How to evaluate $\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x$? Stuck by the integral
$$\int_{0}^{1} \sqrt{\frac{x}{1-x^{3}}} d x,$$
I finally solve the the problem using Beta Function.
Then I generalize the result to
$$
I(m,n)=\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x
$$
can be tackled by the Beta function by letting $$
y=x^{n} \text {, then } x=y^{\frac{1}{n}} \textrm{ followed by} \quad d x=\frac{1}{n} y^{\frac{1}{n}-1} d y
$$
$$
\begin{aligned}
I &=\int_{0}^{1} y^{\frac{m}{2 n}}(1-y)^{-\frac{1}{2}} \cdot\frac{1}{n} y^{\frac{1}{n}-1} d y \\
&=\frac{1}{n} \int_{0}^{1} y^{\frac{m+2}{2 n}-1}{(1-y)^{\frac{1}{2}-1}} d y \\
&=\frac{1}{n} B\left(\frac{m+2}{2 n} , \frac{1}{2}\right) \\
&=\frac{\Gamma\left(\frac{m+2}{2 n}\right) \Gamma\left(\frac{1}{2}\right)}{n\left(\frac{m+2+n}{2 n}\right)}\\&=\frac{\sqrt{\pi} \Gamma\left(\frac{m+2}{2 n}\right)}{n \Gamma\left(\frac{m+2}{2 n}+\frac{1}{2} \right)}
\end{aligned}
$$
For example, $$
\begin{aligned}
I(1,3) &=\frac{1}{3} B\left(\frac{1}{2}, \frac{1}{2}\right) \\
&=\frac{1}{3} \Gamma^{2}\left(\frac{1}{2}\right) \\
&=\frac{\pi}{3}
\end{aligned}
$$
My Question:
Can we simplify the last answer?  Your suggestion or help are highly appreciated.
 A: $$\int \sqrt{\frac{x^{m}}{1-x^{n}}}\, dx=\frac 2{2+m}x^{\frac{m}{2}+1}  \, _2F_1\left(1,\frac{m+n+2}{2 n};\frac{m+2n+2}{2
   n};x^n\right)$$
$$\int_0^1 \sqrt{\frac{x^{m}}{1-x^{n}}}\, dx=\frac{\sqrt{\pi }\, \Gamma \left(\frac{m+2}{2 n}\right)}{n\, \Gamma
   \left(\frac{m+n+2}{2 n}\right)} \quad \text{if} \quad \Re(n)>0\land \Re(m)>-2$$
Making the problem more general
$$\int\left(\frac{x^m}{1-x^n}\right)^p\,dx=\frac 1{1+mp} x^{mp+1}\, _2F_1\left(p,\frac{m p+1}{n};\frac{m p+n+1}{n};x^n\right)$$
$$\int_0^1 \left(\frac{x^m}{1-x^n}\right)^p\,dx=\frac{\Gamma (1-p)\, \Gamma \left(\frac{m p+1}{n}\right)}{n\,\Gamma
   \left(\frac{(m-n)p+n+1}{n}\right)}\quad \text{if} \quad \Re(p)<1\land \Re(m p)>-1\land \Re(n)>0$$
A: Your first integral has an elementary antiderivative.
$$\int_0^1\sqrt{\frac{x}{1-x^3}}=\int_0^1\frac{x^\frac{1}{2}}{\sqrt{1-(x^\frac{3}{2})^2}}dx=\frac{2}{3}\int_0^1\frac{d(x^\frac{3}{2})}{\sqrt{1-(x^\frac{3}{2})^2}}=\frac{2}{3}\arcsin(x^\frac{3}{2})\Bigg|_0^1=\frac{2}{3}\frac{\pi}{2}=\frac{\pi}{3}$$
I doubt there is any simplification from that ratio of gammas, but we could convert it into a cool product form if m and n are integers using the definition of the gamma function
