# Elliptic Integral from Gradshteyn and Ryzhik

In my course of work I came across the following elliptic integral, and found its solution in Gradshteyn and Ryzhik:

$$\int_{b}^{u} \frac{\mathrm{d} x}{\sqrt{(a-x)(x-b)(x-c)(x-d)}} = \frac{2}{\sqrt{(a-c)(b-d)}}\,F(\lambda, r),$$ where $$\lambda = \arcsin \left(\sqrt{\frac{(a-c)(u-b)}{(a-b)(u-c)}}\right), \quad r = \sqrt{\frac{(a-b)(c-d)}{(a-c)(b-d)}},$$ for $a \geq u > b > c > d$.

Now, I trust Gradshteyn and Ryzhik that this is correct, but I am curious how to derive it. I know that the elliptic integral of the first kind can be defined as $$F(\lambda, r) = \int_{0}^{\sin\lambda} \frac{\mathrm{d} t}{\sqrt{(1-r^{2} t^{2})(1-t^{2})}},$$ but I struggle to see how to reconcile this with the formula above, where one has: $$\sqrt{(a-x)(x-b)(x-c)(x-d)} = \sqrt{(-ac+(a+c)x-x^{2})(bd-(b+d)x+x^{2})}.$$

If someone can give me a hint how to proceed with the derivation, I'd be very grateful.

• I would try starting with the substitution $t = \sqrt{\frac{(a-c)(x-b)}{(a-b)(x-c)}}$ and see where that took me. – Michael Biro May 11 '18 at 13:05