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IntegrateOverPowerOfQuadratic​(AngouriMath.​Entity,​AngouriMath.​Entity,​AngouriMath.​Entity,​AngouriMath.​Entity,​System.​Int32,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

∫ k / (a x^2 + b x + c)^n dx for a whole n of two or more, by the reduction that
takes one power off the denominator at a time.

Remarks

Write Q for the quadratic, u for its derivative 2ax + b and
D for 4ac - b^2. Then u^2 is 4aQ - D, so differentiating
u / Q^(m-1) gives 2a(3 - 2m)/Q^(m-1) + (m-1)D/Q^m, and reading that as
an equation for the integral of 1/Q^m leaves
J_m = (u / Q^(m-1) + 2a(2m - 3) J_(m-1)) / ((m - 1) D)

which is unrolled down to J_1, the case the table above already answers. It is
an identity in a, b and c rather than a fact about signs, so one line serves a
positive and a negative discriminant alike; D = 0 is the only exclusion, and
the division by D is why.
 The two shapes the reduction cannot speak for get their own arms, and each is
elementary: with a = 0 the denominator is a linear power, and with
D = 0 the quadratic is u^2/(4a) so the whole integrand is a power of
u. Both are already answered for a numeric quadratic — by the power rule and
by partial fractions — and the arms are here for a symbolic one, where the
discriminant's sign is not known and dropping an arm is not available.
 What was missing was the irreducible case. 1/(x^2 - 1)^2 had an antiderivative
and 1/(x^2 + 2x + 1)^2 had one, because a denominator with real roots comes
apart into linear factors and never reaches here; 1/(x^2 + 1)^2 had none.
https://github.com/asc-community/AngouriMath/issues/180

























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