TITLE: A closed form for $\int_0^\infty\frac{\ln(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx$
QUESTION [35 upvotes]: I need to a evaluate the following integral
$$I=\int_0^\infty\frac{\ln(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx.$$
Both Mathematica  and Maple failed to evaluate it in a closed form, and lookups of the approximate numeric value $4.555919963334436...$ in ISC+ and WolframAlpha did not return plausible closed form candidates either. Does anybody by any chance have an idea if a closed form exists for this integral, and what it could be?

REPLY [5 votes]: This integral in more general form you can find in the tables of integrals by Prudnikov A. P., Brychkov Yu. A. and Marichev O. I.:
$$\int_0^{\infty}\!\!\!\!\frac{x^{\alpha-1}\ln^n(cx+d)}{(ax+b)^\sigma(cx+d)^\rho} \!\mathrm dx=(-1)^n\left(\!\frac{d}{c}\!\right)^\alpha\!\! b^{-\sigma}\!\frac{\partial^n}{\partial \rho^n}\!\!\left[d^{-\rho}\mathbf{B}(\alpha,\sigma+\rho-\alpha) \ _2F_1\!\!\left(\!\sigma,\alpha;\sigma+\rho;1\!-\!\frac{ad}{bc}\!\right)\!\right]$$
when $\Re\{\alpha\}>0; \ \Re\{\alpha-\sigma-\rho\}<0; \ |\arg(ax+b)|,|\arg(cx+d)|<\pi \  \forall x\in[0,\infty)$,  where $\mathbf{B}(\cdot \ , \ \cdot)$ -is beta function and $_2F_1$ - is Gauss hypergeometric function.
Here we have $\alpha=\rho=\sigma=\frac{1}{2}, \ a=c=1,  \ d=4, \ b=3$ so all the conditions are satisfied. 
Then
$$
\begin{eqnarray}
I=\int_0^\infty\frac{\ln(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}\mathrm dx&=&-\!\frac{2}{\sqrt{3}}\!\!\left(\!\frac{\partial}{\partial \rho}\!\!\left[\frac{\mathbf{B}\left(\frac{1}{2},\rho\right)}{4^{\rho}} \ _2F_1\!\left(\frac{1}{2},\frac{1}{2};\rho+\frac{1}{2};-\frac{1}{3}\right)\!\right]\right)_{\rho=\frac{1}{2}}=\\
&=& -\!\frac{2 \ _2F_1\!\left(\frac{1}{2},\frac{1}{2};1;-\frac{1}{3}\right)}{\sqrt{3}}\!\!\left(\!\frac{\partial}{\partial \rho}\!\!\left[\frac{\mathbf{B}\left(\frac{1}{2},\rho\right)}{4^{\rho}}\!\right]\right)_{\rho=\frac{1}{2}}-\\
&-&\!\frac{\mathbf{B}\left(\frac{1}{2},\frac{1}{2}\right)}{\sqrt{3}}\!\!\left(\!\frac{\partial}{\partial \rho}\!\!\left[\ _2F_1\!\left(\frac{1}{2},\frac{1}{2};\rho+\frac{1}{2};-\frac{1}{3}\right)\!\right]\right)_{\rho=\frac{1}{2}}
\end{eqnarray}
$$
Then we perform some simplifications (one can use Wolfram Mathematica ;) ) $$\left(\!\frac{\partial}{\partial \rho}\!\!\left[\frac{\mathbf{B}\left(\frac{1}{2},\rho\right)}{4^{\rho}}\!\right]\right)_{\rho=\frac{1}{2}}=-2\pi\ln 2, \\ \mathbf{B}\left(\frac{1}{2},\frac{1}{2}\right)=\pi, \qquad \ _2F_1\!\left(\frac{1}{2},\frac{1}{2};1;-\frac{1}{3}\right)=\frac{2}{\pi } \mathbf{K}\left(i\frac{1}{\sqrt{3}}\right)$$
where $\mathbf{K}(\cdot)$ is the complete elliptic integral of the first kind.
So $$I=\frac{8\ln 2}{\sqrt{3} }\mathbf{K}\left(i\frac{1}{\sqrt{3}}\right)-\frac{\pi}{\sqrt{3} }\left(\!\frac{\partial}{\partial \rho}\!\!\left[\ _2F_1\!\left(\frac{1}{2},\frac{1}{2};\rho+\frac{1}{2};-\frac{1}{3}\right)\!\right]\right)_{\rho=\frac{1}{2}}\approx 4.55592$$