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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)

Method (no overloads)

Summary

The integral of T(x)/((g + h x)^k S), T a polynomial with the coefficients
t and S a square root of Q = q0 + q1 x + q2 x^2, by
undetermined coefficients: the algebraic terms, which the caller multiplies by
S; a multiple of int 1/S; and with the linear, a multiple of
int 1/((g + h x) S). Null where a coefficient is not written back in the
symbols.

Remarks

All this uses of S is S' = Q'/(2 S), which holds because
sqrt(z)^2 = z for every complex z, so S may be the root of Q or
a product of the roots of its factors.
The polynomial part. In powers of y = g + h x, the terms of T from
y^k up leave a polynomial R(x), and
int R/S = U S + lambda int 1/S. The derivative of U S is
(U' Q + U Q'/2)/S, so R = U' Q + U Q'/2 + lambda, which for U of one
degree less than R is triangular. The coefficient of x^j on the right is
(j + 1) q0 u_(j+1) + (j + 1/2) q1 u_j + j q2 u_(j-1). That is solved for
u_(j-1) from the top down, and lambda is what is left at x^0.
The terms over powers of y. With xi = -g/h and s = x - xi, write
Q = alpha + beta s + gamma s^2. The derivative of S s^(1-j) is
(2 (1 - j) alpha s^(-j) + (3 - 2j) beta s^(1-j) + (4 - 2j) gamma s^(2-j)) / (2 S).
So for j >= 2, I_j = int 1/(s^j S) is given by S s^(1-j) and the two
integrals before it, and all the way down it is algebraic terms and a multiple of
I_1, where alpha, the value of Q at the root of y, is not zero.
Where it is, the linear shares a root with Q, and the derivative of
S s^(-j) is ((1 - 2j) beta s^(-j) + (2 - 2j) gamma s^(1-j)) / (2 S): then
every I_j is algebraic, from I_1 = -2 S/(beta s) up, and there is no
third integral.
A power of Q below the bar too.T/((g + h x)^k Q^n S) for
belowTheRootn >= 1, where alpha is not zero: the
terms over powers of y are those of the series of T/Q^n at the linear's root
below s^k, and what is left is V/Q^n with V a polynomial. Divided by
Qn times, V is a polynomial part, which goes as above, and a
linear r0 + r1 x over each power Q^p. And
int (r0 + r1 x)/Q^(p + 1/2) is (A + B x)/Q^(p - 1/2) and
(2p - 2) B int 1/Q^(p - 1/2), with A and B from two equations whose
determinant is q1^2 - 4 q0 q2: algebraic all the way down, since at p = 1 there is nothing left.
Back in the symbols. A symbol stands for xi until the end, so that what
is computed is polynomial in it, and alpha^(k-1), or beta^k beside a shared
root, is divided by once. Each coefficient is then written homogeneous in g and
h, with the powers of h counted, over the factors the caller writes
h^2 alpha or h beta as: pole's AtThePole. The multiple of int 1/S is divided by
firstKindOver first, where there is one, and the multiple of
int 1/(y S) by the factor thirdKindAlsoOver names: the
caller's closed forms have them below the bar.

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