e^(n i arctan(a x)) to a power that is not whole is one power of a quotient of linears

BREAKING-CHANGES.md

@@ -521,6 +521,21 @@ read them as nonzero. They are now decided over one bar, expanded, and at pinned
 | `"e^(3*acoth(a*x))/(c-a*c*x)^3".Integrate("x")`, and over `(c - a c x)^4` | left unevaluated | an antiderivative in `sqrt((a x + 1)/(a x - 1))` |
 | `"e^(2*acoth(a*x))*sqrt(c-a*c*x)/x".Integrate("x")`, and over `x^2` | left unevaluated | an antiderivative in `sqrt(c - a c x)`, by cases on the sign of `c` |
 
+### `e^(n i arctan(a x))` to a power that is not whole is integrated
+
+**Answers where there were none.** `e^(n i arctan(L))` is written algebraically, as
+`(1 + i L)^n (1 + L^2)^(-n/2)`, and for an `n` that is not whole those are two radicals of different
+orders that nothing reads: `x^2 e^(3/2 i arctan(a x))` was left unevaluated. For such an `n` it is
+written `((1 + i L)/(1 - i L))^(n/2)`, the same on the principal branch for a real `L`, one power of a
+quotient of linears, which the substitution for that reads. Rubi's 5.3.6
+([#718](https://github.com/asc-community/AngouriMath/issues/718)).
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"e^(3/2*i*arctan(a*x))*x^2".ToEntity().Integrate("x")` | `integral(...)` | an antiderivative in `((1 + i a x)/(1 - i a x))^(1/4)` |
+| `"x^3/e^(3/2*i*arctan(a*x))".ToEntity().Integrate("x")` | `integral(...)` | the same |
+| `"e^(1/3*i*arctan(x))*x".ToEntity().Integrate("x")` | `integral(...)` | an antiderivative in `((1 + i x)/(1 - i x))^(1/6)` |
+
 ### `NaN` again, from an exponent that was read as written rather than as a number
 
 **A wrong answer, and a second one of the same kind.** `(a^2 + 2abx^2 + b^2x^4)^3/x^7` came back as

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

@@ -7356,10 +7356,18 @@ node is Powf(var @base, Number.Rational power) && power is not Number.Integer &&
                     return node;
                 rewrote = true;
                 // The tangent's as two powers, `(1 + i L)^n (1 + L^2)^(-n/2)`, which the
-                // radical rules read where a power of the quotient is one node to them.
+                // radical rules read where a power of the quotient is one node to them; for an
+                // n that is not whole, as the power of the quotient,
+                // `((1 + i L)/(1 - i L))^(n/2)`, the same on the principal branch for a real L --
+                // the quotient is `e^(2 i arctan(L))` and `2 arctan(L)` is its principal
+                // argument -- and one power of a quotient of linears, which the substitution
+                // for it reads, where `(1 + i a x)^(3/2) (1 + a^2 x^2)^(-3/4)` is two radicals
+                // of different orders that nothing reads.
                 if (inverse is Arctanf)
                 {
                     var halfPower = Number.Rational.Create(n.ERational.Negate().Divide(2));
+                    if (n is not Number.Integer)
+                        return MathS.Pow((1 + MathS.i * argument) / (1 - MathS.i * argument), Number.Rational.Create(n.ERational.Divide(2)));
                     return (n == Number.Integer.One ? 1 + MathS.i * argument : MathS.Pow(1 + MathS.i * argument, n)) * MathS.Pow(1 + MathS.Sqr(argument), halfPower);
                 }
                 Entity unit = inverse is Arcsinf

Sources/Tests/UnitTests/Calculus/InverseTrigonometricSubstitutionTest.cs

@@ -216,5 +216,20 @@ public void ALinearArgumentAndAMultipleOfTheQuadratic(string integrand, double[]
         [InlineData("x/e^(2*i*arctan(1 + 2*x))", new[] { -1.6, -0.4, 0.3, 1.1, 2.4 })]
         public void AnExponentialOfTheInverseIsAlgebraic(string integrand, double[] points)
             => DifferentiatesBack(integrand, points);
+
+        /// <summary>
+        /// To a power that is not whole, <c>e^(n i arctan(L))</c> is one power of a quotient of
+        /// linears, <c>((1 + i L)/(1 - i L))^(n/2)</c> for a real <c>L</c>, which the substitution
+        /// for such a power reads, where <c>(1 + i L)^n (1 + L^2)^(-n/2)</c> is two radicals of
+        /// different orders: Rubi's 5.3.6.
+        /// </summary>
+        [Theory]
+        [InlineData("e^(3/2*i*arctan(2*x))*x^2", new[] { -1.6, -0.4, 0.3, 1.1, 2.4 })]
+        [InlineData("e^(3/2*i*arctan(2*x))/x^4", new[] { -1.6, -0.4, 0.3, 1.1, 2.4 })]
+        [InlineData("x^3/e^(3/2*i*arctan(2*x))", new[] { -1.6, -0.4, 0.3, 1.1, 2.4 })]
+        [InlineData("e^(1/3*i*arctan(x))*x", new[] { -1.6, -0.4, 0.3, 1.1, 2.4 })]
+        [InlineData("e^(3/2*i*arctan(1 + 2*x))*x^2", new[] { -1.6, -0.4, 0.3, 1.1, 2.4 })]
+        public void ToAPowerThatIsNotWhole(string integrand, double[] points)
+            => DifferentiatesBack(integrand, points);
     }
 }