Evaluate $\int_{x=0}^{2 \pi}\frac{dx}{(1+a \cos{x})^{b^2}}$ 
Evaluate:
  $$\int_{0}^{2 \pi}\frac{dx}{(l^2+r^2+2 l r \cos{x})^{b^2}}$$
  where $b^2$ is a real number, $r>0$, and $l \geq 0$.

I can just simplify it to 
$$c\int_{0}^{2 \pi}\frac{dx}{(1+a \cos{x})^{b^2}}$$
where $a>0$.
Any ideas? simplifications? results?
 A: We want to calculate the integral (set $x\rightarrow 2\pi-x$)
$$
I(a,\beta)=\int_0^{2\pi}\frac{1}{(1+a \cos(x))^{\beta}}=2\int_0^{\pi}\frac{1}{(1+a \cos(x))^{\beta}}=2\int_0^{\pi}\frac{1}{(1+a-2a \sin^2(x/2))^{\beta}}
$$
where $|a|<1$ and $\beta\in \mathbb{R}_+$. 
Let us start with a subsitution $y=\sin(x/2)$. This yields
$$
I(a,\beta)=\frac{4}{(1+a)^\beta}\int_{0}^{1}\frac{1}{\sqrt{1-y^2}}\frac{1}{(1-\frac{2a}{1+a} y^2)^{\beta}}
$$
setting $y=t^{1/2}$ we get
$$
I(a,b)=\frac{2}{(1+a)^\beta}\int_{0}^{1}\frac{t^{-1/2}}{\sqrt{1-t}}\frac{1}{(1-\frac{2a}{1+a} t)^{\beta}}
$$...
which equals by Euler's formula (we use $B(1/2,1)=\Gamma(1/2)^2/\Gamma(1)=\pi$ in the second step)

$$
I(a,\beta)=\frac{2}{(1+a)^\beta}B(1/2,1){_2F_1}\left(\beta,\frac{1}{2};1;\frac{2a}{1+a}\right)=\\\frac{2\pi}{(1+a)^\beta}{_2F_1}\left(\beta,\frac{1}{2};1;\frac{2a}{1+a}\right) \quad 
$$

This reproduces Mathematicas result from the comments by the means of Pfaff's transformation 
Set:
$\frac{2a}{1+a}\rightarrow\frac{2a}{a-1}$ and use that $1-\frac{1}{2}=\frac{1}{2}$ :-)
--
edit: To make the last step more clear:
Due to Pfaff it holds that
$$
_2F_1(a,b;c,z)=\frac{1}{(1-z)^b}{_2F_1}\left(b,c-a;c,\frac{z}{z-1}\right)
$$
now we set $z=\frac{2a}{a-1}$,$a=\frac{1}{2}$, $b=\beta$ and $c=1$. We find
$$
\frac{1}{(1-a)^{\beta}}{_2F_1}\left(\frac{1}{2},\beta;1,\frac{2a}{a-1}\right)=\frac{1}{(1+a)^{\beta}}{_2F_1}\left(\beta,\frac{1}{2};1,\frac{2a}{a+1}\right)
$$
which also means that

$$
I(a,\beta)=\frac{2\pi}{(1-a)^{\beta}}{_2F_1}\left(\frac{1}{2},\beta;1,\frac{2a}{a-1}\right)
$$

which is exactly Mathematicas claim
