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SolveAPolynomialOverAPowerOfALinearBesideARootOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A polynomial over a power of a linear beside a half-odd power of a quadratic,
P(x) Q^(m/2) / (g + h x)^k with m odd and k >= 1, by undetermined
coefficients: algebraic terms times sqrt(Q), and a multiple of each of two
integrals that are not algebraic.

Remarks

Over S = sqrt(Q) the integrand is T(x)/((g + h x)^k S), where
T = P Q^((m + 1)/2) is a polynomial, or below m = -1,
P/((g + h x)^k Q^n S) with n = -(m + 1)/2; and
IntegrateOverARootOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity[],AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,System.Func{AngouriMath.Entity,AngouriMath.Entity},AngouriMath.Functions.Algebra.IndefiniteIntegralSolver.ALinearBesideTheRoot,AngouriMath.Entity,System.Nullable{System.Int32},System.Int32) takes either to algebraic terms and two
integrals.
The two integrals left. With Q = q0 + q1 x + q2 x^2, int 1/S is
ln(2 sqrt(q2) S + 2 q2 x + q1)/sqrt(q2), and
-arcsin((2 q2 x + q1)/sqrt(q1^2 - 4 q0 q2))/sqrt(-q2), real where q2 < 0,
as the root of a quadratic alone is answered. With y = g + h x,
H = q0 h^2 - q1 g h + q2 g^2, which is h^2 Q at the root of y, and
Z = 2 q0 h - q1 g + (q1 h - 2 q2 g) x, int 1/(y S) is
-artanh(Z/(2 sqrt(H) S))/sqrt(H), written as the logarithm the library writes it
as, and arctan(Z/(2 sqrt(-H) S))/sqrt(-H), real where H < 0. The
derivative of each of those two needs only sqrt(z)^2 = z. Each pair is given by
the sign of its quantity, as for the roots of two linears. And since
Z^2 - 4 H Q is (q1^2 - 4 q0 q2) y^2, the artanh is of a number less than
1 in size where Q has no real root, and of one more than 1 where it has two:
there it is written as the logarithm of (z + 1)/(z - 1), which differs from
that of (1 + z)/(1 - z) by a constant and is the real one.
Where it is asked. Only with the linear below the bar. A polynomial beside the
root alone is SolveAPolynomialTimesAnOddHalfPowerOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)'s, which also
answers a leading coefficient that is zero. A first power of the linear with numbers in
it and in Q is SolveALinearBesideTheRootOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)'s, whose answer
is real on both sides of the linear's root, and
SolveARationalFunctionBesideTheRootOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable) takes distinct linears
apart into it. A power past the first is this rule's, and so is a first power with a
symbol in it, which those decline: 1/((g + h x) sqrt(a + c x^2)) was declined.
Rubi's 1.2.1.9, (d + e x + f x^2) sqrt(a + c x^2)/(g + h x)^4 and its like, was
declined or ran past its budget, where this is a triangular system and a recurrence.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

SolveAPolynomialOverAPowerOfALinearBesideARootOfAQuadratic​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

SolveAPolynomialOverAPowerOfALinearBesideARootOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)