Inverse Laplace Transform how to find the exact solution

Using:

$$frac{1}{sinh left(b sqrt{s}right)}=text{csch}left(b sqrt{s}right)=sum _{k=0}^{infty } 2 exp left(-(2 k+1) b sqrt{s}right)$$ then we have:

func = InverseLaplaceTransform[(2 E^(b (-1 - 2 k) Sqrt[s]) Sinh[a Sqrt[s]])/s^2, s, t]
Sum[func[[1]], {k, 0, Infinity}]

Solution only by Infinite Sum:

HoldForm[InverseLaplaceTransform[
Sinh[a Sqrt[s]]/(s^2*Sinh[b Sqrt[s]]), s, t] == 
Sum[1/(2 Sqrt[[Pi]]) (a^2 Sqrt[[Pi]] - 2 a b Sqrt[[Pi]] + 
   b^2 Sqrt[[Pi]] - 4 a b k Sqrt[[Pi]] + 4 b^2 k Sqrt[[Pi]] + 
   4 b^2 k^2 Sqrt[[Pi]] + 
   2 a E^(-((-a + b + 2 b k)^2/(4 t))) Sqrt[t] - 
   2 b E^(-((-a + b + 2 b k)^2/(4 t))) Sqrt[t] + 
   2 a E^(-((a + b + 2 b k)^2/(4 t))) Sqrt[t] + 
   2 b E^(-((a + b + 2 b k)^2/(4 t))) Sqrt[t] - 
   4 b E^(-((-a + b + 2 b k)^2/(4 t))) k Sqrt[t] + 
   4 b E^(-((a + b + 2 b k)^2/(4 t))) k Sqrt[t] + 
   2 Sqrt[[Pi]] t + 
   Sqrt[[Pi]] (a^2 - 2 a (b + 2 b k) + (b + 2 b k)^2 + 2 t) Erf[(
     a - b (1 + 2 k))/(2 Sqrt[t])] - 
   Sqrt[[Pi]] (a^2 + 2 a (b + 2 b k) + (b + 2 b k)^2 + 
      2 t) Erfc[(a + b + 2 b k)/(2 Sqrt[t])]), {k, 0, 
  Infinity}]] // TraditionalForm

I doubt there's a closed form for the Inverse Laplace Transform or Series.