AngouriMath
SolveByTheMoebiusSubstitutionOfAQuotientOfLinears(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A power of a quotient of two linears beside whole powers of the two, under the Moebius
substitutiont = (p + q x)/(r + s x) : then p + q x = t (r + s x) ,
r + s x = D/(q - s t) and dx = D/(q - s t)^2 dt with D = q r - p s , so
t^alpha (p + q x)^j (r + s x)^k dx is D^(j + k + 1) t^(alpha + j) (q - s t)^(-(j + k + 2)) dt ,
a polynomial int times a power of it wherever j + k <= -2 . A whole power of
a quadratic that is a multiple of the two linears' product counts as a power of each.
e^(n acoth(a x)) arrives folded as ((a x + 1)/(a x - 1))^(n/2) , and over
(c - a^2 c x^2)^m it is Rubi's 7.4.2.
https://github.com/asc-community/AngouriMath/issues/233
substitution
a polynomial in
a quadratic that is a multiple of the two linears' product counts as a power of each.
https://github.com/asc-community/AngouriMath/issues/233
Remarks
Only for a power alpha that is not whole: a whole one makes a rational function,
which the partial fractions answer.t^gamma is differentiated to
gamma t^(gamma - 1) t' on the principal branch wherever t is not 0, the
negative half-line included, so the answer written back in the quotient holds where
the integrand is defined. A half-odd power of the quadratic is left alone: it is not
the product of the linears' powers on every side of their roots.
which the partial fractions answer.
negative half-line included, so the answer written back in the quotient holds where
the integrand is defined. A half-odd power of the quadratic is left alone: it is not
the product of the linears' powers on every side of their roots.
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