<s>[INST] Calculating in closed form $\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$ It's not hard to see that for powers like $1,2$, we have a nice closed form. What can be said about
the cubic version, that is 
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$$
What are your ideas on it? Differentiation under the integral sign? Other ways?
Mathematica 9 says that 
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx=\frac{2}{3} \, _5F_4\left(\frac{1}{2},\frac{2}{3},1,1,\frac{4}{3};\frac{5}{6},\frac{7}{6},\frac{3}{2},\frac{3}{2};-1\right).$$
 [/INST] Just exploiting the arctangent Taylor series,
$$\begin{eqnarray*} \int_{0}^{\frac{\pi}{2}} \arctan(\sin^3 x)\,dx &=& \sum_{m\geq 0}\frac{(-1)^m}{2m+1}\int_{0}^{\frac{\pi}{2}}\sin^{6m+3}(x)\,dx\\ &=& \frac{3\sqrt{\pi}}{4}\sum_{m\geq 0}\frac{(-1)^m\,\Gamma\left(3m+2\right)}{(3m+\frac{3}{2})\Gamma\left(3m+\frac{5}{2}\right)}\end{eqnarray*} $$
so, at least in principle, we may compute the RHS by applying a discrete Fourier transform to the power series:
$$ \sum_{m\geq 0}\frac{x^m\, \Gamma(m+2)}{\left(m+\frac{3}{2}\right)\Gamma\left(m+\frac{5}{2}\right)}$$
that is just a $\phantom{}_{3} F_2$ hypergeometric function, namely $\frac{8}{9\sqrt{\pi}}\;\phantom{}_3 F_2\!\left(1,\frac{3}{2},2;\frac{5}{2},\frac{5}{2};x\right).$
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