A novelty integral for $\pi$

My lab friends always play a mentally challenging brain game every month to keep our mind running on all four cylinders and the last month challenge was to find a novelty expression for $\pi$. In order to stick to the rule, of course we must avoid the good old Ramanujan and online available expressions, for instance: the coolest ways of expressing $\pi$ on Quora. The winner of the last month challenge is this integral

$${\large\int_0^\infty}\frac{(1+x)\log(1+x)(2+\log x)\log\left(\!\frac{1+x}{2}\!\right)-2x\log(1+x)\log x}{x^{3/2}(1+x)\log^2x}\ dx={\Large\pi}$$

The equality is precise to at least thousand decimal places. Unfortunately, my friend who proposes this integral keeping the mystery to himself. I tried to crack this integral while waiting for a solution to be offered by one of my friends, but failed to get any.

I have tried to break this integral into two part:

$${\large\int_0^\infty}\frac{\log(1+x)(2+\log x)\log\left(\!\frac{1+x}{2}\!\right)}{x^{3/2}\log^2x}\ dx-2{\large\int_0^\infty}\frac{\log(1+x)}{\sqrt{x}(1+x)\log x}\ dx$$

but each integrals diverges. I have tried many substitutions like $x=y-1$, $x=\frac{1}{y}$, or $x=\tan^2y$ hoping for familiar functions, but couldn't get one. I also tried the method of differentiation under integral sign by introducing

$$I(s)={\large\int_0^\infty}x^{s}\cdot\frac{(1+x)\log(1+x)(2+\log x)\log\left(\!\frac{1+x}{2}\!\right)-2x\log(1+x)\log x}{(1+x)\log^2x}\ dx$$

and differentiating twice with respect to $s$ to get rid of $\log^2x$ couldn't work either. I have a strong feeling that I miss something completely obvious in my calculation. I'm not having much success in evaluating this integral since two weeks ago, so I thought it's about time to ask you for help. Can you help me out to prove it, please?

The integrand can be broken up as $$I=\int_0^{\infty} \left(\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x} +\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln x}-\frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x}\right)dx.$$

But, by integration by parts, $$\int \frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x} dx= \int \frac{2\ln(1+x)}{x^{1/2} \ln x} d\left(\ln\left(\frac{1+x}{2}\right)\right) \\=\small\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{1/2} \ln x}-2\int \left( \frac{\ln\left(\frac{1+x}{2}\right)}{x^{1/2} (1+x) \ln x}-\frac{ \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x}-\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{2 x^{3/2} \ln x}\right)dx$$

That is, $$\int \left(\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x} +\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln x}-\frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x}\right)dx \\= -\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{1/2} \ln x}+2 \int \frac{\ln\left(\frac{1+x}{2}\right)}{x^{1/2} (1+x) \ln x} dx$$

And the claim follows from Jack D'Aurizio's preliminary result.

A preliminary result.

$$I_1=\int_{0}^{+\infty}\frac{2\log\left(\frac{1+x}{2}\right)}{\sqrt{x}(1+x)\log(x)}=\color{red}{\pi}.\tag{1}$$

Proof: through the substitution $x=e^t$, $$\begin{eqnarray*}I_1=\int_{-\infty}^{+\infty}\frac{\log\left(\frac{e^t+1}{2}\right)}{\cosh\left(\frac{t}{2}\right)t}\,dt&=&\color{red}{\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dt}{\cosh\left(\frac{t}{2}\right)}}+\color{blue}{\int_{-\infty}^{+\infty}\frac{\log\cosh\left(\frac{t}{2}\right)}{t\cosh\left(\frac{t}{2}\right)}\,dt}\end{eqnarray*}$$ where the red integral is easy to compute and the blue one vanishes by symmetry.

Thanks to nospoon, this is ultimately everything we need to prove the OP's identity through integration by parts.