Is the value of this integral expressible in closed form and which numerical status does it have? 
Can the value of the integral $$u:=\int_0^1 \ln(x)\tan(x)dx$$
  be expressed in closed form ? 

Wolfram Alpha neither finds an antiderivate nor a closed form expression, but there are cases where no antiderivate exists, but with parameter-integrals , differentiating or integrating with respect to the parameter , gives a closed form of the value of the integral, so there still might be a closed form for $u$.

Is $u$ rational, irrational algebraic or trascendental ?

Using the $lindep-$ and $algdep-$ command of PARI/GP, I could not find any indication that $u$ might be algebraic or even rational. But is there any way to find out its numerical status ?
The numerical value of $u$ is $-0.2756872738004371638897520614\cdots$
Substituting $t=\ln(x)$ gives $$u=\int_{-\infty}^0 te^t \tan(e^t)dt$$
Substituting $t=\tan(x)$ gives $$u=\int_0^{tan(1)} \ln(\arctan(t))\frac{t}{t^2+1}dt$$
Here what Wolfram Alpha gets trying integration by parts :
https://www.wolframalpha.com/input/?i=integrate+log(x)*tan(x)+by+parts
 A: I don't believe a closed form solution (in the conventional sense) exists. It seems to be pretty much impossible that the value is algebraic. In any case, it can be transformed into an infinite sum via
$$\int_0^1\!dx\log x \tan x = \sum_{n=1}^\infty \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!}\overbrace{\int_0^1\!dx \log x\, x^{2n-1}}^{=-(2n)^{-2}}
= \sum_{n=1}^\infty \frac{B_{2n} (-4)^{n-1} (1-4^n)}{n^2(2n)!} $$
with $B_j$ the $j$-th Bernoulli number.
A: As Fabian wrote, probably there is no closed form for this series, but maybe it is interesting to see another series representation. Note that $$I=\int_{0}^{1}\log\left(x\right)\tan\left(x\right)dx
 $$ $$\stackrel{IBP}{=}\int_{0}^{1}\frac{\log\left(\cos\left(x\right)\right)}{x}dx
 $$ and using the Fourier series of $-\log\left(\cos\left(x\right)\right)
 $ we have $$I=\lim_{\epsilon\rightarrow0^{+}}\log\left(2\right)\log\left(\epsilon\right)+\sum_{k\geq1}\frac{\left(-1\right)^{n+1}}{k}\int_{\epsilon}^{1}\frac{\cos\left(2kx\right)}{x}dx
 $$ $$=\lim_{\epsilon\rightarrow0^{+}}\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{\epsilon}^{1}\frac{\cos\left(2kx\right)-1}{x}dx
 $$ $$=\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\frac{\cos\left(2kx\right)-1}{x}dx
 =\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{2k}\frac{\cos\left(u\right)-1}{u}du
 $$ and recalling the definition of the Cosine integral we get $$I=\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\left(\textrm{Ci}\left(2k\right)-\gamma-\log\left(2k\right)\right)
 $$ or $$I=\color{red}{\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\sum_{n\geq1}\frac{\left(-1\right)^{n}\left(2k\right)^{2n}}{2n\left(2n\right)!}}
 $$ unfortunately we can't switch the two series and so we have to stop here. Note that if we had $$S_{2}=\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{\color{blue}{2}}}\sum_{n\geq1}\frac{\left(-1\right)^{n}\left(2k\right)^{2n}}{2n\left(2n\right)!}
 $$ we could exchange the series $$S_{2}=\sum_{n\geq1}\frac{\left(-1\right)^{n}2^{2n-1}}{n\left(2n\right)!}\left(1-2^{2n-1}\right)\zeta\left(2-2n\right)
 $$ actually this is a finite series, due to the fact that $\zeta\left(-2m\right)=0,\,\forall m\geq1$.
