Ways to prove a relation featuring $\int _0^{\infty }\frac{\arctan ^2\left(x\right)\ln ^2\left(1+x^2\right)}{x^2}\:dx$ From a previous post the following integral arised:
$$\int _0^{\infty }\frac{\arctan ^2\left(x\right)\ln ^2\left(1+x^2\right)}{x^2}\:dx$$
And there user178256 linked us to this question where the user M.N.C.E. stated the following:

$$\int _0^{\infty }\frac{\arctan ^2\left(x\right)\ln ^2\left(1+x^2\right)}{x^2}\:dx=\frac{1}{24}\int _0^{\infty }\frac{\ln ^4\left(1+x^2\right)}{x^2}\:dx+\frac{2}{3}\int _0^{\infty }\frac{\arctan ^4\left(x\right)}{x^2}\:dx$$

My question is, how can we prove this relation without evaluating each integral while also avoiding complex numbers?
What I thought of doing is to use the following algebraic identity:
$$a^2b^2=\frac{1}{12}\left(a+b\right)^4+\frac{1}{12}\left(a-b\right)^4-\frac{1}{6}a^4-\frac{1}{6}b^4$$
This means that if we set $a=\arctan\left(x\right)$ and $b=\ln\left(1+x^2\right)$ we have:
$$\int _0^{\infty }\frac{\arctan ^2\left(x\right)\ln ^2\left(1+x^2\right)}{x^2}\:dx=\frac{1}{12}\int _0^{\infty }\frac{\left(\arctan \left(x\right)+\ln \left(1+x^2\right)\right)^4}{x^2}\:dx$$
$$+\frac{1}{12}\int _0^{\infty }\frac{\left(\arctan \left(x\right)-\ln \left(1+x^2\right)\right)^4}{x^2}\:dx-\frac{1}{6}\int _0^{\infty }\frac{\arctan ^4\left(x\right)}{x^2}\:dx-\frac{1}{6}\int _0^{\infty }\frac{\ln ^4\left(1+x^2\right)}{x^2}\:dx$$
Though I'm having a little bit of trouble getting somewhere with the first $2$ integrals, any help will be well received.
 A: A comment regarding a possible way to go proposed by Cornel Ioan Valean (the author of the book, (Almost) Impossible Integrals, Sums, and Series)
There are two key steps in attacking the problem by real methods and avoiding contour integration.
$1)$ The first key step: we consider
$$I(a,b)=\int _0^{\infty }\frac{\arctan(ax)\arctan(bx)\log^2(1+x^2)}{x^2}\textrm{d}x,$$
which then we differentiate with respect to $a$ and $b$ and then use partial fraction decomposition.
$2)$ The second (magical) key step: prove (cleverly) and use that
$$\int_0^{\infty} \frac{\log^2(1+x^2)}{1+a^2 x^2}\textrm{d}x=\frac{1}{a}\left(\frac{\pi^3}{6}-2\log^2(2)\pi+\log^2\left(\frac{2(1+a)}{a}\right)\pi-2\pi \operatorname{Li}_2\left(\frac{a-1}{2a}\right)\right), \ a>0.$$
A BIG NOTE: The strategy described above, with the use of the second step, works smoothly if writing
$$\int_0^{\pi/2} x\cot(x)\log^2(\cos(x))\textrm{d}x= \frac{1}{4}\int_0^{\infty}\frac{\arctan(x)\log^2(1+x^2)}{x(1+x^2)} \textrm{d}x,$$
since all efforts are considered for calculating the trigonometric integral above. Thus, we only need the differentiation with respect to a single parameter, $\displaystyle I(a)=\int_0^{\infty}\frac{\arctan(ax)\log^2(1+x^2)}{x(1+x^2)} \textrm{d}x$. The rest is represented by boring work to do of average difficulty.
And a final note: A solution in large steps to the trigonometric integral above, exploiting results from (Almost) Impossible Integrals, Sums, and Series), is given in large steps here.
End of comment (I hope it helps a little bit!)
