# Simplify product of cosh and sech expressions

I would like to simplify a product of hyperbolic cosine and hyperbolic secant functions, with the key simplifying assumption being that the two are inverses of each other. This sounds like a silly thing to do, but I have some complicated integrals that won't evaluate unless these expressions simplify properly. If I do the following:

Simplify[Cosh[x]^(1/q)*Sech[x]^(2 + 1/q),
Assumptions -> q \[Element] Integers && q > 1]


I would naively expect to get $$\text{sech}^2(x)$$ as the simplified expression. However, Mathematica is unable to simplify this expression further. What additional assumptions do I need to make in order to get this simplification to go through?

• You would get $sech^2(x)$ with the expression Cosh[x]^(1/q)*Sech[x]^(2 + 1/q) (note the + sign). However, Mathematica does not seem to simplify this either. – user64074 Oct 14 '19 at 19:19
• Thanks, fixed that. – Henry Shackleton Oct 14 '19 at 19:31

For some values of x and q your assertion is not true
Cosh[x]^(1/q)*Sech[x]^(2 - 1/q) == Sech[x]^2 /.{q->2,x->3I}

• @HenryShackleton - When doing the integral did you use the option GenerateConditions -> True? – Bob Hanlon Oct 14 '19 at 20:16
• @HenryShackleton Of course! I missed that: if x is complex, then cosh(x) may vanish, and the simplified expression is not exactly equivalent. Substitute x -> I Pi/2 for instance. – user64074 Oct 15 '19 at 6:03