What is $\int_{0}^{\pi} (a^2+1+2a \cos\theta)^{\frac{p-1}{2}}d\theta$? I have tried the problem using $\theta\rightarrow\theta/2$, then taking $z=e^{i\theta}$ we get $\frac{1}{2i}\int_{|z|=1}\frac{(a^2+1+a\sqrt{z}+\frac{a}{\sqrt{z}})^{\frac{p-1}{2}}}{z}dz$. 

With the change of variables $t = 2 \arccos \sqrt u$,
$$\int_0^\pi (a^2 + 2 a \cos t + 1)^{(p - 1)/2} dt = \\
|a - 1|^{p - 1} \int_0^1 u^{-1/2} (1 - u)^{-1/2}
 \left( 1 + \frac {4 a} {(a - 1)^2} u \right)^{(p - 1)/2} du = \\
\pi |a - 1|^{p - 1} {_2 \hspace{-1px} F_{\hspace{-1px} 1}} {\left(
 \frac 1 2, \frac {1 - p} 2; 1; -\frac {4 a} {(a - 1)^2} \right)},
\quad a, p \in \mathbb R, \; a^2 \neq 1.$$
