A root of a square in a trigonometric function is its modulus

BREAKING-CHANGES.md

@@ -740,6 +740,19 @@ inner-simplified now, and declined in seconds where the parts do not separate
 | `"x^4*e^(2*i*atan(a + b*x))".ToEntity().Integrate("x")` | `integral(...)` | a polynomial, logarithms and arctangents of `a + b x` |
 | `"sec(c + d*x)^2/(a + i*a*tan(c + d*x))".ToEntity().Integrate("x")` | `ln(i a d tan(c + d x) + a d)/(i a d)` | `ln(i tan(c + d x) + 1)/(i a d)`, the same up to a constant |
 
+### A root of a square in a trigonometric function is its modulus
+
+**Answers where there were none.** `(a + b sin(x)) sqrt(b^2 + 2 a b sin(x) + a^2 sin(x)^2)` was
+declined or past the budget, with the rest of Rubi's 4.7.7 half-odd powers of a perfect square in a
+sine, tangent or secant. The radicand is `a^2 (f + b/a)^2`, so its root is `sqrt(a^2) |f + b/a|`,
+written with the sign of `f + b/a` in front of the integral, as a root of a perfect square in `x`
+already was ([#718](https://github.com/asc-community/AngouriMath/issues/718)).
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"(a + b*sin(x))*sqrt(b^2 + 2*a*b*sin(x) + a^2*sin(x)^2)".ToEntity().Integrate("x")` | `integral(...)` | `sgn(sin(x) + b/a) sqrt(a^2)` times an expression in `x`, `cos(x)` and `sin(2x)`, `provided a^2 > 0` |
+| `"(a + b*tan(x))/sqrt(b^2 + 2*a*b*tan(x) + a^2*tan(x)^2)".ToEntity().Integrate("x")` | `integral(...)` | `sgn(tan(x) + b/a) a/sqrt(a^2)` times the antiderivative of `(a + b tan(x))/(a tan(x) + b)`, `provided a^2 > 0` |
+
 ### `NaN` was returned as the antiderivative of something that has one
 
 **A wrong answer, not a missing one.** `1/(a*x^2)` came back as `NaN + C`, and `NaN` is this

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

@@ -18029,6 +18029,86 @@ bool MayBeARootOfASquare(Entity node, Entity.Variable x)
             return assumed == Entity.Boolean.True ? answer : answer.Provided(assumed);
         }
 
+        /// <summary>
+        /// A half-odd power of a perfect square in one trigonometric function of <paramref name="x"/>
+        /// is that power of the modulus: <c>sqrt(b^2 + 2 a b sin(y) + a^2 sin(y)^2)</c> is
+        /// <c>sqrt(a^2) |sin(y) + b/a|</c>, which is <c>sqrt(a^2) sgn(sin(y) + b/a) (sin(y) + b/a)</c>,
+        /// the sign constant between the zeros of the linear in the sine.
+        /// </summary>
+        /// <remarks>
+        /// <para>
+        /// <see cref="SolveByTakingARootOfAPerfectSquare"/>'s reading, with the function for the
+        /// variable: Rubi's 4.7.7 has <c>(a + b sin(d + e x)) sqrt(b^2 + 2 a b sin(d + e x) + a^2 sin(d + e x)^2)</c>
+        /// and the same with a tangent or a secant, at the half-odd powers either way, and they
+        /// were declined or past the budget where with the modulus written the rest is rational in
+        /// the function. At the top only, and only where each such root is a factor of the
+        /// integrand, so that its sign goes in front; a leading coefficient a number or one of
+        /// known sign for a real parameter, whose condition travels with the answer, as that
+        /// rule's does.
+        /// https://github.com/asc-community/AngouriMath/issues/718
+        /// </para>
+        /// </remarks>
+        internal static Entity? SolveByTakingARootOfAPerfectSquareInATrigonometricFunction(Entity expr, Entity.Variable x, bool integrateByParts)
+        {
+            if (!Integration.AnsweringTheQuestionAsked)
+                return null;
+            var factors = FactorsOfTheIntegrand(expr).Select(pair => pair.Factor).ToList();
+            var changed = false;
+            var declined = false;
+            Entity assumed = Entity.Boolean.True;
+            Entity signs = Number.Integer.One;
+            var written = expr.Replace(node =>
+            {
+                if (declined || node is not Powf(var radicand, Number.Rational r) || r is Number.Integer
+                    || !r.ERational.Denominator.Equals(EInteger.FromInt32(2)) || !radicand.ContainsNode(x)
+                    || radicand.Complexity > 40 || TreeAnalyzer.TryGetPolynomial(radicand, x, out _))
+                    return node;
+                var functions = radicand.Nodes.Where(inner => inner.ContainsNode(x)
+                    && inner is Sinf or Cosf or Tanf or Cotanf or Secantf or Cosecantf).Distinct().ToList();
+                if (functions.Count != 1)
+                    return node;
+                var function = functions[0];
+                var w = Variable.CreateUnique(expr, "w_sq");
+                var inW = radicand.Replace(inner => inner == function ? w : inner);
+                if (inW.ContainsNode(x) || !TreeAnalyzer.TryGetPolynomial(inW, w, out var monomials)
+                    || monomials.Keys.Max() is not { } degree || !degree.Equals(EInteger.FromInt32(2))
+                    || monomials.Keys.Any(k => k.Sign < 0))
+                    return node;
+                var a = monomials[degree];
+                var b = monomials.TryGetValue(EInteger.One, out var b1) ? b1 : Number.Integer.Zero;
+                var c = monomials.TryGetValue(EInteger.Zero, out var c0) ? c0 : Number.Integer.Zero;
+                if (b.Evaled is Number.Complex and not Number.Real || c.Evaled is Number.Complex and not Number.Real
+                    || !IsAPerfectSquareDiscriminant(a, b, c))
+                    return node;
+                if (a.Evaled is not Number.Real { IsPositive: true })
+                {
+                    if (!IsPositiveForARealParameter(a, x, out var leadingAssumed, assumed))
+                        return node;
+                    assumed = leadingAssumed;
+                }
+                // The sign in front of the integral, which needs the root to be a factor of it.
+                if (factors.Count(factor => factor == node) < expr.Nodes.Count(inner => inner == node))
+                {
+                    declined = true;
+                    return node;
+                }
+                var h = (b / (Number.Integer.Create(2) * a)).InnerSimplified;
+                if (h.Vars.Any())
+                    h = Functions.PartialFractions.Bare(h.Simplify());
+                var linear = h == Number.Integer.Zero ? function : (function + h).InnerSimplified;
+                var twoR = Number.Integer.Create(r.ERational.Numerator);
+                signs = signs * MathS.Signum(linear);
+                changed = true;
+                return MathS.Pow(a, r) * (twoR == Number.Integer.One ? linear : MathS.Pow(linear, twoR));
+            });
+            if (declined || !changed)
+                return null;
+            if (Integration.ComputeAsTheSameQuestion(written, x, integrateByParts) is not { } answer)
+                return null;
+            answer = signs * answer;
+            return assumed == Entity.Boolean.True ? answer : answer.Provided(assumed);
+        }
+
         /// <summary>
         /// A fractional power of a sum whose every term has x to a power in it,
         /// <c>(a x^j + b x^n)^p</c> with <c>0 &lt; j &lt; n</c>, as <c>K x^(j p) (a + b x^(n - j))^p</c>:

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

@@ -702,6 +702,8 @@ private static Entity Normalized(Entity expr, Entity.Variable x) =>
             if ((answer = IndefiniteIntegralSolver.SolveAReciprocalOfAnInverseTrigonometricFunction(expr, x, integrateByParts)) is { }) return answer;
             // A fractional power of a perfect square is the power of the modulus, sgn(P) P^(2r).
             if ((answer = IndefiniteIntegralSolver.SolveByTakingARootOfAPerfectSquare(expr, x, integrateByParts)) is { }) return answer;
+            // And a square in one trigonometric function, at the top: the modulus of the linear in it.
+            if ((answer = IndefiniteIntegralSolver.SolveByTakingARootOfAPerfectSquareInATrigonometricFunction(expr, x, integrateByParts)) is { }) return answer;
             // And any power of a square in any power of x, as the power of its root times a factor
             // constant where the root is not zero: `x^2 (a^2 + 2 a b x^3 + b^2 x^6)^p`.
             if ((answer = IndefiniteIntegralSolver.SolveByWritingAPowerOfASquareAsAPowerOfItsRoot(expr, x, integrateByParts)) is { }) return answer;

Sources/Tests/UnitTests/Calculus/RootOfASquareInATrigonometricFunctionIntegralTest.cs

@@ -0,0 +1,49 @@
+//
+// 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 half-odd power of <c>b^2 + 2 a b f + a^2 f^2</c>, a perfect square in one trigonometric
+    /// function <c>f</c>, is that power of <c>sqrt(a^2) |f + b/a|</c>, the sign in front of the
+    /// integral. 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 = 1.3</c>, <c>b = 0.7</c>, on both sides of the
+    /// zeros of <c>f + b/a</c> and of zero.
+    /// </remarks>
+    [Trait("Area", "Calculus")]
+    public sealed class RootOfASquareInATrigonometricFunctionIntegralTest
+    {
+        [Theory]
+        [InlineData("(a + b*sin(x))*sqrt(b^2 + 2*a*b*sin(x) + a^2*sin(x)^2)")]
+        [InlineData("(a + b*tan(x))/sqrt(b^2 + 2*a*b*tan(x) + a^2*tan(x)^2)")]
+        [InlineData("(a + b*sec(x))*(b^2 + 2*a*b*sec(x) + a^2*sec(x)^2)^(3/2)")]
+        [InlineData("(a + b*sin(g + f*x))/(b^2 + 2*a*b*sin(g + f*x) + a^2*sin(g + f*x)^2)^(3/2)")]
+        public void IsAPowerOfTheModulusOfTheLinear(string integrand)
+        {
+            var integral = integrand.ToEntity().Integrate("x");
+            Assert.DoesNotContain("integral(", integral.Stringize());
+            Entity Pinned(Entity e) => e.Substitute("a", 1.3).Substitute("b", 0.7).Substitute("g", 0.4).Substitute("f", 1.1);
+            var derivative = Pinned(integral.Substitute("C", 0)).Differentiate("x");
+            var original = Pinned(integrand.ToEntity());
+            foreach (var at in new[] { -2.3, -1.1, -0.3, 0.4, 1.1, 2.3 })
+            {
+                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}");
+            }
+        }
+    }
+}