calculus - Tough integral with coth[1/u]

I'm hoping to come up with an analytical solution for the following definite integral:

$$\int_{-2/y}^{2/y} \cfrac{u}{\coth\left( \frac{1}{u}\right) - i \cot\left( \frac{z}{2} \right)} du$$

Here, $y$ and $z$ are strictly positive real constants. $i$ is the imaginary number $\sqrt{-1}$.

I tried doing this on Mathematica but it's been 20 minutes and it hasn't given me anything yet. I was wondering if anybody had any tricks/tips for solving this integral?

Thanks!

EDIT: The imaginary component can be eliminated as follows:

$$\int_{-2/y}^{2/y} \cfrac{u}{\coth\left( \frac{1}{u}\right) - i \cot\left( \frac{z}{2} \right)} du = \int_{-2/y}^{2/y} \cfrac{u\coth\left( \frac{1}{u}\right)}{\coth^2\left( \frac{1}{u}\right) + \cot^2\left( \frac{z}{2} \right)} du + i\int_{-2/y}^{2/y} \cfrac{u\cot\left( \frac{z}{2}\right)}{\coth^2\left( \frac{1}{u}\right) + \cot^2\left( \frac{z}{2} \right)} du$$
The second integral is odd, so it evalutes to zero. Thus,
$$\int_{-2/y}^{2/y} \cfrac{u}{\coth\left( \frac{1}{u}\right) - i \cot\left( \frac{z}{2} \right)} du = \int_{-2/y}^{2/y} \cfrac{u\coth\left( \frac{1}{u}\right)}{\coth^2\left( \frac{1}{u}\right) + \cot^2\left( \frac{z}{2} \right)} du$$
However, I don't think this is any easier to solve...

EDIT2: As noted by the comments, it might be a bit nicer if I set $b = 2/y$ and $c = \cot(z/2)$ so that my integral looks like:

$$\int_{-b}^{b} \cfrac{u}{\coth\left( \frac{1}{u}\right) - i c} du = \int_{-b}^{b} \cfrac{u\coth\left( \frac{1}{u}\right)}{\coth^2\left( \frac{1}{u}\right) + c^2} du$$