Clear a sum with the root of a quadratic below the bar by its conjugate

[Rafael-SOWNet]

Part of #718

(A + B sqrt(Q))/(C + D sqrt(Q)) is (A + B sqrt(Q)) (C - D sqrt(Q))/(C^2 - D^2 Q), a rational function plus one beside the root. Each piece goes back to the chain. The rule runs after every rule that already answers the numeric spellings.

Euler's substitution declined Rubi's 1/(a c + b c x^2 + d sqrt(a + b x^2)), because a symbolic leading coefficient puts sqrt(b) into every coefficient.

Integrand	master ed821bf	2.5.0	this branch
1.3.2:750 x^2/(a c + b c x^2 + d sqrt(a + b x^2))	unsolved	unsolved	1,605 chars
1.3.2:751 1/(a c + b c x^2 + d sqrt(a + b x^2))	unsolved	unsolved	3,464 chars
1.3.2:752 1/(x^2 (a c + b c x^2 + d sqrt(a + b x^2)))	unsolved	unsolved	5,052 chars
1.3.2:1182 x sqrt(2 - x^2)/(x - sqrt(2 - x^2))	unsolved	unsolved	195 chars
1/(c sqrt(a + b x^2) + d)	unsolved	unsolved	2,888 chars

The symbolic answers are large because each piece's answer is piecewise in the signs of the constants, and adding the two pieces multiplies their cases. Master does the same on sums it already answers, so I'll file that separately.

Verification

• Exact check: each answer is differentiated back with exact constants, at 50 and 2,000 digits. The constants were pinned on both signs of b, both signs of a c^2 - d^2, and d < 0; 1182 was checked on both sides of its pole. Every residual is 0.
• Rubi corpus: on the 9,684-problem pocket (1.3.2, 1.3.1, 1.2.1.2–6) and the sample from families 1–8, 4 problems go from unsolved to solved (750–752 and 1182). Nothing moves the other way, and there are no wrong answers.
• The benchmark allocation gate passes.
• The unit suite passes: 15,325 passed, 0 failed.
• The new test, TheConjugateOfASumWithARootBelowTheBarIntegralTest, differentiates each answer back under four pin sets.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura