A polynomial over distinct linears beside the root of a quadratic is taken apart as the rational function it is over that root, and each piece is closed

[Rafael-SOWNet]

N sqrt(Q)/D is N Q/(D sqrt(Q)), and P/D with D a product of
distinct linears is a polynomial plus a constant over each linear, so the
integrand is a polynomial over the root plus one K/((x - p) sqrt(Q)) per
linear. The first is the trigonometric substitution's or Euler's by its
shape and goes to the chain; each of the others is the reciprocal of its
linear, closed. Moses' sqrt(A^2 + B^2 (1 - y^2))/(1 - y^2) is B^2 over
the root -- the table's arcsine -- and A^2/2 over each of 1 - y and
1 + y beside it, and had no antiderivative: the Euler substitution takes
its generic first substitution for a symbolic leading coefficient, with
sqrt(-B^2) in it.

The denominator is factored over the integers where it is numeric and
read as written otherwise, and two factors with a common root are one
repeated, which is not this shape and is declined. Only where there is
something to take apart: a polynomial over the root alone, or one linear
with nothing divided out, is a rule before this one's already, and asking
again would be asking the same question. Before Euler, for the numeric
shapes too: 1/((x - 1)(x + 2) sqrt(x^2 + 1)) is two logarithms of the two
reciprocals where it was a partial-fraction decomposition in Euler's t.

One thing this reached that the answers with a sign in them had not: the
sign of e^x - 1/e^x - 2, which is what Timofeev's
sinh(x)^2 sinh(2x)/(1 - sinh(x)^2)^(3/2) carries once the reciprocal
substitution's u = e^x - 1/e^x hands one of its pieces here. Its
derivative was left unevaluated -- sgn is flat only for a real-valued
argument, and e^x was not shown real -- and no numeric check of the
antiderivative got past that. A positive constant base is real to a real
exponent, and the real-valued analysis says so now.

Rubi sample (463 problems, 5 s budget), against the #1315 master this branch is cut from:

master at #1315     406/463, 0 wrong, 0 timeouts, ~100 s
with this           407/463, 0 wrong, 0 timeouts, 97 s

One more, nothing lost: Moses 112, in 0.4 s. Timofeev 803 -- the hyperbolic integrand under the reciprocal substitution -- now goes through this rule for one of its pieces, in 0.9 s where it was 1.9 s, and is where the sign of e^x - 1/e^x - 2 came up. No other problem moved by more than noise.

The five test suites are green.

Part of #718.

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