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I will be honest about it: the following integral is involved in my current investigations about the interplay between Fourier-Legendre series expansions, hypergeometric functions of the $\phantom{}_3 F_2$ and $\phantom{}_4 F_3$ kind and polylogarithms. Significative contributions will be rewarded through a proper mention of the author.

The value of $\phantom{}_3 F_2\left(\tfrac{1}{2},\tfrac{3}{4},1;\tfrac{5}{4},\tfrac{3}{2};-1\right)$ is related to the integral $$ \mathcal{J}_1 = \int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx, $$ which on its turn (by Feynman's trick) is related to $$ \mathcal{J}_2 = \int_{0}^{1}\frac{1-\phantom{}_2 F_1\left(-\frac{1}{4},1;\frac{1}{4};-x^2\right)}{x^2}\,dx.$$ I would be glad to know how to convert $\mathcal{J}_1$ (or $\mathcal{J}_2$) into a combination of standard mathematical constants and polylogarithms. It most certainly can be done since $\arctan(x)=\text{Im}\log(1+ix)$, but my version of Mathematica is not collaborating with me as I wished.

I will be away from keyboard for a few days (time to work for the Italian most celebrated mathematical contest), I hope to find some nice insight after my return.

Jack D'Aurizio
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    Never mind, I solved it. $$ \mathcal{J}_1 = \tfrac{1}{16},\sqrt{\tfrac{\pi}{2}},\Gamma\left(\tfrac{1}{4}\right)^2 $$ by a nasty symmetry trick, which, as usual, is not recognized by Mathematica. I do not spoil the steps of my solution since you might have some fun with this problem. – Jack D'Aurizio May 02 '18 at 02:08
  • My friend sure did :D – Mr Pie Jun 18 '18 at 04:43

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