I will attempt to simplify the integral. It might take me a while and I might edit this answer a few times, so this is not its final form yet.
Let $D$ be the differential operator.
We know $\displaystyle\int \dfrac{dx}{\sqrt{x^2-1}}=\ln(2(x+\sqrt{x^2-1}))$.
So by using integration by parts we reduce the problem to solving $$\displaystyle\mathcal{A}=\int_1^\infty D\left[\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(2(x+\sqrt{x^2-1}))\, dx.$$
We know that $D[\mathrm{arccot}(x)] = \dfrac{-1}{1+x^2} $ hence by the chain rule for derivatives we get
$$\mathcal{A}=\int_1^\infty \dfrac{D\left[\left(\frac{-2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(2(x+\sqrt{x^2-1}))}{1+\left(1+\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}\right)^2}\,dx.$$
Because we know that $D[\mathrm{arccot}(x)] = \dfrac{-1}{1+x^2} $ and also that $ln(2z)=ln(2)+ln(z)$ we can use integration by parts again and arrive at
$$\mathcal{B}=\int_1^\infty \dfrac{D\left[\left(\frac{-2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(x+\sqrt{x^2-1})}{1+\left(1+\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}\right)^2}\,dx.$$
Now we can turn this into an infinite sum because we set $U=\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}$ and use the Taylor expansion for $\frac{z}{1+(1+z)^2}$.
In other words we substitute $z=U$.
( Remember that $\int \Sigma = \Sigma \int$ )
Finally the core problem is reduced mainly to solving:
$$\displaystyle\mathcal{C_n}=\int_1^\infty \frac{\ln(x+\sqrt{x^2-1})}{(\mathrm{arcoth}(x)-\mathrm{arccsc}(x))^n}\, dx.$$
by induction we get the need to solve for $C_1$ and then are able to get the others.
At this point, I must admit that I have ignored convergeance issues.
Those issues can be solved by taking limits.
For instance $C_1$ does not actually converge by itself.
For all clarity the problem is not resolved.
In fact it might require a new bounty. Still thinking about it ...
(in progress)
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
$$\int_0^\infty \frac{1}{\sqrt{(1+x^2)(a^2+x^2)}}dx=\begin{cases}\displaystyle\frac{1}{a}K' \left(\frac{1}{a} \right)\quad \text{if }a>1 \\ K'(a) \quad \text{if } 0<a<1\end{cases}$$
Best Answer
Sub $u=\log{(2^x-1)}$. Then $x=\log{(1+e^u)}/\log{2}$, $dx = (1/\log{2}) (du/(1+e^{-u})$. The integral then becomes
$$\begin{align}\frac{1}{\log{2}} \int_{-\infty}^{\infty} \frac{du}{1+e^{-u}} e^{-u/2} \frac{\frac{\log{(1+e^u)}}{\log{2}}-1}{u} = \frac{1}{2\log^2{2}} \int_{-\infty}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ = \underbrace{\frac{1}{2\log^2{2}} \int_{-\infty}^{0} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}}_{u\rightarrow -u} \\+ \frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ = \underbrace{-\frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^{-u})}-\log{2}}{u}}_{\log{(1+e^{-u})} = \log{(1+e^u)}-u}\\+ \frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ \end{align}$$
The nasty pieces of the integral cancel, and we are left with
$$ \frac{1}{2\log^2{2}}\int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} = \frac{\pi}{2 \log^2{2}} $$
as correctly conjectured.