SolveByCancellingOnePlusAnImaginaryHyperbolicSineAgainstTheCosineSquared​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

Rubi's 1/(sech(x) ± i tanh(x))^4 in 6.7.1 were declined: the quotient is even in
the hyperbolic cosine, so SolveByBiochesOddHyperbolicSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) does not
read it, and the substitutions spent the budget on a denominator of degree eight in the
exponential, where the quotient the cancellation leaves is answered in a fifth of a
second. Exact on the real line, where 1 ± i sinh(y) is not zero, and the points
off it where it is are removable. Only this shape, read by the multiplicities of
sinh(y) ∓ i on either side once the squares of the cosine are written in the sine:
cancelling where a polynomial or another power of 1 ± i sinh(y) is left made the
answers to cosh(x)^8/(i + sinh(x)) and cosh(x)^2/(1 + i sinh(x))^3 longer,
and writing the cosh(y)^2 a quotient carries on both sides as the two binomials
made (a sech(x) + b tanh(x))^4 a search past the budget.