I have been trying to evaluate this integral: $$\int_0^{\infty} \frac{\arctan^2 x \log^2 (1+x^2)}{x^2}\,dx$$ but have failed despite all my attemps. I tried using the trig substitution $x = \tan(a)$, and through a series of steps, was able to simplify the integral to $$I(b) = 4\displaystyle \int_0^{\pi/2} \frac{(a\log(\cos(a))^b}{(\sin(a))^b}\,da$$ I then tried to differentiate under the integral sign but failed. What should I do now?
Note: b = 2 in the latter integral.
Let's start integrating by parts: $$\int_0^\infty\frac{\arctan^2\left(x\right)\log^2\left(1+x^2\right)}{x^2}dx=$$ $$\underbrace{-\left[\frac{\arctan^2\left(x\right)\log^2\left(1+x^2\right)}{x}\right]^\infty_0}_{0}+\underbrace{4\int_0^\infty\frac{\arctan^2\left(x\right)\log\left(1+x^2\right)}{1+x^2}dx}_{I_1}+\underbrace{2\int_0^\infty\frac{\arctan\left(x\right)\log^2\left(1+x^2\right)}{x(1+x^2)}dx}_{I_2}$$
To solve $I_1$, let $x\rightarrow \frac{y}{\sqrt{1-y^2}}$. Then, let's rewrite $\arctan(x)$ as $\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)$ and apply the series expansion of $\arcsin^2(x)$:
$$I_1=-4\int_0^1\frac{\arcsin^2\left(y\right)\log\left(1-y^2\right)}{\sqrt{1-y^2}}dy=-2\sum_{n=1}^{\infty}\frac{\left(2^nn!\right)^2}{n^2(2n)!}\int_0^1\frac{y^{2n}\log\left(1-y^2\right)}{\sqrt{1-y^2}}dy$$
Bearing in mind that: $$\frac{\partial}{\partial b}\int_0^1t^{a-1}\left(1-t\right)^{b-1}dt=\frac{\partial}{\partial b}\mathfrak{B}(a,b)$$ $$\int_0^1t^{a-1}\left(1-t\right)^{b-1}\log\left(1-t\right)dt=\mathfrak{B}(a,b)\left[\mathfrak{\psi^{(0)}}(b)-\mathfrak{\psi^{(0)}}(a+b)\right]=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}\left[\mathfrak{\psi^{(0)}}(b)-\mathfrak{\psi^{(0)}}(a+b)\right]$$ and $$\Gamma\left(z\right)\Gamma\left(z+\frac{1}{2}\right)=2^{1-2z}\sqrt{\pi}\ \Gamma\left(2z\right)$$
$$I_1=-\sum_{n=1}^{\infty}\frac{\left(2^nn!\right)^2}{n^2(2n)!}\frac{\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(n+1\right)}\left[\mathfrak{\psi^{(0)}}\left(\frac{1}{2}\right)-\mathfrak{\psi^{(0)}}\left(n+1\right)\right]=-\pi\sum_{n=1}^{\infty}\frac{\mathfrak{\psi^{(0)}}\left(\frac{1}{2}\right)-\mathfrak{\psi^{(0)}}\left(n+1\right)}{n^2}$$
$$=-\pi\sum_{n=1}^{\infty}\frac{-2\log(2)-\gamma+\gamma-H_n}{n^2}=2\pi\zeta(2)\log(2)+\pi\sum_{n=1}^{\infty}\int_0^1\frac{dx}{1-x}\frac{1-x^n}{n^2}$$
$$=2\pi\zeta(2)\log(2)+\pi\int_0^1\frac{\zeta(2)-Li_2(x)}{1-x}dx\\=2\pi\zeta(2)\log(2)-\pi\underbrace{\left[\left(\zeta(2)-Li_2(x)\right)\log\left(1-x\right)\right]^1_0}_{0}+\pi\underbrace{\int_{0}^{1}\frac{\log^2\left(1-x\right)}{x}dx}_{2\zeta\left(3\right)}$$ $$\boxed{I_1=2\pi\zeta\left(2\right)\log(2)+2\pi\zeta\left(3\right)}$$
To solve $I_2$, let $x\rightarrow {\frac{1}{x}}$: $$I_2=\underbrace{\int_{0}^{\infty}\frac{2x\arctan\left(\frac{1}{x}\right)\left(\log^2{\left(1+x^2\right)-4\log(x)\log(1+x^2)}\right)}{1+x^2}dx}_{A}+\underbrace{8\int_{0}^{\infty}{\frac{x\arctan\left(\frac{1}{x}\right)\log^2{\left(x\right)}}{1+x^2}dx}}_{B}$$
$$A=\underbrace{\left[\frac{\arctan\left(\frac{1}{x}\right)\log^3\left(1+x^2\right)}{3}-2\arctan\left(\frac{1}{x}\right)\log^2\left(1+x^2\right)\log(x)\right]_{0}^{\infty}}_{0}+\\\frac{1}{3}\int_{0}^{\infty}\frac{\log^3{\left(1+x^2\right)}}{1+x^2}dx-2\int_{0}^{\infty}{\frac{\log^2{\left(1+x^2\right)}\log{\left(x\right)}}{1+x^2}dx}+2\int_{0}^{\infty}{\frac{\arctan\left(\frac{1}{x}\right)\log^2{\left(1+x^2\right)}}{x}dx}$$
Let $x\rightarrow \sqrt\frac{y}{1-y}$:
$$A=\int_{0}^{1}\frac{\frac{1}{3}\log^3{\left(1-y\right)}-\frac{1}{2}\log^2{\left(1-y\right)}\log{\left(y\right)}}{\sqrt{y\left(1-y\right)}}dy+\int_{0}^{1}{\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{y(1-y)}dy}$$ Notice that:
$$\int_{0}^{1}\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{y(1-y)}dy=\int_{0}^{1}\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{y}dy+\int_{0}^{1}\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{1-y}dy\\=\frac{\pi}{2}\underbrace{\int_{0}^{1}\frac{\log^2{\left(1-y\right)}}{y}dy}_{2\zeta(3)}-\underbrace{\int_{0}^{1}\frac{\arcsin(\sqrt{y})\log^2{\left(1-y\right)}}{y}dy}_{y\rightarrow \frac{x^2}{1+x^2}}+\int_{0}^{1}\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{1-y}dy\\=\pi\zeta(3)-\underbrace{2\int_0^\infty\frac{\arctan\left(x\right)\log^2\left(1+x^2\right)}{x(1+x^2)}dx}_{I_2}-\underbrace{\left[\frac{\arccos(\sqrt{y})\log^3\left(1-y\right)}{3}\right]^1_0}_{0}-\frac{1}{6}\int_{0}^{1}\frac{\log^3{\left(1-y\right)}}{\sqrt{y(1-y)}}dy$$
Therefore $$A=\pi\zeta(3)-I_2+\frac{1}{6}\lim_{a\rightarrow 1/2\\ b\rightarrow 1/2}\frac{\partial^3}{\partial^3a}\mathfrak{B}(a,b)-\frac{1}{2}\lim_{a\rightarrow 1/2\\ b\rightarrow 1/2}\frac{\partial^3}{\partial^2a\partial b}\mathfrak{B}(a,b)$$
$$A=\pi\zeta(3)-I_2+\left(-2\pi\zeta(3)-\frac{4}{3}\log^3(2)-2\pi\zeta(2)\log(2)\right)-\left(\pi\zeta(3)-4\pi\log^3(2)\right)$$
$$\boxed{A=-2\pi\zeta(3)-2\pi\zeta(2)\log(2)+\frac{8}{3}\log^3(2)-I_2}$$
$$B=8\int_{0}^{1}dy\int_{0}^{\infty}{\frac{x^2\log^2{\left(x\right)}}{\left(1+x^2\right)\left(1+y^2x^2\right)}dx}=8\int_{0}^{1}\frac{dy}{1-y^2}\int_{0}^{\infty}\left(\frac{\log^2{\left(x\right)}}{1+x^2}-\frac{y^2\log^2{\left(x\right)}}{y^2+x^2}\right)dx$$
$$=8\int_{0}^{1}\frac{\frac{3\pi}{4}\zeta\left(2\right)\left(1-y\right)-\frac{\pi}{2}ylog^2\left(y\right)}{1-y^2}dy\\\boxed{B=6\pi\zeta\left(2\right)\log{\left(2\right)}-\pi\zeta\left(3\right)}$$
$$I_2=-2\pi\zeta(3)-2\pi\zeta(2)\log(2)+\frac{8}{3}\log^3(2)-I_2+6\pi\zeta\left(2\right)\log{\left(2\right)}-\pi\zeta\left(3\right)\\ \boxed{I_2=-\frac{3\pi}{2}\zeta\left(3\right)+2\pi\zeta\left(2\right)\log{\left(2\right)}+\frac{4\pi}{3}\log^3{\left(2\right)}}$$
Hence:
$$\boxed{\int_0^\infty\frac{\arctan^2\left(x\right)\log^2\left(1+x^2\right)}{x^2}dx=\frac{\pi}{2}\zeta\left(3\right)+4\pi\zeta\left(2\right)\log{\left(2\right)}+\frac{4\pi}{3}\log^3{\left(2\right)}}$$
Here is an idea:
From messing around with coefficients on OEIS A049034, it seems that $$ \arctan^2(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2+2n}}{2n+2}\left(H_{n+\frac{1}{2}}+\log(4)\right) $$ then $$ I=\int_0^{\infty} \frac{\arctan^2 x \log^2 (1+x^2)}{x^2}\,dx $$ $$ I=\int_0^{\infty} \frac{\sum_{n=0}^\infty \frac{(-1)^n x^{2+2n}}{2n+2}\left(H_{n+\frac{1}{2}}+\log(4)\right) \log^2 (1+x^2)}{x^2}\,dx $$ $$ I=\int_0^{\infty} \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{2n+2}\left(H_{n+\frac{1}{2}}+\log(4)\right) \log^2 (1+x^2)\,dx $$ $$ I=\log(4)\sum_{n=0}^\infty \frac{(-1)^n }{2n+2}\int_0^{\infty} x^{2n}\log^2 (1+x^2)\,dx \\+ \sum_{n=0}^\infty \frac{(-1)^n H_{n+\frac{1}{2}}}{2n+2} \int_0^{\infty} x^{2n} \log^2 (1+x^2)\,dx $$ the Mellin transform of $\log(x^2+1)^2$ is $$ \int_0^\infty x^{s-1} \log(x^2+1)\;dx = -\frac{2\pi}{s} \csc\left(\frac{\pi s}{2}\right)\left(\gamma + \psi\left(-\frac{s}{2}\right)\right) $$ setting $s=2n+1$ then gives $$ \int_0^{\infty} x^{2n} \log^2 (1+x^2)\,dx = -(-1)^n\frac{2\pi}{1+2n} H_{-n-\frac{3}{2}} $$ $$ I=-\log(4)\sum_{n=0}^\infty \frac{1}{2n+2}\frac{2\pi}{1+2n} H_{-n-\frac{3}{2}} \\-\sum_{n=0}^\infty \frac{H_{n+\frac{1}{2}}}{2n+2}\frac{2\pi}{1+2n} H_{-n-\frac{3}{2}} $$ this gives $$ I=-2\pi\sum_{n=0}^\infty \frac{H_{-n-\frac{3}{2}}\left(\log(4)+H_{n+\frac{1}{2}}\right)}{2+6n+4n^2} $$ but I don't trust the result yet because they don't appear to be numerically the same. There might be an error somewhere, but I'll leave it here and edit it if I work it out.
