Two proportional linears are not divided by their zero determinant

BREAKING-CHANGES.md

@@ -1433,13 +1433,17 @@ polynomial, is integrated now by the recurrence
 `int L1^m L2^n = L1^(m + 1) L2^(n + 1)/((m + 1) D) - (m + n + 2) d/((m + 1) D) int L1^(m + 1) L2^n`,
 `D = b c - a d`, down to `int L2^n/L1`, the polynomial written in powers of `L1` first. The
 answers are the same functions as master's where master gave one, written as powers of the two
-linears ([#718](https://github.com/asc-community/AngouriMath/issues/718)).
+linears ([#718](https://github.com/asc-community/AngouriMath/issues/718)). Beside a multiple of the
+same linear, where `D` is zero only as a value -- `b a c - a b c` for `a + b x` beside
+`c a + c b x` -- the recurrence is not taken: it divided by `D`, and the answer had no value anywhere
+([#1776](https://github.com/asc-community/AngouriMath/issues/1776)).
 
 | Input | Was (2.5.0) | Now |
 |---|---|---|
 | `"1/((a + b*x)^4*sqrt(c + d*x))".ToEntity().Integrate("x")` | `integral(...)` | powers of `a/b + x` beside `sqrt(c + d x)`, and the arctangent or logarithm `1/((a + b x) sqrt(c + d x))` integrates to |
 | `"(p + q*x + r*x^2 + s*x^3)/((a + b*x)^5*sqrt(c + d*x))".ToEntity().Integrate("x")` | `integral(...)` | the same, term by term |
 | `"x/((a + b*x)^4*sqrt(c + d*x))".ToEntity().Integrate("x")` | `integral(...)` | the same |
+| `"(a + b*x)^(-3)/sqrt(c*a + c*b*x)".ToEntity().Integrate("x")` | `integral(...)` | `-2/(5 b) (a + b x)^(-2) (c a + c b x)^(-1/2)` |
 
 ### A polynomial over a power of a linear beside the root of a quadratic is integrated
 

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

@@ -10117,8 +10117,11 @@ void Gather(Entity node, int times)
                 return null;
             if (!TreeAnalyzer.TryGetPolyLinear(first, x, out var b, out var a) || !TreeAnalyzer.TryGetPolyLinear(second, x, out var d, out var c))
                 return null;
+            // Zero as a value, not only as written: for `a + b x` beside `c a + c b x` it is
+            // `b a c - a b c`, and the recurrence divided by it -- an answer with no value anywhere.
+            // https://github.com/asc-community/AngouriMath/issues/1776
             var determinant = (b * c - a * d).InnerSimplified;
-            if (determinant.Evaled is Number.Complex { IsZero: true })
+            if (Functions.PartialFractions.IsZeroAsAValue(determinant))
                 return null;
             // The polynomial in powers of the first linear: x = (t - a)/b.
             var t = Variable.CreateUnique(expr, "t_linear");

Sources/Tests/UnitTests/Calculus/LinearPowerRecurrenceIntegralTest.cs

@@ -27,6 +27,22 @@ public sealed class LinearPowerRecurrenceIntegralTest
         [InlineData("1/(x^3*(c + d*x)^(5/2))")]
         [InlineData("x/((a + b*x)^4*sqrt(c + d*x))")]
         public void ByTheRecurrence(string integrand)
+            => DifferentiatesBack(integrand, new[] { -2.4, -1.5, 0.3, 0.8, 1.5, 2.4 });
+
+        /// <summary>
+        /// Beside a root of a multiple of the same linear, where the determinant the recurrence
+        /// divides by is zero only as a value, <c>b a c - a b c</c>: it was divided by, and the
+        /// answer had no value anywhere.
+        /// <a href="https://github.com/asc-community/AngouriMath/issues/1776">#1776</a>
+        /// </summary>
+        [Theory]
+        [InlineData("(a + b*x)^(-3)/sqrt(c*a + c*b*x)")]
+        [InlineData("1/(c*(a + b*x)^3)^(3/2)")]
+        [InlineData("1/(c*(a + b*x)^3)^(5/2)")]
+        public void BesideAMultipleOfTheSameLinear(string integrand)
+            => DifferentiatesBack(integrand, new[] { -1.2, -0.6, 0.3, 0.8, 1.5, 2.4 });
+
+        private static void DifferentiatesBack(string integrand, double[] points)
         {
             var integral = integrand.ToEntity().Integrate("x");
             Assert.DoesNotContain("integral(", integral.Stringize());
@@ -35,7 +51,7 @@ Entity Pinned(Entity e) => e.Substitute("a", 1.3).Substitute("b", 0.9).Substitut
                 .Substitute("s", 0.6);
             var derivative = Pinned(integral.Substitute("C", 0)).Differentiate("x");
             var original = Pinned(integrand.ToEntity());
-            foreach (var at in new[] { -2.4, -1.5, 0.3, 0.8, 1.5, 2.4 })
+            foreach (var at in points)
             {
                 var want = original.Substitute("x", at).EvalNumerical();
                 var got = derivative.Substitute("x", at).EvalNumerical();