TITLE: Calculating $\int_0^{\infty } \frac{\log (v+1)}{\sqrt{(v+1)^2+1} \sqrt{(v+1)^2+4 \sqrt{(v+1)^2+1} (v+1)+4}} \, dv$
QUESTION [7 upvotes]: What tools would you recommend me for this one? $$\int_0^{\infty } \frac{\log (v+1)}{\sqrt{(v+1)^2+1} \sqrt{(v+1)^2+4 \sqrt{(v+1)^2+1} (v+1)+4}} \, dv$$
It's related to Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ if my so long calculations are correct, and it only represents a tiny bit of the whole story.

REPLY [2 votes]: (Way too long for comment. I wanted to post my progress so far on this problem because I find it interesting. I'll delete this in the event that someone else can provide a complete solution.)

Define $\mathcal{I}$ to be the value of the definite integral
$$\mathcal{I}:=\int_{0}^{\infty}\frac{\ln{\left(v+1\right)}}{\sqrt{\left(v+1\right)^2+1}\sqrt{4+\left(v+1\right)^2+4\left(v+1\right)\sqrt{\left(v+1\right)^2+1}}}\,\mathrm{d}v.$$
As an approximate numerical value, WRA found $\mathcal{I}\approx0.407122.$
Applying the obvious simplifying substitution, $v+1=x$, the integral becomes
$$\mathcal{I}=\int_{1}^{\infty}\frac{\ln{\left(x\right)}}{\sqrt{x^2+1}\sqrt{4+x^2+4x\sqrt{x^2+1}}}\,\mathrm{d}x,$$
and transforming the integral via the reciprocal substitution $\frac{1}{x}=y$ transforms the integral to one over the unit interval:
$$\begin{align}
\mathcal{I}
&=\int_{1}^{\infty}\frac{\ln{\left(x\right)}}{\sqrt{x^2+1}\sqrt{4+x^2+4x\sqrt{x^2+1}}}\,\mathrm{d}x\\
&=\int_{1}^{0}\frac{\ln{\left(\frac{1}{y}\right)}}{\sqrt{\frac{1}{y^2}+1}\sqrt{4+\frac{1}{y^2}+\frac{4}{y}\sqrt{\frac{1}{y^2}+1}}}\cdot\frac{(-1)\,\mathrm{d}y}{y^2};~~~\small{\left[\frac{1}{x}=y\right]}\\
&=-\int_{0}^{1}\frac{\ln{\left(y\right)}}{\sqrt{1+y^2}\sqrt{4y^2+1+4\sqrt{1+y^2}}}\,\mathrm{d}y.\\
\end{align}$$
Next, consider an Euler substitution of the first kind of the form $a\left(\sqrt{1+y^2}-y\right)=z$, where $a\in\mathbb{R}^{+}$. Then,
$$y=\frac{a^2-z^2}{2az}\implies\mathrm{d}y=-\frac{a^2+z^2}{2az^2}\,\mathrm{d}z,$$
and
$$\sqrt{1+y^2}=\frac{a^2+z^2}{2az}\implies\frac{\mathrm{d}y}{\sqrt{1+y^2}}=-\frac{\mathrm{d}z}{z}.$$
Thus,
$$\begin{align}
\mathcal{I}
&=-\int_{0}^{1}\frac{\ln{\left(y\right)}}{\sqrt{4y^2+1+4\sqrt{1+y^2}}}\cdot\frac{\mathrm{d}y}{\sqrt{1+y^2}}\\
&=-\int_{a}^{\left(\sqrt{2}-1\right)a}\frac{\ln{\left(\frac{a^2-z^2}{2az}\right)}}{\sqrt{\frac{a^4-a^2z^2+z^4}{a^2z^2}+4\cdot\frac{a^2+z^2}{2az}}}\cdot\frac{(-1)\,\mathrm{d}z}{z};~~~\small{\left[a\left(\sqrt{1+y^2}-y\right)=z\right]}\\
&=\int_{a}^{\left(\sqrt{2}-1\right)a}\frac{a\,\ln{\left(\frac{a^2-z^2}{2az}\right)}}{\sqrt{z^4+2az^3-a^2z^2+2a^3z+a^4}}\,\mathrm{d}z\\
&=-a\int_{\left(\sqrt{2}-1\right)a}^{a}\frac{\ln{\left(\frac{a^2-z^2}{2az}\right)}}{\sqrt{\left(z^2+3az+a^2\right)\left(z^2-az+a^2\right)}}\,\mathrm{d}z.\\
\end{align}$$
We may as well let $a=1$, in which case
$$\mathcal{I}=-\int_{\sqrt{2}-1}^{1}\frac{\ln{\left(\frac{1-z^2}{2z}\right)}}{\sqrt{\left(z^2+3z+1\right)\left(z^2-z+1\right)}}\,\mathrm{d}z.$$
Substituting the linear fractional transformation, $\frac{1-z}{1+z}=t$, we eliminate the odd terms from the quartic under the radical in the integrand:
$$\begin{align}
\mathcal{I}
&=-\int_{\sqrt{2}-1}^{1}\frac{\ln{\left(\frac{1-z^2}{2z}\right)}}{\sqrt{\left(z^2+3z+1\right)\left(z^2-z+1\right)}}\,\mathrm{d}z\\
&=-\int_{\sqrt{2}-1}^{0}\frac{\ln{\left(\frac{2t}{1-t^2}\right)}}{\sqrt{\left(\frac{5-t^2}{\left(1+t\right)^2}\right)\left(\frac{1+3t^2}{\left(1+t\right)^2}\right)}}\cdot\frac{(-2)\,\mathrm{d}t}{\left(1+t\right)^2};~~~\small{\left[\frac{1-z}{1+z}=t\right]}\\
&=\int_{0}^{\sqrt{2}-1}\frac{2\,\ln{\left(\frac{1-t^2}{2t}\right)}}{\sqrt{\left(5-t^2\right)\left(1+3t^2\right)}}\,\mathrm{d}t\\
&=\frac{2}{\sqrt{5}}\int_{0}^{\sqrt{2}-1}\frac{\ln{\left(\frac{1-t^2}{2t}\right)}}{\sqrt{\left(1-\frac15t^2\right)\left(1+3t^2\right)}}\,\mathrm{d}t.\\
\end{align}$$
Then, rescaling the integrand and applying the technique of trigonometric substitution, the integrand starts to look a lot like that of an elliptic integral:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{\sqrt{2}-1}\frac{2\,\ln{\left(\frac{1-t^2}{2t}\right)}}{\sqrt{\left(5-t^2\right)\left(1+3t^2\right)}}\,\mathrm{d}t\\
&=\frac{2}{\sqrt{5}}\int_{0}^{\sqrt{2}-1}\frac{\ln{\left(\frac{1-t^2}{2t}\right)}}{\sqrt{\left(1-\frac15t^2\right)\left(1+3t^2\right)}}\,\mathrm{d}t\\
&=\frac{2}{\sqrt{15}}\int_{0}^{\sqrt{6}-\sqrt{3}}\frac{\ln{\left(\frac{3-u^2}{2\sqrt{3}\,u}\right)}}{\sqrt{\left(1-\frac{1}{15}u^2\right)\left(1+u^2\right)}}\,\mathrm{d}u;~~~\small{\left[\sqrt{3}\,t=u\right]}\\
&=-2\int_{\frac{\pi}{2}}^{\arccos{\left(\frac{\sqrt{2}-1}{\sqrt{5}}\right)}}\frac{\ln{\left(\frac{3-15\cos^2{\left(\theta\right)}}{6\sqrt{5}\cos{\left(\theta\right)}}\right)}}{\sqrt{\sin^2{\left(\theta\right)}\left(1+15\cos^2{\left(\theta\right)}\right)}}\,\sin{\left(\theta\right)}\,\mathrm{d}\theta;~~~\small{\left[\arccos{\left(\frac{u}{\sqrt{15}}\right)}=\theta\right]}\\
&=2\int_{\arccos{\left(\frac{\sqrt{2}-1}{\sqrt{5}}\right)}}^{\frac{\pi}{2}}\frac{\ln{\left(\frac{1-5\cos^2{\left(\theta\right)}}{2\sqrt{5}\cos{\left(\theta\right)}}\right)}}{\sqrt{1+15\cos^2{\left(\theta\right)}}}\,\mathrm{d}\theta\\
&=2\int_{\arccos{\left(\frac{\sqrt{2}-1}{\sqrt{5}}\right)}}^{\frac{\pi}{2}}\frac{\ln{\left(\frac{1-5\cos^2{\left(\theta\right)}}{2\sqrt{5}\cos{\left(\theta\right)}}\right)}}{\sqrt{16-15\sin^2{\left(\theta\right)}}}\,\mathrm{d}\theta\\
&=\frac12\int_{\arccos{\left(\frac{\sqrt{2}-1}{\sqrt{5}}\right)}}^{\frac{\pi}{2}}\frac{\ln{\left(\frac{1-5\cos^2{\left(\theta\right)}}{2\sqrt{5}\cos{\left(\theta\right)}}\right)}}{\sqrt{1-\frac{15}{16}\sin^2{\left(\theta\right)}}}\,\mathrm{d}\theta.\\
\end{align}$$
To be continued...