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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?

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  • $\begingroup$ 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. $\endgroup$
    – user64074
    Commented Oct 14, 2019 at 19:19
  • $\begingroup$ Thanks, fixed that. $\endgroup$
    – Henry Shackleton
    Commented Oct 14, 2019 at 19:31

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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}
(*False*)
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  • $\begingroup$ Hm, so it seems like I can get the simplification to go through if I assume x is real. This is unusual because in the integral that this expression is a part of, x is integrated over the real axis, so I would have thought that this assumption would have been made. Perhaps the overall problem is different than what I had thought. $\endgroup$
    – Henry Shackleton
    Commented Oct 14, 2019 at 19:33
  • $\begingroup$ @HenryShackleton - When doing the integral did you use the option GenerateConditions -> True? $\endgroup$
    – Bob Hanlon
    Commented Oct 14, 2019 at 20:16
  • $\begingroup$ @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. $\endgroup$
    – user64074
    Commented Oct 15, 2019 at 6:03

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