I am interested in obtaining the asymptotic expansion of $r(\rho)$ (which is the inverse of $\rho$ below). Basically I want to series expand $\rho$ for large $r$ (i.e. as $r\to \infty$) and then invert the series to obtain $r(\rho)$. I have tried some readily available asymptotic expansion of $_2F_1$ but I think I ruined my calculations and I am not really sure if I am doing the right thing. By the way, $b$ is just some positive constant while $-\infty<q<1$.
$$\rho=\frac{2b}{1-q}\left(1-\left(\frac br\right)^{1-q}\right)^{1/2}\left(_2F_1\left(\frac{1}{2},1-\frac{1}{q-1},\frac{3}{2},1-\left(\frac br\right)^{1-q}\right)\right)$$
We need an asymptotic for ${_2\hspace{-1px}F_1}(a, b; c; z)$ with $z \to 1^-$. Since $c - a - b = 1/(q - 1) < 0$, the leading term is $$\frac {\Gamma(c) \Gamma(a + b - c)} {\Gamma(a) \Gamma(b)} (1 - z)^{c - a - b}.$$ This gives $\rho \sim r \sqrt{1 - (b/r)^{1 - q}}$, therefore $\rho \sim r$ and $r \sim \rho$.
For $q < 0$, the next term in the expansion of ${_2\hspace{-1px}F_1}$ is a constant, and we need only the first term from the $(1 - z)^{1/2}$ factor: $${_2\hspace{-1px}F_1} {\left( \frac 1 2, p; \frac 3 2; 1 - z \right)} = \frac {z^{1 - p}} {2 (p - 1)} + \frac {\sqrt \pi \,\Gamma(1 - p)} {2 \Gamma {\left( \frac 3 2 - p \right)}} + \dots, \\ \rho = r + \frac {b \sqrt \pi \,\Gamma(1 - p)} {(1 - q) \Gamma {\left( \frac 3 2 - p \right)}} + \dots, \\ r = \rho - \frac {b \sqrt \pi \,\Gamma(1 - p)} {(1 - q) \Gamma {\left( \frac 3 2 - p \right)}} + \dots,$$ where $p = (2 - q)/(1 - q)$. For $q > 0$ and excluding the logarithmic case $p \in \mathbb N$, the next term in the expansion of ${_2\hspace{-1px}F_1}$ is of order $z^{2 - p}$, and we need two terms from $(1 - z)^{1/2}$: $${_2\hspace{-1px}F_1} {\left( \frac 1 2, p; \frac 3 2; 1 - z \right)} = \frac {z^{1 - p}} {2 (p - 1)} + \frac {(2 p - 3) z^{2 - p}} {4 (p - 1) (p - 2)} + \dots, \\ \rho = r + \frac {b^{1 - q} r^q} {2 q} + \dots, \\ r = \rho - \frac {b^{1 - q} \rho^q} {2 q} + \dots,$$ the last step coming from the Lagrange reversion theorem.