AngouriMath
SolveByDividingOutOnePlusAnImaginaryHyperbolicSine(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A polynomial in sinh(y) and cosh(y) over a ± i a sinh(y) , beside
anything else inx , divided out: cosh(y)^2 = (1 + i sinh(y))(1 - i sinh(y)) ,
so withcosh(y)^2 written as 1 + sinh(y)^2 the numerator is
N0(sinh) + cosh N1(sinh) , and each part divides by the linear
a ± i a sinh(y) to a polynomial and a constant. (e + f x) cosh(c + d x)^2/(a + i a sinh(c + d x)) is (e + f x)(1 - i sinh(c + d x))/a , and (e + f x) sinh(y)^2/(a + i a sinh(y)) is (e + f x)(1 - i sinh(y))/a - (e + f x)/(a + i a sinh(y)) .
https://github.com/asc-community/AngouriMath/issues/233
anything else in
so with
https://github.com/asc-community/AngouriMath/issues/233
Remarks
Rubi's (e + f x)^m cosh(c + d x)^k/(a + i a sinh(c + d x)) and its sinh^k rows were declined or past the budget, where the pieces the division leaves are answered
in a fraction of a second. Exact wherever the denominator is not zero. Only where the remainder beside
the cosine vanishes, sincecosh(y)/(a ± i a sinh(y)) beside a power of x is
a polylogarithm by parts; where the quotient keeps a hyperbolic function, a constant
there being one step from the quotient the rules already answer; and where something
else inx multiplies it, a rational function of the exponential alone being the
exponential substitution's.
in a fraction of a second. Exact wherever the denominator is not zero. Only where the remainder beside
the cosine vanishes, since
a polylogarithm by parts; where the quotient keeps a hyperbolic function, a constant
there being one step from the quotient the rules already answer; and where something
else in
exponential substitution's.
Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 4862 pages online