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 ?

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 

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

