A closed-form expression for $\int_0^\infty \frac{\ln (1+x^\alpha) \ln (1+x^{-\beta})}{x} \, \mathrm{d} x$
Only a comment.
For $\,z> 0\,$ I’ve got:
$$f(1,z) = \frac{3}{4}\zeta(3)\left(z+\frac{1}{z^2}\right) + 2 g(z)$$
with $\enspace\displaystyle g(z):=\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n} \sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k(zn+k)}\enspace$ and $\enspace\displaystyle g(\frac{1}{z})=zg(z)$
Maybe someone can take it from here to create formulas.
Note: $\enspace\displaystyle f(1,\frac{1}{z})=zf(1,z)\enspace$ which is equivalent to $\enspace\displaystyle \frac{1}{\alpha}f(1,\frac{\beta}{\alpha})=\frac{1}{\beta}f(1,\frac{\alpha}{\beta})$
Hint:
$\displaystyle zg(z)=\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\left(H(zn)- H(\frac{zn}{2})\right)=\int\limits_0^1\frac{Li_2(-t^z) - Li_2(-t^{z/2})}{1-t}dt $
$\hspace{1.1cm}\displaystyle = \frac{\pi^2\ln 2}{12} + \int\limits_0^1\frac{Li_2(-x^z)}{1+x}dx = z \int\limits_0^1\frac{\ln(1+x)\ln(1+x^z)}{x}$
with $\enspace\displaystyle H(x):=\sum\limits_{k=1}^\infty \frac{x}{k(x+k)}=\gamma + \psi(1+x) =\int\limits_0^1\frac{1-t^x}{1-t}dt$
It's trivial that $\,f(1,z)\,$ can be split:
$\hspace{1.1cm}\displaystyle \int\limits_0^1 \frac{\ln(1+x)\ln(1+x^{-z})}{x}dx = \frac{3z}{4}\zeta(3) + g(z) = g(-z)$
$\hspace{1.1cm}\displaystyle \int\limits_1^\infty \frac{\ln(1+x)\ln(1+x^{-z})}{x}dx = \frac{3}{4z^2}\zeta(3) + g(z)$
Only a comment. We have
$$ \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta/x)}{x} \, dx = 2\operatorname{Li}_3(\alpha\beta) - \operatorname{Li}_2(\alpha\beta)\log(\alpha\beta) $$
which is valid initially for $\alpha, \beta > 0$ and extends to a larger domain by the principle of analytic continuation. Then for integers $m, n \geq 1$ we obtain
\begin{align*} f(m, n) &=\int_{0}^{\infty} \frac{\log(1+x^m)\log(1+x^{-n})}{x}\,dx \\ &\hspace{6em} = \sum_{j=0}^{m-1}\sum_{k=0}^{n-1} \left[ 2\operatorname{Li}_3\left(e^{i(\alpha_j+\beta_k)}\right) - i(\alpha_j+\beta_k)\operatorname{Li}_2\left(e^{i(\alpha_j+\beta_k)}\right) \right], \end{align*}
where $\alpha_j = \frac{2j-m+1}{n}\pi$ and $\beta_k = \frac{2k-n+1}{n}\pi$. (Although we cannot always split complex logarithms, this happens to work in the above situation.) By the multiplication formula, this simplifies to
\begin{align*} f(m, n) &= \frac{2\gcd(m,n)^3}{m^2n^2}\operatorname{Li}_3\left((-1)^{(m+n)/\gcd(m,n)}\right) \\ &\hspace{2em} - \frac{i}{n} \sum_{j=0}^{m-1} \alpha_j \operatorname{Li}_2\left((-1)^{n-1}e^{in\alpha_j}\right) \\ &\hspace{2em} - \frac{i}{m} \sum_{k=0}^{n-1} \beta_k \operatorname{Li}_2\left((-1)^{m-1}e^{im\beta_k}\right). \end{align*}
Here, $\gcd(m,n)$ is the greatest common divisor of $m$ and $n$.
The following code tests the above formula.
The integral of $x^p$ times a product of two Meijer G-functions gives a Fox H-function: $$\int_0^\infty \frac {\ln(1+x^\alpha) \ln(1+x^{-\beta})} x dx = \frac 1 \alpha \int_0^\infty G_{2,2}^{1,2} \left( x \middle| {1, 1 \atop 1, 0} \right) G_{2,2}^{2,1} \left( x^{\beta/\alpha} \middle| {0, 1 \atop 0, 0} \right) \frac {dx} x = \\ \frac 1 \alpha H_{4,4}^{4,2} \left( 1 \middle| {(0, 1), (0, \frac \beta \alpha), (1, \frac \beta \alpha), (1, 1) \atop (0, 1), (0, 1), (0, \frac \beta \alpha), (0, \frac \beta \alpha)} \right) = \frac 1 {2 \pi i } \int_\mathcal L \frac {\pi^2 \csc \pi s \csc \frac {\pi \beta s} \alpha} {\beta s^2} ds.$$ When $\beta/\alpha$ is rational, the H-function reduces to a G-function.