A closed form for a triple integral with sines and cosines $$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$
I saw this integral $I$ posted on a page on Facebook . The author claims that there is a closed form for it. 

My Attempt
This can be rewritten as 
$$3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz$$
Now consider 
$$F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz$$
Taking the derivative 
$$F'(a) = -3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz}\,dx\,dy\,dz$$
By symmetry we have
$$F'(a) = -3\left(\int^\infty_0 \frac{\sin^2(x)e^{-ax}}{x}\,dx \right)\left( \int^\infty_0 \frac{\sin(x)\cos(x)e^{-ax}}{x}\,dx\right)^2$$
Using W|A I got 
$$F'(a) = -\frac{3}{16} \log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)$$
By integeration we have 
$$F(0) = \frac{3}{16} \int^\infty_0\log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)\,da$$
Let $x = 2/a$
$$\tag{1}I = \frac{3}{8} \int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$
Question
I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ?
 A: Another approach to break down the last integral might be to consider the integral of $\displaystyle \frac{\log^3 (1-iz)}{z^2}$ along a positively oriented semi-circular contour $\gamma_R = [-R,R]\cup Re^{i[0,\pi]}$ in the upper half-plane. (We choose the branch of logarithm $\log (1-iz)$ in the lower half-plane along $[-i,-i\infty)$).
The integral along the arc is $\displaystyle \mathcal{O}\left(\frac{\log^3 R}{R}\right)$, which vanishes as $R \to +\infty$ we have,
\begin{align*}0 = \lim\limits_{R \to \infty} \int\limits_{\gamma_R} \frac{\log^3 (1-iz)}{z^2}\,dz = \int_{-\infty}^{\infty} \frac{\log^3\left((1+x^2)^{1/2} - i\arctan x\right)}{x^2}\,dx\end{align*}
Comparing the real parts on both sides,
\begin{align*} \int_{-\infty}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx &= \frac{1}{12}\int_{-\infty}^{\infty} \frac{\log^3(1+x^2)}{x^2}\,dx \\&= \frac{1}{6}\int_{0}^{\infty} \frac{\log^3(1+x^2)}{x^2}\,dx\\&\underset{\text{(IBP)}}{=} \int_0^{\infty} \frac{\log^2 (1+x^2)}{1+x^2}\,dx \\&= \int_0^{\pi/2} \log^2 (\cos \theta)\,d\theta \\&= \lim\limits_{b \to \frac{1}{2}}\frac{1}{2}\frac{\partial^2}{\partial b^2} B\left(\frac{1}{2},b\right)\\&= \frac{\pi}{2}\left(4\log^2 2 + \frac{\pi^2}{3}\right)\end{align*}
Hence, $$\displaystyle \int_{0}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx = \frac{\pi^3}{12} + \pi\log^2 2$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Given the 'cyclic symmetry' of your integral $\ds{\color{#f00}{\mc{J}}}$, it's equivalent to
\begin{align}
\color{#f00}{\mc{J}} & \equiv 3\int_{0}^{\infty}\!\!\int_{0}^{\infty}\!\!\int_{0}^{\infty}\!\!
\mrm{sinc}\pars{x}\mrm{sinc}\pars{y}\mrm{sinc}\pars{z}\sin\pars{x}\cos\pars{y}
\cos\pars{z}\ \times
\\[3mm] & \phantom{\equiv 3}
\underbrace{\bracks{\int_{0}^{\infty}\expo{-\pars{x + y + z}t}\,\dd t}}
_{\ds{1 \over x + y + z}}\dd x\,\dd y\,\dd z =
3\int_{0}^{\infty}\mrm{f}\pars{t}\mrm{g}^{2}\pars{t}\,\dd t
\\[5mm] & \mbox{where}\qquad
\left\{\begin{array}{rcl}
\ds{\mrm{f}\pars{t}} & \ds{=} &
\ds{\Im\mc{I}\pars{t}}
\\[2mm]
\ds{\mrm{g}\pars{t}} & \ds{=} &
\ds{\Re\mc{I}\pars{t}}
\\[2mm]
\ds{\mc{I}\pars{t}} & \ds{\equiv} &
\ds{\int_{0}^{\infty}\mrm{sinc}\pars{x}\expo{-\pars{t - \ic}x}\dd x}
\end{array}\right.\label{1}\tag{1}
\end{align}

Then
\begin{align}
\left.\mc{I}\pars{t}\right\vert_{\large\ \color{#f00}{t\ >\ 0}} & \equiv
\int_{0}^{\infty}\mrm{sinc}\pars{x}\expo{-\pars{t - \ic}x}\dd x =
\int_{0}^{\infty}
\overbrace{\pars{{1 \over 2}\int_{-1}^{1}\expo{\ic kx}\,\dd k}}
^{\ds{\mrm{sinc}\pars{x}}}\ \expo{-\pars{t - \ic}x}\dd x
\\[5mm] & =
{1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}\expo{-\pars{t - \ic - \ic k}x}
\dd x\,\dd k =
{1 \over 2}\int_{-1}^{1}{\dd k \over t - \ic - \ic k} =
{1 \over 2}\int_{-1}^{1}{t + \pars{k + 1}\ic \over
\pars{k + 1}^{2} + t^{2}}\,\dd k
\\[5mm] & =
{1 \over 2}\int_{0}^{2}{t + k\ic \over k^{2} + t^{2}}\,\dd k =
{1 \over 2}\int_{0}^{2/t}{1 + k\ic \over k^{2} + 1}\,\dd k =
{1 \over 2}\arctan\pars{2 \over t} +
{1 \over 4}\ln\pars{{4 \over t^{2}} + 1 }\ic
\end{align}

$\ds{\color{#f00}{\mc{J}}}$ becomes ( see \eqref{1} ):
\begin{align}
\color{#f00}{\mc{J}} & =
3\int_{0}^{\infty}\bracks{{1 \over 4}\ln\pars{{4 \over t^{2}} + 1}}
\bracks{{1 \over 2}\arctan\pars{2 \over t}}^{2}\dd t
\\[5mm] & \stackrel{2/t\ \mapsto\ t}{=}\,\,\,
{3 \over 16}\int_{\infty}^{0}\ln\pars{t^{2} + 1}
\arctan^{2}\pars{t}\pars{-\,{2\,\dd t \over t^{2}}} =
{3 \over 8}\ \underbrace{\int_{0}^{\infty}
{\ln\pars{t^{2} + 1}\arctan^{2}\pars{t} \over t^{2}}\,\dd t}
_{\ds{{\Large\color{#f00}{\S}}: {\pi^{3} \over 12} + \pi\ln^{2}\pars{2}}}
\end{align}


$\ds{{\Large\color{#f00}{\S}}}$: The integral was already evaluated in the
  $\texttt{@Zaid Alyafeai}$ fine answer.


Finally, the answer to the proposed OP integral is given by
$$
\bbox[15px,#ffe,border:1px dotted navy]{\ds{{\color{#f00}{\mc{J}} =
{\pi^{3} \over 32} + {3 \over 8}\,\pi\ln^{2}\pars{2}}}} \approx 1.5350
$$
A: Ok I was able to find the integral 
$$\int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$
First note that 
$$\int \frac{\log(1+x^2)}{x^2}\,dx = 2 \arctan(x) - \frac{\log(1 + x^2)}{x}+C$$
Using integration by parts
$$I = \frac{\pi^3}{12}+2\int^\infty_0\frac{\arctan(x)\log(1 + x^2)}{(1+x^2)x}\,dx$$
For the integral let
$$F(a) = \int^\infty_0\frac{\arctan(ax)\log(1 + x^2)}{(1+x^2)x}\,dx$$
By differentiation we have 
$$F'(a) = \int^\infty_0 \frac{\log(1+x^2)}{(1 + a^2 x^2)(1+x^2)}\,dx $$
Letting $1/a = b$ we get
$$\frac{1}{(1 + a^2 x^2)(1+x^2)} = \frac{1}{a^2} \left\{ \frac{1}{((1/a)^2+x)(1+x^2)}\right\} =\frac{b^2}{1-b^2}\left\{ \frac{1}{b^2+x^2}-\frac{1}{1+x^2} \right\}$$
We conclude that 
$$\frac{b^2}{1-b^2}\int^\infty_0 \frac{\log(1+x^2)}{b^2+x^2}-\frac{\log(1+x^2)}{1+x^2} \,dx = \frac{b^2}{1-b^2}\left\{ \frac{\pi}{b}\log (1+b)-\pi\log(2)\right\}$$
Where we used that 
$$\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+g^2x^2}\,dx = \frac{\pi}{cg}\log \frac{ag+bc}{g}$$
By integration we deduce that 
$$\int^1_0 \frac{\pi}{a^2-1}\left\{ a\log \left(1+\frac{1}{a} \right)-\log(2)\right\}\,da = \frac{\pi}{2}\log^2(2)$$
For the last one I used wolfram alpha, however it shouldn't be difficult to prove.
Finally we have 

$$\int^\infty_0\frac{\log\left(x^2+1
 \right)\arctan^2\left(x\right)}{x^2}\,dx = \frac{\pi^3}{12}+\pi
\log^2(2)$$

