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!
 A: 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 $$I=\int_{-\infty}^{\infty} \frac{du}{(4u^4+1)(u^2+1)}=\frac{2\pi}{5}$$
A: You can indeed set $x=u^2$ as Mostafa Ayaz did. However sooner or later you need to know how to deal with $\sqrt[\alpha]{z}$ with $z\in\mathbb{C}$ which is not even that difficult. From your words I understand that you were using the keyhole contour. The pole at zero was not inside the contour. So you actually had $z=1$, $\pm \frac{1}{2}i$ as simple poles inside your contour. 
Moreover you have two kinds of pathes that look like circles, the large "circle" which will have no contribution as the radius goes to infinity, and the small "circle" which will have no contribution as well as the radius of it goes to zero. Check these, please! 
Assuming you were using the Principal Logarithm to define the square root, I'll show you how to do the residue at $\frac{1}{2}i$:
\begin{align}
\text{Res}_{z=i/2}\frac{1}{\sqrt[]{z}(4z^2+1)(z+1)}=\frac{1}{4\sqrt[]{i/2}(i/2+i/2)(i/2+1)}
\end{align} 
The only problematic thing here is the square root right? 
\begin{align}
\sqrt[]{i/2}=e^{\frac{1}{2}\text{Log}(i/2)}
\end{align}
That is how we defined it (I think you too? Since that is the standard way). 
\begin{align}
\text{Log}(i/2)=\ln|i/2| +i\text{Arg}(i/2)=-\ln(2)+i\frac{\pi}{2}
\end{align} 
So:
\begin{align}
\sqrt[]{i/2}=e^{\frac{1}{2}\left(-\ln(2)+i\frac{\pi}{2} \right)}=\frac{1}{\sqrt[]{2}}e^{i\pi/4}
\end{align}
I hope this was helpful.
