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we need more integral solvers #233
Description
There are a lot of stuff to do, but to begin with, I will put some unsolvable by AM integrals here:
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cos(x^2)*xvariable substitution method, WolframAlpha -
sin(x)*e^xintegrating by parts and solving integral equation, WolframAlpha -
1/(a^2 + x^2)standard integral for arctan, WolframAlpha -
x^2 / (x^4 + 1)partial fractioning WolframAlpha -
sqrt(tan(x))very painful, requires different solvers, WolframAlpha
I do not mention special integrals here, like e^(x^2) or 1/ln(x)
Activity
Add five families of standard integrals (#233) (#675)
Integrate a quotient whose numerator is not constant (#233) (#681)
Split a rational function at a rational root of its denominator (#233) (
Integrate a constant over x squared times the root of a quadratic (#691)
Rafael-SOWNet commentedon Aug 16, 2026
Three of the five are done. Measured on
masterata45a7256, each verified by differentiating the answer back rather than by eye.cos(x^2)*xsin(x^2) / 2 + Csin(x)*e^xe^x * (sin(x) - cos(x)) / 2 + C1/(a^2 + x^2)piecewiseon the sign ofa^2, witharctan(x / sqrt(a^2)) / sqrt(a^2) + Con the positive branchx^2 / (x^4 + 1)sqrt(tan(x))So the substitution case, the by-parts-and-solve-the-equation case, and the arctan standard integral are all answered. The two left are the two the body itself flags as hardest — partial fractioning over an irreducible quartic, and
sqrt(tan(x)), which it calls "very painful, requires different solvers". Both are also the twocasbenchreports unsolved, so this agrees with the corpus.On the arctan one, a correction to my own first reading. Differentiating it back and simplifying does not reduce to zero, and I nearly reported that as a wrong answer. It is not — the
piecewisecarries branch conditions that block symbolic cancellation. Checked numerically instead:1/51/501/131/1304/54/501/411/410Worth recording because it will catch the next person too: an antiderivative that comes back as a
Piecewisecannot be verified by simplifyingd/dx F - fto zero, and a harness that does so will report a false defect.Suggest ticking the three and leaving this open for the remaining two.
A power of the variable is a substitution even where it is written no…
A biquadratic denominator is decomposed over the reals (#1145)
A fractional power, and the tangent, are substitutions (#1146)
21 remaining items
Rafael-SOWNet commentedon Oct 19, 2026
Moved to 2.7.0: the integrator work goes on past 2.6.0. This release's part of it is closed: #1469, #1501 and the closed sub-issues.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Rafael-SOWNet commentedon Oct 19, 2026
@Happypig375, should this close with 2.6.0 or carry on into 2.7.0? I moved it to 2.7.0 earlier today and have put it back on 2.6.0 until you answer, because you moved it there for calculus first.
What 2.6.0 delivers, measured on the full Rubi corpus at master
3d5d1461:Close at 2.6.0, with a 2.7.0 successor for the remaining families and #1486? Or keep this open into 2.7.0? #1486 goes wherever this does, since both touch the integrator's code.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Happypig375 commentedon Oct 19, 2026
@Rafael-SOWNet Your choice - 2.7 might go too fast while 2.6 too packed, but all integration being in the same release looks more organised.
Integrate an exponential of the reciprocal of a linear beside another…
Integrate an exponential of a quadratic over the linear through its v…
Integrate a power of x times an artanh whose derivative is a root, by…
Integrate a power of a quotient of linears beside whole powers of them (