Evaluating $\int x^2 \sqrt{x^2-1} dx$ How do I evaluate the following indefinite integral?
$$\int x^2 \sqrt{x^2-1} dx$$
Through integration of parts, I have obtained 
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int (x^2-1)^{3/2} dx $$
I've attempted evaluating the second term through substitution, where
$$ x =  \sec(u)$$
However, I am stuck with 
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int \tan^4(u) \sec (u) du $$
What would be my next step?

$$\int x^{2}\sqrt{x^{2}-1}\,dx\\
x=\cosh{y}\Rightarrow dx=\sinh{y}dy\\
\int x^{2}\sqrt{x^{2}-1}\,dx=\int \cosh^{2}{y}\sinh^{2}{y}\,dy=\frac{1}{4}\int\sinh^{2}{2y}\,dy\\
=\frac{1}{4}\int\frac{\cosh{4y}-1}{2}\,dy=\frac{\sinh{4y}}{2}-\frac{y}{8}+C\\
=\sinh{2y}\cosh{2y}-\frac{\cosh^{-1}x}{8}+C\\
=2(\cosh{y})(\cosh{2y})(\sinh{y})-\frac{\cosh^{-1}x}{8}+C\\
=2x(2x^{2}-1)\sqrt{x^{2}-1}-\frac{\cosh^{-1}x}{8}+C\\
$$
