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]