Solve the integral $\int{\frac{x^2dx}{\sqrt[n]{x^2(3-x)}}}$

My attempt: $$\int{\frac{x^2dx}{\sqrt[n]{x^2(3-x)}}}=\int{\frac{x^2dx}{\sqrt[n]{\left( \frac{3-x}{x} \right)x^3 }}}$$ Let $$t^n = \frac{3-x}{x} ~~\rightarrow~~ nt^{n-1}dt=-3\frac{dx}{x^2}$$ or $$x = \frac{3}{t^n+1} ~~ \rightarrow~~dx=-\frac{3nt^{n-1}}{(t^n+1)^2}dt$$ But this substitution is not helping me.

Let rearrenge the integrand as $$x^{2-\frac{2}{k}}(3-x)^{-1/k}$$. As we know, the integral of this binomial differentian can be done only in $4$ cases. Considering the link, when $p$ in not an integer so we think of $\frac{m+1}{n}$ or $\frac{m+1}{n}+p$. These two should be an integer otherwise the integral cannot be expressed as elementary functions. Now put $m=2-2/k$ and $n=1$ and $p=-1/k$ in the formula inside the link page and think of possible way of solving the integral.
Let $x=3z$