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.
 A: 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
$$
\begin{split} 
\int (x^2 -1)(x^3 - 3x)^{4/3}dx &= \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx\cr
&= \frac{1}{3} \int u^{4/3} du \cr
&= \frac{1}{3} \times \frac{3u^{7/3}}{7} + C \cr
&= \frac{1}{7} (x^3 - 3x)^{7/3} + C \cr
\end{split}
$$
A: Let $x^3-3x=t\implies (3x^2-3)dx=dt$ or $(x^2-1)dx=\frac{dt}{3}$
$$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\int t^{4/3}\frac{dt}{3}$$
$$=\frac13\frac{t^{7/3}}{7/3}+C$$$$=\frac{(x^3-3x)^{7/3}}{7}+C$$
or alternatively,
$$\int (x^2-1)(x^3-3x)^{4/3}\ dx=\frac13\int (3x^2-3)(x^3-3x)^{4/3}\ dx$$
$$=\frac13\int (x^3-3x)^{4/3}\ d(x^3-3x)$$
$$=\frac13\frac{(x^3-3x)^{7/3}}{7/3}+C$$
$$=\frac{(x^3-3x)^{7/3}}{7}+C$$
A: I agree with the other answers.  My response is long-winded so...
Often when attacking indefinite integrals, you will immediately suspect that a substitution [i.e. $u = g(x)$] is needed, but won't be sure which substitution to try.
I have to ask the OP:
Why did you think that $x = \sec \theta$ was the right substitution?  Had you recently been exposed to problems that seemed similar where $x = \sec \theta$ was the right substitution?
The point of my response/rant is to develop the OP's intuition. Since the integral contains $(x^3 - 3x)^{(4/3)},$ my first guess as to the right substitution to try would be $u = (x^3 - 3x).$  This would convert this portion of the integral to $u^{(4/3)}.$
The idea is that (as a first guess for the right substitution), I would be hoping that (except for the $u^{(4/3)}$ factor), the remainder of the integral would be a polynomial in $u$, where each term has an integer exponent.
As I say, the point of my response is simply to expand the OP's intuition (and perspective).
A: $$t=x^{3}-3x\\
I=\frac{1}{3}\int t^{\frac{4}{3}}dt=\frac{1}{7}t^{2}\sqrt[3]{t}+c=\frac{1}{7}(x^{3}-3x)^{2}\sqrt[3]{x^{3}-3x}+c$$
