This is an integral from a chat group: $$ \int_{0}^{1} \frac{(1+5x^2)\arctan(x)^2}{x^{5/4}(1-x^2)^{1/4}} \text{d}x =\frac{\sqrt{\sqrt{2}+1 } }{4}\left( \left ( \sqrt{2 }+1 \right )\pi^2+2\left ( \sqrt{2}-1 \right ) (8-3\ln(2))\pi-\frac{4\sqrt{2}\,\Gamma\left ( \frac34 \right )^2 }{\sqrt{\pi} } \left ( 2\ln(2)+\left ( 2\sqrt{2} -1\right ) \pi\right ) \right ). $$ What I already know is that(integrating by parts) $$ \int_{0}^{1} \frac{(1+5x^2)\arctan(x)^2}{x^{5/4}(1-x^2)^{1/4}} \text{d}x=8\int_{0}^{1} \frac{(1-x^2)^{3/4}\arctan(x)}{x^{1/4}(1+x^2)} \text{d}x, $$ but I don't know how to proceed. So any help would be highly appreciated.
(June 5th,23) Expanding $\arctan(x)/(1+x^2)$ into power series, denoting $$ \mathscr{H}_n=\sum_{k=1}^n\frac1{2k-1} $$, we have $$ \int_{0}^{1} \frac{(1-x^2)^{3/4}\arctan(x)}{x^{1/4}(1+x^2)} \text{d}x=\frac{3}{8}\Gamma\left ( \frac34 \right ) \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\mathscr{H}_n\,\Gamma\left ( n-\frac18 \right ) }{ \Gamma\left ( n+\frac{13}8 \right ) }. $$