Correction terms in the asymptotic expansion of hypergeometric function I am interested in obtaining the asymptotic expansion of $r(\rho)$ (which is the inverse of $\rho$ below),
$$\rho=\frac{2b}{1-q}\left(1-\left(\frac br\right)^{1-q}\right)^{1/2}\left(_2F_1\left(\frac{1}{2},1-\frac{1}{q-1},\frac{3}{2},1-\left(\frac br\right)^{1-q}\right)\right)$$
where $b$ is just some positive constant, while $-\infty<q<1$.
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 ${}_2F_1$.
Basically I got,
$$\rho\sim r\sqrt{1-(b/r)^{1-q}}$$
as the leading term in the expansion. I am not that sure what to pick as the next two correction terms from the link given the allowed values $-\infty<q<1$. Is there a detailed easy way of finding the next two correction terms in the expansion, given the allowed values of $q$? Thank you.
 A: Since for all $\alpha\in\mathbb{R}$ and $z\in(0,1)$, by Euler integral representation,
\begin{align*}
f(\alpha,z)&:=(1-z)^{1/2} {_2F_1}\left(\frac{1}{2},1+\alpha;\frac{3}{2};1-z\right)\\
&=\frac{(1-z)^{1/2}}{2}\int_{0}^1x^{-1/2}(1-(1-z)x)^{-1-\alpha}\,dx\\
&=\frac{1}{2}\int_{z}^{
1}\frac{v^{-1-\alpha}}{\sqrt{1-v}}\,dv=\frac{1}{2}\sum_{k\ge 0}\frac{(1/2)_k}{k!}\int_{z}^1\frac{\,dv}{v^{1-k+\alpha}}\\
&=\sum_{\substack{k\ge 0\\ k\neq \alpha}}\frac{(1/2)_k}{2k!(k-\alpha)}\left(1-z^{k-\alpha}\right)-{\bf 1}_{\alpha\in\mathbb{N}}\frac{(1/2)_{\alpha}}{2\alpha!}\log z.
\end{align*}
Namely, for all $z\in(0,1)$,
$$f(\alpha,z)=C_{\alpha}-{\bf 1}_{\alpha\in\mathbb{N}}\frac{(1/2)_{\alpha}}{2\alpha!}\log z-\sum_{\substack{k\ge 0\\ k\neq \alpha}}\frac{(1/2)_k}{2k!(k-\alpha)}z^{k-\alpha}$$
with
$C_{\alpha}=\sum_{\substack{k\ge 0\\ k\neq \alpha}}\frac{(1/2)_k}{2k!(k-\alpha)}$; it is a convergence series expansion. 
In particular, if $\alpha\in(0,1)$,
\begin{align}
f(\alpha,z)&=\sum_{k\ge 0}\frac{(1/2)_k}{2k!(k-\alpha)}-\sum_{k\ge 0}\frac{(1/2)_k}{2k!(k-\alpha)}z^{k-\alpha}\\
&=\frac{1}{2\alpha}z^{-\alpha}+\sum_{k\ge 0}\frac{(1/2)_k}{2k!(k-\alpha)}-\frac{1}{4(1-\alpha)}z^{1-\alpha}+O(z^{2-\alpha}).
\end{align}
Now by set $\alpha=1/(1-q)$ and $z=(b/r)^{1-q}$ with $b>0$, and if $q<0$ then
\begin{align}
\rho
&=\frac{2b}{1-q}\left(\frac{1-q}{2b}r+\frac{1-q}{2}\sum_{k\ge 0}\frac{(1/2)_k}{k!((1-q)k-1)}+\frac{1-q}{4q}\left(\frac{b}{r}\right)^{-q}+O\left(\left(\frac{b}{r}\right)^{1-2q}\right)\right)\\
&=r+b\sum_{k\ge 0}\frac{(1/2)_k}{k!((1-q)k-1)}+\frac{b^{1-q}}{2q}r^q+O(r^{2q-1}).
\end{align}
