The closed form of $\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$ What tools or ways would you propose for getting the closed form of this integral?
$$\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$
EDIT: It took a while since I made this post. I'll give a little bounty for the solver of the problem, 500 points bounty.
Supplementary question:
Calculate 
$$\int_0^{\pi/4}\frac{\log(1-x)\log(x)\log(1+x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$
 A: Just a few notes for a series development, because a "closed" formula is very unlikely.
$$\int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = -\sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{1}{n-k+1} \int\limits_0^{\pi/4} x^{n+1} (\tan x)^{2k+2} dx$$
Or using $\int\limits_0^{\pi/4} \frac{\ln(1-x)}{x (1-x\tan^2 x)} dx$ it becomes a little more handsomely:
$$ \int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = \text{Li}_2\left(\frac{\pi}{4}\right) -\sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{1}{n-k+1} \int\limits_0^{\pi/4} x^n (\tan x)^{2k} dx$$
Then we have: $$ \int\limits_0^{\pi/4} x^n (\tan x)^{2k} dx = (-1)^k \left(\frac{\pi^{n+1}}{(n+1)4^{n+1}} + A_{n,k} - B_{n,k} - C_{n,k} \right) \enspace$$ for:
$$A_{n,k} := \left(\sin \frac{\pi n}{2}\right)\frac{n!}{2^{n+1}} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} c_{2k-1,j}~\eta(n-j+1)\\
B_{n,k} := \frac{1}{2^{2n+2}}\sum\limits_{~v=0 \\ v~\text{odd}}^n \left(\sin \frac{\pi v}{2}\right)\frac{n!\pi^{n-v}}{(n-v)!} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} 2^j c_{2k-1,j}~\eta(v-j+1)\\
 C_{n,k} := \frac{1}{2^{2n+1}}\sum\limits_{~v=0 \\ v~\text{even}}^n \left(\cos \frac{\pi v}{2}\right)\frac{n!\pi^{n-v}2^v}{(n-v)!} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} c_{2k-1,j}~\beta(v-j+1)\\
 c_{n,j} := \frac{1}{n!} \sum\limits_{v=0}^{n+1} {\binom {n+1} v} \sum\limits_{l=j}^n \begin{bmatrix}{n+1}\\{l+1}\end{bmatrix}{\binom l j}(-v)^{l-j}$$ 
and the Stirling numbers of the first kind $\begin{bmatrix}n\\k\end{bmatrix}$ defined by: $$\sum\limits_{k=0}^n\begin{bmatrix}n\\k\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k),$$ 
so that $ c_{ \text{odd},\text{even} }=0 $ and $ c_{ \text{even},\text{odd} }=0$. Some values $\,c_{n,j}\,$ can be seen here .
The needed analytical continuation for Dirichlet eta function $\eta(s)$ and Dirichlet beta function $\beta(s)$ can be seen in my answer of the question here . With the additional information 
$$ \sum\limits_{n=0}^\infty\frac{\pi^{n+1}}{(n+1) 4^{n+1}}\sum\limits_{k=0}^n \frac{(-1)^k}{n-k+1} = \\ \text{Li}_2\left(\frac{\pi}{4}\right) + \text{Li}_2 \left(\frac{1}{2} -\frac{\pi}{8} \right) + \left(\ln\left(\frac{1}{2} +\frac{\pi}{8} \right)\right)\ln\left(1-\frac{\pi}{4}\right)$$
we get: 
$$ \int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = \\
-\text{Li}_2 \left(\frac{1}{2} -\frac{\pi}{8} \right) - \left(\ln\left(\frac{1}{2} +\frac{\pi}{8} \right)\right)\ln\left(1-\frac{\pi}{4}\right) - \sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{(-1)^k}{n-k+1}(A_{n,k} - B_{n,k} - C_{n,k} )$$
