<s>[INST] Evaluate integral $\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}$ How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,
$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3} 
  \sec\theta\ \tan\theta d\theta $$
$$=\int \tan^2\theta \sec^4\theta(1-3\cos^2\theta)^{4/3} \sec\theta\ \tan\theta d\theta $$
$$=\int \tan^3\theta \sec^5\theta(1-3\cos^2\theta)^{4/3}\ d\theta $$
$$=\int\dfrac{ \sin^3\theta}{ \cos^8\theta}(1-3\cos^2\theta)^{4/3}\ d\theta $$
I can't see if this substitution will work or not. This has become so complicated.
Please help me solve this integral.
 [/INST] If you multiply and divide by $3$, you get
$$ \int (x^2 -1)(x^3 - 3x)^{4/3}dx = \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx $$
changing variable to $u = x^3 - 3x$ you have $du = (3x^2 - 3x)dx$ so
$$
\begin{split} 
\int (x^2 -1)(x^3 - 3x)^{4/3}dx &= \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx\cr
&= \frac{1}{3} \int u^{4/3} du \cr
&= \frac{1}{3} \times \frac{3u^{7/3}}{7} + C \cr
&= \frac{1}{7} (x^3 - 3x)^{7/3} + C \cr
\end{split}
$$
</s>