inverse Laplace transform for exponential function How does the answer to the inverse Laplace transform of this function get? 
I solved this Laplace reversal by using binomial extensions and extending the exponential function. But I want to know if there is a closed form for this series, like the Wright function or  Mainardi function or other function? Is there a solution for this reversal of Laplace in other ways?The function shown in the photograph is attached.enter image description here
 A: Well, in general:
$$\text{y}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{n}_1\cdot\sqrt{\text{s}^{\text{n}_2}+\text{n}_3}\right)\right]_{\left(t\right)}\tag1$$
Using, that:


*

*$$\exp\left(x\right):=e^x=\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\tag2$$

*When $\left|\text{a}\right|<2$:
$$\left(\text{a}+\text{b}\right)^\frac{\text{c}}{2}=\sum_{\text{p}=0}^\infty\text{a}^\text{p}\cdot\text{b}^{\frac{1}{2}\cdot\left(\text{c}-2\cdot\text{p}\right)}\cdot\binom{\frac{\text{c}}{2}}{\text{p}}\tag3$$


So, we get:
$$\text{y}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{n}_1\cdot\sqrt{\text{s}^{\text{n}_2}+\text{n}_3}\right)\right]_{\left(t\right)}=\mathscr{L}_\text{s}^{-1}\left[\sum_{\text{k}=0}^\infty\frac{\left(\text{n}_1\cdot\sqrt{\text{s}^{\text{n}_2}+\text{n}_3}\right)^\text{k}}{\text{k}!}\right]_{\left(t\right)}=$$
$$\sum_{\text{k}=0}^\infty\frac{\text{n}_1^\text{k}}{\text{k}!}\cdot\mathscr{L}_\text{s}^{-1}\left[\left(\text{s}^{\text{n}_2}+\text{n}_3\right)^\frac{\text{k}}{2}\right]_{\left(t\right)}=$$
$$\sum_{\text{k}=0}^\infty\frac{\text{n}_1^\text{k}}{\text{k}!}\cdot\left\{\mathscr{L}_\text{s}^{-1}\left[\sum_{\text{p}=0}^\infty\left(\text{s}^{\text{n}_2}\right)^\text{p}\cdot\left(\text{n}_3\right)^{\frac{1}{2}\cdot\left(\text{k}-2\cdot\text{p}\right)}\cdot\binom{\frac{\text{k}}{2}}{\text{p}}\right]_{\left(t\right)}\right\}=$$
$$\sum_{\text{k}=0}^\infty\frac{\text{n}_1^\text{k}}{\text{k}!}\cdot\left\{\sum_{\text{p}=0}^\infty\text{n}_3^{\frac{1}{2}\cdot\left(\text{k}-2\cdot\text{p}\right)}\cdot\binom{\frac{\text{k}}{2}}{\text{p}}\cdot\mathscr{L}_\text{s}^{-1}\left[\left(\text{s}^{\text{n}_2}\right)^\text{p}\right]_{\left(t\right)}\right\}=$$
$$\sum_{\text{k}=0}^\infty\frac{\text{n}_1^\text{k}}{\text{k}!}\cdot\left\{\sum_{\text{p}=0}^\infty\text{n}_3^{\frac{1}{2}\cdot\left(\text{k}-2\cdot\text{p}\right)}\cdot\binom{\frac{\text{k}}{2}}{\text{p}}\cdot\mathscr{L}_\text{s}^{-1}\left[\text{s}^{\text{p}\cdot\text{n}_2}\right]_{\left(t\right)}\right\}\tag4$$
Well, we know that:
$$\mathscr{L}_\text{s}^{-1}\left[\text{s}^{\text{p}\cdot\text{n}_2}\right]_{\left(t\right)}=\frac{1}{\Gamma\left(-\text{p}\cdot\text{n}_2\right)}\cdot\frac{1}{t^{1+\text{p}\cdot\text{n}_2}}\tag5$$
So, we end up with:
$$\text{y}\left(t\right)=\sum_{\text{k}=0}^\infty\frac{\text{n}_1^\text{k}}{\text{k}!}\cdot\left\{\sum_{\text{p}=0}^\infty\text{n}_3^{\frac{1}{2}\cdot\left(\text{k}-2\cdot\text{p}\right)}\cdot\binom{\frac{\text{k}}{2}}{\text{p}}\cdot\frac{1}{\Gamma\left(-\text{p}\cdot\text{n}_2\right)}\cdot\frac{1}{t^{1+\text{p}\cdot\text{n}_2}}\right\}\tag6$$
