If we make the substitution $z=e^{i\theta}$ and use the trigonometric Euler identities then we can convert the integral into a contour integral over the unit circle in the complex plane:
$$I=\oint_C\frac{1}{3+z+\frac{1}{z}+\frac{1}{2i}(z-\frac{1}{z})}\frac{-i}{z}dz=-2i\oint_C\frac{dz}{z^2(2-i)+6z+(2+i)}$$
Using the quadratic formula to find the roots of the denominator, which we will call $\alpha^+$ and $\alpha^-$:
$$\alpha^+=-\frac{1}{5}(2+i)$$ $$\alpha^-=-(2+i)$$
$$I=-2i\oint_C\frac{dz}{(z-\alpha^+)(z-\alpha^-)}$$
$\alpha^+$ is a simple pole that lies within the unit circle. It clearly has residue $\frac{1}{\alpha^+-\alpha^-}=\frac{5}{8+4i}$. Therefore by residue theorem:
$$I=\frac{20\pi}{8+4i}$$
However, this is clearly incorrect because it is not a real number, whereas the integrand originally was real. Furthermore, a calculator will give that $I=\pi$, so something has gone wrong somewhere, but my reasoning seems solid. Any help will be greatly appreciated.
Your computations are fine, but not your formula. Let$$f(x,y)=\frac1{3+2x+y};$$then your integral is $\int_0^{2\pi}f\bigl(\cos(\theta),\sin(\theta)\bigr)\,\mathrm d\theta$. Now, let$$g(z)=\frac1zf\left(\frac{z+z^{-1}}2,\frac{z-z^{-1}}{2i}\right).$$The integral that you're after is equal to $2\pi$ times the sum of the residues of$$g(z)=\frac{2 i}{(1+2 i) z^2+6 i z-(1-2 i)}\tag1$$ within the unit circle. There is a single singularity there (located at $-\frac25(1+i)$, as you know), and the residue of $(1)$ there is $\frac12$. Therefore, your integral is equal to $\pi$.