Yes, there is an elementary closed form for this integral:
$$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx=\frac{\pi}{2\,\sqrt2}\cdot\exp\left(\frac1{\sqrt2}\right)\cdot\frac{\sin\left(\frac1{\sqrt2}\right)-\cos\left(\frac1{\sqrt2}\right)+2\,\exp\left(\frac1{\sqrt2}\right)}{1-4\,\exp\left(\frac1{\sqrt2}\right)\cos\left(\frac1{\sqrt2}\right)+4\,\exp\left(\sqrt2\right)}\tag1$$
Proof:
Let us denote the integral in question as
$$\mathcal{I}=\int_0^\infty\frac{2-\cos x}{\left(x^4+1\right)\,\left(5-4\cos x\right)}dx\tag2$$
Note that the trigonometric part of the integrand is a periodic function and can be expanded to a Fourier series with particularly simple coefficients:
$$\frac{2-\cos x}{5-4\cos x}=\sum_{n=0}^\infty\frac{\cos(n\,x)}{2^{n+1}}\tag3$$
(this can be easily checked by expressing cosines via exponents of an imaginary argument).
Now we can integrate it term-wise:
$$\mathcal{I}=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\int_0^\infty\frac{\cos(n\,x)}{x^4+1}dx\right)=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\cdot\frac{\pi}{2\,\sqrt2}\cdot\exp\left(-\frac{n}{\sqrt2}\right)\cdot\left(\sin\left(\frac{n}{\sqrt2}\right)+\cos\left(\frac{n}{\sqrt2}\right)\right)\right)\tag4$$
(for the integral, see DLMF 1.14, vii, Table 1.14.2, $4^{th}$ row).
Trig functions in the last sum can again be expressed via exponents of an imaginary argument, and then the sum is easily evaluated. Converting exponents back to trig functions and getting rid of complex numbers, we get the final result $(1)$.
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
$$I=\frac\pi{20}\ln^23+\frac\pi4\operatorname{Li_2}\left(\tfrac13\right)-\frac15\operatorname{Ti}_3\left(\sqrt3\right),$$ where $$\operatorname{Ti}_3\left(\sqrt3\right)=\Im\Big[\operatorname{Li}_3\left(i\sqrt3\right)\Big]=\frac{\sqrt3}8\Phi\left(-3,3,\tfrac12\right)=\frac{5\sqrt3}4\,{_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac34\right)-\frac{5\pi^3}{432}-\frac\pi{16}\ln^23.$$