A log arctan integral $\int_0^1 \log x \arctan^2 x \, dx$
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\mathcal{J}} & = \int_{0}^{1}\ln\pars{x}\arctan^{2}\pars{x}\,\dd x = \int_{x\ =\ 0}^{x\ =\ 1}\arctan^{2}\pars{x}\,\dd\bracks{x\ln\pars{x} - x} \\[3mm] & = -\,{\phantom{^{2}}\pi^{2} \over 16} - \int_{0}^{1}\bracks{x\ln\pars{x} - x} \bracks{2\arctan\pars{x}\,{1 \over x^{2} + 1}}\,\dd x \\[3mm] & = -\,{\phantom{^{2}}\pi^{2} \over 16}\ +\ \underbrace{2\int_{0}^{1}{x\,\arctan\pars{x} \over x^{2} + 1}\,\dd x} _{\ds{\mathcal{J}_{1}}}\ -\ \underbrace{2\int_{0}^{1}{x\,\ln\pars{x}\arctan\pars{x} \over x^{2} + 1}\,\dd x} _{\ds{\mathcal{J}_{2}}} \\[3mm] & = -\,{\phantom{^{2}}\pi^{2} \over 16} + \mathcal{J}_{1} - \mathcal{J}_{2}\tag{1} \end{align}
- $\ds{\large \mathcal{J}_{1} =\, ?}$
\begin{align}
\mathcal{J}_{1} & =
2\int_{0}^{1}{x\,\arctan\pars{x} \over x^{2} + 1}\,\dd x\
\stackrel{x\ \equiv\ \tan\pars{t}}{=}\
2\int_{0}^{\pi/4}\tan\pars{t}t\,\dd t =
-2\int_{t\ =\ 0}^{t\ =\ \pi/4}t\,\,\dd\bracks{\ln\pars{\cos\pars{t}}}
\\[3mm] & =
{\pi \over 4}\,\ln\pars{2}\ +\ 2\
\underbrace{\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t}
_{\ds{\half\,G - {\pi \over 4}\,\ln\pars{2}}}\,,\qquad
G\ \mbox{is the}\ Catalan\ Constant.
\end{align}
The last integral is straightforward evaluated by using the
well known '$\ds{\cos}$-expansion' $\ds{\pars{~Fourier~}}$ of$\ $
$\ds{\ln\pars{\cos\pars{t}}}$.
\begin{align}
\fbox{$\ds{%
\quad\mathcal{J}_{1} =
2\int_{0}^{1}{x\arctan\pars{x} \over x^{2} + 1}\,\dd x =
G - {\pi \over 4}\,\ln\pars{2}\quad}$}\tag{2}
\end{align}
- $\ds{\mathcal{J}_{2} = \, ?}$ \begin{align} \mathcal{J}_{2} & = 2\int_{0}^{1}{x\,\ln\pars{x}\arctan\pars{x} \over x^{2} + 1}\,\dd x = 2\,\Im\int_{0}^{1}{x\,\ln\pars{x}\ln\pars{1 + x\ic} \over x^{2} + 1}\,\dd x \\[3mm] & = \overbrace{\Im\int_{0}^{1}{x\,\ln^{2}\pars{1 + x\ic} \over x^{2} + 1}\,\dd x} ^{\ds{t\ \equiv\ 1 + x\ic}}\ -\ \overbrace{% \Im\int_{0}^{1}{x\,\ln^{2}\pars{x/\bracks{1 + x\ic}} \over x^{2} + 1}\,\dd x} ^{\ds{t\ \equiv\ {x \over 1 + x\ic}}} \\[3mm] & = \underbrace{\Im\int_{1}^{1 + \ic}{t - 1 \over t\pars{t - 2}}\,\ln^{2}\pars{t}\,\dd t}_{\ds{\mathcal{J}_{21}}}\ +\ \underbrace{\Im\int_{0}^{\pars{1 - \ic}/2}{t \over 2t^{2} + 3\ic t - 1}\,\ln^{2}\pars{t}\,\dd t}_{\ds{\mathcal{J}_{22}}}\tag{2.a} \\[3mm] & = \mathcal{J}_{21} + \mathcal{J}_{22} \end{align} Those integrals are typically evaluated in a standard fashion where the results involve $PolyLogarithms$ $\ds{\,\mathrm{Li}_{s}}$ expressions. Namely, \begin{align} \mathcal{J}_{21} & = \half\,G\ln\pars{2} - {5 \over 32}\,\pi\ln^{2}\pars{2} + {7 \over 384}\,\pi^{3} - \Im\,\mathrm{Li}_{3}\pars{\half + \half\,\ic} \\[3mm] \mathcal{J}_{22} & = -\,\half\,G\ln\pars{2} + {9 \over 32}\,\pi\ln^{2}\pars{2} + {17 \over 384}\,\pi^{3} -3\,\Im\,\mathrm{Li}_{3}\pars{\half + \half\,\ic} \end{align} such that $\pars{~\mathcal{J}_{2} = \mathcal{J}_{21} + \mathcal{J}_{22}~}$ \begin{equation}\fbox{$\ds{\ \mathcal{J}_{2} = 2\int_{0}^{1}{x\ln\pars{x}\arctan\pars{x} \over x^{2} + 1} \,\dd x = {1 \over 8}\,\pi\ln^{2}\pars{2} + {1 \over 16}\,\pi^{3} -4\,\Im\,\mathrm{Li}_{3}\pars{\half + \half\,\ic}\ }$}\tag{3} \end{equation}
With the results $\pars{1}$, $\pars{2}$ and $\pars{3}$: \begin{align} \color{#f00}{\mathcal{J}} & \equiv \int_{0}^{1}\ln\pars{x}\arctan^{2}\pars{x}\,\dd x \\[3mm] & = \color{#f00}{% -\,{1 \over 16}\,\pi^{2} + G - {1 \over 4}\,\pi\ln\pars{2} - {1 \over 8}\,\pi\ln^{2}\pars{2} - {1 \over 16}\,\pi^{3} +4\,\Im\,\mathrm{Li}_{3}\pars{\half + \half\,\ic}} \approx -0.0915 \end{align}
Integrals in $\ds{\pars{2.a}}$ are reduced to an evaluation of: \begin{align} &\color{#f00}{\int_{a}^{b}{\ln^{2}\pars{t} \over c - t}\,\dd t} = \int_{a}^{b}{\ln^{2}\pars{t} \over 1 - t/c}\,{\dd t \over c} = \int_{a/c}^{b/c}{\ln^{2}\pars{ct} \over 1 - t}\,\dd t \\[3mm] = &\ \left.\vphantom{\Large A} -\ln\pars{1 - t}\ln\pars{ct}\right\vert_{\ a/c}^{\ b/c} + 2\int_{a/c}^{b/c}\ \overbrace{{\ln\pars{1 - t} \over t}} ^{\ds{-\,\mathrm{Li}_{2}'\pars{t}}}\ \,\ln\pars{ct}\,\dd t \\[3mm] = &\ -\ln\pars{1 - {b \over c}}\ln\pars{b} + \ln\pars{1 - {a \over c}}\ln\pars{a} -2\,\mathrm{Li}_{2}\pars{b \over c}\ln\pars{b} + 2\,\mathrm{Li}_{2}\pars{a \over c}\ln\pars{a} \\[3mm] +&\ 2\,\mathrm{Li}_{3}\pars{b \over c} - 2\,\mathrm{Li}_{3}\pars{a \over c} \end{align}
We have $$\frac{d}{dx}\arctan^2(x) = 2\cdot\frac{\arctan x}{1+x^2}\tag{1}$$ hence it follows that the Taylor series of $\arctan(x)^2$ is given by $$\arctan^2(x)=2\sum_{n\geq 0}(-1)^n \frac{x^{2n+2}}{2n+2} \sum_{m=0}^{n}\frac{1}{2m+1}\tag{2} $$ and: $$ \mathcal{J} = \color{red}{\sum_{n\geq 1}\frac{(-1)^{n}\left(2H_{2n}-H_{n}\right)}{(2n)(2n+1)^2}}.\tag{3}$$ Now Mathematica provides: $$ \sum_{n\geq 1}\frac{H_n}{(2n)(2n+1)^2}=\frac{2\pi^2\log(2)+8\log^2(2)-14\,\zeta(3)}{8}\tag{4}$$ but the alternating sign in $(3)$ gives an extra level of complexity. Anyway, $$ \frac{1}{(2n)(2n+1)^2}=\frac{1}{2n}-\frac{1}{2n+1}-\frac{1}{(2n+1)^2}\tag{5}$$ hence the crucial value to compute is $\color{red}{2\,S_6-S_3}$, where:
$$ S_1 = \sum_{n\geq 1}\frac{(-1)^n H_n}{2n}=\frac{\log^2(2)-\zeta(2)}{4},\qquad S_2 = \sum_{n\geq 1}\frac{(-1)^n H_n}{2n+1}=K-\frac{\pi}{2}\log 2$$ $$ \color{red}{S_3} = \sum_{n\geq 1}\frac{(-1)^n H_n}{(2n+1)^2}=\,\color{red}{???}\qquad S_4=\sum_{n\geq 1}\frac{(-1)^n H_{2n}}{2n}=\frac{\log^2(2)}{8}-\frac{5\pi^2}{96} $$ $$ S_5=\sum_{n\geq 1}\frac{(-1)^n H_{2n}}{2n+1}=-\frac{\pi}{8}\log(2),\qquad \color{red}{S_6}=\sum_{n\geq 1}\frac{(-1)^n H_{2n}}{(2n+1)^2}=\,\color{red}{???} $$
We have: $$ S_3 = \int_{0}^{1}\frac{\log(x)\log(1+x^2)}{1+x^2}\,dx $$ that Mathematica is able to convert into a complicated expression involving $\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)$ (besides $\pi^3,K\log(2)$ and $\pi\log^2(2)$) but I am stil clueless about $S_6$.