diff --git a/BREAKING-CHANGES.md b/BREAKING-CHANGES.md
index 3547f0dd3..62e2ea2ab 100644
--- a/BREAKING-CHANGES.md
+++ b/BREAKING-CHANGES.md
@@ -302,6 +302,24 @@ rules' already
 | `"asin(sqrt(1 + x) - sqrt(x))".ToEntity().Integrate("x")` | `integral(...)` | an antiderivative in `arcsin(sqrt(1 + x) - sqrt(x))` |
 | `"x^3*atan(-sqrt(x)+sqrt(1+x))".ToEntity().Integrate("x")` | `integral(...)` | an antiderivative in the arctangent |
 
+### A sum of two squares below the bar, beside a repeated factor, is split over its conjugates
+
+**Answers where there were none.** `(c + d tan(x))^(3/2)/(a + b tan(x))^2` was declined in 2.5.0 and ran
+past two minutes since. Under `u = tan(x)` and `t = sqrt(c + d u)` the `1 + u^2` the tangent leaves is
+`(t^2 - c)^2 + d^2`, a quartic beside the square of `a d + b (t^2 - c)`, and the Hermite reduction
+solved for the numerators over it with the symbols in every entry. A written sum of two squares
+`P^2 + Q^2`, the greater of their degrees two, beside a repeated factor, is `(P - i Q)(P + i Q)` below
+the bar now, and over those every factor is one the split by residues reads -- under a root of the
+quotient of two such linears the sum is `(b - d t^2)^2 + (c t^2 - a)^2`. Rubi's 4.3.2.1
+([#718](https://github.com/asc-community/AngouriMath/issues/718)).
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"(c + d*tan(x))^(3/2)/(a + b*tan(x))^2".ToEntity().Integrate("x")` | `integral(...)`; past two minutes on the unreleased master | arctangents and logarithms of `sqrt(c + d tan(x))`, in two seconds |
+| `"(c + d*tan(x))^(5/2)/(a + b*tan(x))^2".ToEntity().Integrate("x")` | `integral(...)` | the same |
+| `"(c + d*tan(x))^(3/2)/(a + b*tan(x))^3".ToEntity().Integrate("x")` | `integral(...)` | the same |
+| `"1/((a + b*tan(x))^(3/2)*(c + d*tan(x))^(3/2))".ToEntity().Integrate("x")` | `integral(...)`; past a minute on the unreleased master | in the root of the quotient of the two, in a second |
+
 ### A function of `x^n` for a symbolic `n` beside a power of `x` is integrated in a power of `x`
 
 **Answers where there were none.** `x^(-1 + 4n)/(a + b x^n + c x^(2n))` and the rest of Rubi's
diff --git a/Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs b/Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs
index 31ec42351..a5964c8bd 100644
--- a/Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs
+++ b/Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs
@@ -847,6 +847,19 @@ is var (multiple, leftover)
                     ?? Integration.ComputeIndefiniteIntegral(inY, y, integrateByParts)) is { } inTheShiftedVariable)
                 return inTheShiftedVariable.Substitute(y, (x + shift).InnerSimplified);
 
+            // A written factor that is a sum of two squares, `P^2 + Q^2` with the greater of their
+            // degrees two, beside a repeated factor: it is `(P - i Q)(P + i Q)`, and over those two
+            // the split by residues above reads every factor. `1 + u^2` beside a root of a
+            // linear is `(t^2 - c)^2 + d^2` under `t = sqrt(c + d u)`, a quartic nothing splits, and
+            // the Hermite reduction below solved for its numerators with the symbols in every entry:
+            // `(c + d tan(x))^(3/2)/(a + b tan(x))^2` ran past two minutes. Once: what this writes
+            // has no such factor left.
+            if (Mulf.LinearChildren(denominator).Any(f => f.ContainsNode(x) && f is Powf(_, Number.Integer { EInteger.Sign: > 0 } e) && e != Number.Integer.One)
+                && OverTheConjugatesOfASumOfTwoSquares(denominator, x) is { } overTheConjugates
+                && (SolveByPartialFractions(numerator / overTheConjugates, x, integrateByParts)
+                    ?? Integration.ComputeIndefiniteIntegral(numerator / overTheConjugates, x, integrateByParts)) is { } overConjugates)
+                return overConjugates;
+
             // A denominator with a written repeated factor takes the Hermite reduction first:
             // the rational part of the answer in one linear solve, and what is left is a proper
             // fraction over a squarefree denominator for the splits below. `(1 + x^2)/(x (1 + x^3)^2)`
@@ -14363,6 +14376,43 @@ private static bool IsASumOfMonomials(Entity expr, Entity.Variable x)
             return changed ? product : null;
         }
 
+        /// <summary>
+        /// <paramref name="denominator"/> with every written factor <c>P^2 + Q^2</c>, <c>P</c> and
+        /// <c>Q</c> polynomials in <paramref name="x"/> the greater of whose degrees is two, written
+        /// as <c>(P - i Q)(P + i Q)</c>, with its power; <see langword="null"/> where there is none,
+        /// or where nothing in the quotient is a symbol: with numbers alone the split over the reals
+        /// reads a biquadratic as it is. Not of the first degree: <c>(a + b x)^2 + k^2</c> is a
+        /// quadratic every rule reads, by an arctangent.
+        /// </summary>
+        private static Entity? OverTheConjugatesOfASumOfTwoSquares(Entity denominator, Entity.Variable x)
+        {
+            static int? DegreeOfAPolynomial(Entity expr, Entity.Variable x)
+                => !expr.ContainsNode(x) ? 0
+                : TreeAnalyzer.TryGetPolynomial(expr, x, out var read) && read.Count > 0 && read.Values.All(coefficient => !coefficient.ContainsNode(x))
+                    && read.Keys.All(power => power.Sign >= 0 && power.CanFitInInt32())
+                ? read.Keys.Max()!.ToInt32Unchecked() : null;
+            if (!denominator.Vars.Any(symbol => symbol != x))
+                return null;
+            var changed = false;
+            Entity product = Number.Integer.One;
+            foreach (var factor in Mulf.LinearChildren(denominator))
+            {
+                var (@base, power) = factor is Powf(var b, Number.Integer { EInteger.Sign: > 0 } p) ? (b, p) : (factor, Number.Integer.One);
+                if (@base is Sumf(Powf(var first, Number.Integer(2)), Powf(var second, Number.Integer(2)))
+                    && (first.ContainsNode(x) || second.ContainsNode(x))
+                    && DegreeOfAPolynomial(first, x) is { } firstDegree && DegreeOfAPolynomial(second, x) is { } secondDegree
+                    && System.Math.Max(firstDegree, secondDegree) == 2)
+                {
+                    var (minus, plus) = (first - MathS.i * second, first + MathS.i * second);
+                    product *= power == Number.Integer.One ? minus * plus : MathS.Pow(minus, power) * MathS.Pow(plus, power);
+                    changed = true;
+                    continue;
+                }
+                product *= factor;
+            }
+            return changed ? product : null;
+        }
+
         /// <summary>
         /// <paramref name="denominator"/> with every quadratic factor that has a root off the real line
         /// in common with a linear factor beside it written as its leading coefficient times the
diff --git a/Sources/Tests/UnitTests/Calculus/SumOfTwoSquaresBelowTheBarIntegralTest.cs b/Sources/Tests/UnitTests/Calculus/SumOfTwoSquaresBelowTheBarIntegralTest.cs
new file mode 100644
index 000000000..ef3a1a731
--- /dev/null
+++ b/Sources/Tests/UnitTests/Calculus/SumOfTwoSquaresBelowTheBarIntegralTest.cs
@@ -0,0 +1,55 @@
+//
+// 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 sum of two squares below the bar beside a repeated factor, written over its two
+    /// conjugates. Under <c>u = tan(x)</c> and <c>t = sqrt(c + d u)</c>, the <c>1 + u^2</c> of
+    /// Rubi's 4.3.2.1 <c>(c + d tan(x))^(3/2)/(a + b tan(x))^2</c> is <c>(t^2 - c)^2 + d^2</c>,
+    /// a quartic beside the square of <c>a d + b (t^2 - c)</c>; the Hermite reduction solved for
+    /// its numerators with the symbols in every entry and ran past two minutes. Written as
+    /// <c>(t^2 - c - i d)(t^2 - c + i d)</c>, every factor is one the split by residues reads.
+    /// <a href="https://github.com/asc-community/AngouriMath/issues/718">#718</a>
+    /// </summary>
+    /// <remarks>
+    /// Under a root of the quotient of two such linears the sum is <c>(b - d t^2)^2 + (c t^2 - a)^2</c>,
+    /// both parts in <c>t</c>, and its conjugates are quadratics as well. The integrands are real
+    /// where <c>a + b tan(x)</c> and <c>c + d tan(x)</c> are positive, which the points are.
+    /// </remarks>
+    [Trait("Area", "Calculus")]
+    public sealed class SumOfTwoSquaresBelowTheBarIntegralTest
+    {
+        [Theory]
+        [InlineData("(c + d*tan(x))^(3/2)/(a + b*tan(x))^2")]
+        [InlineData("(c + d*tan(x))^(5/2)/(a + b*tan(x))^2")]
+        [InlineData("(c + d*tan(x))^(3/2)/(a + b*tan(x))^3")]
+        [InlineData("x^4/((a*d + (x^2 - c)*b)^2*((x^2 - c)^2 + d^2))")]
+        [InlineData("1/((a + b*tan(x))^(3/2)*(c + d*tan(x))^(3/2))")]
+        [InlineData("1/((a + b*tan(x))^(5/2)*(c + d*tan(x))^(3/2))")]
+        public void OverItsConjugates(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("c", 1.1).Substitute("d", 0.6);
+            var derivative = Pinned(integral.Substitute("C", 0)).Differentiate("x");
+            var original = Pinned(integrand.ToEntity());
+            foreach (var at in new[] { 0.2, 0.5, 0.9, 1.2 })
+            {
+                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)),
+                    $"d/dx of the antiderivative of {integrand} is {got} at x = {at}, where the integrand is {want}");
+            }
+        }
+    }
+}