$$\int_0^{\infty } \frac{\sinh (x)}{\left(h^2-x^2\right) (\cosh (b)+\cosh (x))} \, dx=\\\int_0^1 \frac{e^b \left(-1+t^2\right)}{t \left(e^b+t+e^{2 b} t+e^b t^2\right) \left(-h^2+\log ^2(t)\right)} \, dt=\\\int_0^1 \left(\frac{1}{t (h-\log (t)) (h+\log (t))}-\frac{1}{\left(e^b+t\right) (h-\log (t)) (h+\log (t))}-\frac{e^b}{\left(1+e^b t\right) (h-\log (t)) (h+\log (t))}\right) \, dt=\\\int_0^1 \left(-\frac{1}{\left(e^b+t\right) \left(h^2-\log ^2(t)\right)}-\frac{e^b}{\left(1+e^b t\right) \left(h^2-\log ^2(t)\right)}\right) \, dt=\\\Re\left(\sum _{m=0}^{\infty } \frac{(-1)^m e^{-((b+h) (1+m))} \left(\Gamma (0,-h (1+m))-e^{2 h (1+m)} \Gamma (0,h+h m)\right)}{2 h}\right)+\Re\left(\sum _{m=0}^{\infty } -\frac{(-1)^{2 m} h^{2 m} \left(-i \pi \text{Li}_{-1-2 m}\left(-e^b\right)+2 \log (h) \text{Li}_{-1-2 m}\left(-e^b\right)-2 \psi ^{(0)}(2+2 m) \text{Li}_{-1-2 m}\left(-e^b\right)-2 \text{PolyLog}^{(1,0)}\left(-1-2 m,-e^b\right)\right)}{2 (1+2 m)!}\right)$$

Maybe exist closed form I can't find them.

Numerically is easy to compute

I doubt there's a closed-form for the Integral. Another maybe simple solution:

Regards M.I.

Setting values for $b$ and $h$ results in warnings that the integration doesn't converge on $(0,\infty)$. Here's an example:

Integrate[Sinh[x]/((Cosh[x] + Cosh[b])*(h^2 - x^2)) /. {h -> 1, b -> 1}, {x, 0, \[Infinity]}]

Integrate::idiv: Integral of -(Sinh[x]/((-1+x^2) (Cosh[1]+Cosh[x]))) does not converge on {0,[Infinity]}.

Do you have some reason to think the integral exists for either specific values of $h$ and $b$ or on a different interval?