I am away from Laplace transform for years, and now I have to solve $$\mathcal{L}^{-1} \left\{ s^{-\frac32}\sqrt{\frac{as+b}{cs+b}} \right\}$$a,b,c are real positive numbers.I can find inverse Laplace for $s^{-k}$ but really I got stuck on this. Please guide me to solve this form. Thanks in advance. Implicitly it has the condition $\lim_{s\to \infty}F(s)=0$
Honestly I don't know what should I do for $\sqrt{\frac{as+b}{cs+b}}$

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    $\begingroup$ Do you know about the inversion formula? $\endgroup$ – Allawonder Jan 17 at 11:05
  • $\begingroup$ @Allawonder: Yes, I can find ordinary Laplace inverse, but this problem was not simple. As far as I search in tables, I did not see this form. I did not have an Idea about $\sqrt{\frac{as+b}{cs+b}}$ $\endgroup$ – Khosrotash Jan 17 at 13:05
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    $\begingroup$ Looks like: $$\mathcal{L}_s^{-1}\left[\sqrt{\frac{a s+b}{c s+b}}\right](t)=-\frac{b (a-c) e^{-\frac{b (a+c) t}{2 a c}} \left(I_0\left(\frac{b (a-c) t}{2 a c}\right)+I_1\left(\frac{1}{2} b \left(\frac{1}{a}-\frac{1}{c}\right) t\right)\right)}{2 \sqrt{\frac{a}{c}} c^2}$$,Mathematica code:-((b (a - c) E^(-((b (a + c) t)/( 2 a c))) (BesselI[0, (b (a - c) t)/(2 a c)] + BesselI[1, 1/2 b (1/a - 1/c) t]))/(2 Sqrt[a/c] c^2)). $\endgroup$ – Mariusz Iwaniuk Jan 24 at 11:23
  • $\begingroup$ @MariuszIwaniuk:Thank you for code and solution, Do you understand what does functionality of $$\sqrt{\frac{as+b}{cs+b}}$$? behind this problem $\endgroup$ – Khosrotash Jan 24 at 14:42
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    $\begingroup$ I have only fast coverages series: $$\mathcal{L}_s^{-1}\left[\frac{\sqrt{\frac{a s+b}{c s+b}}}{s^{3/2}}\right](t)=-\frac{\left(2 \sqrt{a t}\right) \sum _{n=0}^{\infty } \frac{\left(-\frac{1}{a c}\right)^n (b (-a+c) t)^n \, _1F_1\left(n;\frac{3}{2}+n;-\frac{b t}{c}\right)}{\left(-1+4 n^2\right) \Gamma (1+n)}}{\sqrt{c} \sqrt{\pi }}$$ for: a>0,b>0,c>0 $\endgroup$ – Mariusz Iwaniuk Feb 1 at 10:01

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