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.
 A: 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.
A: Let $x=3z$
\begin{align}
\int \frac{x^{2-2/n}}{(3-x)^{1/n}} dx &= 3^{-1/n} \int \frac{x^{2-2/n}}{(1-x/3)^{1/n}} dx \\
&= 3^{3-3/n} \int z^{2-2/n} (1-z)^{-1/n} dz \\
&= 3^{3-3/n} \mathrm{B}_{z} \left( 3-\frac{2}{n}, 1-\frac{1}{n} \right) \\
&= 3^{3-3/n} \frac{n}{3n-2} z^{3-2/n} {}_{2}\mathrm{F}_{1}\left(3-\frac{2}{n},\frac{1}{n};4-\frac{2}{n};z \right) \\
&= \frac{1}{3^{1/n}} \frac{n}{3n-2} x^{3-2/n} {}_{2}\mathrm{F}_{1}\left(3-\frac{2}{n},\frac{1}{n};4-\frac{2}{n};\frac{x}{3} \right)
\end{align}
Note:
\begin{align}
\mathrm{B}_{z}(p,q) &= \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t \\
&= \frac{z^{p}}{p} \,{}_{2}\mathrm{F}_{1}(p,1-q;p+1;z)
\end{align}
The incomplete beta function and hypergeometric function.
