New $_2F_1$ identity?

Question

The following identity arises in (a new) derivation of the distribution of the estimate $r$ of the binormal correlation coefficient $\rho$. Here formulated in terms of $x = r\rho$.

For $-1 < x < 1$, we have:

$\qquad{}_2\mathrm{F}_1\big(a,b ; \mbox{$\frac{a+b+1}{2} ; \frac{x+1}{2}$}\big) = $

$\qquad\qquad\large \frac{\sqrt{\pi}\hspace{3pt}\Gamma\big(\mbox{$\normalsize\frac{a+b+1}{2}$}\big)}{\Gamma\big(\mbox{$\normalsize\frac{a+1}{2}$}\big)\Gamma\big(\mbox{$\normalsize\frac{b+1}{2}$}\big)}\normalsize \hspace{1pt} \mbox{${}_2\mathrm{F}_1\big(\frac{a}{2},\frac{b}{2} ; \mbox{$\frac{1}{2} ; x^2$}\big)$} \hspace{2pt} \hspace{2pt} + \hspace{2pt} \large \frac{2\sqrt{\pi}\hspace{3pt}\Gamma\big(\mbox{$\normalsize\frac{a+b+1}{2}$}\big)}{\Gamma\big(\mbox{$\normalsize\frac{a}{2}$}\big)\Gamma\big(\mbox{$\normalsize\frac{b}{2}$}\big)}\hspace{1pt}\normalsize x\hspace{2pt}\mbox{${}_2\mathrm{F}_1\big(\frac{a+1}{2},\frac{b+1}{2} ; \mbox{$\frac{3}{2} ; x^2$}\big)$} $

This can be written:

$\qquad\frac{1}{B\big(\mbox{$\frac{1}{2}$},\mbox{$\frac{a+b+1}{2}$}\big)}\hspace{1pt} \normalsize{}_2\mathrm{F}_1\big(a,b ; \mbox{$\frac{a+b+1}{2} ; \frac{x+1}{2}$}\big) = $ $\qquad\qquad\qquad\frac{1}{B\big(\mbox{$\frac{a+1}{2}$},\mbox{$\frac{b+1}{2}$}\big)} \hspace{1pt} \mbox{${}_2\mathrm{F}_1\big(\frac{a}{2},\frac{b}{2} ; \mbox{$\frac{1}{2} ; x^2$}\big)$} \hspace{2pt} + \hspace{2pt} \frac{\Large a+b}{B\big(\mbox{$\frac{a}{2},\frac{b}{2}$}\big)} \hspace{1pt}\hspace{1pt}x\hspace{2pt}\mbox{${}_2\mathrm{F}_1\big(\frac{a+1}{2},\frac{b+1}{2} ; \mbox{$\frac{3}{2} ; x^2$}\big)$} $

where $B(a,b)$ is the Beta function.

Is this a known identity? It doesn't appear in A&S or G&R. From the identities listed in Abramowitz and Stegun, it seems to be a combination of a linear and quadratic transformation. Can it be derived by such transformations?

I think the identity can be shown by making a linear change of variable in the hypergeometric differential equation, showing that $f(x)={}_2\mathrm{F}_1\big(a,b ; \mbox{$\frac{a+b+1}{2} ; \frac{x+1}{2}$}\big)$ satisfies:

$$(1-x^2)f''(x) - (a+b+1)\,x f'(x) - a\hspace{1pt}b f(x) = 0$$

for $-1<x<1$. Then showing that the right hand side also satisfies the differential equation, and has the same value and derivative at $x=0$ as the left hand side.

It follows immediately that for $-1 < x < 1$, we have:

$\qquad{}_2\mathrm{F}_1\big(a,b ; \mbox{$\frac{a+b+1}{2} ; \frac{1+x}{2}$}\big) \hspace{2pt}+\hspace{2pt} {}_2\mathrm{F}_1\big(a,b ; \mbox{$\frac{a+b+1}{2} ; \frac{1-x}{2}$}\big) \hspace{2pt} = \hspace{2pt}\large \frac{2\hspace{0.5pt}B\big(\mbox{$\normalsize\frac{1}{2}$},\mbox{$\normalsize\frac{a+b+1}{2}$}\big)}{B\big(\mbox{$\normalsize\frac{a+1}{2}$},\mbox{$\normalsize\frac{b+1}{2}$}\big)} \normalsize \hspace{1pt} \mbox{${}_2\mathrm{F}_1\big(\frac{a}{2},\frac{b}{2} ; \mbox{$\frac{1}{2} ; x^2$}\big)$}$

which can be used to simplify the distribution $r^2$.