Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$ 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.
 A: 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.
A: \begin{align}
-\frac{1}{4}\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x
&=-\frac{1}{4}\int^\infty_0\frac{\ln{\sinh{x}}}{1+\sinh{x}}e^{x/2}{\rm d}x\\
&=-\frac{1}{2}\int^\infty_1\frac{\ln\left(\frac{x^2-x^{-2}}{2}\right)}{1+\frac{x^2-x^{-2}}{2}}{\rm d}x\\
&=\int^1_0\frac{\ln(1-x^4)-\ln(2x^2)}{x^4-2x^2-1}{\rm d}x\\
\end{align}
For simplicity's sake, let $\displaystyle\frac{1}{x^4-2x^2-1}=\sum^4_{k=1}\frac{c_k}{x-r_k}$. The integral becomes
$$I=-4\sum^4_{k=1}\int^1_0\frac{c_k\ln(1-x^4)-c_k\ln{2}-2c_k\ln{x}}{x-r_k}{\rm d}x$$
The second integral is
\begin{align}
-c_k\ln{2}\int^1_0\frac{1}{x-r_k}{\rm d}x=-c_k\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)
\end{align}
The third integral is
\begin{align}
-2c_k\int^1_0\frac{\ln{x}}{x-r_k}{\rm d}x
&=2c_k\int^{1/r_k}_0\frac{\ln(r_kx)}{1-x}{\rm d}x\\
&=-2c_k{\rm Li}_2\left(\frac{1}{r_k}\right)
\end{align}
Pluck these results back in.
$$I=4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)-\int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\right]$$
The remaining integral is
\begin{align}
&\ \ \ \ \ \int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\\
&=\sum_{j=1,-1,i,-i}\int^1_0\frac{\ln(1+jx)}{x-r_k}{\rm d}x\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\ln\left(1+\frac{j\lambda}{y}+jr_k\right)\frac{{\rm d}y}{y}\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\left[\ln\left(1+\frac{1+jr_k}{j\lambda}y\right)-\ln\left(\frac{y}{j\lambda}\right)\right]\frac{{\rm d}y}{y}\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{1+jr_k}{j-jr_k}}_{-\frac{1+jr_k}{jr_k}}\frac{\ln(1+y)}{y}-\frac{1}{y}\ln\left(\frac{y}{1+jr_k}\right){\rm d}y\\
&=-\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right]
\end{align}
Final Result:
\begin{align}
\color\purple{\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x
=4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)+\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right]\right]}
\end{align}
where
\begin{align}
\frac{c_1}{x-r_1}&=\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x-\sqrt{1+\sqrt{2}}\right)}\\
\frac{c_2}{x-r_2}&=-\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x+\sqrt{1+\sqrt{2}}\right)}\\
\frac{c_3}{x-r_3}&=-\frac{1}{4\sqrt{2}\left(x+i\sqrt{\sqrt{2}-1}\right)}\\
\frac{c_4}{x-r_4}&=-\frac{1}{4\sqrt{2}\left(x-i\sqrt{\sqrt{2}-1}\right)}\\
\end{align}
A: Okay, first of all: it is not easy to answer this question, and I'm not able to show the truth in an elegant way. I did the job with Maple computer algebra system (CAS).
In this form there is no chance to evaulate to solution even with CAS, so we transform into the form, what @Lucian offered. Let $f$ be
$$f:=\frac{\ln(\sinh x)}{1+ \sinh x} \cdot e^{x/2}$$
I calculated for you the indefinite integral $\int f(x) \ dx$. It is too long to render with $\LaTeX$, so I paste here the raw Maple output:

After that I calculated the definite integral $\int_0^{\infty} f(x) \ dx$. Here is the closed form:

Which is by 100 digits precision: $4.253149825368695481030630266176679795147658282 \\ 387249226057297960587772473334190569825379100016048097$
A good approximation for this integral for example
$$\frac{5}{9} \cdot 3^{4/5} \cdot 5^{2/3} \cdot \ln(3)^{8/9}.$$
The value of this one is $4.2531498232523271$ which is correct for $9$ digits, and also has quite an easy form.
If you need the closed form in raw text I can paste here. For a more accurate answer I think we need here @Cleo or @Tunk-Fey.
