By replacing $\phi$ with $\arctan(t)$, then using integration by parts, we have:
$$ I = \int_{0}^{1}\frac{1}{1+t^2}\,\arctan\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)\,dt =\frac{\pi^2}{8}-\int_{0}^{1}\frac{3\sqrt{2}\, t \arctan(t)}{(3-t^2)\sqrt{1-t^2}}\,dt.$$
Now comes the magic. Since:
$$\int \frac{3\sqrt{2}\,t}{(3-t^2)\sqrt{1-t^2}}\,dt = -3\arctan\sqrt{\frac{1-t^2}{2}}\tag{1}$$
integrating by parts once again we get:
$$ I = \frac{\pi^2}{8}-3\int_{0}^{1}\frac{1}{1+t^2}\arctan\sqrt{\frac{1-t^2}{2}}\,dt \tag{2}$$
hence we just need to prove that:
$$ \int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\arctan\sqrt{1-2t^2}}{1+t^2}\,dt=\color{red}{\frac{\pi^2}{24}}\tag{3}$$
and this is not difficult since both
$$\int_{0}^{1}\frac{dt}{1+t^2}(1-t^2)^{\frac{2m+1}{2}},\qquad \int_{0}^{\frac{1}{\sqrt{2}}}\frac{(1-2t^2)^{\frac{2m+1}{2}}}{1+t^2}\,dt $$
can be computed through the residue theorem or other techniques. For instance:
$$\int_{0}^{1}\frac{(1-t)^{\frac{2m+1}{2}}}{t^{\frac{1}{2}}(1+t)}\,dt = \sum_{n\geq 0}(-1)^n \int_{0}^{1}(1-t)^{\frac{2m+1}{2}} t^{n-\frac{1}{2}}\,dt=\sum_{n\geq 0}(-1)^n\frac{\Gamma\left(m+\frac{3}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(m+n+2)}$$
or just:
$$\int_{0}^{1}\frac{\sqrt{\frac{1-t^2}{2}}}{(1+t^2)\left(1+\frac{1-t^2}{2}u^2\right)}\,dt = \frac{\pi}{2(1+u^2)}\left(1-\frac{1}{\sqrt{2+u^2}}\right)\tag{4}$$
from which:
$$\int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\frac{\pi}{2}\int_{0}^{1}\frac{du}{1+u^2}\left(1-\frac{1}{\sqrt{2+u^2}}\right) =\color{red}{\frac{\pi^2}{24}} $$
as wanted, since:
$$ \int \frac{du}{(1+u^2)\sqrt{2+u^2}}=\arctan\frac{u}{\sqrt{2+u^2}}.$$
Another approach:
Using the Fourier series $$1 + 2 \sum_{k=1}^{\infty} (-1)^{n} a^{n}\cos(nx) = \frac{1-a^{2}}{1+2a \cos (x) +a^{2}}, \quad |a| <1, $$ we have
$$ \begin{align} \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\frac{1-a^{2}}{1+2a \cos (x) +a^{2}} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\left(1+2 \sum_{n=1}^{\infty} (-1)^{n} a^{n} \cos(nx) \right) \, \mathrm dx \\ &= \pi + 2 \sum_{n=1}^{\infty} (-1)^{n}a^{n}\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \cos(nx) \, \mathrm dx \\ &= \pi - 2a \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \, \cos (x) \, \mathrm dx + 2 \sum_{n=2}^{\infty} (-1)^{n} a^{n} \int_{-\infty}^{\infty}\frac{\sin (x)}{x} \, \cos(nx) \mathrm dx \\ &= \pi - 2a \left(\frac{\pi}{2} \right)+2\sum_{n=2}^{\infty} (-1)^{n}a^{n}(0) \\ &= \pi \left(1-a \right) . \end{align}$$
Rewriting the integral as
$$\frac{1}{1+a^{2}}\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1-a^{2}}{1+\frac{2a}{1+a^{2}}\cos(x)} \, \mathrm dx,$$ and letting $a= 2-\sqrt{3}$, we get $$\sqrt{3} \int_{-\infty}^{\infty}\frac{\sin (x)}{x} \frac{1}{2+\cos(x)} \, \mathrm dx = \pi \left(\sqrt{3}-1 \right). $$
Therefore, $$ \begin{align} \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1+\cos (x)}{2+ \cos (x)} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \, \mathrm dx - \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1}{2+ \cos (x)} \, \mathrm dx \\ &= \pi - \frac{\pi}{\sqrt{3}} \left(\sqrt{3} - 1 \right) \\ &= \frac{\pi}{\sqrt{3}}. \end{align} $$
The one issue with this approach is justification for switching the order of integration and summation. Fubini's theorem is not satisfied.
Fortunately, we can use Sangchul Lee's result from the addendum of this answer to justify the switching.
Best Answer
For any $a>0$ we have that $f_a(x)=\frac{a}{x^2+a^2}$ is an even function with an absolute maximum at the origin such that $\int_{-\infty}^{+\infty}f_a(x)\,dx=\pi$. In terms of the Laplace transform $$(\mathcal{L}^{-1} f_a)(s)=\sin(as)$$ holds, and if we assume that $$ g(x) \stackrel{L^2}{=} c_0 + \sum_{n\geq 1}\left( c_n \cos(nx) + s_n \sin(nx)\right) $$ has a uniformly convergent Fourier series (which is granted, for instance, by $\max(s_n,c_n)=O\left(\frac{1}{n^2}\right)$) we have $$\begin{eqnarray*} \int_{-\infty}^{+\infty}g(x)f_a(x)\,dx &=& \int_{0}^{+\infty}(g(x)+g(-x))f_a(x)\,dx\\&=&\pi c_0+2\sum_{n\geq 1}c_n\int_{0}^{+\infty}\cos(nx)\frac{a}{a^2+x^2}\,dx\\ &=&\pi c_0+2\sum_{n\geq 1}c_n \int_{0}^{+\infty}\frac{s}{n^2+s^2}\sin(as)\,ds \\&=&\pi c_0+2\sum_{n\geq 1}c_n \int_{0}^{+\infty}\frac{s}{1+s^2}\sin(nas)\,ds\\&=&\pi c_0+\pi\sum_{n\geq 1}c_n e^{-na}\end{eqnarray*} $$ by the self-adjointness of the Laplace transform. Since $c_n=O\left(\frac{1}{n^2}\right)$, by the dominated convergence theorem $$ \lim_{a\to 0^+}\int_{-\infty}^{+\infty}g(x)f_a(x)\,dx = \pi\sum_{n\geq 0}c_n = \pi g(0)$$ and by translation $$ \lim_{a\to 0^+}\int_{-\infty}^{+\infty}g(x-\pi/3)f_a(x-\pi/3)\,dx = \pi g(\pi/3).\tag{1}$$ In order to apply $(1)$, it only remains to prove that $g(x)=\frac{\cos^4(x+\pi/3)}{2+\cos(x+\pi/3)}$ meets the wanted constraints. In order to do that, it is sufficient to estimate the Fourier coefficients of $\frac{1}{2+\cos(x)}$ or $\frac{1}{2-\cos(x)}$. For any $R>1$ we have
$$ \frac{1}{1-\frac{1}{R}e^{ix}}\cdot \frac{1}{1-\frac{1}{R}e^{-ix}}=\frac{R^2}{(1+R^2)-2R\cos(x)}=\sum_{n\geq 0}\frac{e^{nix}}{R^n}\sum_{m\geq 0}\frac{e^{-mix}}{R^m} \tag{2}$$ hence by picking $R=2+\sqrt{3}$ we have $$\frac{2+\sqrt{3}}{2}\cdot\frac{1}{2-\cos(x)}=\sum_{n\geq 0}\frac{e^{nix}}{(2+\sqrt{3})^n}\sum_{m\geq 0}\frac{e^{-mix}}{(2+\sqrt{3})^m} \tag{3}$$ and the coefficients of the Fourier series of $\frac{1}{2\pm\cos(x)}$ are given by explicit convolutions.
They decay like $\frac{1}{(2+\sqrt{3})^n}$, i.e. way faster than $\frac{1}{n^2}$, and this proves that $$ \lim_{a\to 0^+}\int_{-\infty}^{+\infty}\frac{\cos^4(x)}{2+\cos(x)}\cdot\frac{a}{(x-\pi/3)^2+a^2}\,dx = \pi\cdot\frac{\cos^4(\pi/3)}{2+\cos(\pi/3)}=\frac{\pi}{40}.\tag{4}$$