Integrate a linear below a cube root of a quadratic in closed form

BREAKING-CHANGES.md

@@ -396,6 +396,23 @@ in Rubi's 1.1.3.4 come out shorter as well, and those that took two seconds take
 | `"1/(x^8*(a + b*x^3)*(c + d*x^3))".ToEntity().Integrate("x")` | `integral(...)` | 1,217 characters |
 | `"1/(x^6*(a + b*x^4)*(c + d*x^4))".ToEntity().Integrate("x")` | `integral(...)` | 795 characters |
 
+### A linear below the cube root of a quadratic is integrated in closed form
+
+**Answers where there were none.** `1/((d + e x) (d^2 + 3 e^2 x^2)^(1/3))` was declined, and so were six more of Rubi's
+1.2.1.2. With `y = x + b/(2c)`, `1/((d + e x) (a + b x + c x^2)^(1/3))` is a linear in `y` below the cube root of
+`a' + c y^2`, and where `c d'^2 = 3 a' e^2` or `c d'^2 = -9 a' e^2` it has a closed form of two or three logarithms and
+an arctangent ([#718](https://github.com/asc-community/AngouriMath/issues/718)).
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"1/((d+k*x)*(d^2+3*k^2*x^2)^(1/3))".ToEntity().Integrate("x")` | `integral(...)` | 773 characters |
+| `"1/((2+3*i*x)*(4-27*x^2)^(1/3))".ToEntity().Integrate("x")` | `integral(...)` | 398 characters |
+| `"1/((1+x^2)^(1/3)*(x+sqrt(3)))".ToEntity().Integrate("x")` | `integral(...)` | 290 characters |
+| `"1/((1+x^2)^(1/3)*(-x+sqrt(3)))".ToEntity().Integrate("x")` | `integral(...)` | 302 characters |
+| `"1/((d+k*x)*(d^2-9*k^2*x^2)^(1/3))".ToEntity().Integrate("x")` | `integral(...)` | 397 characters |
+| `"1/((d+k*x)*(c^2*d^2-b*c*d*k+b^2*k^2+3*b*c*k^2*x+3*c^2*k^2*x^2)^(1/3))".ToEntity().Integrate("x")` | `integral(...)` | 1,934 characters |
+| `"1/((d+k*x)*(-c^2*d^2+b*c*d*k+2*b^2*k^2+9*b*c*k^2*x+9*c^2*k^2*x^2)^(1/3))".ToEntity().Integrate("x")` | `integral(...)` | 1,274 characters |
+
 ### A linear and a quadratic to one power are integrated as one binomial
 
 **Answers where there were none.** `1/((1 + x)^(1/3) (1 - x + x^2)^(1/3))` was declined. A linear and a quadratic to

Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs

@@ -19988,6 +19988,83 @@ Piecewise piecewise when piecewise.Cases.Any() => piecewise.Cases.First().Expres
             return null;
         }
 
+        /// <summary>
+        /// A linear below the cube root of a quadratic, <c>1/((d + e x) (a + b x + c x^2)^(1/3))</c>,
+        /// in closed form where the shifted quadratic <c>a' + c y^2</c>, <c>y = x + b/(2 c)</c>, and
+        /// the linear <c>d' + e y</c> satisfy <c>c d'^2 = 3 a' e^2</c> or <c>c d'^2 = -9 a' e^2</c>.
+        /// </summary>
+        /// <remarks>
+        /// With <c>q^3 = 6 c^2 e^2/d'^2</c> and <c>R = (a + b x + c x^2)^(1/3)</c> the integral is
+        /// <c>-sqrt(3) c e/(d'^2 q^2) arctan(1/sqrt(3) + 2 c (d' - e y)/(sqrt(3) d' q R))
+        /// - 3 c e/(2 d'^2 q^2) ln(d + e x) + 3 c e/(2 d'^2 q^2) ln(c d' - c e y - d' q R)</c>, Rubi's
+        /// 1.2.1.2 rule. Where <c>c d'^2 = -9 a' e^2</c>, with <c>t = e y/d'</c>, <c>u = 1 - 3 t</c>
+        /// and <c>k^3 = a'</c>, it is <c>(-3/8 ln(2 R + k u) + 1/4 ln(d + e x) + 1/8 ln(u)
+        /// - sqrt(3)/4 arctan(sqrt(3) k u/(4 R - k u)))/(e k)</c>: on the curve
+        /// <c>w^3 = 1 - 9 t^2</c> the tangent at each point over <c>t = -1</c> meets it again at
+        /// <c>(1/3, 0)</c>, so <c>w + r (1 - 3 t)/2</c> for the cube roots of unity <c>r</c> are the
+        /// functions whose logarithms carry the residues. Otherwise the integral is elliptic. The
+        /// answer is differentiated back.
+        /// https://github.com/asc-community/AngouriMath/issues/718
+        /// </remarks>
+        internal static Entity? SolveALinearBelowACubeRootOfAQuadraticInClosedForm(Entity expr, Entity.Variable x)
+        {
+            var (numerator, denominator) = Functions.SingleQuotient.Of(expr);
+            if (numerator.ContainsNode(x))
+                return null;
+            Entity? linear = null, quadratic = null;
+            foreach (var factor in Mulf.LinearChildren(denominator))
+            {
+                if (!factor.ContainsNode(x))
+                    numerator /= factor;
+                else if (quadratic is null && factor is Powf(var @base, Number.Rational power) && power == Number.Rational.Create(1, 3))
+                    quadratic = @base;
+                else if (linear is null && TreeAnalyzer.TryGetPolyLinear(factor, x, out _, out _))
+                    linear = factor;
+                else
+                    return null;
+            }
+            if (linear is null || quadratic is null
+                || !TreeAnalyzer.TryGetPolyLinear(linear, x, out var e, out var d)
+                || !TreeAnalyzer.TryGetPolynomial(quadratic, x, out var read)
+                || read.Keys.Any(key => key.Sign < 0 || key.CompareTo(EInteger.FromInt32(2)) > 0)
+                || !read.TryGetValue(EInteger.FromInt32(2), out var c) || VanishesIdentically(c.Expand()))
+                return null;
+            var a = read.TryGetValue(EInteger.Zero, out var a0) ? a0 : Number.Integer.Zero;
+            var b = read.TryGetValue(EInteger.One, out var b0) ? b0 : Number.Integer.Zero;
+            var shift = (b / (2 * c)).InnerSimplified;
+            var dShifted = (d - e * shift).InnerSimplified;
+            var aShifted = (a - b * b / (4 * c)).InnerSimplified;
+            if (VanishesIdentically(dShifted.Expand()) || VanishesIdentically(e.Expand()))
+                return null;
+            var three = Number.Integer.Create(3);
+            var root3 = MathS.Sqrt(three);
+            if (VanishesIdentically((c * dShifted * dShifted + 9 * aShifted * e * e).Expand().InnerSimplified.Expand()))
+            {
+                var k = MathS.Pow(aShifted, Number.Rational.Create(1, 3));
+                var cubeRoot = MathS.Pow(quadratic, Number.Rational.Create(1, 3));
+                var u = 1 - 3 * e * (x + shift) / dShifted;
+                var closed = numerator / (e * k) * (
+                    -Number.Rational.Create(3, 8) * MathS.Ln(2 * cubeRoot + k * u)
+                    + Number.Rational.Create(1, 4) * MathS.Ln(linear)
+                    + Number.Rational.Create(1, 8) * MathS.Ln(u)
+                    - root3 / 4 * MathS.Arctan(root3 * k * u / (4 * cubeRoot - k * u)));
+                closed = closed.InnerSimplified;
+                return Functions.PartialFractions.DerivativeHoldsAtSampledPoints(closed, expr, x) ? closed : null;
+            }
+            if (!VanishesIdentically((c * dShifted * dShifted - 3 * aShifted * e * e).Expand().InnerSimplified.Expand()))
+                return null;
+            var q = MathS.Pow(6 * c * c * e * e / (dShifted * dShifted), Number.Rational.Create(1, 3));
+            var r = MathS.Pow(quadratic, Number.Rational.Create(1, 3));
+            var y = x + shift;
+            var scale = c * e / (dShifted * dShifted * q * q);
+            var answer = numerator * (
+                -root3 * scale * MathS.Arctan(1 / root3 + 2 * c * (dShifted - e * y) / (root3 * dShifted * q * r))
+                - three / 2 * scale * MathS.Ln(linear)
+                + three / 2 * scale * MathS.Ln(c * dShifted - c * e * y - dShifted * q * r));
+            answer = answer.InnerSimplified;
+            return Functions.PartialFractions.DerivativeHoldsAtSampledPoints(answer, expr, x) ? answer : null;
+        }
+
         /// <summary>
         /// A power of a linear that is not whole, times a linear, over a whole power of a quadratic,
         /// with the power of the quadratic lowered one at a time:

Sources/AngouriMath/Functions/Continuous/Integration/Integration.Definition.cs

@@ -668,6 +668,8 @@ private static Entity WithItsPiecewiseSumsCombined(Entity expr)
             // Two linears to one power that is not whole, as one quadratic to that power, before each
             // linear is written once: `(2 + 3x)^4/((1 - 2x)^(5/2) (3 + 5x)^(5/2))` took forty-five
             // seconds there, and over `(3 - x - 10x^2)^(5/2)` it takes half of one.
+            // A linear below the cube root of a quadratic, in closed form where Rubi has one.
+            if ((answer = IndefiniteIntegralSolver.SolveALinearBelowACubeRootOfAQuadraticInClosedForm(expr, x)) is { }) return answer;
             if ((answer = IndefiniteIntegralSolver.SolveTwoLinearsToOnePowerAsOneQuadratic(expr, x, integrateByParts)) is { }) return answer;
             // A linear and a quadratic to one power as one binomial, `(1 + x) (1 - x + x^2)` as
             // `1 + x^3`, which the rules for a power of a binomial answer.

Sources/Tests/UnitTests/Calculus/ALinearBelowACubeRootOfAQuadraticIntegralTest.cs

@@ -0,0 +1,75 @@
+//
+// Copyright (c) 2019-2026 Angouri.
+// AngouriMath is licensed under MIT.
+// Details: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.
+// Website: https://am.angouri.org.
+//
+
+using System;
+using AngouriMath.Extensions;
+using Xunit;
+
+namespace AngouriMath.Tests.Calculus
+{
+    /// <summary>
+    /// A linear below the cube root of a quadratic, <c>1/((d + e x) (a + b x + c x^2)^(1/3))</c>,
+    /// where the shifted quadratic and the linear satisfy <c>c d^2 = 3 a e^2</c> or
+    /// <c>c d^2 = -9 a e^2</c>: Rubi's 1.2.1.2 closed forms, an arctangent and logarithms. Sampled on both sides of the linear's root,
+    /// and where the quadratic is negative, so the cube root is complex.
+    /// <a href="https://github.com/asc-community/AngouriMath/issues/718">#718</a>
+    /// </summary>
+    [Trait("Area", "Calculus")]
+    public sealed class ALinearBelowACubeRootOfAQuadraticIntegralTest
+    {
+        private static void DifferentiatesBack(string integrand, params (string, double)[] pins)
+        {
+            var integral = integrand.ToEntity().Integrate("x");
+            Assert.DoesNotContain("integral(", integral.Stringize());
+
+            var pinned = integral.Substitute("C", 0);
+            Entity original = integrand.ToEntity();
+            foreach (var (name, value) in pins)
+            {
+                pinned = pinned.Substitute(name, value);
+                original = original.Substitute(name, value);
+            }
+            var derivative = pinned.Differentiate("x");
+            foreach (var at in new[] { -2.3, -0.7, 0.4, 1.9 })
+            {
+                var got = derivative.Substitute("x", at).EvalNumerical();
+                var want = original.Substitute("x", at).EvalNumerical();
+                var difference = Math.Abs((double)(got - want).RealPart) + Math.Abs((double)(got - want).ImaginaryPart);
+                var scale = Math.Max(1.0, Math.Abs((double)want.RealPart) + Math.Abs((double)want.ImaginaryPart));
+                Assert.True(difference / scale < 1e-8,
+                    $"d/dx of the antiderivative of {integrand} is {got} at x = {at}, where the integrand is {want}");
+            }
+        }
+
+        [Theory]
+        [InlineData("1/((1+x^2)^(1/3)*(x+sqrt(3)))")]
+        [InlineData("1/((1+x^2)^(1/3)*(-x+sqrt(3)))")]
+        [InlineData("1/((2+3*i*x)*(4-27*x^2)^(1/3))")]
+        [InlineData("1/((1+x)*(1-9*x^2)^(1/3))")]
+        public void WithNumbersIsAnsweredAndDifferentiatesBack(string integrand)
+            => DifferentiatesBack(integrand);
+
+        [Theory]
+        [InlineData("1/((d+k*x)*(d^2+3*k^2*x^2)^(1/3))")]
+        [InlineData("1/((d+k*x)*(c^2*d^2-b*c*d*k+b^2*k^2+3*b*c*k^2*x+3*c^2*k^2*x^2)^(1/3))")]
+        [InlineData("1/((d+k*x)*(d^2-9*k^2*x^2)^(1/3))")]
+        [InlineData("1/((d+k*x)*(-c^2*d^2+b*c*d*k+2*b^2*k^2+9*b*c*k^2*x+9*c^2*k^2*x^2)^(1/3))")]
+        public void WithSymbolsIsAnsweredAndDifferentiatesBack(string integrand)
+        {
+            DifferentiatesBack(integrand, ("b", 2), ("c", 3), ("d", 0.7), ("k", 0.5));
+            DifferentiatesBack(integrand, ("b", -2), ("c", 3), ("d", -0.7), ("k", 1.5));
+        }
+
+        [Fact]
+        public void IsNotAnsweredInClosedFormWithoutTheRelation()
+        {
+            // c d^2 = 2 a e^2: the integral is elliptic, and no closed form is claimed.
+            var integral = "1/((1+x)*(1+2*x^2)^(1/3))".ToEntity().Integrate("x");
+            Assert.DoesNotContain("arctan", integral.Stringize());
+        }
+    }
+}