Asymptotic expansion at infinity given a branch cut

Question

Basically, I have obtained the function $\rho (r)$ below as a result of integrating

$$\rho(r)=\int_{b_0}^{r}\frac{dx}{\sqrt{1-(b_{0}/x)^{1-q}}}$$ which results to

 b0 = 1;
 rho[r_] := (2 b0)/(1 - q) Sqrt[1 - (b0/r)^(1 - q)] Hypergeometric2F1[1/2, (q - 2)/(q - 1), 3/2, 1 - (b0/r)^(1 - q)];

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

Answer 1

The Mathematica 12.2 code

b0 = 1; AsymptoticIntegrate[1/Sqrt[1-(b0/x)^(1 - q)],{x, b0, r},{r,Infinity,1}, Assumptions->q<1]

performs

$$\frac{r^q}{2 q}+\frac{3 (q-1) r^{2 q-1}}{8 q (2 q-1)}+\frac{3 r^{2 q-1}}{8 q (2 q-1)}-\frac{\sqrt{\pi } \Gamma \left(\frac{q}{q-1}\right)}{\Gamma \left(\frac{1}{2}+\frac{1}{q-1}\right)}+r $$