sinh(z) = 0 $\Rightarrow z = ik\pi$ and we have to find the distance between $5i+1$ and $i\pi$ then between $5i+1$ and $-i\pi$ then with $2i\pi, -2i\pi, 3i\pi, -3i\pi...$ to see which poles are inside the contour in order to apply the Residue theorem.
Is there any fast way to solve this integral, especially when you're an exam and you do not have a calculator to compute each distance ?
The poles are at multiples of $\pi$ along the imaginary axis. Thus, with the equation of the circle being
$$(x-1)^2+(y-5)^2=192$$
plug in $x=0$ and solve for $y$ to get the bounds of the poles that are within the contour and thus count toward the integral. Here,
$$y_{\pm} = 5 \pm \sqrt{191}$$
are the bounds on the imaginary axis. Unfortunately, it becomes a bit unfair if you do not have a calculator and you are in an exam situation. Really, the poles inside the contour are in the set $\{-i 2 \pi, -i \pi, i \pi, \ldots, i 5 \pi\}$ - note that we exclude zero because that singularity is removed by the $z$ in the numerator. That is the hard part of this calculation.
The rest is a simple application of the residue theorem. The residues at the above poles $z_k = i k \pi$ are $i (-1)^k k \pi$. Thus, the value of the integral is
$$i 2 \pi (i \pi) (-2 +1 - 1 + 2 - 3 + 4 - 5) = 8 \pi^2$$