Let us denote:
\begin{equation}
{\mathcal I}^{(3,2)}:=\int\limits_0^1 \frac{\log(1+x)^3}{x} \cdot [\log(x)]^2 dx
\end{equation}
We have:
\begin{eqnarray}
&&{\mathcal I}^{(3,2)}=\\
&&-\frac{2}{3} i \pi \left(-12 \text{Li}_5\left(\frac{1}{2}\right)-12 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^2 \zeta (3)}{2}+\frac{99 \zeta (5)}{16}-\frac{21}{4} \zeta (3) \log ^2(2)-\frac{2 \log
^5(2)}{5}+\frac{1}{3} \pi ^2 \log ^3(2)\right)-\\
&&\frac{2}{3} \int\limits_0^1 \frac{6\left(\text{Li}_4(x+1)-\frac{\pi ^4}{90}\right)+3 \text{Li}_2(x+1) \log ^2(x+1)-6 \text{Li}_3(x+1) \log (x+1)}{x} \cdot \log(x)dx
\end{eqnarray}
In the above we used the knowledge of the anti-derivative of the fraction in the integrand and we integrated by parts once. Indeed we have:
\begin{eqnarray}
\int \frac{\log(1+x)^k}{x} dx = \sum\limits_{l=1}^{k+1} (-1)^l \binom{k}{l-1} (l-1)! Li_l(1+x) \log(1+x)^{k+1-l}
\end{eqnarray}
Now it is actually fairy easy to construct the anti-derivative of the fraction in the remaining integrand above and then to do another integration by parts.As a matter of fact we have:
\begin{eqnarray}
&&\int \frac{6\left(\text{Li}_4(x+1)-\frac{\pi ^4}{90}\right)+3 \text{Li}_2(x+1) \log ^2(x+1)-6 \text{Li}_3(x+1) \log (x+1)}{x} dx=\\
&&\log (-x) \left(6 \left(\text{Li}_4(x+1)-\frac{\pi ^4}{90}\right)+3 \text{Li}_2(x+1) \log ^2(x+1)-6 \text{Li}_3(x+1) \log (x+1)\right)+\\
&&3 \int \frac{Li_1(1+x)^2}{(1+x)} \cdot \log(1+x)^2 dx
\end{eqnarray}
In doing that it turns out that the boundary term vanishes and then what we are left with are integrals of the kind $\int\limits_0^{1/2} \log(x)^p \log(1-x)^q/x dx$ for $p+q \le 5$. All those integrals have allready been dealt with and are expressed thorough poly-logarithms with one exception only , namely when $(p,q)=(3,2)$. In this case a new quantity ${\bf H}^{(1)}_5(1/2)$ enters the result. Then the final result reads:
\begin{eqnarray}
&&{\mathcal I}^{(3,2)}=\\
&&-108 \text{Li}_6\left(\frac{1}{2}\right)-36 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{429 \zeta (2)^3}{35}+12 \zeta (3)^2-\frac{3}{2} \zeta (2) \log ^4(2)+6 \zeta (3) \log ^3(2)-\frac{9}{10} \zeta (2)^2 \log
^2(2)-18 \zeta (3) \zeta (2) \log (2)+\frac{9}{8} \zeta (5) \log (2)+\frac{3 \log ^6(2)}{20}
+ 36 {\bf H}^{(1)}_5(1/2)
\end{eqnarray}
Below I include the Mathematica code that verifies the results:
M = 2000; Clear[H];
H[p_, q_, x_] := N[Sum[ HarmonicNumber[n, p]/n^q x^n, {n, 1, M}], 50];
k = 3;
NIntegrate[Log[1 + x]^k/x Log[x]^2, {x, 0, 1}, WorkingPrecision :> 30]
(*The border term is equal to Int Log[1+x]^3 Log[x]/x,{x,0,1}]*)
-2 I Pi/3 (Pi^2/3 Log[2]^3 - 2/5 Log[2]^5 + Pi^2/2 Zeta[3] +
99/16 Zeta[5] - 21/4 Zeta[3] Log[2]^2 -
12 PolyLog[4, 1/2] Log[2] - 12 PolyLog[5, 1/2]) -
2/3 NIntegrate[(3 Log[1 + x]^2 PolyLog[2, 1 + x] -
6 Log[1 + x] PolyLog[3, 1 + x] +
6 (-PolyLog[4, 1] + PolyLog[4, 1 + x])) Log[x]/x, {x, 0, 1},
WorkingPrecision :> 30]
(3 Log[2]^6)/20 - 36 Log[2] PolyLog[5, 1/2] - 108 PolyLog[6, 1/2] +
6 Log[2]^3 Zeta[3] + 12 Zeta[3]^2 + 9/8 Log[2] Zeta[5] -
3/2 Log[2]^4 Zeta[2] - 18 Log[2] Zeta[3] Zeta[2] -
9/10 Log[2]^2 Zeta[2]^2 + (429 Zeta[2]^3)/35 + 36 H[1, 5, 1/2]