How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$? I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$

This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ we will use the substitution: $$ a+bx^k = u^{\frac{1}{n}}$$
Therefore,
$$ u^3 = 1 + x^2 \iff x = \sqrt{u^3 +1} \text{ (Mistake Here. Check the comments) } $$ $$\iff dx = \frac{3u^2}{2\sqrt{u^3}+1}$$
So the new integral is,
$$ \int x^5 (1+x^2)^{\frac{2}{3}} dx = \frac{3}{2} \int u^4 (u^3+1)^{\frac{7}{6}} du$$

Instead of simplifying the integral, the substitution did nothing by keeping it at the same form, with different values on the variables $m,k,n$.
I tried to substitute once again and it doesn't seem to lead in any known paths, anytime soon.
Any ideas on how this could be evaluated?

HINT:
Let $1+x^2=t^3\implies 2xdx=3t^2dt$ or $xdx=\frac{3}{2}t^2dt$
$$\int x^5(1+x^2)^{2/3}dx=\int (x^2)^2(1+x^2)^{2/3}xdx$$
$$=\int (t^3-1)^2(t^3)^{2/3}\ \frac{3t^2}{2}dt$$
$$=\frac32\int(t^3-1)^2t^4 dt $$
$$=\frac{3}{2}\int (t^{10}-2t^7+t^4)dt$$
