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AngouriMath

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

 Method (no overloads)

Summary

A product of powers of quotients of polynomials in x whose roots
are rational in the symbols, where one root stands in two of the bases, written over
each linear once: K (x - r_1)^(e_1) ... (x - r_k)^(e_k), with e_i the sum
of the exponents r_i has in the bases. K is the integrand over that
product, and its logarithmic derivative is zero -- each base is its leading coefficient
times its linears, so both sides have the same sum e_i/(x - r_i) -- so it is
constant wherever it is defined and stands in front of the answer, the way Rubi writes
x^p (c + d/x)^p/(1 + c x/d)^p in front of its.

Remarks

e^atanh(a x) sqrt(c - c/(a x)) is sqrt((1 + a x)/(1 - a x)) sqrt(c (a x - 1)/(a x)),
and the root 1/a stands in both bases, with exponents -1/2 and 1/2.
Written once it is gone: the integrand is K sqrt(a x + 1)/sqrt(x), which the
rule for two linear radicals answers in milliseconds, where the substitution search ran
out of time. Rubi's 7.3.6 and 7.4.2. https://github.com/asc-community/AngouriMath/issues/718
 Only where it helps: a root stands in two bases under powers that are not whole --
sqrt(c - a c x)/sqrt(1 - a x) -- or in a root of a quotient with x below the bar,
sqrt(c - c/(a x)), and in another base; and at most two linears are left under a
power that is not whole, which is what the rule for two linear radicals reads. A root
beside a whole power of one of its linears -- sqrt(x)/(x (x + 1)), and
(1 - a^2 x^2)^(3/2)/((1 - a x)^2 (c + d x)), which the rules behind this one write
over the radicand -- is read by the rules as it is written, and is answered without a
K.

























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