Integrals identity By numerical integration I found the identity
$$ \int_{-\infty}^{0}\frac{db}{\sqrt{R_4(b,W)}} = 2\int_{(W+1)^2}^{\infty}\frac{db}{\sqrt{R_4(b,W)}} $$
where $R_4(b,W)= b \ (b-4) \ (b-(W-1)^2) \ (b-(W+1)^2) $ and $W>3$.
Now I would like to prove that formally but I can't find the right way. I started by substituting
$$ b \rightarrow t=-b $$ and
$$ t \rightarrow x=t+(W+1)^2 $$
to have the same integration range on both sides, but $R_4(b,W)$ becomes quite strange.

Let the roots of $R_4$ be $a=(W+1)^2,b=(W-1)^2,c=4,d=0$. Byrd and Friedman 251.00 and 258.00 resolves both sides in terms of elliptic integrals as follows:
$$gF(\varphi,m)=2gF(\psi,m)$$
where $g=\frac2{\sqrt{(a-c)(b-d)}}$, $m=k^2=\frac{(b-c)(a-d)}{(a-c)(b-d)}=\frac{(W-3)(W+1)^3}{(W+3)(W-1)^3}$, $\sin^2\varphi=\frac{a-c}{a-d}=1-\frac4{(W+1)^2}$ (hence $\cos\varphi=\frac2{W+1}$) and $\sin^2\psi=\frac{b-d}{a-d}=\frac{(W-1)^2}{(W+1)^2}$ (hence $\cos\psi=\frac{2\sqrt W}{W+1}$). Thus it remains to show $F(\varphi,m)=2F(\psi,m)$. Applying the $F$ addition formula to the RHS (see DLMF 19.11.14) gives
$$\sin\varphi=\frac{2\sin\psi\cos\psi\sqrt{1-m\sin^2\psi}}{1-m\sin^4\psi}=\frac{\sqrt{(W-1)(W+3)}}{W+1}$$
$$\sin^2\varphi=\frac{(W-1)(W+3)}{(W+1)^2}=1-\frac4{(W+1)^2}$$
which matches the expression we obtained for $\sin^2\varphi$ earlier. So the equality is true.
