A closed form for $\int_0^\infty\ln x\cdot\ln\left(1+\frac1{2\cosh x}\right)dx$

Define

$$ I(\alpha) = \int_{0}^{\infty} \log x \log(1 - e^{-\alpha x}) \, dx. $$

Integrating by parts, followed by the substitution $\alpha x \mapsto x$, we have

\begin{align*} I(\alpha) &= \alpha \int_{0}^{\infty} \frac{x - x\log x}{e^{\alpha x} - 1} \, dx \\ &= \frac{1}{\alpha} \int_{0}^{\infty} \frac{(1+\log \alpha) x - x \log x}{e^{x} - 1} \, dx\\ &= \frac{1}{\alpha} \left\{ (1+\log\alpha)\zeta(2) - \left.\frac{d \zeta(s)\Gamma(s)}{s}\right|_{s=2} \right\}\\ &= \frac{1}{\alpha} \left\{ (\gamma+\log\alpha)\zeta(2) - \zeta'(2) \right\}. \end{align*}

Then it follows that

\begin{align*} \int_{0}^{\infty} \log x \log \left( 1 + \frac{1}{2\cosh x} \right) \, dx &= \int_{0}^{\infty} \log x \log \left( \frac{1 - e^{-3x}}{1 - e^{-x}} \cdot \frac{1 - e^{-2x}}{1 - e^{-4x}} \right) \, dx \\ &= I(2) + I(3) - I(1) - I(4) \\ &= \frac{5}{12} \zeta'(2) - \frac{5}{72}\gamma\pi^{2} + \frac{1}{18}\pi^{2} \log (3). \end{align*}

Plugging some identities relating $\zeta'(2)$ and the Glaisher-Kinkelin constant $A$, this reduces to Vladimir's answer.

Addendum - Something you might want to know:

The following identity played the key role in this proof.

$$ \int_{0}^{\infty} \frac{x^{s-1}}{e^{x} - 1} \, dx = \Gamma(s)\zeta(s), $$

which holds for $\Re s > 1$.