How is it possible to prevent separation of negative and positive exponents when symbolic simplifying?

Question

As parts of my computations, I use InverseLaplaceTransform for some symbolic expressions. As an example, when I have the following transformation for sum of two similar expressions:

InverseLaplaceTransform[(b  E^-(b x)^n (b x)^(-1 + n))/ s + (b (b x)^(-1 + n))/s, s, t]

then, Mathematica simplifies and converts it into a form which the negative numeric exponent is separated from the symbolic part of exponent and is brought into the denominator as:

(E^-(b x)^n (1 + E^(b x)^n) (b x)^n)/x

and this becomes a challenge in next step, once evaluating the expression for some values of the parameters at x=0, which has become now the root of the denominator. I think this is not really related to the InverseLaplaceTransform, since similar results could also appear when using Simplify, for example, the following simplification:

Simplify[b (b x)^(-1 + n), n>1]

would also result in:

(b x)^n/x

Regarding that simplifying is necessary in my computations, is there a way to prevent only this type of simplification? I was thinking of using some options similar to those used in HoldForm, but I don’t know if it is possible and how to do it.

Answer 1

Restricting the appearance of $x^{-1}$ with ComplexityFunction option in FullSimplify helps here:

invlapl = InverseLaplaceTransform[(b E^-(b x)^n (b x)^(-1 + n))/
    s + (b (b x)^(-1 + n))/s, s, t];
f[e_] := LeafCount[e] + 100*Boole[ ! FreeQ[e, x^(-1)]]
FullSimplify[invlapl, ComplexityFunction -> f]

gives

$$ b e^{-(b x)^n} \left(e^{(b x)^n}+1\right) (b x)^{n-1}. $$

Answer 2

Try the following simple approach. Here is your expression:

expr1 = (E^-(b x)^n (1 + E^(b x)^n) (b x)^n)/x

The way to remove x from the denominator and include it into the expression in the numerator is to apply PowerExpand:

expr2 = MapAt[PowerExpand, expr1, 4]

(*  b^n E^-(b x)^n (1 + E^(b x)^n) x^(-1 + n)  *)

However, if you want to protect some combination (say, (b x)^(-1 + n)) from further changes you may, of course use a command from the Hold group, but there is also a more simple and convenient way. You may simply rename this combination. For example, here I give it the name c:

 expr3 = expr2 /. x^(-1 + n) -> c/b^(-1 + n)

(*  b c E^-(b x)^n (1 + E^(b x)^n) *)

Now you may do any transformation you need, and AFTER that replace c back: c -> b^(-1 + n) x^(-1 + n).

Have fun!