AngouriMath
SolveByPartsAgainstAnArtanhWhoseDerivativeIsARoot(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A power of x times artanh(A x/sqrt(d + B x^2)) with A^2 = B , by parts:
the logarithmln((1 + u)/(1 - u)) that artanh(u) is written as has the
derivative2 u'/(1 - u^2) , and for this u , 1 - u^2 = d/(d + B x^2) and
u' = A d/(d + B x^2)^(3/2) , so it is 2 A/sqrt(d + B x^2) . So
integral x^m ln(...) = x^(m + 1)/(m + 1) ln(...) - 2 A/(m + 1) integral x^(m + 1)/sqrt(d + B x^2) ,
for every wholem but -1 . Rubi's 7.3.7, x^m artanh(sqrt(e) x/sqrt(d + e x^2)) .
https://github.com/asc-community/AngouriMath/issues/233
the logarithm
derivative
for every whole
https://github.com/asc-community/AngouriMath/issues/233
Remarks
The search takes it by parts too, and differentiates the logarithm as written: a quotient
of two sums with a root in each, which it never brought down to the root alone, and the
remainder it asked about ran past the budget. The identity is algebraic, with no branch in
it: it holds whereverd is not 0 and the root is.
of two sums with a root in each, which it never brought down to the root alone, and the
remainder it asked about ran past the budget. The identity is algebraic, with no branch in
it: it holds wherever
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