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We seek to construct a Möbius transformation $x = L(u)$ which transforms the integral below into Weierstrass form. \begin{equation} \label{eq:integrals} \bigint_{C} \frac 1 {\sqrt {ax^4 + bx^3 + cx^2 + dx + e}} dx = \bigint_{C'} \frac 1 {\sqrt {4u^3 - g_2u - g_3}} du \end{equation} Doing so is also equivalent to constructing a birational mapping \begin{equation} \label{eq:birational} x = L(u),\quad y = vL'(u) \end{equation} between the curves \begin{equation} \label{eq:curves} y^2 = ax^4 + bx^3 + cx^2 + dx + e,\quad v^2 = 4u^3 - g_2u - g_3 \end{equation} and the holomorphic differentials \begin{equation} \label{eq:differentials} \frac {dx} {y} = \frac {du} {v} \end{equation} It is also equivalent to solving the differential equation \begin{equation} \label{eq:de} f'^2 = af^4 + bf^3 + cf^2 + df + e \end{equation} using the Weierstrass $\wp$ function \begin{equation} \label{eq:uniformisers} f(z) = L(\wp(z, g_2, g_3)) \end{equation}

HISTORICAL NOTE

Invariant Integral

Consider the integral whose integrand is modelled on the cross-ratio formula: \begin{equation} \label{eq:model} \bigint_{C} \sqrt {\frac {(e_1 - e_2)(e_3 - e_4)} {(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx \end{equation} Using the Möbius difference formula we see that it is invariant under Möbius transformations. That is if $L$ is a Möbius transformation mapping $e_1',e_2',e_3',e_4'$ to $e_1,e_2,e_3,e_4$ and the path of integration $C\thinspace'$ to $C$ then the change of variables $x = L(u)$ gives \begin{equation} \label{eq:invariant} \bigint_{C} \sqrt {\frac {(e_1 - e_2)(e_3 - e_4)} {(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C\thinspace'} \sqrt {\frac {(e_1' - e_2')(e_3' - e_4')} {(u - e_1')(u - e_2')(u - e_3')(u - e_4')}} du \end{equation} The cross-ratio's of the two sets of roots are by necessity equal. \begin{equation} \label{eq:crossratio} \crossratio{e_1,e_2,e_3,e_4}=\crossratio{e_1',e_2',e_3',e_4'} \end{equation}

Equation \eqref{eq:invariant} remains true if we introduce arbitrary leading coefficients $K$ and $K'$ into the polynomials \begin{equation} \label{eq:invariantK} \bigint_{C} \sqrt {\frac {K(e_1 - e_2)(e_3 - e_4)} {K(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C\thinspace'} \sqrt {\frac {K'(e_1' - e_2')(e_3' - e_4')} {K'(u - e_1')(u - e_2')(u - e_3')(u - e_4')}} du \end{equation} If we let $e_4' \rightarrow \infty$ then \eqref{eq:invariantK} becomes \begin{equation} \label{eq:invariantcubic} \bigint_{C} \sqrt {\frac {K(e_1 - e_2)(e_3 - e_4)} {K(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C\thinspace'} \sqrt {\frac {K'(e_1' - e_2')} {K'(u - e_1')(u - e_2')(u - e_3')}} du \end{equation} and the cross-ratio formula \eqref{eq:crossratio} becomes \begin{equation} \label{eq:crossratiocubic} \crossratio{e_1,e_2,e_3,e_4}=\crossratio{e_1',e_2',e_3',\infty} \end{equation} We now have the formulae needed to carry out the reduction.

Reduction to Weierstrass Form

Let \begin{equation} \label{eq:ABC} A = K(e_1 - e_2)(e_3 - e_4),\quad B = K(e_3 - e_1)(e_2 - e_4),\quad C = K(e_2 - e_3)(e_1 - e_4) \end{equation} then it is easily confirmed that \begin{equation*} A + B + C = 0 \end{equation*} Put \begin{equation} \label{eq:proots} \epsilon_1 = \tfrac 1 {12} (A - B),\quad \epsilon_2=\tfrac 1 {12} (C - A),\quad \epsilon_3=\tfrac 1 {12} (B - C) \end{equation} so that \begin{equation*} \epsilon_1 + \epsilon_2 + \epsilon_3 = 0 \end{equation*} and \begin{equation} \label{eq:ABCP} K(e_1 - e_2)(e_3 - e_4) = 4(\epsilon_1 - \epsilon_2), \quad\quad K(e_3 - e_1)(e_2 - e_4) = 4(\epsilon_3 - \epsilon_1), \quad\quad K(e_2 - e_3)(e_1 - e_4) = 4(\epsilon_2 - \epsilon_3) \end{equation} and \begin{equation*} \crossratio{e_1,e_2,e_3,e_4} = \crossratio{\epsilon_1,\epsilon_2,\epsilon_3,\infty} \end{equation*} Then under the change of variables $x = L(u)$ given implicitly by \begin{equation} \label{eq:mobius} \crossratio{x,e_1,e_2,e_3} = \crossratio{u,\epsilon_1,\epsilon_2,\epsilon_3} \end{equation} we get using \eqref{eq:invariantcubic} and \eqref{eq:ABCP} \begin{equation} \label{eq:reduction} \bigint_{C} \frac 1 {\sqrt {ax^4 + bx^3 + cx^2 + dx + e}} dx = \bigint_{C} \frac 1 {\sqrt {K(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C'} \frac 1 {\sqrt {4(u - \epsilon_1)(u - \epsilon_2)(u - \epsilon_3)}} du = \bigint_{C'} \frac 1 {\sqrt {4u^3 - g_2u - g_3}} du \end{equation}

Quartic Invariants

Using \eqref{eq:ABC}, \eqref{eq:proots} and \eqref{eq:reduction} we can obtain expressions for $g_2$ and $g_3$ in terms of the other coefficients. \begin{equation} \label{eq:g2} g_2 = -4 (\epsilon_1 \epsilon_2 + \epsilon_2 \epsilon_3 + \epsilon_3 \epsilon_1) = \tfrac {1} {24} (A^2 + B^2 + C^2) = \tfrac {1} {12} (12ae - 3bd + c^2) \end{equation} and \begin{equation} \label{eq:g3} g_3 = 4 \epsilon_1 \epsilon_2 \epsilon_3 = \tfrac {1} {432} (A - B)(B - C)(C - A) = \tfrac {1} {432} (72ace - 27ad^2 - 27b^2e + 9bcd - 2c^3) \end{equation} The derivation of the last terms in \eqref{eq:g2} and \eqref{eq:g3} is not obvious. The expressions involving $A, B, C$ are symmetric polynomials in $e_1,e_2,e_3,e_4$, a fact which is also not entirely obvious. Therefore by a well known theorem they can be written as polynomials in the coefficients $a,b,c,d,e$. In fact more than that. Because they are symmetric root differences, they are invariants of the quartic polynomial, in the sense of Invariant Theory.

Modular Discriminant

In addition the modular discriminant is defined by \begin{equation} \label{eq:delta} \Delta = g_2^3 - 27 g_3^2 = 16 (\epsilon_1 - \epsilon_2)^2 (\epsilon_2 - \epsilon_3)^2 (\epsilon_3 - \epsilon_1)^2 = \tfrac {1} {256} A^2 B^2 C^2 = \tfrac {1} {256} \discrim(a,b,c,d,e) \end{equation} where $\discrim(a,b,c,d,e)$ is the discriminant of the quartic polynomial. \begin{multline} \label{eq:discriminant} \discrim(a,b,c,d,e) = 256a^3e^3 - 192a^2bde^2 - 128a^2c^2e^2 + 144a^2cd^2e - 27a^2d^4 + 144ab^2ce^2 - 6ab^2d^2e \\ - 80abc^2de + 18abcd^3 + 16ac^4e - 4ac^3d^2 - 27b^4e^2 + 18b^3cde - 4b^3d^3 - 4b^2c^3e + b^2c^2d^2 \end{multline} Observe that \begin{equation} \discrim(a,b,c,d,e) = \discrim(0,4,0,-g_2,-g_3) \end{equation}

Finally we have achieved our goal of transforming the general integral to Weierstrass form via a Möbius transformation \eqref{eq:mobius}.

Cubic Case

Well not quite. Equation \eqref{eq:ABC} needs an adjustment to deal with the cubic case. Letting $e_4 \rightarrow \infty$ in \eqref{eq:invariantcubic} reveals the appropriate formula is \begin{equation} \label{eq:ABC3} A = K(e_1 - e_2),\quad B = K(e_3 - e_1),\quad C = K(e_2 - e_3) \end{equation} From there the rest of the reduction is the same with $a = 0$ and $e_4 = \infty$ and the $e_4$ terms dropping out of \eqref{eq:ABCP} and \eqref{eq:reduction}.

12th Root Formula

Equation \eqref{eq:invariant} is a pretty formula but its numerator appears to lack symmetry. The explanation for this is simple. The equality of the cross-ratios \eqref{eq:crossratio} implies the following identity \begin{equation} \label{eq:crossratiotriple} \frac {(e_1 - e_2)(e_3 - e_4)} {(e_1' - e_2')(e_3' - e_4')} = \frac {(e_3 - e_1)(e_2 - e_4)} {(e_3' - e_1')(e_2' - e_4')} = \frac {(e_2 - e_3)(e_1 - e_4)} {(e_2' - e_3')(e_1' - e_4')} \end{equation} and so \eqref{eq:invariant} also holds for any permutation of the roots in the numerator. It can even be written symmetrically as \begin{equation} \label{eq:invariantdelta} \bigint_{C} \frac {\sqrt[12] {(e_1 - e_2)^2(e_1 - e_3)^2(e_1 - e_4)^2(e_2 - e_3)^2(e_2 - e_4)^2(e_3 - e_4)^2}} {\sqrt{(x - e_1)(x - e_2)(x - e_3)(x - e_4)}} dx = \bigint_{C\thinspace'} \frac {\sqrt[12] {(e_1' - e_2')^2(e_1' - e_3')^2(e_1' - e_4')^2(e_2' - e_3')^2(e_2' - e_4')^2(e_3' - e_4')^2}} {\sqrt{(u - e_1')(u - e_2')(u - e_3')(u - e_4')}} du \end{equation} The symmetry in this formula allows it to be written purely in terms of the coefficients of the polynomials. If the roots of two quartic polynomials can be mapped to one another by a Möbius transform $L$ then under the change of variables $x = L(u)$ we get

Formula \eqref{eq:invariantdeltaK} remains valid when one or both of the polynomials are cubic ie. $a=0$ or $p=0$. In such cases the "missing" 4th root used to define the Möbius transform is $\infty$.

FUN FACT

Jacobian and Legendre Curves

We can now apply the reduction formulae to compute the $g_2$ and $g_3$ invariants for other standard curves.

The curve associated with Jacobi's elliptic sine function is \begin{equation} y^2 = (1-x^2)(1-k^2x^2) \end{equation} Put $K = k^2,\space e_1=1,\space e_2=-1,\space e_3=1/k,\space e_4= -1/k$ then using \eqref{eq:ABC} \begin{equation} \label{eq:ABCjacobi} A=4k,\quad B=(k-1)^2,\quad C=-(k+1)^2 \end{equation}

The Legendre curve is \begin{equation} y^2 = x(x-1)(x-\lambda) \end{equation} Put $K = 1,\space e_1=0,\space e_2=\lambda,\space e_3=1,\space e_4= \infty$ then using \eqref{eq:ABC3} \begin{equation}\label{eq:ABClegendre} A=-\lambda,\quad B=1,\quad C=\lambda - 1 \end{equation}

In summary

$ \dagger \space e_1 + e_2 + e_3 = 0$

The $j$ invariants are given by \begin{equation} j = 1728 {g_2^3 \over g_2^3 - 27g_3^2} = 32 {(A^2 + B^2 + C^2)^3 \over A^2 B^2 C^2} = 16 {(k^4 + 14k^2 + 1)^3 \over k^2(k^2 - 1)^4} = 256 {(\lambda^2 - \lambda + 1)^3 \over \lambda^2(\lambda - 1)^2} \end{equation}

Invariants of the holomorphic differential

Just to emphasize the point. Because of \eqref{eq:reduction}, \eqref{eq:g2} and \eqref{eq:g3} we have calculated $\sum {\frac 1 {\omega^4}}$ and $\sum {\frac 1 {\omega^6}}$ where $\omega$ ranges over all the non-zero periods of the differential $\frac {dx} {y}$. Explicitly let \begin{equation} \label{eq:y2r4} y^2 = ax^4 + bx^3 + cx^2 + dx + e \end{equation} and let $\omega_C$ be the period of the differential associated with closed path $C$ \begin{equation*} \omega_C = \bigint_{C} \frac {dx} {y} \end{equation*} then \begin{aligned} 60 \sum_{C} {\frac 1 {\omega_C^4}} &= \tfrac {1} {12} (12ae - 3bd + c^2) \\\\ 140 \sum_{C} {\frac 1 {\omega_C^6}} &= \tfrac {1} {432} (72ace - 27ad^2 - 27b^2e + 9bcd - 2c^3) \end{aligned} where $C$ ranges over all non-trival closed paths in the homology group of the curve.

Curve as a differential equation

To solve the differential equation below for $f$ in terms of the Weierstrass $\wp$ function: \begin{equation} \label{eq:fde} f'(z)^2 = a f(z)^4 + b f(z)^3 + c f(z)^2 + d f(z) + e \end{equation} Let $e_1,e_2,e_3,e_4,\epsilon_1,\epsilon_2,\epsilon_3,g_2,g_3$ be defined as in \eqref{eq:ABC} thru \eqref{eq:g3}. Assume a boundary condition of $f(0) = e_4$. Then $f(z)$ is given implicitly by the cross ratio formula \begin{equation} \label{eq:solution} \crossratio{f(z),e_1,e_2,e_3} = \crossratio{\wp(z,g_2,g_3),\epsilon_1,\epsilon_2,\epsilon_3} \end{equation} In fact \eqref{eq:solution}, and the invariance of cross-ratio's under the Möbius transforms, implies for any $w,x,y,z$ \begin{equation} \label{eq:crossratioidentity} \crossratio{f(w),f(x),f(y),f(z)} = \crossratio{\wp(w),\wp(x),\wp(y),\wp(z)} = \frac {\sigma(w-x)\sigma(w+x)\sigma(y-z)\sigma(y+z)} {\sigma(w-y)\sigma(w+y)\sigma(x-z)\sigma(x+z)} \end{equation} The general solution of the \eqref{eq:fde} is given by \begin{equation} \label{eq:gensolution} f(z) = L(\wp(z+z_0,g_2,g_3)) \end{equation} where $z_0$ is effectively the constant of integration. For the boundary condition $f(0) = x_0$ there are in general two solutions corresponding to the two possible values of $f'(0) = y_0$. It might appear from \eqref{eq:solution} that this solution would explicitly involve the roots $e_1,e_2,e_3,\epsilon_1,\epsilon_2,\epsilon_3$. But that is not the case because, for a fixed value of $x_0$ and $y_0$, such a solution must be unchanged when the roots are permuted, and therefore only involves the rational functions of the coefficients $a,b,c,d,e$ along with $x_0, y_0$. We could attempt to find this solution by expanding \eqref{eq:gensolution} using the addition formula for $\wp(z+z_0)$. But using this power series method is more direct.

Addition Formula In 4 Variables

Similar to the well known symmetric addition formula for $\wp$ there is a symmetric addition formula for the general elliptic function $f$ of order 2.

If $z_1 + z_2 + z_3 + z_4 \equiv 2(a_1 + a_2) \mod \Omega$ where $a_1$ and $a_2$ are the location of the poles of $f$, then:

EXAMPLE

To get a symmetric 3 variable formula put $z_4 = 2(a_1 + a_2)$.

Addition Formula For The Integral

The natural way to express the above addition formula in terms of integrals is to say that if $R$ is a polynomial, of degree 3 or 4, and \begin{equation*} \bigint_{l_1}^u \frac {dt} {\sqrt{R(t)}} \space + \space \bigint_{l_1}^v \frac {dt} {\sqrt{R(t)}} \space + \space \bigint_{l_2}^w \frac {dt} {\sqrt{R(t)}} \space + \space \bigint_{l_2}^x \frac {dt} {\sqrt{R(t)}} \space \equiv \space 0 \mod \Omega \end{equation*} then $u,v,w,x$ satisfy the algebraic relation

The congruence relation, modulo the period lattice $\Omega$, is to take into account different paths of integration. There is also some complexity in the interpretation of the lower bounds. The congruence is only true when $l_1$ and $l_2$ are a pair of conjugate points. That is points which have the same complex $t$ value but correspond to two different points (or a branch point) on the underlying Riemann surface where the integration takes place.

As we will see below, many of the standard addition formula for elliptic integrals and functions are an expansion or specialisation of this formula.

ALTERNATIVE EXPRESSION

Level 1 Addition Formula In 3 Variables For Degree 3

Assume $R(t) = K(t-a)(t-b)(t-c)$.

We can obtain a 3 variable addition formula by letting $x \rightarrow \infty$ in \eqref{eq:addvar4} giving \begin{equation} \label{eq:addvar3deg3lev1} \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 0 & 0 & 1 & 0 \\ \end{vmatrix} \space = \space - \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & \sqrt{R(w)} \\ \end{vmatrix} \space = \space (v-w)\sqrt{\smash[b]{R(u)}} \space + \space (w-u)\sqrt{\smash[b]{R(v)}} \space + \space (u-v)\sqrt{\smash[b]{R(w)}} \space = \space 0 \end{equation} In order to distinguish this from the closely related formulae below, call it the level 1 formula.

Level 2 Addition Formula In 3 Variables For Degree 3

We can partialy rationalise the level 1 formula by multiplying by a conjugate expression to give \begin{equation*} \label{eq:addvar3deg3lev2} \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & \sqrt{R(w)} \\ \end{vmatrix} \space \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & -\sqrt{R(w)} \\ \end{vmatrix} \space = \space (w-u)(w-v)\left(K\left[(u + v + w) - (a + b + c)\right]\left(u - v\right)^2 - \left(\sqrt{\smash[b]{R(u)}} - \sqrt{\smash[b]{R(v)}}\right)^2\right) \space = \space 0 \end{equation*} Call this the level 2 addition formula. It can be solved for $w$ to give the non-trivial solution \begin{equation} w = (a + b + c) + \frac 1 K \left[\frac {\sqrt{R(u)} - \sqrt{R(v)}} {u - v}\right]^2 - (u + v) \end{equation} When $K=4$ and $a+b+c=0$ this is the standard addition formula for the Weierstrass integral.

Level 3 Addition Formula in 3 Variables For Degree 3

We can repeat this process once more to produce level 3 addition formulae. \begin{multline} \label{eq:deg3lev3} \left. \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & \sqrt{R(w)} \end{vmatrix} \space \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & \sqrt{R(v)} \\ 1 & w & -\sqrt{R(w)} \end{vmatrix} \space \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & -\sqrt{R(v)} \\ 1 & w & \sqrt{R(w)} \end{vmatrix} \space \begin{vmatrix} 1 & u & \sqrt{R(u)} \\ 1 & v & -\sqrt{R(v)} \\ 1 & w & -\sqrt{R(w)} \end{vmatrix} \space \middle/ \space \begin{vmatrix} 1 & u & u^2 \\ 1 & v & v^2 \\ 1 & w & w^2 \\ \end{vmatrix}^2 \right. \\\\ \space = \space K^2 \Big[\big((uv + uw + vw) - (ab + ac + bc)\big)^2 - 4\big((u + v + w) - (a + b + c)\big)(uvw - abc)\Big] \space = \space 0 \end{multline} This is an example of Abel's classic addition theorem for algebraic integrals. Rather mysteriously, in this case it is equivalent to the discriminant expression \begin{equation*} \discriminant_{t}\big( R(t) - K(t-u)(t-v)(t-w) \big) \space = \space 0 \end{equation*} FUN FACT

Differential Equation Equivalent

The three variable addition formulae can also be interpreted as the solution of a different equation. For example the integral of this differential equation \begin{equation*} \frac {dx} {\sqrt{(x-a)(x-b)(x-c)}} = \frac {dy} {\sqrt{(y-a)(y-b)(y-c)}} \end{equation*} is equation \eqref{eq:deg3lev3}. Explicitly \begin{equation*} \left[(xy + xC + yC) - (ab + ac + bc)\right]^2 - 4\left[(x + y + C) - (a + b + c)\right]\left[xyC - abc\right] \space = \space 0 \end{equation*} where $C$ is the constant of integration.

HISTORICAL NOTE

Addition Formulae In 3 Variables For Degree 4

Assume $R(t) = K(t-a)(t-b)(t-c)(t-d)$.

By letting $x$ take some specific value in equation \eqref{eq:addvar4} we can get an addition formula for degree 4 in the 3 variables. However, without extra requirements, these formulae are not rational in the coefficients of $R$. For example $x = \infty, 0\space \text{or}\space d$ gives \begin{equation*} \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 0 & 0 & 1 & \sqrt{K} \\ \end{vmatrix} \space = \space 0, \qquad \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & 0 & 0 & \sqrt{Kabcd} \\ \end{vmatrix} \space = \space 0 \qquad \text{or} \qquad \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & d & d^2 & 0 \\ \end{vmatrix} \space = \space 0 \end{equation*} The corresponding level 2 addition formulae are not rational unless $\sqrt{K}, \sqrt{Kabcd} \space \text{or} \space d$, respectively, are rational. For this reason it's better to focus on the 4 variable addition formula in the degree 4 case.

Level 2 Addition Formula In 4 Variables For Degree 4

Similar to above, computation of the 4 variable level 2 addition formula gives \begin{multline*} \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & x & x^2 & \sqrt{R(x)} \\ \end{vmatrix} \space \begin{vmatrix} 1 & u & u^2 & \sqrt{R(u)} \\ 1 & v & v^2 & \sqrt{R(v)} \\ 1 & w & w^2 & \sqrt{R(w)} \\ 1 & x & x^2 & -\sqrt{R(x)} \\ \end{vmatrix} \frac 1 {(u-x)(v-x)(w-x)} \\\\ \space = \space K(u - v)^2(v - w)^2(w - u)^2\big[(u + v + w + x) - (a + b + c + d)\big] \space + \space (u-x)S(u,v,w) \space + (v-x)S(v,u,w) \space + \space (w-x)S(w,u,v) \space = \space 0 \end{multline*} where \begin{equation*} S(u,v,w) = (u - v)(u - w)\left(\sqrt{\smash[b]{R(v)}} - \sqrt{\smash[b]{R(w)}}\right)^2 \end{equation*} solving for $x$ gives \begin{equation} \label{eq:addvar4lev2} x = \frac {K(u - v)^2(v - w)^2(w - u)^2\big[(u + v + w) - (a + b + c + d)\big] \space + \space u S(u,v,w) \space + \space v S(v,u,w) \space + \space w S(w,u,v)} {-K(u - v)^2(v - w)^2(w - u)^2 \space + \space S(u,v,w) \space + \space S(v,u,w) \space + \space S(w,u,v)} \end{equation} There are many ways of expressing this formula, see Quartic Addition Formulae.

Level 4 Addition Formula In 4 Variables For Degree 4

Computation of the 4 variable level 4 addition formula, using CAS, gives \begin{equation} \label{eq:addvar4lev4} \prod_{\text{all signs}} \begin{vmatrix} 1 & u & u^2 & \hphantom{\pm}\sqrt{R(u)} \\ 1 & v & v^2 & \pm\sqrt{R(v)} \\ 1 & w & w^2 & \pm\sqrt{R(w)} \\ 1 & x & x^2 & \pm\sqrt{R(x)} \\ \end{vmatrix} \space = \space \begin{vmatrix} 1 & u & u^2 & u^3 \\ 1 & v & v^2 & v^3 \\ 1 & w & w^2 & w^3 \\ 1 & x & x^2 & x^3 \\ \end{vmatrix}^4 \space K^4 \space P(u,v,w,x,a,b,c,d) \space = \space 0 \end{equation} where $P$ is a polynomial with several hundred terms.

For reasons which are not at all obvious it has the following symmetry $P(u,v,w,x,a,b,c,d) = P(a,b,c,d,u,v,w,x)$ which implies another

FUN FACT

References

[1] Euler, Leonhard (1761) Observationes de comparatione arcuum curvarum irrectificibilium Euler Archive - All Works. 252.

[2] Barrios, Jose (2009) A Brief History of Elliptic Integral Addition Theorems Rose-Hulman Undergraduate Mathematics Journal: Vol. 10 : Iss. 2 , Article 2.

[3] Trotti, Manuela (2016) Applications of elliptic functions to solve differential equations [Laurea magistrale], Università di Bologna, Corso di Studio in Matematica [LM-DM270]