Complete Elliptic Integral Of The First Kind?

First apply the following substitution $\theta=\arctan{\left(\frac{1}{2}\left(\frac{x^{-1/8}-x^{1/8}}{2^{1/4}}-\frac{2^{1/4}}{x^{-1/8}-x^{1/8}}\right)\right)}$ It is easy to check that x runs from 0 to 1 and the integral becomes $\frac{\sqrt{\sqrt{2}+1}}{2^{13/4}}\int_0^1\frac{x^{-7/8}+x^{-5/8}}{\sqrt{x+1}}dx=\frac{\sqrt{\sqrt{2}+1}}{2^{13/4}}\int_0^\infty\frac{x^{-7/8}}{\sqrt{x+1}}dx$ Then the claim follows by the following formula $\int_0^\infty \frac{x^{a-1}}{(1+x)^b}dx=\frac{\Gamma(a)\Gamma(b-a)}{\Gamma(b)}$ then $\frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{d\theta}{\sqrt{1-(\sqrt{2}-1)^2\sin^2{\theta}}}=\frac{\sqrt{\sqrt{2}+1}}{2^{13/4}\sqrt{\pi}}\Gamma\left(\frac{1}{8}\right)\Gamma\left(\frac{3}{8}\right)$ and $A+B+C+D+E+F=2+1+13+4+8+3=\fbox{31}$