AngouriMath
SolveByWritingAConstantMultipleOfARadicandOverIt(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A power of a constant multiple of a radicand that a logarithm holds under a root,
written over that radicand:acosh(c x) is ln(c x + sqrt(c^2 x^2 - 1)) ,
and beside itd - c^2 d x^2 is -d (c^2 x^2 - 1) . A whole power is the
product of the powers; a root(lambda M)^(k/2) is K^k M^(k/2) with
K = sqrt(lambda M)/sqrt(M) , which is constant wherever it is defined -- plus or
minusi , or plus or minus sqrt(lambda) -- so it stands in front of the
answer the way Rubi writes it.
written over that radicand:
and beside it
product of the powers; a root
minus
answer the way Rubi writes it.
Remarks
The substitution u = acosh(c x) needs the root its derivative carries,
sqrt(c^2 x^2 - 1) ; written over 1 - c^2 x^2 the integrand is the same
function timesi and nothing reads it: acosh(a x)^2/sqrt(1 - a^2 x^2) ran
out of time whereacosh(a x)^2/sqrt(a^2 x^2 - 1) takes a third of a second, and
x (a + b acosh(c x))/(d - c^2 d x^2)^3 -- a whole power, no root at all -- the same.
K is carried through the integration as a symbol: it is constant, so an
antiderivative in it is one inx once it is written out, whatever the branches.
https://github.com/asc-community/AngouriMath/issues/718
function times
out of time where
antiderivative in it is one in
https://github.com/asc-community/AngouriMath/issues/718
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