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?

Let $x = \sec \theta$. Hence, $dx = \sec \theta \tan \theta d\theta$.
$$
\int x^2\sqrt{x^2 - 1}dx = \int \sec^2\theta \tan \theta \sec \theta \tan \theta d\theta = \int \sec^3 \theta \tan^2 \theta d\theta 
$$
$$
= \int \sec^3d\theta(\sec^2\theta - 1)d\theta = \int \sec^5\theta d\theta - \int \sec^3 \theta d\theta
$$
In this link 
$$
K_n(\theta) := \int \sec^n \theta d\theta = \frac{1}{n-1}\sec^{n-2}\theta\tan \theta + \frac{n-2}{n-1}K_{n-2}(\theta)
$$
Thus,
$$
\int \sec^5\theta d\theta = \dfrac{1}{4}\sec^3d\theta \tan \theta d\theta + \dfrac{3}{4}K_3(\theta)
$$
and
$$
K_3(\theta) = \dfrac{1}{2}\sec \theta \tan \theta + \dfrac{1}{2}\ln(\sec \theta + \tan \theta)
$$
