Hypergeometric functions & integral I'm having difficulty re-deriving a result a calculation from a paper. The integral is
$$\int_0^{2\pi} \int_0^{2\pi} \frac{\sinh\eta}{(\cosh\eta-\cos\theta)^2}\left(1-c\sinh^2\eta\sin\phi\right)^\frac12d\theta d\phi,$$
where $\eta$ and $c$ are parameters such that $\sinh^2\eta = 2$ and $c\sinh^2\eta < 1.$
From what I've read about hypergeometric functions, one could potentially evaluate this using elliptic integrals of the first and second kind,
\begin{align*}
K(z) &= \int_0^{\pi/2} (1-z^2\sin^2 t)^{-1/2} dt \\
E(z) &= \int_0^{\pi/2} (1-z^2\sin^2 t)^{1/2} dt.
\end{align*}
These can be written in hypergeometric form as 
\begin{align*}
K(z) &= \frac{\pi}{2}F(\frac{1}{2},\frac{1}{2};1;z^2)\\
E(z) &= \frac{\pi}{2}F(-\frac{1}{2},\frac{1}{2};1;z^2).
\end{align*}
The expected result of the integral is 
$$S = 8\pi^2 \frac{b^2}{a} G_1(c/a).$$
where $a = \sqrt{c^2+b^2}$ and
$$G_1(x) = F(3/2,1/2,1;x^2)+x^2/2F(3/2,3/2,2;x^2).$$
I've tried doing this, but I am inexperienced in these calculations and it's going to take me a while. I'll keep working at it, but I was wondering if anybody could take a look and tell me whether I am even going about this in the right way.
 A: Going off Peter's comment, note that your integral is separable, and can thus be factored into a product of two one-dimensional integrals:
$$\begin{split}&\int_0^{2\pi} \int_0^{2\pi} \frac{\sinh\eta}{(\cosh\eta-\cos\theta)^2}\left(1-c\sinh^2\eta\sin\phi\right)^\frac12\mathrm d\theta\mathrm d\phi=\\&\quad\left(\color{green}{\sinh\eta\int_0^{2\pi} \frac{\mathrm d\theta}{(\cosh\eta-\cos\theta)^2}}\right)\left(\color{blue}{\int_0^{2\pi} \sqrt{1-c\sinh^2\eta\sin\phi} \mathrm d\phi}\right)\end{split}$$
The first integral is elementary:
$$\color{green}{\sinh\eta\int_0^{2\pi} \frac{\mathrm d\theta}{(\cosh\eta-\cos\theta)^2}}=\frac{2\pi\cosh\,\eta}{\sinh^2\eta}$$
while the second requires the services of the incomplete elliptic integral of the second kind $E(\phi\mid m)$ (see Byrd and Friedman, formula 288.01 for the required identity):
$\displaystyle\begin{split}&\color{blue}{\int_0^{2\pi} \sqrt{1-c\sinh^2\eta\sin\phi} \mathrm d\phi}=\\&\Tiny 2\left(\sqrt{\frac{1+c\sinh^2\eta}{1-c\sinh^2\eta}}\;E\left(\arcsin\left(\sqrt{\frac{1+c\sinh^2\eta}{2}}\right)\mid\frac{2c\sinh^2\eta}{1+c\sinh^2\eta}\right)-\frac{c\sinh^2\eta}{\sqrt{1-c\sinh^2\eta}}+2\sqrt{1+c\sinh^2\eta}\;E\left(\frac{\pi}{4}\mid\frac{2c\sinh^2\eta}{1+c\sinh^2\eta}\right)\right)\end{split}$
