Integrating $\int_0^{\frac{\pi}{2}} x (\log\tan x)^{2n+1}\;dx$

By setting $x=\arctan u$ we are left with

$$ \mathcal{I}(n) = \int_{0}^{+\infty}\frac{\arctan u}{1+u^2}\left(\log u\right)^{2n+1}\,du =\left.\frac{d^{2n+1}}{d\alpha^{2n+1}}\int_{0}^{+\infty}\frac{u^\alpha \arctan u}{1+u^2}\,du\right|_{\alpha=0}$$ but the integral in the RHS is related to the Beta function. By un-doing the previous substitution, $$ \int_{0}^{+\infty}\frac{u^\alpha \arctan u}{1+u^2}\,du = \int_{0}^{\pi/2}x\left(\sin x\right)^{\alpha}\left(\cos x\right)^{-\alpha}\,dx $$ where we may write $x$ as $$ \arcsin(\sin x)= \sum_{n\geq 0}\frac{(\sin x)^{2n+1}}{(2n+1)4^n}\binom{2n}{n}$$ leading to: $$ \int_{0}^{+\infty}\frac{u^{\alpha}\arctan u}{1+u^2}\,du=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n}\cdot\frac{\Gamma\left(\frac{1}{2}-\frac{a}{2}\right)\, \Gamma\left(1+\frac{a}{2}+n\right)}{2\,\Gamma\left(\frac{3}{2}+n\right)}. $$ Now "it is enough" to differentiate both sides with respect to $\alpha$ the correct number of times and perform an evaluation at $\alpha=0$ in order to convert the original integral in a "twisted hypergeometric series", whose terms depend both on hypergeometric terms and generalized harmonic numbers (arising from the differentiation of the Beta function).