Calculating integrals of the form $\int\frac{dx}{(1+x^{q})^\frac{p}{r}}$

Mathematica gives the following nasty result $$\int \frac{dx}{(1+x^q)^{p/r}}=x\cdot {_2}F_1\left[\frac{1}{q},\frac{p}{r};1+\frac{1}{q};-x^q\right]+C,$$ where ${_2}F_1$ is a hypergeometric function. With inspiration from @TymaGaidash's comment on my previous edit containing the derivation for the definite integral on $[0,\infty)$, one can obtain a neater result in terms of the incomplete beta function $$\int \frac{dx}{(1+x^q)^{p/r}}=\frac{1}{q}B\left(\frac{x^q}{x^q+1};\frac{1}{q},\frac{p}{r}-\frac{1}{q}\right)+C.$$

Note that by the substitution $y=x^q$ one has $$\int \frac{1}{(1+x^q)^{p/r}}~dx=\frac{1}{q}\int \frac{y^{1/q-1}}{(1+y)^{p/r}}~dy.$$ By the substitution $t=\frac{y}{y+1}$, one has $$\begin{align} \frac{1}{q}\int \frac{y^{1/q-1}}{(1+y)^{p/r}}~dy&=\frac{1}{q}\int t^{1/q-1} (1-t)^{p/r-1/q-1}~dt\\&=\frac{1}{q}B\left(t;\frac{1}{q},\frac{p}{r}-\frac{1}{q}\right)+C\\&=\frac{1}{q}B\left(\frac{x^q}{x^q+1};\frac{1}{q},\frac{p}{r}-\frac{1}{q}\right)+C,\end{align}$$ which is what we wanted to prove. In the case that the integral is on $[0,\infty)$ one has the following neat closed form in terms of the $\Gamma$ function (under suitable conditions for convergence) $$\int_0^{\infty} \frac{dx}{(1+x^q)^{p/r}}=\frac{1}{q}B\left(\frac{1}{q},\frac{p}{r}-\frac{1}{q}\right)=\frac{\Gamma(1+1/q)\Gamma(p/r-1/q)}{\Gamma(p/r)}.$$