What's about $-\int_0^1\frac{\log(1+x^{10})\log x}{x}dx$? Playing with Wolfram Alpha and inspired in [1] (I refers it if someone know how relates my problem with some of problems involving the Apéry constant in this reference, but the relation doesn't seem explicit), defining
$$I_n:=-\int_0^1\frac{\log(1+x^{2n})\log x}{x}dx$$
for integers $n\geq 1$, I can calculate, as I am saying with Wolfram Alpha (but I don't know how get the indefinite integrals) $I_1$, $I_2$, $I_3$ and $I_4$. And as a conjecture $$I_8=\frac{6\zeta(3)}{8\cdot 16^2}.$$
Motivation. I would like to do a comparison with the sequence  $I_1$, $I_2$, $I_3$, $I_4$ and $I_8$.

Question. If do you know that this problem was solved in the literature please add a comment: can you evaluate in a closed-form $I_5$? Many thanks.

References:
[1] Walther  Janous , Around's Apéry's constant, J. Ineq. Pure and Appl. Math. 7(1) Art. 35 (2006).
 A: By Taylor series expansion ,
For $0<x<1$ and $n\geq 1$,
$\displaystyle -\ln(1+x^{2n})=\sum_{k=1}^{+\infty} \dfrac{(-1)^kx^{2kn}}{k}$
For $k \geq 0$,
$\displaystyle \int_0^1 x^k\ln x dx=-\dfrac{1}{(k+1)^2}$
(integration by parts)
Therefore,
$\begin{align}
I_n&=\int_0^1 \left(\sum_{k=1}^{+\infty} \dfrac{(-1)^kx^{2kn-1}\ln x}{k}\right) dx\\
&=\sum_{k=1}^{+\infty} \left(\int_0^1 \dfrac{(-1)^kx^{2kn-1}\ln x}{k} dx\right)\\
&=-\sum_{k=1}^{+\infty} \dfrac{(-1)^k}{k(2kn)^2}\\
&=-\dfrac{1}{4n^2}\sum_{k=1}^{+\infty} \dfrac{(-1)^k}{k^3}\\
&=-\dfrac{1}{4n^2}\left(\sum_{k=1}^{+\infty} \dfrac{1}{(2k)^3}-\sum_{k=0}^{+\infty}\dfrac{1}{(2k+1)^3}\right)\\
&=-\dfrac{1}{4n^2}\left(\dfrac{1}{8}\zeta(3)-\left(\zeta(3)-\dfrac{1}{8}\zeta(3)\right)\right)\tag{1}\\
&=-\dfrac{1}{4n^2}\times -\dfrac{3}{4}\zeta(3)\\
&=\boxed{\dfrac{3\zeta(3)}{16n^2}}
\end{align}$
For (1) observe that,
$\begin{align}\sum_{k=0}^{+\infty}\dfrac{1}{(2k+1)^3}&=\sum_{k=1}^{+\infty}\dfrac{1}{k^3}-\sum_{k=1}^{+\infty}\dfrac{1}{(2k)^3}\\
&=\zeta(3)-\dfrac{1}{8}\zeta(3)
\end{align}$
A: Integration by part leads to$$ I_n=-I_n+2n\int_0^1\frac{x^{2n-1}\log^2 x}{1+x^{2n}}\,dx$$
Then $$ I_n=n\int_0^1\frac{x^{2n-1}\log^2 x}{1+x^{2n}}\,dx$$
and Maple integrates:$$I_n=\frac{3\zeta(3)}{16n^2}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&-\int_{0}^{1}{\ln\pars{1 + x^{2n}}\ln\pars{x} \over x}\,\dd x \,\,\,\,\stackrel{x^{2n}\ \mapsto\ x}{=}\,\,
-\,{1 \over 4n^{2}}\int_{0}^{1}{\ln\pars{1 + x}\ln\pars{x} \over x}\,\dd x
\\[5mm]= &\
-\,{1 \over 4n^{2}}\int_{0}^{-1}{\ln\pars{1 - x}\ln\pars{-x} \over x}\,\dd x =
{1 \over 4n^{2}}\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln\pars{-x}\,\dd x =
-\,{1 \over 4n^{2}}\int_{0}^{-1}{\mrm{Li}_{2}\pars{x} \over x}\,\dd x
\\[5mm] = &\
-\,{1 \over 4n^{2}}\int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\,\dd x =
-\,{1 \over 4n^{2}}\,\mrm{Li}_{3}\pars{-1} =
-\,{1 \over 4n^{2}}\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k^{3}}
\\[5mm] = &\
-\,{1 \over 4n^{2}}\pars{\sum_{k = 1\ \mrm{even}}^{\infty}{1 \over k^{3}} -
\sum_{k = 1\ \mrm{odd}}^{\infty}{1 \over k^{3}}} =
-\,{1 \over 4n^{2}}\pars{2\sum_{k = 1}^{\infty}{1 \over \pars{2k}^{3}} -
\sum_{n = 1}^{\infty}{1 \over n^{3}}} =
\bbx{\ds{{3 \over 16n^{2}}\,\zeta\pars{3}}}
\end{align}
