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 ?

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, $$\displaystyle \int_{0}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx = \frac{\pi^3}{12} + \pi\log^2 2$$
