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.
 A: By transforming the variable from $x$ to $t$ using the relation,
\begin{equation}
\begin{split}
t=\sqrt{\frac{(a-c)(x-b)}{(a-b)(x-c)}}
\end{split}
\end{equation}
you will obtain
\begin{equation}
\begin{split}
&x=\frac{-(a-c)b+(a-b)c\,t^2}{-(a-c)+(a-b)\,t^2} \\
&dx=\frac{2(a-b)(a-c)(b-c)\,t}{[-(a-c)+(a-b)\,t^2]^2}\,dt
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)
\end{split}
\end{equation}
Inserting the above relation to the integral
\begin{equation}
\begin{split}
&\int^u_b\frac{dx}{\sqrt{(a-x)(x-b)(x-c)(x-d)}}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{split}
\end{equation}
gives
\begin{equation}
\begin{split}
&\int^u_b\frac{dx}{\sqrt{(a-x)(x-b)(x-c)(x-d)}} \\
&=\frac{2}{\sqrt{(a-c)(b-d)}}
\int^{a_1}_{a_2}\frac{dt}{\sqrt{(1-t^2)\left[1-\frac{(a-b)(c-d)}{(a-c)(b-d)}t^2\right]}}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)
\end{split}
\end{equation}
Transforming from $t$ to $\theta$ using the relation $t=\sin\theta$, the integral in (3) is expressed as
\begin{equation}
\begin{split}
&\frac{2}{\sqrt{(a-c)(b-d)}}
\int^{\alpha}_{\beta}\frac{d\theta}{\sqrt{1-k^2\,\sin^2\theta}}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)
\end{split}
\end{equation}
Here
\begin{equation}
\begin{split}
&a_1=\sqrt{\frac{(a-c)(u-b)}{(a-b)(u-c)}} \\
&a_2=0 \\
&\alpha=\sin^{-1}\left[\sqrt{\frac{(a-c)(u-b)}{(a-b)(u-c)}}\right] \\
&\beta=0 \\
&k^2=\frac{(a-b)(c-d)}{(a-c)(b-d)} \\
\end{split}
\end{equation}
Therefore the integral becomes
\begin{equation}
\begin{split}
&\frac{2}{\sqrt{(a-c)(b-d)}}
\int^{\alpha}_{\beta}\frac{d\theta}{\sqrt{1-k^2\,\sin^2\theta}}=\frac{2}{\sqrt{(a-c)(b-d)}}\,F[\alpha,k]
\end{split}
\end{equation}
