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SolveAPowerOfXOverAPowerOfAQuadraticTimesAnInverseTrigonometricFunctionByParts​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

x^m (d + e x^2)^p (a + b F(c x)), F an inverse trigonometric function and
m + 2p + 3 = 0, by parts against the antiderivative of x^m (d + e x^2)^p that is
one term, v = x^(m + 1) (d + e x^2)^(p + 1)/(d (m + 1)):
integral = v (a + b F(c x)) - b integral v F(c x)', and F(c x)' is algebraic.

Remarks

v' = x^m (d + e x^2)^p ((m + 1)(d + e x^2) + 2 (p + 1) e x^2)/(d (m + 1)), and the bracket
is (m + 1) d + (m + 2p + 3) e x^2, which is (m + 1) d for every e: an
identity of rational functions, so the parts are exact wherever the integrand is defined, on
any branch of F. It is Rubi's choice of v, and the one that vanishes at
x = 0.
 The integrator's own antiderivative of x^3/(d + e x^2)^3 is a piecewise on e = 0,
and by parts against it spent the budget on every arm: Rubi's 5.1.4, 5.3.4, 5.5.1 and 5.6.1
x^3 (a + b F(c x))/(d + e x^2)^3, for F the inverse sine, tangent, secant and
cosecant, ran out of time, where the remainder against this v is answered in under a
second.
 Narrow on purpose. p whole and at most -3, so that v is rational and
m = -2p - 3 at least 3: m = 1, x/(d + e x^2)^2, is answered by parts as it
was, and a half-odd p is answered as it was beside the inverse sine and tangent and is
elliptic beside the secant. And not for e = +-c^2 d, the quadratic a multiple of
F's own radicand, which the substitution under F answers.
https://github.com/asc-community/AngouriMath/issues/233

























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