Little help with a residue integration with branch point 
$$\int_0^{+\infty} \frac{dx}{\sqrt{x}(4x^2+1)(x+1)}$$

Now I transformed it into
$$\int_0^{+\infty} \frac{z\ dz}{z(4z^2+1)(z+1)}$$
To obtain the form
$$\frac{z^{\alpha}F(z)}{G(z)}$$
In such a wa I can apply residue calculus to get the answer as
$$\frac{2\pi i}{1 - e^{2\pi i \alpha}}\text{Res}[f, z]$$
Where a sum of residues is understood.
The poles are at $z = 0, -1, \pm \frac{1}{2}i$
The result shall be $\frac{2\pi }{5}$ but now the question is: how to deal with the poles at $\pm \frac{1}{2}i$? I would have a cumbersome square root of an imaginary unit.
Is there some "nicer" or smart way to deal with the integral? 
Thank you!

Substitute $x=u^2$ therefore$$\int_0^{+\infty} \frac{dx}{\sqrt{x}(4x^2+1)(x+1)}=\int_{0}^{\infty} \frac{2du}{(4u^4+1)(u^2+1)}=\int_{-\infty}^{\infty} \frac{du}{(4u^4+1)(u^2+1)}$$using the same famous contour for complex integration we have:$$\int_{C} \frac{dz}{(4z^4+1)(z^2+1)}=\int_{-R}^{R} \frac{du}{(4u^4+1)(u^2+1)}+\int_{0}^{\pi} \frac{iRe^{i\theta}d\theta}{(4R^4e^{4i\theta}+1)(R^2e^{2i\theta}+1)}$$also $f(z)=\dfrac{1}{(4z^4+1)(z^2+1)}$ has 6 roots among those 3 contained by the contour and all are of order 1. Those 3 are:$$r_1=i\\r_2=\dfrac{1}{\sqrt2}e^{i\dfrac{\pi}{4}}\\r_3=\dfrac{1}{\sqrt2}e^{i\dfrac{3\pi}{4}}$$if we denote the residue of $f(z)$ in these points by $R_1$, $R_2$ and $R_3$ we have:$$R_1=R_2+R_3=\dfrac{1}{10i}$$therefore$$R_1+R_2+R_3=\dfrac{1}{5i}$$and the integral would be$$\int_{C} \frac{dz}{(4z^4+1)(z^2+1)}=2\pi i(R_1+R_2+R_3)=\frac{2\pi}{5}$$which leads to $$I=\int_{-\infty}^{\infty} \frac{du}{(4u^4+1)(u^2+1)}=\frac{2\pi}{5}$$
