Basically, I have obtained the function $\rho (r)$ below as a result of integrating
$$\rho(r)=\int_{b}^{r}\frac{dx}{\sqrt{1-(b_{0}/x)^{1-q}}}$$ which results to $$\rho=\frac{2b}{1-q}\sqrt{1-\left(\frac br\right)^{1-q}}\,_2F_1\left(\frac{1}{2},\frac{q-2}{q-1},\frac{3}{2},1-\left(\frac br\right)^{1-q}\right)$$ where $b$ is just some positive constant while $-\infty<q<1$.
What I need is $r(\rho)$ which is the inverse of the function above (which I can implement with numerics). I am also interested in the series expansion of $r(\rho)$ at $\infty$ but unknowingly, the hypergeometric function above has a branch cut at $1-\left(\frac br\right)^{1-q}=1$ (i.e. $r\rightarrow \infty$). My question would be, am I allowed to expand $r(\rho)$ at $\infty$ given that there is a branch cut at infinity? What are the possible complications? What can I possibly do to safely obtain an asymptotic expansion of $r(\rho)$ at $\infty$? I am also thinking of converting the integral above into a differential equation for $r(\rho)$ and expand it at infinity to bypass the hypergeometric function branch cut problem but I am not really sure of this. Thanks for the help
$$\rho(r)=r+a+\sum_{n\ge 1} {-1/2\choose n} \frac{b_0^{n(1-q)}}{1+n(q-1)} r^{1+n(q-1)}\in r \ \Bbb{R}[[r^{-1},r^q]]^\times$$
If $q <-1$ then $$\rho(r)^{-1}=r^{-1} u, \qquad u\in \Bbb{R}[[r^{-1},r^q]]^\times$$ Note that $u^{-q}=\sum_{k\ge 0} {-q\choose k} (u-1)^k\in \Bbb{R}[[r^{-1},r^q]]$.
Thus $$\phi(f(r^{-1},r^q))=f(\rho(r)^{-1},\rho(r)^q)$$ is an automorphism of $\Bbb{R}[[r^{-1},r^{q-1}]]$, there is some $u\in \Bbb{R}[[r^{-1},r^q]]$ such that $\phi(u)=r^{-1}$.
$u^{-1}$ is your asymptotic expansion of $\rho^{-1}(r)$.