Principal value of Carlson elliptic integral $R_C$ $$R_C(x,y)={1\over2}\int_0^\infty\frac{dt}{(t+y)\sqrt{t+x}},\quad(x\ge0)$$
$R_C$ has a singularity in the denominator when $y$ is negative. Wikipedia says in that case $R_C$ can be evaluated as follows
$$
PV\;R_C(x,-y)=\sqrt{\frac{x}{x+y}}R_C(x+y,y),\quad(y>0)
$$

Q: How can we prove the above equality?

The integrand has an easily computable anti-derivative but I don't know if it's useful in this context. 

$$\newcommand{\PV}{\operatorname{PV}}
\begin{align}
R(x,y)
&=\frac12\,\PV\!\!\int_0^\infty\frac{\mathrm{d}z}{(z+y)\sqrt{z+x}}\tag{1}\\[6pt]
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}}^\infty\frac{2\,\mathrm{d}z}{z^2-1}\tag{2}\\[3pt]
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}}^\infty\left(\frac1{z-1}-\frac1{z+1}\right)\mathrm{d}z\tag{3}\\
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}\,-1}^{\sqrt{\frac{x}{x-y}}\,+1}\frac{\mathrm{d}z}z\tag{4}\\
&=\frac1{2\sqrt{x-y}}\int_{\left|\sqrt{\frac{x}{x-y}}\,-1\right|}^{\sqrt{\frac{x}{x-y}}\,+1}\frac{\mathrm{d}z}z\tag{5}\\
&=\bbox[5px,border:2px solid #C0A000]{\frac1{2\sqrt{x-y}}\log\left|\frac{\sqrt{x}+\sqrt{x-y}}{\sqrt{x}-\sqrt{x-y}}\right|}\tag{6}\\[6pt]
&=\sqrt{\frac{u-v}{u}}\frac1{2\sqrt{u-v}}\log\left|\frac{\sqrt{u-v}+\sqrt{u}}{\sqrt{u-v}-\sqrt{u}}\right|\tag{7}\\[6pt]
&=\sqrt{\frac{u-v}{u}}\,R(u,v)\tag{8}\\[9pt]
&=\sqrt{\frac{x}{x-y}}\,R(x-y,-y)\tag{9}
\end{align}
$$
Explanation:
$(2)$: substitute $z\mapsto(x-y)z^2-x$
$(3)$: partial fractions
$(4)$: substitute $z\mapsto z+1$ and $z\mapsto z-1$ and subtract integrals
$(5)$: Principal Value is easy with an odd function
$(6)$: evaluate integral
$(7)$: $u=x-y$ and $v=-y$
$(8)$: apply $(6)$ to $(7)$
$(9)$: undo the substitution in $(7)$
Substituting $y\mapsto-y$ in $(9)$ yields
$$
R(x,-y)=\sqrt{\frac{x}{x+y}}\,R(x+y,y)\tag{10}
$$
