Problem with InverseLaplaceTransform[] using Mathematica 11.3

Questions
1

Hello,
I have problems with one special InverseLaplaceTransform.
Mathematica 11.3 doesn’t solve it.
Have I done any mistakes?
Thanks
Leo

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Carla.nb (21 KB)

2

I don’t know whether there are mistakes, but perhaps Mathematica simply cannot solve it.
Do it piecewise:

Your LPT is ( without factors)

lap = ( e^(-u p) (-1 + e^(u p)))/(a^2 + ac p + p^2) // Expand

You can do the first term

t1= InverseLaplaceTransform[1/(a^2 + ac p + p^2), p,` x]

For the 2nd term split the denominator to get aa and bb to be used later

nn = a^2 + ac p + p^2;
lsg = p /. Solve[nn == 0, p]
nn1 = (p - lsg[[1]]) (p - lsg[[2]])
% // Expand

Now the denominator can be written as

(p - aa) (p-bb)

and define

t2 = Apart[1/((p - aa) (p - bb))]

and get

t3 = InverseLaplaceTransform[Exp[ -p u] t2[[1]], p, x]
t4 = InverseLaplaceTransform[Exp[ -p u] t2[[2]], p, x]

Then you get the inverse of your LPT

invLPT = t1 + t3 + t4 /. {aa -> lsg[[1]], bb -> lsg[[2]]}

and check the result for u > 0

FullSimplify[LaplaceTransform[invLPT, x, p] // Expand, u > 0]
3

Thanks Hans - Yes, I did it so.
Yesterday I was to lazy to do it piecewise and I demand Mathematica to do it for me.

Now it is solved.

Regards.