I encountered the following relation when studying elliptic integrals:
$$\int_{a_2}^{a_3} \dfrac{dx}{\sqrt{(x-a_1)(x-a_2)(x-a_3)(x-a_4)}}\propto K(k)$$
where $K(k)=\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k t^2)}}$ is an elliptic integral of first kind and $k=\dfrac{(a_1-a_4)(a_2-a_3)}{(a_1-a_3)(a_2-a_4)}$ is the modulus of the elliptic curve defined by the quartic polynomial equation:
$$y^2=(x-a_1)(x-a_2)(x-a_3)(x-a_4)$$
where we order the (real) roots $a_1<a_2<a_3<a_4$ and pick the the branch $\sqrt{(x-a_1)(x-a_2)(x-a_3)(x-a_4)}=+1 \sqrt{|(x-a_1)(x-a_2)(x-a_3)(x-a_4)|}$ in the interval $a_2<x\leq a_3$.
My question is on which substitution to make to translate from the LHS to the RHS of the first equation.
Thanks!
Equation 254.00 in the Byrd-Friemann book says $$ \int_{a_2}^y \frac{dt}{\sqrt{(a_1-t)(a_3-t)(t-a_2)(t-a_1)}} = gF(\varphi,k) $$ with $$ g\equiv \frac{2}{\sqrt{(a_4-a_2)(a_3-a_1)}}. $$ $$ \varphi\equiv \arcsin\sqrt{\frac{(a_3-a_1)(y-a_2)}{(a_3-a_2)(y-a_1)}}. $$ $$ k^2\equiv \frac{(a_3-a_2)(a_4-a_1)}{(a_4-a_2)(a_3-a_1)}. $$ where $F$ is an incomplete Elliptic Integral of the First Kind. For $y\to a_3$ this turns with $\varphi\to 90^\circ$ into the Complete Elliptic Integral.