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?

Take $x^5 \times (1 + x^2)^{\frac{2}{3}} = (x^{15} \times (1+x^2)^2)^{\frac{1}{3}}$ or some other form of that sort and integrate.
Thus, you'll be integrating $(x^{19} + 2x^{17} + x^{15})^{\frac{1}{3}}$
Applying substitution method here may work.
