Two very advanced harmonic series of weight $5$
Update: the paper mentioned below is the preprint On the calculation of two essential harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$ by Cornel Ioan Valean
A solution in large steps by Cornel Ioan Valean:
Considering $\displaystyle -\log(1+y)\log(1-y)=\sum_{n=1}^{\infty} y^{2n} \frac{H_{2n}-H_n}{n}+\frac{1}{2}\sum_{n=1}^{\infty} \frac{y^{2n}}{n^2}$ where we divide both sides by $y$ and then integrate from $y=0$ to $y=x$, we have $\displaystyle -\int_0^x \frac{\log(1+y)\log(1-y)}{y}\textrm{d}y=\sum_{n=1}^{\infty} x^{2n} \frac{H_{2n}-H_n}{2n^2}+\frac{1}{4}\sum_{n=1}^{\infty} \frac{x^{2n}}{n^3}$. Now, if we multiply both sides of this last result by $\log(1+x)/x$ and then integrate from $x=0$ to $x=1$, using the fact that $\displaystyle \int_0^1 x^{2n-1}\log(1+x) \textrm{d}x=\frac{H_{2n}-H_n}{2n}$, we get
{A specific note: one can multiply both sides of the relation above by $\log(1-x)/x$ instead of $\log(1+x)/x$ and use the integral, $\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-H_n/n$, but later in the process one might like to use though the version $\int_0^1 x^{2n-1}\log(1+x) \textrm{d}x$ to nicely get the calculations.}
$$\underbrace{-\int_0^1 \frac{\log(1+x)}{x}\left(\int_0^x \frac{\log(1+y)\log(1-y)}{y}\textrm{d}y\right)\textrm{d}x}_{\displaystyle I}=\frac{5}{4}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}+\frac{7}{8}\sum_{n=1}^{\infty}\frac{H_n}{n^4}-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{n}}{n^4}-\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^2}{n^3}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^3}.$$
Integrating by parts, the integral $I$ may be written as $5/16\zeta(2)\zeta(3)-\underbrace{\int_0^1 \frac{\text{Li}_2(-x) \log (1+x) \log (1-x)}{x} \textrm{d}x}_{J}$, and then we may write that $$\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^3}=2\int_0^1 \frac{\text{Li}_2(-x) \log (1+x) \log (1-x)}{x} \textrm{d}x-\frac{5}{8}\zeta(2)\zeta(3)+\frac{5}{2}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}+\frac{7}{4}\sum_{n=1}^{\infty}\frac{H_n}{n^4}-2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{n}}{n^4}-2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^2}{n^3}\tag1 .$$
Now, the last magical part comes from considering expressing the integral $J$ in a different way, and by using the Cauchy product, $\displaystyle \operatorname{Li}_2(-x)\log(1+x)=3\sum_{n=1}^{\infty}(-1)^n \frac{x^n}{n^3}-2 \sum_{n=1}^{\infty}(-1)^n x^n\frac{H_n}{n^2}-\sum_{n=1}^{\infty}(-1)^nx^n\frac{H_n^{(2)}}{n}$, we get that
$$\int_0^1 \frac{\operatorname{Li}_2(-x) \log (1+x) \log (1-x)}{x} \textrm{d}x= -\sum _{n=1}^{\infty }(-1)^{n-1}\frac{ H_n H_n^{(2)}}{n^2}+3\sum _{n=1}^{\infty }(-1)^{n-1}\frac{ H_n}{n^4}\\-2\sum _{n=1}^{\infty }(-1)^{n-1}\frac{H_n^2}{n^3}.\tag2$$
Combining $(1)$ and $(2)$ and collecting the values of the series from the book, (Almost) Impossible Integrals, Sums, and Series, we are done with the first series.
To get the value of the second series we might use the relation:
\begin{equation*} \sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n-1)^3}-\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n)^3} \end{equation*} \begin{equation*} =6 \log (2)-2 \log ^2(2)-\frac{1}{12}\log ^4(2)+\frac{3}{20} \log ^5(2)-\frac{3}{2} \zeta (2)-\frac{21}{8} \zeta (3)+\frac{173}{32} \zeta (4) \end{equation*} \begin{equation*} +\frac{55}{32} \zeta (5)-\frac{5 }{4}\zeta (2) \zeta (3)+\frac{3}{2} \log (2) \zeta (2)-\frac{7}{2}\log (2)\zeta (3)-4\log (2)\zeta (4)+\frac{1}{2} \log ^2(2) \zeta (2) \end{equation*} \begin{equation*} -\frac{5}{6} \log ^3(2)\zeta (2)+\frac{21}{8}\log ^2(2)\zeta (3)-2 \operatorname{Li}_4\left(\frac{1}{2}\right)+4 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right)+2 \operatorname{Li}_5\left(\frac{1}{2}\right), \end{equation*}
and this is obtained by using a very similar strategy to the one given in Section 6.59, pages $530$-$532$, from the book, (Almost) Impossible Integrals, Sums, and Series. The critical identity here is given in (6.289).
A detailed solution will appear soon in a new paper.
UPDATE (September $30$, $2019$)
A magical way to the series $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$ by Cornel Ioan Valean
By the Cauchy product, we have $\operatorname{Li}_2(x^2) \log(1-x^2)= 3\sum _{n=1}^{\infty } \frac{x^{2 n}}{n^3}-2\sum _{n=1}^{\infty } x^{2n}\frac{H_n}{n^2}-\sum _{n=1}^{\infty } x^{2n}\frac{H_n^{(2)}}{n}$, and if we multiply both sides by $\log(1-x)/x$, and integrate from $x=0$ to $x=1$, using that $\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-H_n/n$, and doing all the reductions, we arrive at
$$2\sum _{n=1}^{\infty } \frac{H_{2 n} H_n^{(2)}}{(2 n)^2}-12\sum _{n=1}^{\infty } \frac{H_n}{n^4}+12\sum _{n=1}^{\infty }(-1)^{n-1} \frac{H_n}{n^4}+\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log \left(1-x^2\right) \log (1-x)}{x} \textrm{d}x$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x} \textrm{d}x+2 \int_0^1 \frac{\text{Li}_2(-x) \log ^2(1-x)}{x} \textrm{d}x\\+2 \int_0^1 \frac{\text{Li}_2(x) \log ^2(1-x)}{x} \textrm{d}x$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x} \textrm{d}x+2 \sum _{n=1}^{\infty } \frac{H_n^2}{n^3}-2 \sum _{n=1}^{\infty } \frac{(-1)^{n-1}H_n^2}{n^3}+2 \sum _{n=1}^{\infty } \frac{H_n^{(2)}}{n^3}\\-2 \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_n^{(2)}}{n^3},$$ where the last integral is given here Two very advanced harmonic series of weight $5$, and all the last resulting harmonic series are given in the book (Almost) Impossible Integrals, Sums, and Series. The reduction to the last series has been achieved by using the identity, $\displaystyle \int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x=\frac{H_n^2+H_n^{(2)}}{n}$. The series $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$ maybe found calculated in the paper On the calculation of two essential harmonicseries with a weight 5 structure, involving harmonic numbers of the type H_{2n} by Cornel Ioan Valean. Thus, we have
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3).$$
All the details will appear in a new paper.
UPDATE (October $30$, $2019$) The details with respect to the evaluation of the previous series may be found in the preprint The evaluation of a special harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$
UPDATE (July $05$, $2020$) The two essential series stated above may be found now as an article here.
A very simple solution to one of the key logarithmic integrals stated above, $ \displaystyle \int_{0}^{1}x^{2n-1}\ln(1+x)\textrm{d}x=\frac{H_{2n}-H_n}{2n}$
Solution (by Cornel): \begin{equation*} \int_{0}^{1}x^{2n-1}\ln(1+x)\textrm{d}x= \int_0^1 \left(\frac{x^{2n}}{2n}-\frac{1}{2n}\right)' \log(1+x)\textrm{d}x=\frac{1}{2n}\int_0^1 \frac{1-x^{2n}}{1+x}\textrm{d}x \end{equation*} \begin{equation*} =\frac{1}{2n}\sum _{k=1}^{2 n} (-1)^{k-1} \int_0^1 x^{k-1} \textrm{d}x=\frac{1}{2n}\sum _{k=1}^{2 n}\frac{ (-1)^{k-1}}{k}=\frac{H_{2n}-H_n}{2n}. \end{equation*}
BONUS: A very simple solution to a logarithmic integral strongly related to the key integral above, $ \displaystyle \int_{0}^{1}x^{2n}\ln(1+x)\textrm{d}x=\frac{2\log(2)}{2n+1}-\frac{1}{(2n+1)^2}+\frac{H_n-H_{2n}}{2n+1}$
Solution (by Cornel): \begin{equation*} \int_{0}^1 x^{2n}\ln(1+x)\textrm{d}x= \int_{0}^1 \left(\frac{x^{2n+1}}{2n+1}+\frac{1}{2n+1}\right)' \ln(1+x)\textrm{d}x \end{equation*} \begin{equation*} =\frac{2\log(2)}{2n+1}-\frac{1}{2n+1}\int_0^1\frac{1+x^{2n+1}}{1+x}\textrm{d}x=\frac{2\log(2)}{2n+1}-\frac{1}{2n+1}\sum_{k=1}^{2n+1} (-1)^{k-1}\int_0^1x^{k-1}\textrm{d}x \end{equation*} \begin{equation*} =\frac{2\log(2)}{2n+1}-\frac{1}{2n+1}\sum_{k=1}^{2n+1} \frac{(-1)^{k-1}}{k}=\frac{2\log(2)}{2n+1}-\frac{1}{(2n+1)^2}+\frac{H_n-H_{2n}}{2n+1}. \end{equation*}
We have
$$\frac{\ln^2(1-y)}{1-y}=\sum_{n=1}^\infty y^n(H_n^2-H_n^{(2)})\tag{1}$$
integrate both sides of (1) from $y=0$ to $y=x$ to get
$$-\frac13\ln^3(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^2-H_n^{(2)}\right)=\sum_{n=1}^\infty\frac{x^{n}}{n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)\tag{2}$$
Now replace $x$ with $x^2$ in (2) then multiply both sides by $-\frac{\ln(1-x)}{x}$ and integrate from $x=0$ to $x=1$, also note that $\int_0^1 -x^{2n-1}\ln(1-x)\ dx=\frac{H_{2n}}{2n}$ we get
$$\frac13\underbrace{\int_0^1\frac{\ln^3(1-x^2)\ln(1-x)}{x}\ dx}_{\large I}=\sum_{n=1}^\infty\frac{H_{2n}}{2n^2}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$
Rearranging the terms to get
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^2}{(2n)^2}=\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}+4\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n)^3}-8\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}+\frac16I\tag{3}$$
@nospoon mentioned in equation (3) of his solution that he found
$$\sum_{n=1}^{\infty} \frac{H_{n-1}^{(2)}\,H_{2n}}{n^2} =\frac{11}{4}\zeta(2)\,\zeta(3)-\frac{47}{16}\zeta(5)$$
Or
$$\boxed{\sum _{n=1}^{\infty } \frac{H_{2n} H_{n}^{(2)}}{(2 n)^2}=\frac{101 }{64}\zeta (5)-\frac{5 }{16}\zeta (2) \zeta (3)}$$
Also Cornel elegantly calculated the second sum above
$$\boxed{\small{\sum _{n=1}^{\infty } \frac{H_{2 n}H_n }{(2 n)^3}=\frac{307}{128}\zeta(5)-\frac{1}{16}\zeta (2) \zeta (3)+\frac{1}{3}\ln ^3(2)\zeta (2) -\frac{7}{8} \ln ^2(2)\zeta (3)-\frac{1}{15} \ln ^5(2) -2 \ln (2) \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right)}}$$
For the third sum: $$\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac12\sum_{n=1}^\infty\frac{H_{n}}{n^4}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}$$
plugging the common results:
$$\sum_{n=1}^\infty\frac{H_{n}}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$
$$\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$
we get
$$\boxed{\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac{37}{64}\zeta(5)-\frac14\zeta(2)\zeta(3)}$$
For the remaining integral $I$, we use the magical identity
$$(a+b)^3a=a^4-b^4+\frac12(a+b)^4-\frac12(a-b)^4-(a-b)^3b$$
with $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write
$$I=\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac12\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}\\-\underbrace{\frac12\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}-\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}$$
$$I=\frac54\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{4!\zeta(5)}-\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{K}-\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}{4}\zeta(5)}+\underbrace{2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx}_{J}$$
$$I=\frac{27}{4}\zeta(5)-K+J\tag{4}$$
we have
\begin{align} K&=\int_0^1\frac{\ln^4(1+x)}{x}=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\ &=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right) \end{align}
and
$$J=2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx=\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx+\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1+x}\ dx$$
using the rule
$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{a+1}}{n}$$
allows us to write
\begin{align} J&=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}\ dx-\ln2\int_0^1\frac{\ln^3x}{1+x}\ dx\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-\sum_{n=1}^\infty(-1)^n H_n\int_0^1x^n\ln^3x\ dx-\ln2\left(-\frac{21}4\zeta(4)\right)\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+6\sum_{n=1}^\infty\frac{(-1)^n H_n}{(n+1)^4}+\frac{21}{4}\ln2 \zeta(4)\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-6\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^4}-\frac{45}{8}\zeta(5)+\frac{21}{4}\ln2 \zeta(4) \end{align}
Plugging
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}=\frac78\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)-2\zeta(5)$$
we get
$$J=\frac{279}{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)$$
Plugging the results of $K$ and $J$ in (4) we get
$$\boxed{\small{I=24\operatorname{Li}_5\left(\frac12\right)+24\ln2\operatorname{Li}_4\left(\frac12\right)+\frac3{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)+\frac{21}2\ln^22\zeta(3)-4\ln^32\zeta(2)+\frac45\ln^52}}$$
and finally by substituting the boxed results in (3) we get
$$\sum _{n=1}^{\infty } \frac{H_{2 n}H_n^2 }{(2 n)^2} =\frac{9 }{16}\zeta (2) \zeta (3)+\frac{421 }{64}\zeta (5)+\frac{2}{3} \ln ^3(2)\zeta (2) -\frac{7}{4} \ln ^2(2)\zeta (3)\\ -\frac{2}{15} \ln^5(2) -4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$
Note:
$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ can be found here and $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}$ can be found here.