Closed form of $ \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { x } } \log { (\cos { x } ) }\ dx }$ Does there exists a closed form of$$ \displaystyle \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { x }  } \log { (\cos { x } ) }\ dx }$$
If exists can someone find a way to tackle this integral and provide a closed-form of it. Many similar integrals have closed form and I believe this one, too. 

Substitute $t = \sqrt{\tan x}$
\begin{align}
I=\int_{0}^{\pi/2}x\sqrt{\tan{x}}\ln({\cos x})\ dx=J(1)\\
\end{align}
where $J(a)= -\int_0^\infty \frac{t^2 \ln(1+t^4) \tan^{-1}(a^2t^2)}{1+t^4}dt$
\begin{align}
J’(a)=&-\int_0^\infty \frac{2at^2 \ln(1+t^4)}{(1+t^4)(1+a^4t^4)}dt\\
=& \ \frac{2a}{1-a^4}\int_0^\infty
\left(\frac{\ln(1+t^4)}{1+t^4}- \frac{\ln(1+t^4)}{1+a^4t^4} \right)dt
\end{align}
Let $K(a,b)=\int_0^\infty \frac{\ln(1+b^4t^4)}{1+a^4t^4}dt$
\begin{align}
K_b’(a,b)=&\int_0^\infty \frac{4b^3 t^4}{(1+a^4t^4)(1+b^4t^4)}dt
=\frac{\sqrt2\pi b^2}{a(a+b)(a^2+b^2)}
\end{align}
Then
\begin{align}
K(a,b)= &\int_0^b K_b’(a,s)ds
= \frac\pi{\sqrt2 a}\left(\frac12\ln\frac{a^2+b^2}{a^2}
 +\ln\frac{a+b}a-\tan^{-1}\frac ba\right)\\
J’(a)=&\ \frac{2a}{1-a^4}\left[K(1,1)-K(a,1) \right]\\
=& \frac{\sqrt2\pi}{1-a^4}
\left[a\left(\frac32\ln2-\frac\pi4 \right)-\frac12 \ln\frac{1+a^2}{a^2} -\ln\frac{1+a}{a}+\cot^{-1}a \right]
\end{align}
and
\begin{align}
I=&\ J(1)=\int_0^1 J’(a)da\\
=& -\sqrt2\pi \bigg[
\left(\frac32\ln2-\frac\pi4 \right)\int_0^1 \frac{1-a}{1-a^4}da
+\frac12\int_0^1 \frac{\ln\frac{1+a^2}{2a^2}}{1-a^4}da\\
&\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>
+\int_0^1 \frac{\ln\frac{1+a}{2a}}{1-a^4}da
+\int_0^1 \frac{\frac\pi4-\cot^{-1}a}{1-a^4}da\bigg]\\
=& -\sqrt2\pi \bigg[
\left(\frac32\ln2-\frac\pi4 \right)\left(\frac\pi8+\frac14\ln2\right)
+\frac12\left(\frac12G+\frac{3\pi^2}{32} +\frac\pi{8}\ln2 \right)\\
&\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>
+\left(\frac12G+\frac{\pi^2}{24} -\frac\pi{16}\ln2\right)
+\left(-\frac14G-\frac{\pi^2}{64}\right)\bigg]\\
 =&-\frac\pi{\sqrt2}\left(G+\frac{\pi^2}{12}+\frac\pi4\ln2+\frac34\ln^22 \right)
\end{align}
