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$\rm Lemma.$
$\D \int_0^x \frac{t^{2m}}{\sqrt{1-t^2}}\,dt=\frac{\binom{2m}{m}}{2^{2m}}\sum_{m< n}\frac{2^{2n}x^{2n-1}\sqrt{1-x^2}}{2n\binom{2n}{n}} $
$\BA\D
\sum_{0\le m< n}\frac{\binom{2m}{m}^2}{2^{4m}}\frac{2^{2n}x^{2n-1}}{2n\binom{2n}{n}}
&=\sum_{0\le m}\frac{\binom{2m}{m}}{2^{2m}}\frac{1}{\sqrt{1-x^2}}\int_0^x \frac{t^{2n}}{\sqrt{1-t^2}}\,dt\\
&=\frac{1}{\sqrt{1-x^2}}\int_0^x \frac{dt}{1-t^2}\\
&=\frac{\tanh^{-1}x}{\sqrt{1-x^2}}
\EA$
$\bf Theorem.$
$\D \frac{\tanh^{-1}x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty \frac{2^{2n}x^{2n-1}}{2n\binom{2n}{n}}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}}$
おまけ
$\BA\D
\frac{\pi}{8}\sum_{n=1}^\infty \frac{1}{n^3}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}}
&=\int_0^1 \frac{1}{\sqrt{1-z^2}}\int_0^z \frac{1}{x}\int_0^x \frac{\tanh^{-1}t}{\sqrt{1-t^2}}\,dt\,dx\,dz\\
&=\int_0^\frac{\pi}{2} \int_0^z\frac{1}{\tan x}\int_0^x \tanh^{-1}\sin t\,dt\,dx\,dz\\
&=\sum_{n=0}^\infty \frac{2(-1)^n}{2n+1}\int_0^\frac{\pi}{2} \int_0^z\frac{1}{\tan x}\int_0^x \sin(2n+1)t\,dt\,dx\,dz\\
&=\sum_{n=0}^\infty \frac{2(-1)^n}{2n+1}\int_0^\frac{\pi}{2} \int_0^z\frac{1}{\tan x}\frac{1-\cos(2n+1)x}{2m+1}\,dx\,dz\\
&=\sum_{n=0}^\infty \frac{2(-1)^n}{(2n+1)^2}\int_0^\frac{\pi}{2} \int_0^z \L(-\tan\frac{x}{2}+\sin(2n+1)x+2\sum_{m=0}^{n-1}\sin(2m+1)x \R)dx\,dz\\
&=\sum_{n=0}^\infty \frac{2(-1)^n}{(2n+1)^2}\int_0^\frac{\pi}{2}\L(2\ln\cos\frac{z}{2}+\frac{1-\cos(2n+1)z}{2n+1}+2\sum_{m=0}^{n-1}\frac{1-\cos(2m+1)z}{2m+1} \R)dz\\
&=\sum_{n=0}^\infty \frac{2(-1)^n}{(2n+1)^2}\L(2\beta(2)-\pi\ln2+\frac{1}{2n+1}\L(\frac{\pi}{2}-\frac{(-1)^n}{2n+1}\R)+2\sum_{m=0}^{n-1}\frac{1}{2m+1}\L(\frac{\pi}{2}-\frac{(-1)^m}{2m+1}\R)\R)\\
&=4\beta(2)^2-2\pi\beta(2)\ln2+\pi\beta(3)-2t(4)+2\pi\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}\sum_{m=0}^{n-1}\frac{1}{2m+1}-4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}\sum_{m=0}^{n-1}\frac{(-1)^m}{(2m+1)^2}\\
&=4\beta(2)^2-2\pi\beta(2)\ln2+\frac{\pi^4}{32}-2t(4)+2\pi\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}\sum_{m=0}^{n-1}\frac{1}{2m+1}-2\L(\beta(2)^2-t(4)\R)\\
&=2\beta(2)^2-2\pi\beta(2)\ln2+\frac{\pi^4}{32}+2\pi\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}\sum_{m=0}^{n-1}\frac{1}{2m+1}\\
\EA$