$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx$

I need help with calculating this integral: $$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx,$$ where $_pF_q$ is a generalized hypergeometric function.

I was told it has a closed-form representation in terms of elementary functions and integers.

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@Laila Maybe this hypergeometric function can be expressed in terms of elementary functions? Have you tried to find such representation? –  Liu Jin Tsai May 11 '13 at 22:47
Look here, starting from "An infinite family of rational values"... –  Myself May 11 '13 at 22:59

This hypergeometric function is not an elementary function, but its inverse is - see Bring radical.

$$\int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac12,\frac34,\frac54;\frac{x}{64}\right)\,dx=\\\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right),$$ where $\alpha$ is the positive root of the polynomial $625\alpha^4-500\alpha^3-100\alpha^2-20\alpha-4$. It can be expressed in radicals as follows:

$$\alpha=\frac15+\sqrt\beta+\sqrt{\frac15-\beta +\frac1{25\sqrt\beta}},$$ where $$\beta=\frac1{30}\left(\frac\gamma5-\frac4\gamma+2\right),$$ where $$\gamma=\sqrt[3]{15\sqrt{105}-125}.$$

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Of course, you can express $\pi$ as $4 \arctan 1$ to get rid of non-integer constants. –  Vladimir Reshetnikov May 11 '13 at 23:24
Commiserations to anyone who attempted this by hand. –  L. F. May 11 '13 at 23:32
That's so obvious, I an kicking myself (very softly) for not thinking of it. –  marty cohen May 12 '13 at 0:53
@L.F. I recalculated the integral completely by hand, and now the result looks much nicer. –  Vladimir Reshetnikov May 12 '13 at 5:52
This is a true marvel...! –  sos440 May 28 '13 at 2:15