Laplace Inverse of the problem. What is the Laplace Inverse of the given question?
$$\frac{\sqrt{4+s^3}}{s^3}$$
I tried solving it by expanding $\sqrt{1+\frac{s^3}{4}}$ but the terms will not have Laplace Inverse.
If I expand it like  $\sqrt{1+\frac{4}{s^3}}$i.e.$$s^{-3/2}* \sqrt{1+\frac{4}{s^3}}$$
It will still have many terms of $t^n$ .
Are there other methods? 

With CAS help I have:
$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{4+s^3}}{s^3}\right](t)=\frac{2 \sqrt{t} \,
   _1F_3\left(-\frac{1}{2};\frac{1}{2},\frac{5}{6},\frac{7}{6};-\frac{4 t^3}{27}\right)}{\sqrt{\pi }}$$
For general a:
$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{a+s^3}}{s^3}\right](t)=\\\mathcal{L}_s^{-1}\left[\sum _{n=0}^{\infty } a^n
   \left(s^3\right)^{-\frac{1}{2}-n} \binom{\frac{1}{2}}{n}\right](t)=\\\sum _{n=0}^{\infty }
   \mathcal{L}_s^{-1}\left[a^n \left(s^3\right)^{-\frac{1}{2}-n} \binom{\frac{1}{2}}{n}\right](t)=\sum
   _{n=0}^{\infty } \frac{a^n \sqrt{\pi } t^{\frac{1}{2}+3 n}}{2 \Gamma \left(\frac{3}{2}-n\right) \Gamma
   (1+n) \Gamma \left(\frac{3}{2}+3 n\right)}=\\\frac{2 \sqrt{t} \,
   _1F_3\left(-\frac{1}{2};\frac{1}{2},\frac{5}{6},\frac{7}{6};-\frac{1}{27} \left(a
   t^3\right)\right)}{\sqrt{\pi }}$$
