Evaluate $\int^\pi_{-\pi}(1-a\cos\theta)^{-b-2}\log(1-a\cos\theta)d\theta$ 
$$ \int^\pi_{-\pi} \left(1-a\cos\theta\right)^{-b-2}
 \log\left(1-a\cos\theta\right)d\theta$$
  Under the condition  $0<a<1$ and $b>0$.

Mathematica found the following form.
$$ 2\pi\left(a+1\right)^{-b-2}
 \bigg({}_2F_1\left(\frac{1}{2},b+2;1;\frac{2a}{a+1}\right)
 \left(\log\left(a+1\right)+\Psi\left(b+2\right)\right)
+\Gamma\left(b+2\right){}_3\tilde{F}_2^{\left(\{0,0,0\},
 \{0,1\},0\right)}
 \left(\frac{1}{2},b+2,b+2;1,b+2;\frac{2a}{a+1}\right)\bigg)$$
I need to use this in an algorithm, but calculation of $\displaystyle {}_3\tilde{F}_2^{\left(\{0,0,0\}, \{0,1\},0\right)}$ takes time. So I am looking for some simpler expression.
 A: Consider $\int_{-\pi}^\pi(1-a\cos\theta)^{-b-2}~d\theta$ ,
$\int_{-\pi}^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_{-\pi}^0(1-a\cos\theta)^{-b-2}~d\theta+\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_\pi^0(1-a\cos(-\theta))^{-b-2}~d(-\theta)+\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta+\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=2\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=2\int_1^{-1}(1-ax)^{-b-2}~d(\cos^{-1}x)$
$=2\int_{-1}^1\dfrac{(1-ax)^{-b-2}}{\sqrt{1-x^2}}dx$
$=2\int_0^2\dfrac{(1-a(x-1))^{-b-2}}{\sqrt{1-(x-1)^2}}d(x-1)$
$=2\int_0^2x^{-\frac{1}{2}}(2-x)^{-\frac{1}{2}}(a+1-ax)^{-b-2}~dx$
$=2\int_0^1(2x)^{-\frac{1}{2}}(2-2x)^{-\frac{1}{2}}(a+1-2ax)^{-b-2}~d(2x)$
$=\dfrac{2}{(a+1)^{b+2}}\int_0^1x^{-\frac{1}{2}}(1-x)^{-\frac{1}{2}}\left(1-\dfrac{2ax}{a+1}\right)^{-b-2}~dx$
$=\dfrac{2\pi}{(a+1)^{b+2}}~_2F_1\left(b+2,\dfrac{1}{2};1;\dfrac{2a}{a+1}\right)$
$\therefore$ $\int_{-\pi}^\pi(1-a\cos\theta)^{-b-2}\log(1-a\cos\theta)d\theta$
$=\int^\pi_{-\pi}\left(-\dfrac{d}{db}\left(1-a\cos\theta\right)^{-b-2}\right)d\theta$
$=-2\pi\dfrac{d}{db}\left(\dfrac{1}{(a+1)^{b+2}}~_2F_1\left(b+2,\dfrac{1}{2};1;\dfrac{2a}{a+1}\right)\right)$
