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?
Here is my solution:
Part 1
Let $$I(a)=\int_0^\infty\frac{\log(1+x^2 )}{\sqrt{(1+x^2 )(a^2+x^2 )}}dx\tag{1}$$ for $a>1$. Then rewrite the integral as $$ \begin{align*} I(a) &= \frac{1}{2}\int_1^\infty\frac{\log x}{\sqrt{x(x-1)(x+a^2-1)}}dx\quad x \mapsto \sqrt{x-1} \\ &= \frac{1}{2}\text{Re}\int_0^\infty\frac{\log x}{\sqrt{x(x-1)(x+a^2-1)}}dx \\ &= 2 \text{Re}\int_0^\infty \frac{\log x}{\sqrt{(x^2-1)(x^2+a^2-1)}}dx\quad x\mapsto x^2 \\ &\stackrel{\color{blue}{[1]}}{=} \text{Re} \left( -i K(a)\log\left(i\sqrt{a^2-1}\right)\right) \\ &= \text{Re} \left( K(a)\left( \frac{\pi}{2}-\frac{i}{2}\log(a^2-1)\right)\right) \end{align*} $$ where $K(k)$ denotes the complete elliptic integral of the first kind. Furthermore, it is known that $$K(k)=\frac{K\left(\frac{1}{k} \right)+iK'\left(\frac{1}{k} \right)}{k} \quad k>1$$ Thus, we obtain $$ \begin{align*} I(a)&= \text{Re} \left\{ \frac{K\left(\frac{1}{a} \right)+iK'\left(\frac{1}{a} \right)}{a}\left( \frac{\pi}{2}-\frac{i}{2}\log(a^2-1)\right)\right\} \\ &= \frac{\pi}{2a}K\left(\frac{1}{a} \right)+\frac{\log(a^2-1)}{2a}K'\left(\frac{1}{a}\right)\tag{2} \end{align*} $$
Part 2
Now, we turn our attention back to the original problem.
$$ \begin{align*} \int_0^\infty\frac{\log(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx &=2 \int_0^\infty \frac{\log(4+x^2)}{\sqrt{(3+x^2)(4+x^2)}}dx \quad x\mapsto x^2 \\ &= 4\int_0^\infty \frac{\log(4+4x^2)}{\sqrt{(3+4x^2)(4+4x^2)}}dx \quad x\mapsto 2x \\ &= 2\int_0^\infty \frac{2\log(2)+\log(1+x^2)}{\sqrt{(4x^2+3 )(1+x^2)}}dx \tag{3} \end{align*} $$ The first integral is straightforward and it's value is $$\int_0^\infty \frac{1}{\sqrt{(4x^2+3)(1+x^2)}}dx\stackrel{\color{blue}{[2]}}{=} \frac{1}{2}K\left(\frac{1}{2} \right)$$ And the second integral can be evaluated as follows: $$ \begin{align*} &\; \int_0^\infty \frac{\log(1+x^2)}{\sqrt{(4x^2+3 )(1+x^2)}}dx \\ &= \int_0^\infty \frac{\log(1+x^2)-2\log(x)}{\sqrt{(4+3x^2 )(1+x^2)}}dx\quad x\mapsto 1/x \\ &\stackrel{\color{blue}{[1]}}{=}\frac{1}{\sqrt{3}}\int_0^\infty \frac{\log(1+x^2)}{\sqrt{\left(x^2+\frac{4}{3} \right)(1+x^2)}}dx+\frac{1}{2}\log\left(\frac{\sqrt{3}}{2} \right)K\left( \frac{1}{2}\right) \\ &= \frac{1}{\sqrt{3}}I\left(\frac{2}{\sqrt{3}} \right)+\frac{1}{2}\log\left(\frac{\sqrt{3}}{2} \right)K\left( \frac{1}{2}\right) \\ &\stackrel{\color{blue}{\text{eq }(2)}}{=} \frac{1}{\sqrt{3}}\left(\frac{\pi\sqrt{3}}{4}K\left(\frac{\sqrt{3}}{2}\right) -\frac{\log(3)\sqrt{3}}{4}K\left(\frac{1}{2}\right)\right)+\frac{1}{2}\log\left(\frac{\sqrt{3}}{2} \right)K\left( \frac{1}{2}\right) \\ &= \frac{\pi}{4}K\left(\frac{\sqrt{3}}{2}\right)-\frac{\log 2}{2}K\left(\frac{1}{2} \right) \end{align*} $$ Substituting everything in equation $(3)$, we get $$\int_0^\infty\frac{\log(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx=K\left(\frac{\sqrt3}2\right)\frac\pi2+K\left(\frac12\right)\log(2)$$
Explanations
$\color{blue}{[1]}$ Refer to equation $(7.13)$ of this paper.
$\color{blue}{[2]}$ In general, we have $$\int_0^\infty \frac{1}{\sqrt{(1+x^2)(a^2+x^2)}}dx=\begin{cases}\displaystyle\frac{1}{a}K' \left(\frac{1}{a} \right)\quad \text{if }a>1 \\ K'(a) \quad \text{if } 0<a<1\end{cases}$$
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 $$I=\frac{8\ln 2}{\sqrt{3} }\mathbf{K}\left(i\frac{1}{\sqrt{3}}\right)-\frac{\pi}{\sqrt{3} }\left(\!\frac{\partial}{\partial \rho}\!\!\left[\ _2F_1\!\left(\frac{1}{2},\frac{1}{2};\rho+\frac{1}{2};-\frac{1}{3}\right)\!\right]\right)_{\rho=\frac{1}{2}}\approx 4.55592$$
Too long for a comment: Apparently, not even Feynman can make this integral any easier :
$$I(a)=\int_0^\infty\frac{(x+4)^a}{\sqrt{x(x+3)}}dx\quad\iff\quad I'\Big(\!\!-\tfrac12\!\Big)=\int_0^\infty\frac{\ln(x+4)}{\sqrt{x(x+3)(x+4)}}dx$$
Now, expanding the denominator into binomial series, then switching the order of summation and integration, and rewriting the result in terms of the hypergeometric function $_2F_1$ , we have
$$I(a)=\frac{\sqrt\pi}{\cos a\pi}\bigg[3^a\,\Gamma(-a)\,_{_2}F_{_1}\bigg(\begin{align}-a\ ,-a\\\tfrac12-a\end{align}\ ;\ \frac43\bigg)-\frac{2^{2a+1}\cdot\pi\cdot\ _{_2}F_{_1}\bigg(\ \begin{align}\tfrac12\ ,\tfrac12\\\tfrac32+a\end{align}\ ;\ \dfrac43\bigg)}{\sqrt3\cdot\Gamma(-a)}\bigg]$$
Notice how for $a=-\frac12$ we have for $I(a)$ an indeterminate form, $\frac00$. Also, notice how for the same value of a, the argument of both hypergeometric functions becomes $\Big(\frac12,\frac12;1;\frac43\Big)$, making them equivalent to two elliptic integrals, as has already been mentioned by Vladimir earlier. Now all that is left to do is to take the derivative, itself no trivial task. Nevertheless, such differentiation formulas for the Gaussian function are known to exist, with regard to any of its four parameters, as can be seen, for instance, on the Wolfram MathWorld site. And all of them will inevitably feature the $\psi$ or polygamma function, when taken with regard to any of the first three parameters, as is clearly the case here. The same function will also appear when differentiating $\Gamma(-a)$.
