<s>[INST] 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?
 [/INST] For the domain $|x|>1$,
\begin{align}
\int x^2 \sqrt{x^2-1}\ dx=& \int\frac{x}{4\sqrt{x^2-1}}d\left[(x^2-1)^2\right]\\
\overset{ibp}=&\ \frac14 x(x^2-1)^{3/2}
 +\frac18\int \frac{\sqrt{x^2-1}}x d(x^2)\\
\overset{ibp}= &\ 
\frac14 x(x^2-1)^{3/2}+\frac18 x(x^2-1)^{1/2}-\frac18
\tanh^{-1}\frac x{\sqrt{x^2-1}}
\end{align}
</s>