A solution using Abel's summation as suggested by Cornel.
Let $\ \displaystyle S\ $ denote $\ \displaystyle \sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}\ $
and by using Abel's summation:
$\displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)\ $ where $\ \displaystyle A_n=\sum_{i=1}^n a_i\ $
and letting let $\ \displaystyle a_k=\frac{1}{(2k+1)^2}\ $ , $\ \displaystyle b_k=H_k^{(2)}$, we get
\begin{align}
\sum_{k=1}^n\frac{H_k^{(2)}}{(2k+1)^2}&=\sum_{i=1}^n\frac{H_{n+1}^{(2)}}{(2i+1)^2}-\sum_{k=1}^n\frac{1}{(k+1)^2}\left(\sum_{i=1}^k\frac{1}{(2i+1)^2}\right)\\
&=\sum_{i=1}^n\frac{H_{n+1}^{(2)}}{(2i+1)^2}-\sum_{k=1}^n\frac{1}{(k+1)^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}+\frac{1}{(2k+1)^2}-1\right)
\end{align}
Letting $n$ approach $\infty$, we get
\begin{align}
S&=\zeta(2)\sum_{i=1}^\infty\frac{1}{(2i+1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)\\
&\quad-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}+\sum_{k=1}^\infty\frac1{(k+1)^2}\\
&=\zeta(2)\left(\frac34\zeta(2)-1\right)-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}-\frac{1}{(2k-1)^2}\right)\\
&\quad-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}+\zeta(2)-1\\
&=\frac{15}8\zeta(4)-1-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)+\sum_{k=1}^\infty\frac{1}{k^2(2k-1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}\\
&=\frac{15}8\zeta(4)-1-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)+1\\
&\quad+\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}\\
&=\frac{15}8\zeta(4)-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)\\
&=\frac{15}8\zeta(4)-4\sum_{k=1}^\infty\frac{H_{2k}^{(2)}}{(2k)^2}+\frac14\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}\\
&=\frac{15}8\zeta(4)-4\left(\frac12\sum_{k=1}^\infty\frac{H_{k}^{(2)}}{k^2}+\frac12\sum_{k=1}^\infty\frac{(-1)^kH_k^{(2)}}{k^2}\right)+\frac14\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}\\
&=\frac{15}8\zeta(4)-\frac74\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}-2\sum_{k=1}^\infty\frac{(-1)^kH_k^{(2)}}{k^2}
\end{align}
By plugging $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^nH_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42\ $
( proved here ) and $\ \displaystyle\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}=\frac74\zeta(4)\ $, we get the closed form of $\ S$
Considering the algebraic identity
\begin{align*}
&(a-b)^3b = a^3b - 3a^2b^2 + 3ab^3 - b^4 = -2a^3b +3(a^3b+ab^3) -3a^2b^2 -b^4\\
&\Longrightarrow \ \ \ 2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b
\end{align*} with $a = \ln(1-x)$ and $b= \ln (1+x)$ it follows that
\begin{align*}
2\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =& - \frac 1 2\int_0^1 {\ln^4(1+x)\over x}d x \\
&-\frac 12 \int_0^1 \frac{\ln^4(1+x) + 6\ln^2(1-x)\ln^2(1+x)}{x}dx\\
&+3\int_0^1 \frac{\ln^3(1-x)\ln(1+x) + \ln(1-x)\ln^3(1+x)}{x}dx\\
&- \int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}dx\\
=:& -I_1 - I_2 + I_3 -I_4.
\end{align*}
For $I_1$, make substitution $y = \frac x {1+x}$ to get:
\begin{align*}
I_1 =& \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{y(1-y)} dy \\
=& \frac 1 2\underbrace{ \int_0^{\frac 12} \frac{\ln^4(1-y)}{y} dy}_{z=1-y}+ \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{1-y} dy\\
=& \frac 1 2 \int_{\frac 1 2 }^1 \frac{\ln^4 z} {1-z} dz + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \int_{\frac 1 2}^1 z^{n-1}\ln^4 z\ dz + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \frac{\partial^4}{\partial n^4}\left[\frac 1 n - \frac 1 {n2^n}\right] + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \left[\frac{24}{n^5} - \frac {24}{n^52^n} - \frac{24 \ln 2}{n^42^n}-\frac{12\ln^2 2}{n^3 2^n}-\frac{4\ln^3 2}{n^2 2^n} - \frac{\ln^4 2}{n2^n}\right] + \frac {\ln^5 2}{10}\\
=&12\zeta(5) - 12\text{Li}_5(1/2) - 12\ln 2 \text{Li}_4(1/2) -6\ln^2 2 \text{Li}_3(1/2) -2\ln^3 2\text{Li}_2(1/2)-\frac {2}{5}\ln^5 2\\
=&\boxed{-12\Big(\text{Li}_5(1/2) + \ln 2\text{Li}_4(1/2)-\zeta(5)\Big)-{21 \over 4}\zeta(3)\ln^2 2 +{1\over 3} \pi^2 \ln^3 2-{2 \over 5} \ln^5 2}
\end{align*} where the well-known values
\begin{align*}\text{Li}_2(1/2) = {\pi^2 \over 12}-{\ln^2 2\over 2} , \qquad \text{Li}_3(1/2) ={7\zeta(3) \over 8} -{\pi^2 \ln 2\over 12} + {\ln^3 2 \over 6}
\end{align*} are used.
Actually, $I_2$ was already evaluated by the OP here using the algebraic identity $$b^4 + 6a^2b^2 = \frac {(a-b)^4} 2+\frac{(a+b)^4}{2} -a^4.$$
It holds that
$$
\boxed{I_2 = \frac {21}{8} \zeta(5).}
$$
In fact, the value of $I_3$ can also be found in the previous answer of @Przemo's. For $I_3$, one can use the algebraic relation $3(a^3b + ab^3) =\frac 3 8 \left[ (a+b)^4 - (a-b)^4\right]$.
This gives
\begin{align*}
I_3=& \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4(1-x^2)}{x} dx}_{x^2 = y} - \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x} dx}_{\frac{1-x}{1+x} = y}\\
=&\frac 3 {16}\underbrace{\int_0^1 \frac{\ln^4(1-y)}{y} dy }_{1-y\mapsto y}- \frac 3 4 \int_0^1 \frac{\ln^4 y}{1-y^2} dy\\
=&\frac 3 {16}\int_0^1 \frac{\ln^4 y}{1-y} dy - \frac 3 4 \sum_{n=0}^\infty \int_0^1 y^{2n} \ln^4 y \ dy\\
=&\frac 3 {16}\sum_{n=1}^\infty \int_0^1 y^{n-1}\ln^4 y \ dy - \frac 3 4 \sum_{n=0}^\infty \frac {24}{(2n+1)^5}\\
=&\frac 3 {16}\sum_{n=1}^\infty \frac{24}{n^5} - 18 \sum_{n=0}^\infty \frac {1}{(2n+1)^5}\\
=&\frac {9}{2} \zeta(5)- 18\cdot \frac {31}{32}\zeta(5)\\
=&\boxed{-\frac{207}{16}\zeta(5)}
\end{align*} as can be found in @Przemo's answer.
For $I_4$, make substitution $ \frac{1-x}{1+x}\mapsto x$ to get
\begin{align*} I_4 = &2\int_0^1 \frac{\ln^3 x \ln\left(\frac 2 {1+x}\right)}{1-x^2} dx \\
=&2\ln 2 \int_0^1 \frac{\ln^3 x}{1-x^2} dx - \underbrace{2\int_0^1\frac{\ln^3 x \ln(1+x)}{1-x^2} dx }_{=:J}\\
=& 2\ln 2\sum_{n=0}^\infty \int_0^1 x^{2n} \ln^3 x\ dx - J\\
=& - 12\ln 2 \underbrace{\sum_{n=0}^\infty \frac 1 {(2n+1)^4}}_{\frac{15}{16}\zeta(4) = \frac{\pi^4}{96}} - J \\
=& -\frac{\pi^4 \ln 2}{8} - J.
\end{align*}
\begin{align*}
J = &\int_0^1\frac{2\ln^3 x \ln(1+x)}{1-x^2} dx \\
=& \underbrace{\int_0^1 \frac{\ln^3 x \ln(1+x)}{1+x}dx}_{=:A} + \int_0^1 \frac{\ln^3 x \ln(1+x)}{1-x}dx\\
=& A + \int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + \int_0^1 \frac{(1+x)\ln^3 x \ln(1-x^2)}{1-x^2}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + \underbrace{\int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x^2}dx }_{=:B}+\underbrace{\int_0^1 \frac{x\ln^3 x \ln(1-x^2)}{1-x^2}dx}_{x^2 \mapsto x}-\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + B - \underbrace{\frac {15}{16} \int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx}_{=:C}\\
=&A + B - C.
\end{align*}
For $A$, we can use the McLaurin series of
$$
\frac{\ln (1+x)}{1+x} = \sum_{n=0}^\infty (-1)^{n-1}H_n x^n
$$ ($H_0= 0$) to get
\begin{align*}
A = & \sum_{n=0}^\infty (-1)^{n-1}H_n \int_0^1 x^n\ln^3 x \ dx \\
=&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_n}{(n+1)^4}\\
=&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_{n+1}}{(n+1)^4} - 6\sum_{n=0}^\infty \frac{(-1)^{n}}{(n+1)^5}\\
=&6 \sum_{n=1}^\infty \frac{(-1)^{n-1}H_{n}}{n^4} - 6\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^5}\\
=& 6\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-6\cdot \frac{15}{16}\zeta(5)\\
=& \frac{87}{16}\zeta(5) - \frac{\pi^2 \zeta(3)}{2}.
\end{align*}
Here, the known value of $ \sum_{n=1}^\infty (-1)^{n-1}{H_n \over n^4}$ is used.
For $B$, make substitution $u = x^2$ to get
\begin{align*}
B =& \frac 1 {16} \int_0^1 \frac{\ln^3 u \ln(1-u)}{\sqrt u (1-u)} du \\
=& \frac 1 {16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=\frac 1 2, y = 0^+}
\end{align*} where $\text{B}(\cdot,\cdot)$ is Euler's Beta function. We can use the fact that
\begin{align*}
\lim_{y\to 0^+}\frac{\partial^2}{\partial x\partial y} \text{B}(x,y) = -\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]
\end{align*} to get
\begin{align*}
B =& \frac 1 {16}\frac{d^2}{dx^2}\left[-\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]\right]_{x=\frac 1 2}\\
=&\frac 1 {16} \left[-\frac 1 2 \psi''''(1/2) + \psi'''(1/2)\big[\psi(1/2) + \gamma\big] + 3\psi'(1/2)\psi''(1/2)\right]\\
=& \frac 1 {16}\left[-21\pi^2 \zeta(3) + 372\zeta(5) - 2\pi^4 \ln 2\right]
\end{align*} which can be evaluated using the series representations of polygamma functions $$\psi(x) +\gamma = - \frac 1 x +\sum_{n=1}^\infty \frac 1 n - \frac 1 { n+x},\\
\psi'(x) = \sum_{n=0}^\infty \frac 1 {(n+x)^2}$$ and the derived fact that $\psi(\tfrac 1 2 )+\gamma = -2\ln 2$ and $\psi^{(k)}(\tfrac 1 2)=(-1)^{k+1}k!(2^{k+1}-1)\zeta(k+1)$ for $k\ge 1$.
For $C$, we can use the same method as used in the evaluation of $B$. It holds that
\begin{align*}
C =& \frac {15}{16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=1, y = 0^+}\\
=&\frac {15} {16}\left[-\frac 1 2 \psi''''(1) + \psi'''(1)\big[\psi(1) + \gamma\big] + 3\psi'(1)\psi''(1)\right]\\
=&\frac{15}{16}\left[12\zeta(5) -6\zeta(2)\zeta(3)\right]\\
=&\frac {45}{4}\zeta(5) -\frac {15\pi^2 \zeta(3)}{16}
\end{align*} where $\psi(1) +\gamma = 0$, $\psi'(1) = \zeta(2)$, $\psi''(1) = -2\zeta(3)$ and $\psi''''(1) = -24\zeta(5)$ are used.
Combining $A,B,C$, we have that $$J =A+B-C= \frac{279}{16}\zeta(5) -\frac{7\pi^2\zeta(3)}{8} - \frac{\pi^4 \ln 2}{8}$$ and
$$
\boxed{I_4 = -\frac{\pi^4 \ln 2}{8} - J = -\frac{279}{16}\zeta(5)+\frac{7\pi^2\zeta(3)}{8}}
$$
Finally, these evaluate $\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =\frac 1 2\big[-I_1-I_2+I_3-I_4\big]$ as follows.
\begin{align*}
\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =&\ 6\text{Li}_5(1/2) + 6\ln 2\ \text{Li}_4(1/2)-\frac{81}{16}\zeta(5)-{7\pi^2 \over 16}\zeta(3)\\
&+\frac{21\ln^2 2}{8}\zeta(3)- \frac{1}{6}\pi^2\ln^3 2+\frac{1}{5}\ln^5 2.
\end{align*}
Using the identity given in the OP, we get the desired integral $I$
\begin{align*}
\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\
&-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}.
\end{align*}
Best Answer
First we start with subbing $ x=\frac{y}{2+y}\Longrightarrow y=\frac{2x}{1-x}\Longrightarrow dy=\frac{2}{(1-x)^2}$
$$\mathcal{I}=\int_0^\infty\frac{\ln(2+y)\operatorname{Li}_2(-y)}{y(2+y)}dy=\frac12\int_0^1\frac{\ln\left(\frac2{1-x}\right)\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx$$
$$=\frac12\ln2\int_0^1\frac{\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx-\frac12\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx$$
$$=\frac12\ln2\cdot\mathcal{I}_1+\frac12\mathcal{I}_2$$
Evaluating the first integral $\mathcal{I}_1$:
By integration by parts we get
$$\mathcal{I}_1=\int_0^1\frac{\ln(x)\ln\left(\frac{1+x}{1-x}\right)}{x(1-x)}dx=\int_0^1\frac{\ln(x)\ln(1+x)}{x(1-x)}dx-\int_0^1\frac{\ln(x)\ln(1-x)}{x(1-x)}dx\\=\mathcal{A}-\mathcal{B}$$
$$\mathcal{A}=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^{n-1}\ln(x)}{1-x}dx=-\sum_{n=1}^\infty \frac{(-1)^n}{n}[H_{n-1}^{(2)}-\zeta(2)]$$
$$=-\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{n}+\sum_{n-1}^\infty\frac{(-1)^n}{n^3}+\ln2\sum_{n=1}^\infty\frac{(-1)^n}{n}$$
$$=-\left(\frac12\ln2\zeta(2)-\zeta(3)\right)-\frac34\zeta(3)-\ln2\zeta(2)$$
$$=\frac14\zeta(3)-\frac32\ln2\zeta(2)$$
By symmetry, we have
$$\mathcal{B}=2\int_0^1\frac{\ln x\ln(1-x)}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1x^{n-1}\ln xdx=2\sum_{n=1}^\infty\frac{1}{n^3}=2\zeta(3)$$
Collect $\mathcal{A}$ and $\mathcal{B}$ we get
Evaluating the second integral $\mathcal{I}_2$:
Use Landen's identity $-\operatorname{Li}_2(z)=\operatorname{Li}_2\left(\frac{z}{z-1}\right)+\ln^2(1-z)$
Set $z=\frac{2x}{x-1}$ we get $-\operatorname{Li}2\left(\frac{2x}{x-1}\right)=\operatorname{Li}2\left(\frac{2x}{x+1}\right)+\frac12\ln^2\left(\frac{1-x}{1+x}\right)$
Then we can write
$$\mathcal{I}_2=\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{2x}{x+1}\right)}{x}dx+\frac12\int_0^1\frac{\ln(1-x)\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx\\=\mathcal{J}+\frac12\mathcal{K}$$
Apply integration by parts for $\mathcal{J}$ we have
$$\mathcal{J}=-\frac52\zeta(4)+\int_0^1\frac{\operatorname{Li}_2(x)\ln\left(\frac{1+x}{1-x}\right)}{x(1+x)}dx$$
$$=-\frac52\zeta(4)+\int_0^1\frac{\operatorname{Li}_2(x)\ln(1+x)}{x(1+x)}dx-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{x(1+x)}dx\\=-\frac52\zeta(4)+\mathcal{J}_1-\mathcal{J}_2$$
$$\mathcal{J}_1=\int_0^1\frac{\operatorname{Li}_2(x)\ln(1+x)}{x(1+x)}dx=-\sum_{n=1}^\infty (-1)^nH_n\int_0^1 x^{n-1}\operatorname{Li}_2(x)dx$$
$$=-\sum_{n=1}^\infty (-1)^nH_n\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\sum_{n=1}^\infty \frac{(-1)^nH_n^2}{n^2}-\zeta(2)\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}$$
You can find here:
$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$
Substitute this result along with $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}=\frac12\ln^22-\frac12\zeta(2)$ we get
$$\boxed{\mathcal{J}_1=2\operatorname{Li}_4\left(\frac12\right)-\frac{21}{16}\zeta(4)+\frac74\ln2\zeta(3)-\ln^22\zeta(2)+\frac1{12}\ln^42}$$
For $\mathcal{J}_2$, break the denominator first
$$\mathcal{J}_2=\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{x}dx-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{1+x}dx$$
$$=-\frac54\zeta(4)-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{1+x}dx$$
Now set $1-x\to x$ then use the reflection formula $\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)$
$$\mathcal{J}_2=-\frac54\zeta(4)-\int_0^1\frac{\ln x}{2-x}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx$$
write $\frac1{2-x}=\sum_{n=1}^\infty \frac1{2^n}x^{n-1}$ to get
$$\mathcal{J}_2=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\int_0^1\ln x\ x^{n-1}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx$$
$$=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\frac{\partial}{\partial n}\int_0^1x^{n-1}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx, \quad IBP$$
$$=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\frac{\partial}{\partial n}\left(\frac{\zeta(2)}{n}-\frac{H_n^{(2)}}{n}\right)$$
$$=-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{1}{2^n}\left(\frac{H_n^{(2)}}{n^2}+\frac{2H_n^{(3)}}{n}-\frac{\zeta(2)}{n^2}-\frac{2\zeta(3)}{n}\right)$$
The first sum is already proved here
$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42$$
For the second sum, we can just set $x=\frac12$ in the generating function to get
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$
Collect these two results along with $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$ we get
$$\boxed{\mathcal{J}_2=-3\operatorname{Li_4}\left(\frac12\right)+\frac{9}{16}\zeta(4)+\frac14\ln^22\zeta(2)-\frac{1}{8}\ln^42}$$
Now collect the result of $\mathcal{J}_1$ and $\mathcal{J}_2$ we get
$$\boxed{\mathcal{J}=5\operatorname{Li_4}\left(\frac12\right)-\frac{35}{8}\zeta(4)+\frac74\ln2\zeta(3)-\frac54\ln^22\zeta(2)+\frac{5}{24}\ln^42}$$
The integral $\mathcal{K}$ is already calculated here
$$\boxed{\mathcal{K}=-4\operatorname{Li}_4\left(\frac12\right)-\frac{41}{8}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac1{6}\ln^42}$$
Thus
and finally