AngouriMath
SolveAPolynomialTimesAnInverseHyperbolicTangentAndTheLogarithmOfItsQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A polynomial in x times a linear combination of 1 , T , B and T B ,
T = artanh(c x) or arcoth(c x) and B = ln(K (1 - c^2 x^2)) , in closed form.
Withy = c x , T' = 1/(1 - y^2) and B' = -2 y/(1 - y^2) , the integral of
y^j T B is p0 T B + p1 T + p2 B + q1 T^2 + q2 B^2 + r , where
p0 = (y^(j + 1) - [j odd])/(j + 1) , q1 = p0(1) , q2 = p0(1)/4 , and
p1 , p2 and r are the integrals of (2 y p0 - 2 q1)/(1 - y^2) ,
(4 y q2 - p0)/(1 - y^2) and (2 y p2 - p1)/(1 - y^2) , polynomials.
With
Remarks
Matching the coefficients of T B , T , B and 1 asks that the three
numerators vanish aty = 1 and y = -1 : the constant of p0 makes the first
two do, and the constants ofp1 and p2 the third. Only the derivatives of
T and B are read, which are rational and the same on every branch, so the
answer holds for either inverse function and everyK , in the generic case
c != 0 . The coefficients are rational in y, and y^k is c^k x^k .
The rules after it ran out of time on Rubi's 7.3.4 and 7.4.1
x^m (a + b artanh(c x))(d + e ln(1 - c^2 x^2)) and the arcoth twins.
Read only whereT B is there: without it the rules after it answer.
https://github.com/asc-community/AngouriMath/issues/233
numerators vanish at
two do, and the constants of
answer holds for either inverse function and every
The rules after it ran out of time on Rubi's 7.3.4 and 7.4.1
Read only where
https://github.com/asc-community/AngouriMath/issues/233
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