Contour integral of rational function to some power I am interested in the following integral $$\int_{0}^{2\pi}d\theta \left(\frac{1-|x|^2}{|1-x e^{-i\theta}|^2}\right)^{y}$$
Does it has a nice closed form expression?
I know that $$\int_{0}^{2\pi}d\theta \left(\frac{1-|x|^2}{|1-x e^{-i\theta}|^2}\right)=2\pi$$
by contour integration.
Any hints are welcome.

Assuming $|x|=r<1$, the integral is $$\int_0^{2\pi}\left(\frac{1-r^2}{1-2r\cos\theta+r^2}\right)^y d\theta.$$ It reduces to the hypergeometric function. A contour integration approach is possible here too; for $0<\Re y<1$, we integrate $\big(z(1-rz)^y(1-r/z)^y\big)^{-1}$ along the circle $|z|=1$, and squeeze the contour down to just encircling the segment $(0,r)$ of the real line (the cut for the integrand); the result is analytically continued w.r.t. $y$. The same result is obtained directly from the binomial series (let $y=1+\alpha$) $$(1-z)^{-1-\alpha}=\sum_{n=0}^\infty\binom{n+\alpha}{n}z^n\qquad(|z|<1)$$ writing $1-2r\cos\theta+r^2=(1-re^{i\theta})(1-re^{-i\theta})$: $$\int_0^{2\pi}\frac{d\theta}{(1-2r\cos\theta+r^2)^{1+\alpha}}=2\pi\sum_{n=0}^\infty\binom{n+\alpha}{n}^2 r^{2n}.$$ This has a (true) closed form only when $\alpha$ (i.e. $y$) is an integer (and when it is half an integer, the result is a sort of complete elliptic integrals).
