A tangent times that of its double is a secant less one

BREAKING-CHANGES.md

@@ -654,6 +654,20 @@ that `sec(x)^2/(a + b sin(x))` is rational in them rather than declined at once
 | `"tan(x)^4/(a + a*cos(x))".ToEntity().Integrate("x")` | `integral(...)` | an antiderivative in `tan(x/2)` |
 | `"csc(x)^2/(a + a*cos(x))".ToEntity().Integrate("x")` | `integral(...)` | `(tan(x/2)^3/12 + tan(x/2)/2 - 1/(4 tan(x/2)))/a` |
 
+### A tangent times that of its double is a secant less one
+
+**Answers where there were none.** `sec(2(a + b x))^2 sqrt(c tan(a + b x) tan(2(a + b x)))` was
+declined, with the rest of Rubi's 4.7.7 powers of `c tan(y) tan(2y)` beside the secant or cosine
+of `2y`, some after searches past the budget. `tan(y) tan(2y)` is `sec(2y) - 1`, and written so the
+integrand is in the one argument `2y`, which the half-angle tangent answers; the answer is exact
+where the integrand is real ([#718](https://github.com/asc-community/AngouriMath/issues/718)).
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"sec(2*(a + b*x))^2*sqrt(c*tan(a + b*x)*tan(2*(a + b*x)))".ToEntity().Integrate("x")` | `integral(...)` | `sgn(tan(a + b x))` times powers of `sqrt(1 - tan(a + b x)^2)` |
+| `"1/sqrt(c*tan(a + b*x)*tan(2*(a + b*x)))".ToEntity().Integrate("x")` | `integral(...)` | logarithms in `tan(a + b x)`, `sgn(tan(a + b x))` in front |
+| `"tan(x)*tan(2*x)".ToEntity().Integrate("x")` | `integral(...)` | `ln((1 + sin(2x))/(1 - sin(2x)))/4 - x` |
+
 ### A power of `a + i a tan` beside a power of the secant is integrated as an exponential
 
 **Answers where there were none.** `sqrt(a + i a tan(x))/sqrt(c sec(x))` is `sqrt(a/c) e^(i x/2)`

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

@@ -4152,6 +4152,56 @@ Entity Integral(ERational power)
             return roots < 2 ? answer : answer.Provided((secantKind ? MathS.Cos(argument) : MathS.Sin(argument)) >= Number.Integer.Zero);
         }
 
+        /// <summary>
+        /// <c>tan(y) tan(2y)</c> written <c>sec(2y) - 1</c>, and the integrand asked again in the
+        /// one argument <c>2y</c>.
+        /// </summary>
+        /// <remarks>
+        /// <para>
+        /// <c>tan(y)</c> is <c>(1 - cos(2y))/sin(2y)</c>, so <c>tan(y) tan(2y)</c> is
+        /// <c>(1 - cos(2y))/cos(2y)</c>, which is <c>sec(2y) - 1</c> wherever both sides are
+        /// defined; at the zeros of <c>cos(y)</c> the left is <c>-2</c> as a limit, which the right
+        /// is. Rubi's 4.7.7 has <c>sec(2(a + b x))^k sqrt(c tan(a + b x) tan(2(a + b x)))</c> and
+        /// its kin, a half-odd power of <c>c sec(2y) - c</c> beside powers of the secant or cosine
+        /// of <c>2y</c>, which <see cref="SolveByTheHalfAngleTangentBesideAHalfOddPowerOfOnePlusASecant"/>
+        /// answers in that one argument; with two arguments nothing read them.
+        /// </para>
+        /// <para>
+        /// The same question in another spelling, so asked as the same question. The answer is
+        /// what that rule gives, exact where the integrand is real.
+        /// https://github.com/asc-community/AngouriMath/issues/718
+        /// </para>
+        /// </remarks>
+        internal static Entity? SolveByWritingATangentTimesThatOfItsDoubleThroughTheSecant(Entity expr, Entity.Variable x, bool integrateByParts)
+        {
+            static bool IsTwice(Entity twice, Entity once)
+                => (twice - 2 * once).Expand().InnerSimplified.Evaled is Number.Complex { IsZero: true };
+            var found = false;
+            var rewritten = expr.Replace(node =>
+            {
+                if (node is not Mulf || !node.ContainsNode(x))
+                    return node;
+                var factors = Mulf.LinearChildren(node).ToList();
+                var single = factors.FindIndex(factor => factor is Tanf(var y) && y.ContainsNode(x)
+                    && factors.Any(other => other is Tanf(var z) && z != y && IsTwice(z, y)));
+                if (single < 0)
+                    return node;
+                var once = ((Tanf)factors[single]).Argument;
+                var twice = factors.FindIndex(other => other is Tanf(var z) && z != once && IsTwice(z, once));
+                var doubled = ((Tanf)factors[twice]).Argument;
+                Entity others = Number.Integer.One;
+                for (var i = 0; i < factors.Count; i++)
+                    if (i != single && i != twice)
+                        others = others == Number.Integer.One ? factors[i] : others * factors[i];
+                found = true;
+                // Spread over the difference, so that a root of it reads as `c sec(2y) - c`.
+                return others == Number.Integer.One ? MathS.Sec(doubled) - 1 : others * MathS.Sec(doubled) - others;
+            });
+            if (!found)
+                return null;
+            return Integration.ComputeAsTheSameQuestion(rewritten, x, integrateByParts);
+        }
+
         /// <summary>
         /// <paramref name="expr"/> as what it is a product of beside <paramref name="bases"/>, and
         /// the sum of the exponents of each base in it, read through products, quotients and

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

@@ -867,6 +867,8 @@ private static Entity Normalized(Entity expr, Entity.Variable x) =>
             if ((answer = IndefiniteIntegralSolver.SolveSymbolicPowersOfOnePlusMinusASineThroughTheSine(expr, x, integrateByParts)) is { }) return answer;
             // And a half-odd power of a +- a sec(y), which is that square over cos(y): by the half-angle
             // tangent, in which the whole is rational beside one root.
+            // And `tan(y) tan(2y)` written `sec(2y) - 1` first, so that the rule reads one argument.
+            if ((answer = IndefiniteIntegralSolver.SolveByWritingATangentTimesThatOfItsDoubleThroughTheSecant(expr, x, integrateByParts)) is { }) return answer;
             if ((answer = IndefiniteIntegralSolver.SolveByTheHalfAngleTangentBesideAHalfOddPowerOfOnePlusASecant(expr, x, integrateByParts)) is { }) return answer;
             // And `a ± a cosh(y)` under a fractional power: `2a cosh(y/2)^2`, `-2a sinh(y/2)^2`.
             if ((answer = IndefiniteIntegralSolver.SolveByTheHalfAngleWhereOnePlusAHyperbolicCosineIsASquare(expr, x, integrateByParts)) is { }) return answer;

Sources/Tests/UnitTests/Calculus/TangentTimesThatOfItsDoubleIntegralTest.cs

@@ -0,0 +1,51 @@
+//
+// 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>
+    /// <c>tan(y) tan(2y)</c> is <c>sec(2y) - 1</c>, and written so a power of
+    /// <c>c tan(y) tan(2y)</c> beside the secant or cosine of <c>2y</c> is in one argument, which
+    /// the half-angle tangent answers. Rubi's 4.7.7.
+    /// <a href="https://github.com/asc-community/AngouriMath/issues/718">#718</a>
+    /// </summary>
+    /// <remarks>
+    /// Checked by differentiating back with <c>a = 0.4</c>, <c>b = 1.1</c>, <c>c = 1.3</c> where the
+    /// integrand is real, <c>cos(2(a + b x))</c> positive, on both sides of the zero of
+    /// <c>tan(a + b x)</c>: the rule the rewriting reaches is exact there.
+    /// </remarks>
+    [Trait("Area", "Calculus")]
+    public sealed class TangentTimesThatOfItsDoubleIntegralTest
+    {
+        [Theory]
+        [InlineData("sec(2*(a + b*x))^2*sqrt(c*tan(a + b*x)*tan(2*(a + b*x)))")]
+        [InlineData("cos(2*(a + b*x))*(c*tan(a + b*x)*tan(2*(a + b*x)))^(3/2)")]
+        [InlineData("1/sqrt(c*tan(a + b*x)*tan(2*(a + b*x)))")]
+        [InlineData("sec(2*(a + b*x))/(c*tan(a + b*x)*tan(2*(a + b*x)))^(3/2)")]
+        [InlineData("tan(x)*tan(2*x)")]
+        public void IsAPowerOfASecantLessOne(string integrand)
+        {
+            var integral = integrand.ToEntity().Integrate("x");
+            Assert.DoesNotContain("integral(", integral.Stringize());
+            Entity Pinned(Entity e) => e.Substitute("a", 0.4).Substitute("b", 1.1).Substitute("c", 1.3);
+            var derivative = Pinned(integral.Substitute("C", 0)).Differentiate("x");
+            var original = Pinned(integrand.ToEntity());
+            foreach (var at in new[] { -0.9, -0.7, -0.5, 0.05, 0.2 })
+            {
+                var want = original.Substitute("x", at).EvalNumerical();
+                var got = derivative.Substitute("x", at).EvalNumerical();
+                Assert.True(Math.Abs((double)(got - want).RealPart) + Math.Abs((double)(got - want).ImaginaryPart)
+                            < 1e-9 * Math.Max(1, Math.Abs((double)want.RealPart) + Math.Abs((double)want.ImaginaryPart)),
+                    $"d/dx of the antiderivative of {integrand} is {got} at x = {at}, where the integrand is {want}");
+            }
+        }
+    }
+}