An exponential of a polynomial beside the polynomial's derivative is integrated under u = P

BREAKING-CHANGES.md

@@ -1710,6 +1710,18 @@ since they are these exponentials
 | `"(p + q*x)^2*f^(a + b*x + c*x^2)".Integrate("x")` | `integral((p + q * x) ^ 2 * f ^ (a + b * x + c * x ^ 2), x)` | the antiderivative |
 | `"x^2*sinh(a + b*x + c*x^2)".Integrate("x")` | `integral(x ^ 2 * (e ^ (a + b * x + c * x ^ 2) - e ^ (-(a + b * x + c * x ^ 2))) / 2, x)` | the antiderivative |
 
+### An exponential of a polynomial beside the polynomial's derivative is integrated
+
+`e^(a + b x + c x^2) (b + 2 c x) sqrt(a + b x + c x^2)` was left unintegrated. `G^P k P' f(P)`, with
+`P` a polynomial of degree two or more and `P'` a factor of its own up to a constant, is now
+integrated under `u = P`, as `k G^u f(u)`
+([#1501](https://github.com/asc-community/AngouriMath/issues/1501)).
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"e^(a + b*x + c*x^2)*(b + 2*c*x)*(a + b*x + c*x^2)^(1/2)".Integrate("x")` | `integral(e ^ (a + b * x + c * x ^ 2) * (b + 2 * c * x) * sqrt(a + b * x + c * x ^ 2), x)` | the antiderivative, with `erfi` |
+| `"e^(a + b*x + c*x^2)*(b + 2*c*x)/(a + b*x + c*x^2)^(3/2)".Integrate("x")` | `integral(e ^ (a + b * x + c * x ^ 2) * (b + 2 * c * x) / (a + b * x + c * x ^ 2) ^ (3/2), x)` | the antiderivative, with `erfi` |
+
 ### An exponential of a quadratic in `1/x` or `1/(c + d x)` is integrated
 
 `e^(-1/x^2)` was left unintegrated. `L^m F^(A/L^2 + B/L + C)`, with `L = c + d x` and a whole `m`,

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

@@ -17193,6 +17193,63 @@ private static bool IsARationalFunctionOfExponentials(Entity expr, Entity.Variab
             }
         }
 
+        /// <summary>
+        /// An exponential of a polynomial beside the polynomial's derivative, with the rest a
+        /// function of the polynomial: <c>G^P k P' f(P)</c> is <c>k G^u f(u)</c> under <c>u = P</c>.
+        /// Rubi's 2.3, <c>e^(a + b x + c x^2) (b + 2 c x) (a + b x + c x^2)^(n/2)</c>.
+        /// https://github.com/asc-community/AngouriMath/issues/1501
+        /// </summary>
+        /// <remarks>
+        /// <see cref="SolveBySubstitution"/> reads this <c>u</c> and does not reach it: for its
+        /// own search it writes an exponential of a sum as a product of exponentials, so that
+        /// <c>e^(e^x) e^x</c> is <c>e^u du</c>, and <c>e^(a + b x + c x^2)</c> is then
+        /// <c>e^a e^(b x) e^(c x^2)</c>, with no <c>P</c> left in it to replace. Declining took
+        /// it 24 s. Only a polynomial of degree two or more, whose derivative is a factor of its
+        /// own: an exponential of a linear is the table's.
+        /// </remarks>
+        internal static Entity? SolveByTheExponentAsTheVariable(Entity expr, Entity.Variable x, bool integrateByParts)
+        {
+            if (expr is not (Mulf or Divf) || AnExponentialOfAPolynomial(expr, x) is not { } exponential)
+                return null;
+            var polynomial = exponential.Exponent;
+            var derivative = polynomial.Differentiate(x);
+            Entity? scale = null;
+            Entity rest = Number.Integer.One;
+            foreach (var (factor, underneath) in FactorsOfTheIntegrand(expr))
+            {
+                // The derivative, up to a constant, above the bar: a sum, as it is written.
+                if (scale is null && !underneath && factor is Sumf or Minusf && factor.ContainsNode(x)
+                    && Functions.PartialFractions.Bare((factor / derivative).Simplify()) is var ratio && !ratio.ContainsNode(x)
+                    && ratio.Evaled is not Number.Complex { IsZero: true })
+                {
+                    scale = ratio;
+                    continue;
+                }
+                rest = underneath ? rest / factor : rest * factor;
+            }
+            if (scale is null)
+                return null;
+            var u = Variable.CreateUnique(expr, "u");
+            var inU = (scale * rest.Replace(node => node == polynomial ? u : node)).InnerSimplified;
+            if (inU.ContainsNode(x))
+                return null;
+            return Integration.ComputeAsAQuestionOfItsOwn(inU, u, integrateByParts)?.Substitute(u, polynomial);
+        }
+
+        /// <summary>
+        /// A factor of <paramref name="expr"/>, read through products and quotients, that is a
+        /// constant to a polynomial in <paramref name="x"/> of degree two or more, or <see langword="null"/>.
+        /// </summary>
+        private static Powf? AnExponentialOfAPolynomial(Entity expr, Entity.Variable x) => expr switch
+        {
+            Mulf(var left, var right) => AnExponentialOfAPolynomial(left, x) ?? AnExponentialOfAPolynomial(right, x),
+            Divf(var numerator, var denominator) => AnExponentialOfAPolynomial(numerator, x) ?? AnExponentialOfAPolynomial(denominator, x),
+            Powf(var @base, Sumf or Minusf) power when !@base.ContainsNode(x) && power.Exponent.ContainsNode(x)
+                && TreeAnalyzer.TryGetPolynomial(power.Exponent, x, out var monomials)
+                && monomials.Keys.All(degree => degree.Sign >= 0) && monomials.Keys.Any(degree => degree.CompareTo(EInteger.One) > 0) => power,
+            _ => null
+        };
+
         /// <summary>
         /// Attempts to solve an integral using u-substitution.
         /// Looks for patterns where f(g(x)) * g'(x) can be integrated as F(g(x)).

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

@@ -579,6 +579,9 @@ private static Entity Normalized(Entity expr, Entity.Variable x) =>
             // An exponential of a multiple of a logarithm is a power of the argument, which is
             // how every inverse hyperbolic function under an exponential arrives.
             if ((answer = IndefiniteIntegralSolver.SolveByFoldingAnExponentialOfALogarithm(expr, x, integrateByParts)) is { }) return answer;
+            // An exponential of a polynomial beside the polynomial's derivative, under u = P,
+            // which the substitution search does not reach: it writes the exponential apart.
+            if ((answer = IndefiniteIntegralSolver.SolveByTheExponentAsTheVariable(expr, x, integrateByParts)) is { }) return answer;
             // `A + i A tan(z)` is `A e^(i z)/cos(z)`, which beside a polynomial is a shape the
             // closed rules answer, where the imaginary unit in the coefficient is read by none.
             if ((answer = IndefiniteIntegralSolver.SolveByWritingAnImaginaryTangentAsAnExponential(expr, x, integrateByParts)) is { }) return answer;

Sources/Tests/UnitTests/Calculus/GaussianIntegralTest.cs

@@ -146,6 +146,19 @@ public void ASymbolicGaussianWithALinearTermBesideAPolynomial(string integrand)
         public void ASymbolicGaussianInAReciprocal(string integrand)
             => DifferentiatesBack(integrand, ("f", "2"), ("a", "1/3"), ("b", "-2/3"), ("c", "5"), ("d", "3/2"));
 
+        /// <summary>
+        /// An exponential of a quadratic beside the quadratic's derivative and a half-odd power of
+        /// the quadratic, under <c>u = a + b x + c x^2</c>: <c>e^u u^(n/2)</c>. Rubi's 2.3. The
+        /// quadratic is positive at every point for these pins.
+        /// </summary>
+        [Theory]
+        [InlineData("e^(a + b*x + c*x^2)*(b + 2*c*x)*(a + b*x + c*x^2)^(1/2)")]
+        [InlineData("e^(a + b*x + c*x^2)*(b + 2*c*x)*(a + b*x + c*x^2)^(5/2)")]
+        [InlineData("e^(a + b*x + c*x^2)*(b + 2*c*x)/(a + b*x + c*x^2)^(3/2)")]
+        [InlineData("3*e^(a + b*x + c*x^2)*(2*b + 4*c*x)/sqrt(a + b*x + c*x^2)")]
+        public void TheExponentAsTheVariable(string integrand)
+            => DifferentiatesBack(integrand, ("a", "1/3"), ("b", "-2/3"), ("c", "5/4"));
+
         /// <summary>
         /// An odd negative power ends at <c>int e^(A x^2)/x</c>, which is the exponential
         /// integral, so it is not taken.