Closed form for ${\large\int}_0^1\frac{\ln^4(1+x)\ln x}x \, dx$ Can someone compute
$$ \int_0^1\frac{\ln^4(1+x)\ln x}x \,dx$$
in closed form?  
I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the form $\operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $\mathrm{Li}_n$ is the $n$th polylogarithm.  
The reason for my conjecture is that 
$$ \int_0^1\frac{\ln^2(1+x)\ln x}x \; dx =\frac{\pi^4}{24}-\frac16\ln^42+\frac{\pi^2}6\ln^22-\frac72\zeta(3)\ln2-4\operatorname{Li}_4\!\left(\frac12\right) $$
and as shown here on Math StackExchange
$$ \int_0^1\frac{\ln^3(1+x)\ln x}x \; dx = \frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac{21}4\zeta(3)\ln^22\\-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right).$$
The Riemann zeta function obeys
$$ \zeta_n = \operatorname{Li}_n(1) $$
and $\pi^{2n}$ is a rational number times $\zeta_{2n}$.  Also, 
$$ \operatorname{Li}_1(x) = -\ln(1 - x) .$$
So, the two integrals above are polynomials with rational coefficients in constants of the form $\operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $\mathrm{Li}_n$ is the $n$th polylogarithm.  Maybe this pattern continues!
If my conjecture is true, next I'll ask about 
$$ \int_0^1\frac{\ln^k(1+x)\ln x}x \,dx$$
for $k = 4, 5, 6, \dots $
 A: The Stirling numbers of the first kind $\left[ \begin{array}{c} n \\ k \end{array} \right]$ are usually defined by: $$ \sum\limits_{k=0}^n \left[ \begin{array}{c} n \\ k \end{array} \right] x^k := x(x+1)…(x+n-1) $$ 
The definition of  $\enspace\eta_n(m)\enspace$ in $\enspace$ Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $ is
$$\eta_n(m):=\sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$ 
with $\enspace m>0$, $\enspace n\in\mathbb{N}_0$ , $\enspace\eta_0(m)=\eta(m)$ 
and $\enspace\displaystyle \frac{1}{(k-1)!} \left[ \begin{array}{c} k \\ {n+1} \end{array} \right]= \sum\limits_{i_1=1}^{k-1}\sum\limits_{i_2=i_1+1}^{k-1}…\sum\limits_{i_n=i_{n-1}+1}^{k-1}\frac{1}{i_1 i_2 … i_n}$ .
To combine this series with the polylogarithm is a separate problem.
Now we can write:
$\displaystyle \sum\limits_{k=1}^\infty \frac{z^k}{k!}\int\limits_0^1 \frac{\ln^k(1+x)\ln x}{x} dx = \int\limits_0^1 \frac{((1+x)^z-1)\ln x}{x} dx = - \sum\limits_{k=1}^\infty \binom z k \frac{1}{k^2}$ 
$\hspace{5.3cm}\displaystyle = - \sum\limits_{k=1}^\infty z^k \sum\limits_{v=k}^\infty \frac{(-1)^{k-v}}{v^2 v!} \left[ \begin{array}{c} v \\ k \end{array} \right]= \sum\limits_{k=1}^\infty \frac{(-z)^k}{(k-1)!} \eta_{k-1}(3)$ 
It follows
$$\int\limits_0^1 \frac{\ln^k(1+x)\ln x}{x} dx = (-1)^k\,k\,\eta_{k-1}(3)$$
for $k\in\mathbb{N}$ .
A: It's not a complete answer but it is too lengthy for a comment.
@user14717: 
Program GP PARI contains a routine to perform something like PSQL stuff.
Here a script
\p 200
A1=Pi^6
A2=Pi^4*log(2)^2
A3=Pi^2*log(2)^4
A4=polylog(6,1/2)
A5=polylog(5,1/2)*log(2)
A6=polylog(4,1/2)*log(2)^2
A7=polylog(3,1/2)*log(2)^3
A8=polylog(2,1/2)*log(2)^4
A9=log(2)^6
A10=zeta(3)*log(2)^3
A11=zeta(5)*log(2)
A12=zeta(3)^2
A13=Pi^2*log(2)*zeta(3)
A14=polylog(3,1/2)*Pi^2*log(2)
A15=polylog(4,1/2)*Pi^2
A16=polylog(3,1/2)^2
A17=polylog(2,1/2)^2*log(2)^2
A18=polylog(2,1/2)^2*Pi^2
A19=polylog(2,1/2)^3
A20=polylog(2,1/2)*Pi^2*log(2)^2
A21=polylog(2,1/2)*Pi^4
A22=polylog(3,1/2)*zeta(3)
J=intnum(x=0,1,log(1+x)^4*log(x)/x)
lindep([J,A1,A2,A4,A5,A6,A7,A8,A9,A11,A12,A13,A15,A22])
Last command returns an integer relation of Ai's equals to 0.
Notice that some Ai's are linearly dependant on integers.
Anyway i wasn't able to find out such integer relation using these constants.
