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

 Method (no overloads)

Summary

A rational function of tan(z) beside roots of linears in it, integrated in
t = tan(z) term by term over the partial fractions of its rational part, with the
1 + t^2 that dz = dt/(f (1 + t^2)) puts below the bar written as the linears
of its roots, (1 + i t)(1 - i t): each term is the roots over one power of a linear,
which the rules for a root of a linear over a linear answer.

Remarks

Rubi's 4.3.4.2 (c + d tan(z))^(5/2) (A + B tan(z) + C tan(z)^2)/sqrt(a + b tan(z)) and
4.3.2.1 1/((a + i a tan(z))^3 (c + d tan(z))^(3/2)) ran past the budget, as their
neighbours with smaller powers did not: in t, under the root of the quotient of the two
linears or the root of the one, 1 + t^2 is a quartic with symbols in every coefficient.
Over its linears the same integrands are three terms, or a few more, of a second or less each.
Only two roots of linears with distinct roots beside a polynomial or a linear below the
bar, or one root below the bar beside a whole power of a +- i a tan(z) below it, which
shares a root with 1 + t^2; neither root a root of t or of 1 +- i t, which
the rules for a root of the tangent and for the imaginary sum read, and below the bar linears
only. Two roots alone, and a root above the bar beside the imaginary sum, the substitution in
the root answers in as few terms and as soon, and are left to it. The decomposition is exact,
its coefficients constants in the symbols, and the sum is checked by its derivative in
t before it is returned: each term's answer is its own rule's.
https://github.com/asc-community/AngouriMath/issues/718

























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