TITLE: Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$
QUESTION [38 upvotes]: Please help me to evaluate this integral:
$$
I={\large\int}_{0}^{\infty}{\ln\left(x\right) \over 1 + x}\,
\,\sqrt{\,x + \sqrt{\,1 + x^{2}\,}\, \over 1 + x^{2}\,}\,\,{\rm d}x.\tag1
$$
Mathematica could not evaluate it in a closed form. A numerical integration returned
$$I \approx 4.25314982536869548103063\ldots\,,\tag2$$
but neither
WolframAlpha nor ISC+ could find a plausible closed form for this.

REPLY [27 votes]: With some help from a CAS I got this result with only one dilogarithm term:
$$I=\frac{\sqrt[4]2\sqrt{2-\sqrt2}}{16}\\\left\{\left[16\ln2+\left(1+\sqrt2\right)\left(4\pi-16\arctan\sqrt{\sqrt2-1}\right)\right]\ln\left(1+\sqrt2+\sqrt{2+2\sqrt2}\right)\\-\Big(32\left(1+\sqrt2\right)\ln2+4\pi\Big)\arctan\sqrt{\sqrt2-1}-3\pi\arctan\frac{2\sqrt{2+10\sqrt2}}7\\-16\arctan^2\sqrt{\sqrt2-1}+4\ln^2\!\left(1+\sqrt2+\sqrt{2+2\sqrt2}\right)\\+3\pi^2-\frac{32}{\sqrt{2-\sqrt2}}\,\Re\left[\left(-1\right)^{5/8}\operatorname{Li}_2\!\left(2i\left( 1+\sqrt{1+i}\right)\right)\right]\right\}.$$

Update: It was suggested in comments that this expression gives a numerically incorrect result when evaluated with Maple. Here is what I got with Maple 18:

This value numerically matches the integral. I also checked it with Mathematica 10 and it gave the same numeric result.