Finding $\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx$ 
Finding $\displaystyle \int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx$

What I tried
Put $\displaystyle I =\int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx$
\begin{align*}
I&=\int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx\\
&=\int^{\frac{\pi}{4}}_{0}\ln\bigg(2\cos^2\frac{x}{2}\bigg)\mathrm dx\\
&=\frac{\pi}{4}\ln(2)+4\int^{\frac{\pi}{8}}_{0}\ln(\cos x)\mathrm dx\\
&=\frac{\pi}{4}\ln(2)+4\left[\left[x\ln(\cos x)\right]^{\frac{\pi}{8}}_{0}+\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx\right]\\
&=\frac{\pi}{8}\ln(2)+\frac{\pi}{2}\ln\left(\cos \frac{\pi}{8}\right)+4\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx
\end{align*}
How do I solve it? Help me, please.
 A: Using the Clausen Function $\operatorname{Cl}_2(z)$ we can directly write down an anti-derivative for your integral. Since we got that

$$\int_0^t \log(\tan x)\mathrm dx=-\frac12\operatorname{Cl}_2(2t)-\frac12\operatorname{Cl}_2(\pi-2t)\tag1$$

This formula can be shown by using the Fourier Series Expansions of $\log(\sin x)$ and $\log(\cos x)$ under the utilisation of the series representation of the Clausen Function. Using formula $(1)$ we may write your integral as
$$\int_0^{\pi/8} \log(\tan x)\mathrm dx=-\frac12\operatorname{Cl}_2\left(\frac\pi4\right)-\frac12\operatorname{Cl}_2\left(\frac{3\pi}4\right)$$
However, this does not help at all since there are no closed-form representations known for these values of the Clausen Function. By the Duplication Formula of the Clausen Function we can at least eliminate one of the unknown values.

$$\therefore~\int_0^{\pi/8} \log(\tan x)\mathrm dx~=~-\operatorname{Cl}_2\left(\frac\pi4\right)+\frac G4~=~-\operatorname{Cl}_2\left(\frac{3\pi}4\right)-\frac G4$$

I have doubts that we can get any closer to an actual closed-form. Anyway, these terms are still series in disguise. Note that the Clausen Function can be expressed as Dilogarithm aswell, precisely $\operatorname{Cl}_2(z)=\Im\operatorname{Li}_2(e^{iz})$, which moves this problem in the conetxt of Polylogarithms too.
A: \begin{align}J=\int_0^{\frac{\pi}{8}}\ln\left(\tan x\right)\,dx\end{align}
Perform the change of variable $y=\left(\sqrt{2}+1\right)\tan x$,
\begin{align}J&=(\sqrt{2}-1)\int_0^{1}\frac{\ln\big((\sqrt{2}-1)x\big) }{1+(\sqrt{2}-1)^2x^2}\,dx\\
&=(\sqrt{2}-1)\int_0^{1}\frac{\ln\big(\sqrt{2}-1\big) }{1+(\sqrt{2}-1)^2x^2}\,dx+(\sqrt{2}-1)\int_0^{1}\frac{\ln x }{1+(\sqrt{2}-1)^2x^2}\,dx\\
&=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+(\sqrt{2}-1)\int_0^{1}\frac{\ln x }{1+(\sqrt{2}-1)^2x^2}\,dx\\
&=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+\frac{1}{2}\big(\sqrt{2}-1\big)\int_0^1\frac{\ln x}{1-i\big(1-\sqrt{2}\big)x}\,dx+\frac{1}{2}\big(\sqrt{2}-1\big)\int_0^1\frac{\ln x}{1-i\big(\sqrt{2}-1\big)x}\,dx\\
&=\boxed{\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+\frac{1}{2}i\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)-\frac{1}{2}i\text{Li}_2\left(i\big(1-\sqrt{2}\big)\right)}
\end{align}
NB:
For $0<|z|<1$ complex,
\begin{align}\int_0^1 \frac{\ln x}{1-zx}\,dx&=\int_0^1 \left(\sum_{n=0}^{\infty}(xz)^n\right)\ln x\,dx\\
&=\sum_{n=0}^{\infty} z^n\left(\int_0^1 x^n\ln x\,dx\right)\\
&=-\sum_{n=0}^{\infty} \frac{z^n}{(n+1)^2}\\
&=-\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^{n+1}}{(n+1)^2}\\
&=-\frac{1}{z}\text{Li}_2(z)
\end{align}
PS:
if you prefer,
\begin{align}\boxed{J=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)-\Im{\left(\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)\right)}}\end{align}
