Consider this integral
Motivated by this question
$$\int_{0}^{1}{\sqrt{x^n\over 1-x^m}}\mathrm dx=F(n,m)\tag1$$ Where $n\ge -1$ and $m\ge 1$
We note the following values for $F(n,m):$
$$F(-1,1)=\pi$$
$$F(1,1)={\pi\over 2}$$
$$F(1,3)={\pi\over 3}$$
Where $k $ is an integer, what are other values of $ n$ and $m$ that will give ${\pi\over k}$?
I guess to answer this question we have to find the closed form for $(1)$
Hint. This is the Euler beta function in disguise, by the change of variable $$u=x^m, \quad x=u^{1/m}, \quad dx=\frac1m\cdot u^{1/m-1}du, $$ one gets, for $n\ge-1,\, m>0,$
$$ \begin{align} \int_{0}^{1}{\sqrt{x^n\over 1-x^m}}\:\mathrm dx&=\int_{0}^{1}(1-u)^{-\large\frac12}u^{\large\frac{n+2}{2m}-1}\:du=\frac{\sqrt{\pi }\:\Gamma \left(\frac{n+2}{2m}\right)}{m\: \Gamma \left(\frac{m+n+2}{2 m}\right)},\quad \end{align} $$
then by properties of the gamma function one may simplify the preceding expression.
Mathematica says:
$$ \text{ConditionalExpression}\left[\frac{\sqrt{\pi } \Gamma \left(\frac{m+2}{2 n}\right)}{n \Gamma \left(\frac{m+n+2}{2 n}\right)},\Re(n)>0\land \Re(m)>-2\right] $$
