Integrate $\int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}$

Hint. By the change of variable $$ \sqrt{\tan x}=t,\quad x=\arctan (t^2),\quad dx=\frac{2t\:dt}{1+t^4}, $$ one gets

$$ \int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}=\int_0^\infty \frac{4t^4\ln t}{\left(1+t^4\right)^2}\:dt=\frac{\pi \sqrt{2}}{4}-\frac{\pi ^2\sqrt{2}}{16}. \tag1 $$

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Addendum. One may recall the identity $$ \int_0^\infty\frac{x^\alpha}{1+x}\:dx=-\frac{\pi}{\sin \alpha \pi}, \quad -1<\alpha<0, \tag2 $$ giving, with an integration by parts, $$ \int_0^\infty\frac{x^\alpha}{(1+x)^2}\:dx=\frac{\alpha \pi}{\sin \alpha \pi}, \quad -1<\alpha<1 \tag3 $$ differentiating $(3)$ yields $$ \int_0^\infty\frac{x^\alpha\ln x}{(1+x)^2}\:dx=\frac{\pi}{\sin \alpha \pi}-\frac{\pi^2\alpha \cos \alpha \pi}{\sin^2 \alpha \pi}, \quad -1<\alpha<1, \tag4 $$ then putting $x=t^4$, $\alpha=\frac14$ gives $(1)$.