Is there a way to calculate $\int \limits_0^1\frac{x^3}{\sqrt{x^2-1}}\frac{1}{1-a^2x^2}\frac{1}{1-b^2x^2}\frac{1}{c-x}\mathrm dx$ I want to calculate
$\displaystyle \int \limits_0^1\dfrac{x^3}{\sqrt{x^2-1}}\dfrac{1}{1-a^2x^2}\dfrac{1}{1-b^2x^2}\dfrac{1}{c-x}\mathrm dx$
$a$ and $b$ are real parameters, c could be complex and is the solution of a cubic equation. 
I tried to find an appropriate contour in the complex plane but failed because it seems impossible to go around the cut (integration from $-1$ to $1$ would be easier). Have anyone an idea? 
Or is any other integration technique better suited?

Let us denote$A:=\arccos(a)$, $B:=\arccos(b)$ and $C:=\arccos(1/c)$.
By firstly substituting $x=\sin(y)$ and then $u=\tan(y/2)$ we get:
\begin{eqnarray}
&&\int\limits_0^1 \frac{x^3}{\sqrt{1-x^2}} \frac{1}{1-a^2 x^2} \frac{1}{1-b^2 x^2} \frac{1}{c-x} dx= \\
&&\frac{16}{c} \int\limits_0^{\tan(\pi/4)} \frac{u^3+u^5}{\prod\limits_{\xi_2=\pm}(u-\exp(\xi_2 \imath C))\prod\limits_{\xi_1=\pm}(u+\xi_1 \exp(\xi_2 \imath A))(u+\xi_1 \exp(\xi_2 \imath B))} du
\end{eqnarray}
Now the only thing we need to do is to use the following partial fraction decomposition:
\begin{eqnarray}
\frac{u^p}{\prod\limits_{j=1}^d (u-u_j)} = \sum\limits_{i=1}^d \frac{u_i^p}{(u-u_i)} \cdot \prod\limits_{j\ne i} \frac{1}{u_i-u_j}
\end{eqnarray}
which is valid for $d\ge 1$ and $p=0,1,2,\cdots,d-1$ and the computation is completed.
