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?

Do the substitution $u^{3/5} = 1+ x^2$. Then $x = \sqrt{u^{3/5}-1}$, so $\mathrm{d}x = \frac{3\,u^{-2/5}}{5\,\sqrt{u^{3/5}-1}}\,\mathrm d u$ and
$$
∫ x^5 \,(1+x^2)^{2/3}\,\mathrm d x = \frac{3}{5}∫ (u^{3/5}-1)^2 \,\mathrm d x.
$$
