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.

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,
$$\int (x^2-1)(x^3-3x)^{4/3}\ dx=\frac13\int (3x^2-3)(x^3-3x)^{4/3}\ dx$$
$$=\frac13\int (x^3-3x)^{4/3}\ d(x^3-3x)$$
$$=\frac13\frac{(x^3-3x)^{7/3}}{7/3}+C$$
$$=\frac{(x^3-3x)^{7/3}}{7}+C$$
