A constant of integration is matched to a linear divisor with a symbol in it

BREAKING-CHANGES.md

@@ -2315,6 +2315,26 @@ no integrand without one changes: Rubi's independent suites and a sample of its
 | `"Ei(b*x)^2".Integrate("x")` | `Ei * (b * x) ^ 3 / 3 / b + C`, with `Ei` a variable | an antiderivative in `Ei(b x)` and `Ei(2 b x)` |
 | `"x*erf(b*x)^2".Integrate("x")` | `UnrecognizedFunctionParseException`: there is no function `erf` | an antiderivative in `erf(b x)` |
 
+### A constant of integration is matched to a linear divisor with a symbol in it
+
+By parts against a polynomial chooses the polynomial's antiderivative so that what the other
+factor's derivative divides by divides it too, and that was done only where the division left a
+number over. `a + b x` leaves a symbol: `x^2/2` over `a + b x` leaves `a^2/(2 b^2)`, so the
+remainder of `x Shi(a + b x)^2` kept `x^2/(a + b x)` and nothing read it. The antiderivative less
+its value at `-a/b` is taken now, written as `a + b x` times the quotient, so that the linear
+cancels in the remainder ([#1501](https://github.com/asc-community/AngouriMath/issues/1501)).
+`x Shi(a + b x)^2`, `x Chi(a + b x)^2` and `x Ei(a + b x)^2` are answered. An integrand holding one of
+these functions had no reading in 2.5.0, which the entries for the functions themselves record, and
+no integrand without one changes: Rubi's independent suites and a sample of its families 1 to 7
+are answered alone as they were. Where by parts already answered a special function of `a + b x`
+beside a power of `x`, the answer is written with the same constant: `x Ei(a + b x)` now reads
+`Ei(a + b x) (a + b x)(x/(2b) - a/(2b^2)) - ...`, the same function as before.
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"x*Shi(a+b*x)^2".Integrate("x")` | `a ^ 2 * Shi * x ^ 2 / 2 + a * b * 2 * Shi * x ^ 3 / 3 + b ^ 2 * Shi * x ^ 4 / 4 + C`, with `Shi` a variable | an antiderivative in `Shi(a + b x)` and `Ei(2 (a + b x))`, `Ei(-2 (a + b x))` |
+| `"x*Ei(a+b*x)^2".Integrate("x")` | `a ^ 2 * Ei * x ^ 2 / 2 + a * b * 2 * Ei * x ^ 3 / 3 + b ^ 2 * Ei * x ^ 4 / 4 + C`, with `Ei` a variable | an antiderivative in `Ei(a + b x)` and `Ei(2 (a + b x))` |
+
 ### An exponential or a hyperbolic function over several linears is split into partial fractions over them
 
 `e^x/(x (x + 1))` was left unevaluated, where `e^x/x` and `e^x/(x + 1)` were each answered with

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

@@ -1691,6 +1691,19 @@ static Entity WithTheConstantMatchedTo(Entity antiderivative, Entity derivativeO
                     && Functions.SingleQuotient.Of(leftOver) is var (leftOverTop, _) && !leftOverTop.ContainsNode(x)
                     && leftOverTop.Evaled is Number.Complex { IsZero: false } constant)
                     return (antiderivative + (-constant).Evaled).InnerSimplified;
+                // A linear divisor with a symbol in it, `a + b x`, leaves a symbol over, which the
+                // case above does not read: the antiderivative less its value at the root, -a/b, is
+                // divisible by the linear, and it is written as the linear times the quotient, so
+                // that the linear the derivative divides by cancels as written. `x Shi(a + b x)^2`
+                // is two rounds of parts that way, where with `x^2/2` the remainder kept
+                // `x^2/(a + b x)` and nothing read it.
+                // https://github.com/asc-community/AngouriMath/issues/1501
+                if (TreeAnalyzer.TryGetPolyLinear(divisors[0], x, out var slope, out var offset) && !TreeAnalyzer.IsZero(slope)
+                    && antiderivative.Substitute(x, (-offset / slope).InnerSimplified).InnerSimplified is var atTheRoot
+                    && !atTheRoot.ContainsNode(x) && !TreeAnalyzer.IsZero(atTheRoot)
+                    && TreeAnalyzer.PolynomialLongDivision((antiderivative - atTheRoot).Expand().InnerSimplified, divisors[0], genericCase: true, inTermsOf: x) is var (quotient, _)
+                    && !quotient.ContainsNode(MathS.NaN))
+                    return divisors[0] * Functions.PartialFractions.Bare(quotient);
                 return antiderivative;
             }
 

Sources/Tests/UnitTests/Calculus/SpecialFunctionsByPartsTest.cs

@@ -133,5 +133,16 @@ public void BesideAPowerAndTheElementaryFactorOfItsDerivative(string integrand)
         [InlineData("Chi(b*x)^2")]
         public void ASquareIsTwoRoundsOfParts(string integrand)
             => DifferentiatesBack(integrand, Parameters);
+
+        /// <summary>
+        /// Of <c>a + b x</c>, the first round against <c>x</c> takes <c>x^2/2</c> less its value at
+        /// <c>-a/b</c>, written as <c>a + b x</c> times the quotient, so the linear the derivative
+        /// divides by cancels and what is left is the case above.
+        /// </summary>
+        [Theory]
+        [InlineData("x*Shi(a + b*x)^2")]
+        [InlineData("x*Ei(a + b*x)^2")]
+        public void ASquareOfAShiftedArgumentIsTwoRoundsOfParts(string integrand)
+            => DifferentiatesBack(integrand, Parameters);
     }
 }