How to calculate $ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx $?

Substitute $t=\tan\frac x2$

\begin{align} &\int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx \\ =&\>2\int_0^{1}\frac{\ln\frac{(1+t)^2}{1+t^2}\ln \frac{1-t^2}{1+t^2} }{1+t^2}\,dt =\>4I_1 +4 I_2 -2I_3- 6I_4+2I_5 \end{align} where, per the results

\begin{align} I_1 &= \int_0^1 \frac{\ln (1+t)\ln(1-t)}{1+t^2} dt = -G \ln 2-K+\frac{3 \pi ^3}{128}+\frac{3\pi}{32} \ln ^22\\ I_2 &= \int_0^1 \frac{\ln^2(1+t)}{1+t^2} dt = -2 G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{16} \ln ^22 \\ I_3 &= \int_0^1 \frac{\ln (1+t^2)\ln(1-t)}{1+t^2} dt = -\frac{1}{2} G \ln 2+4 K -\frac{5 \pi ^3}{64}+\frac{\pi}{8} \ln ^22 \\ I_4 &= \int_0^1 \frac{\ln (1+t^2)\ln(1+t)}{1+t^2} dt = -\frac{5}{2} G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{8} \ln ^22\\ I_5 &= \int_0^1 \frac{\ln^2(1+t^2)}{1+t^2} dt = -2 G \ln 2+4 K-\frac{7 \pi ^3}{96}+\frac{7\pi}{8} \ln ^22 \end{align}

with $K= \Im\text{Li}_3\left(\frac{1+i}{2}\right)$. Together $$ \int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx =4\Im\text{Li}_3\left(\frac{1+i}{2}\right)-\frac{11\pi^3}{96}+\frac{3\pi}8\ln^22 $$