An Euler squared sum Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Evaluate the following sum
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1}$$
Here $\mathcal{H}_n^2$ is the square harmonic number, e.g $\left ( \sum \limits_{k=1}^{n} \frac{1}{k} \right )^2$. I don't know how to tackle this. One idea of mine was the following:
\begin{align*}
\mathcal{S} &=\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n^2 \int_{0}^{1}x^n \, {\rm d}x \\ 
 &= \int_{0}^{1}\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n^2 x^n \, {\rm d}x
\end{align*}
and the last sum evaluates to ? 
Motivation: The series comes from this question . I tried a different approach. This is what I tried:
\begin{align*}
\int_{0}^{1} \frac{\log(1+x) \log(1-x)}{1+x} \, {\rm d}x&= \int_{0}^{1}\log(1-x) \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n x^n \, {\rm d}x\\ 
 &=\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \int_{0}^{1}x^n \log(1-x) \, {\rm d}x \\ 
 &= -\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \cdot \frac{\mathcal{H}_{n+1}}{n+1} \\ 
 &= -\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \left ( \frac{\mathcal{H}_n + \frac{1}{n+1}}{n+1} \right )\\ 
 &= -\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2}
\end{align*}
The right hand Euler sum appears to be elementary. I have not evaluated it though. But now we known that:
$$\bbox[blue, 2pt]{\color{white}{-\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2} = \frac{1}{3}\log^3 2-\frac{\pi^2}{12}\log 2+\frac{\zeta(3)}{8}}}$$
from the linked question. So, I expect both of these series to have a closed form. Any help how to proceed with the first series? If anyone wishes to give a go for the second (right hand Euler sum) be my guest.
Addendum: For the second Euler sum we have that:
\begin{align*}
-\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2} &= \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \int_{0}^{1}x^n \log x \, {\rm d}x\\ 
 &=\int_{0}^{1}\log x  \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n x^n \, {\rm d}x \\ 
 &= \int_{0}^{1}\frac{\log (1+x) \log x}{1+x}\, {\rm d}x\\ 
 &=-\frac{\zeta(3)}{8}
\end{align*}
The integral is elementary. Thus, it appears that the sum I seek evaluates to
$$\frac{1}{3} \log^3 2 - \frac{\pi^2}{12} \log 2 + \frac{\zeta(3)}{8} = -\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \frac{\zeta(3)}{8}  \Rightarrow \\\\\\
\Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1}= -\frac{1}{3} \log^3 2 +\frac{\pi^2}{12} \log 2 - \frac{\zeta(3)}{4}$$
Of course this is an indirect way of evaluating the sum. Any other way of tackling it?
 A: Some preliminary lemma.
Lemma 1.
$$ \sum_{n\geq 1} H_n x^n = \frac{\log(1-x)}{1-x}. $$
Lemma 2. By Lemma $1$,
$$ \sum_{n\geq 1}\frac{H_n}{n+1} x^{n+1} = \frac{1}{2}\log^2(1-x),\qquad \sum_{n\geq 1}\frac{H_n+H_{n+1}}{n+1}x^{n}=\frac{-x+\log^2(1-x)+\text{Li}_2(x)}{x}.$$
Lemma 3. Since $H_{n+1}^2-H_n^2 = \frac{H_n+H_{n+1}}{n+1}$,
$$ \sum_{n\geq 1}H_{n}^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x}.$$
Lemma 4. By Lemma 3,
$$ \sum_{n\geq 1}\frac{(-1)^{n+1} H_{n}^2}{n+1} = -\int_{0}^{1}\frac{\log^2(1+x)+\text{Li}_2(-x)}{1+x}\,dx=-\frac{\log^3(2)}{3}-\color{red}{\int_{0}^{1}\frac{\text{Li}_{2}(-x)}{1+x}\,dx}.$$
The problem boils down to the evaluation of the last integral. By integration by parts, it is:
$$ \color{red}{\int_{0}^{1}\frac{\text{Li}_{2}(-x)}{1+x}\,dx}=-\frac{\pi^2}{12}\log(2)+\color{blue}{\int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx}\tag{1} $$
but:
$$ \color{blue}{\int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx} = -2\int_{0}^{1}\frac{\log(1+x)\log(x)}{1+x}\,dx=\color{blue}{\frac{\zeta(3)}{4}}\tag{2}$$
and the proof is complete.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

$\underline{\mbox{Just one of the integrals you didn't evaluate}\ }$ :

With $\ds{0 < \mu < 1}$:
\begin{align}
&\color{#f00}{%
\int_{0}^{1 - \mu}{\ln\pars{1 + x}\ln\pars{1 - x} \over 1 + x}\,\dd x} =
\half\,\ln^{2}\pars{2 - \mu}\ln\pars{\mu} +
\half\,\int_{0}^{1 - \mu}{\ln^{2}\pars{1 + x} \over 1 - x}\,\dd x
\\[3mm] = &\
\half\,\ln^{2}\pars{2 - \mu}\ln\pars{\mu} +
\half\,\int_{1}^{2 - \mu}{\ln^{2}\pars{x} \over 2 - x}\,\dd x
\\[3mm] = &\
\half\,\ln^{2}\pars{2 - \mu}\ln\pars{\mu} + 
\half\,\int_{1/2}^{1 - \mu/2}{\ln^{2}\pars{2x} \over 1 - x}\,\dd x
\\[3mm] = &\
\half\,\ln^{2}\pars{2 - \mu}\ln\pars{\mu} -
\half\ln\pars{{\mu \over 2}}\ln^{2}\pars{2 - \mu} + 
\int_{1/2}^{1 - \mu/2}{\ln\pars{1 - x} \over x}\,\ln\pars{2x}\,\dd x
\end{align}

When $\ds{\mu \to 0^{+}}$:
\begin{align}
&\color{#f00}{%
\int_{0}^{1}{\ln\pars{1 + x}\ln\pars{1 - x} \over 1 + x}\,\dd x} =
\half\,\ln^{3}\pars{2} -\int_{1/2}^{1}\mathrm{Li}_{2}'\pars{x}\ln\pars{2x}
\,\dd x
\\[3mm] = &\
\half\,\ln^{3}\pars{2} - \mathrm{Li}_{2}\pars{1}\ln\pars{2} +
\int_{1/2}^{1}\mathrm{Li}_{3}'\pars{x}\,\dd x
\\[3mm] = &\
\color{#f00}{\half\,\ln^{3}\pars{2} - \mathrm{Li}_{2}\pars{1}\ln\pars{2} +
\mathrm{Li}_{3}\pars{1} - \mathrm{Li}_{3}\pars{\half}}
\end{align}

$$
\mbox{With}\quad
\left\lbrace\begin{array}{rcl}
\ds{\mathrm{Li}_{2}\pars{1}} & \ds{=} & \ds{{\pi^{2} \over 6}}
\\[2mm]
\ds{\mathrm{Li}_{3}\pars{1}} & \ds{=} & \ds{\zeta\pars{3}}
\\[2mm]
\ds{\mathrm{Li}_{3}\pars{\half}} & \ds{=} &
\ds{{1 \over 24}\bracks{21\zeta\pars{3} + 4\ln^{3}\pars{2} - 2\pi^{2}\ln\pars{2}}}
\end{array}\right.
$$

we'll get
$$
\color{#f00}{%
\int_{0}^{1}{\ln\pars{1 + x}\ln\pars{1 - x} \over 1 + x}\,\dd x} =
\color{#f00}{{1 \over 24}\bracks{3\zeta\pars{3} + 8\ln^{3}\pars{2} -2\pi^{2}\ln\pars{2}}} \approx -0.3028
$$
