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Write a trinomial with conjugate roots over its roots in the square #1948

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80 changes: 29 additions & 51 deletions Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs
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Original file line number Diff line number Diff line change
Expand Up @@ -19153,27 +19153,28 @@ static Entity WithoutConditions(Entity coefficient)

/// <summary>
/// A rational function with a trinomial in the square below the bar, beside a root of a
/// quadratic, taken apart over the trinomial's factors in the square: where its roots in
/// x^2 are real, <c>a + b x^2 + c x^4</c> is <c>c (x^2 - r) (x^2 - s)</c> with
/// <c>r, s = (-b -+ sqrt(b^2 - 4 a c))/(2c)</c>, and where they are a conjugate pair it is
/// <c>c (x^2 + p x + m) (x^2 - p x + m)</c> with <c>m = sqrt(a/c)</c> and
/// <c>p = sqrt(2 m - b/c)</c>. Each factor is a quadratic the rule before this one splits
/// the integrand over and closes beside the root, and the answer is the piecewise of the
/// two on the sign of the discriminant. Rubi's 1.2.2.4
/// <c>sqrt(d + e x^2)/(a + b x^2 + c x^4)</c> was declined.
/// quadratic, taken apart over the trinomial's roots in the square: <c>a + b x^2 + c x^4</c>
/// is <c>c (x^2 - r) (x^2 - s)</c> with <c>r, s = (-b -+ sqrt(b^2 - 4 a c))/(2c)</c>, and each
/// factor is a quadratic the rule before this one splits the integrand over and closes
/// beside the root. Rubi's 1.2.2.4 <c>sqrt(d + e x^2)/(a + b x^2 + c x^4)</c> was declined.
/// </summary>
/// <remarks>
/// Asked with <c>r</c>, <c>s</c>, <c>m</c> and <c>p</c> symbols of their own and written
/// back after: the integrand so written is an integrand for every value of them, and the
/// one asked for at theirs. Named, each stays one atom through the split: with
/// Asked with <c>r</c> and <c>s</c> symbols of their own and written back after: the
/// integrand so written is an integrand for every value of them, and the one asked for at
/// theirs. Named, each stays one atom through the split: with
/// <c>(sqrt(b^2 - 4 a c) - b)/(2c)</c> in place of <c>r</c> the split declined
/// <c>x^2 sqrt(d + e x^2)/(a + b x^2 + c x^4)</c>, and over <c>x^2 - r</c> it is 1,458
/// characters (<a href="https://github.com/asc-community/AngouriMath/issues/1877">#1877</a>
/// was the same with a quadratic). The real roots' writing alone is not enough: with
/// <c>r</c> complex the pieces' cases read the sign of a number off the real line, and the
/// answer was NaN wherever <c>b^2 &lt; 4 a c</c>. Not where the discriminant vanishes,
/// the generic case's. With a symbol in the trinomial only, and in a function of x^2 only.
/// https://github.com/asc-community/AngouriMath/issues/718
/// <c>x^2 sqrt(d + e x^2)/(a + b x^2 + c x^4)</c>
/// (<a href="https://github.com/asc-community/AngouriMath/issues/1877">#1877</a> was the
/// same with a quadratic). The same writing where the roots are a conjugate pair: each
/// piece over <c>x^2 - r</c> is an arctangent of <c>w x/sqrt(d + e x^2)</c> with
/// <c>w^2 = (d + e r)/(-r)</c>, which holds on either branch of <c>w</c>, and for <c>r</c>
/// off the real line <c>w x/sqrt(d + e x^2)</c> stays off the arctangent's cuts, so the
/// two conjugate pieces add to a real, continuous antiderivative. Written instead over
/// the two real quadratics <c>x^2 -+ p x + m</c>, the answer to
/// <c>x^2 sqrt(d + e x^2)/(a + b x^2 + c x^4)</c> was 29,139 characters where this one is
/// under 2,000. Not where the discriminant vanishes, the generic case's. With a symbol in
/// the trinomial only, and in a function of x^2 only.
/// https://github.com/asc-community/AngouriMath/issues/1885
/// </remarks>
internal static Entity? SolveATrinomialInTheSquareBesideTheRootOfAQuadratic(Entity expr, Entity.Variable x)
{
Expand Down Expand Up @@ -19209,43 +19210,20 @@ static Entity WithoutConditions(Entity coefficient)
var discriminant = (middle * middle - 4 * constant * leading).InnerSimplified;
if (VanishesIdentically(discriminant))
return null;
// Two writings of the trinomial, each real where it is taken, its roots and
// coefficients named so that each stays one atom through the split. Two real roots
// in x^2: c (x^2 - r) (x^2 - s). None: the roots are a conjugate pair and the pieces' cases
// read the sign of an imaginary q, so it is the product of two real quadratics
// c (x^2 + p x + m) (x^2 - p x + m) with m = sqrt(a/c) and p = sqrt(2 m - b/c), both
// real there since b^2 < 4 a c makes a/c positive and |b/c| < 2 m. Each is an
// identity in its symbols, so each answer is an antiderivative wherever its symbols
// are what they stand for. A root twice, where the discriminant vanishes, is the
// case the generic answer leaves, as `ln(a x + b)/a` leaves a = 0.
// c (x^2 - r) (x^2 - s), its roots named so that each stays one atom through the split.
// An identity in r and s, so the answer is an antiderivative wherever they are the
// roots, a conjugate pair included. A root twice, where the discriminant vanishes, is
// the case the generic answer leaves, as `ln(a x + b)/a` leaves a = 0.
var square = MathS.Pow(x, Number.Integer.Create(2));
var r = Variable.CreateUnique(expr, "r");
var s = Variable.CreateUnique(expr + r, "s");
var m = Variable.CreateUnique(expr + r + s, "m");
var p = Variable.CreateUnique(expr + r + s + m, "p");
var q = MathS.Sqrt(discriminant);
var writings = new (Entity Written, Entity[] Named, Entity[] Values, Entity Condition)[]
{
(leading * (square - r) * (square - s),
new Entity[] { r, s }, new[] { (q - middle) / (2 * leading), (-q - middle) / (2 * leading) }, discriminant > 0),
(leading * (square + p * x + m) * (square - p * x + m),
new Entity[] { m, p }, new[] { MathS.Sqrt(constant / leading), MathS.Sqrt(2 * MathS.Sqrt(constant / leading) - middle / leading) }, discriminant < 0),
};
var cases = new List<Providedf>();
foreach (var (overTheRoots, named, values, condition) in writings)
{
var written = expr.Replace(node => node == trinomial ? overTheRoots : node);
if (SolveARationalFunctionBesideTheRootOfAQuadratic(written, x) is not { } inNamed)
return null;
var answered = inNamed;
for (var i = named.Length - 1; i >= 0; i--)
answered = answered.Substitute(named[i], values[i]);
answered = answered.InnerSimplified;
if (answered.Nodes.Any(node => node is Number.Complex { IsNaN: true }))
return null;
cases.Add(new Providedf(answered, condition));
}
var answer = MathS.Piecewise(cases);
var written = expr.Replace(node => node == trinomial ? leading * (square - r) * (square - s) : node);
if (SolveARationalFunctionBesideTheRootOfAQuadratic(written, x) is not { } inNamed)
return null;
var answer = inNamed.Substitute(s, (-q - middle) / (2 * leading)).Substitute(r, (q - middle) / (2 * leading)).InnerSimplified;
if (answer.Nodes.Any(node => node is Number.Complex { IsNaN: true }))
return null;
return Functions.PartialFractions.DerivativeHoldsAtSampledPoints(answer, expr, x) ? answer : null;
}

Expand Down
56 changes: 56 additions & 0 deletions Sources/Tests/UnitTests/Calculus/ATrinomialWithConjugateRootsInTheSquareIntegralTest.cs
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@@ -0,0 +1,56 @@
//
// 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 rational function over <c>a + b x^2 + c x^4</c> beside a root of <c>d + k x^2</c>, taken
/// apart over the trinomial's roots in x^2 whether they are real or a conjugate pair. Over the
/// two real quadratics a conjugate pair also factors into, the answer to
/// <c>x^2 sqrt(d + k x^2)/(a + b x^2 + c x^4)</c> was 29,139 characters. Rubi's 1.2.2.4.
/// <a href="https://github.com/asc-community/AngouriMath/issues/1885">#1885</a>
/// </summary>
[Trait("Area", "Calculus")]
public sealed class ATrinomialWithConjugateRootsInTheSquareIntegralTest
{
[Theory]
// b^2 < 4 a c: the roots in x^2 are a conjugate pair.
[InlineData("x^2*sqrt(d + k*x^2)/(a + b*x^2 + c*x^4)", 2, 1, 3)]
[InlineData("sqrt(d + k*x^2)/(a + b*x^2 + c*x^4)", 2, 1, 3)]
[InlineData("x^4/((a + b*x^2 + c*x^4)*sqrt(d + k*x^2))", 2, -1, 3)]
[InlineData("(d + k*x^2)^(3/2)/(a + b*x^2 + c*x^4)", 2, 1, 3)]
// b^2 > 4 a c: two real roots, the same writing.
[InlineData("x^2*sqrt(d + k*x^2)/(a + b*x^2 + c*x^4)", 1, 5, 2)]
public void OverItsRootsInTheSquare(string integrand, double a, double b, double c)
{
var integral = integrand.ToEntity().Integrate("x");
var text = integral.Stringize();
Assert.DoesNotContain("integral(", text);
Assert.True(text.Length < 5000, $"{text.Length} characters of answer for {integrand}");
Entity Pinned(Entity e) => e.Substitute("a", a).Substitute("b", b).Substitute("c", c).Substitute("d", 1.9).Substitute("k", 0.7);
var antiderivative = Pinned(integral.Substitute("C", 0));
var derivative = antiderivative.Differentiate("x");
var original = Pinned(integrand.ToEntity());
foreach (var at in new[] { -1.5, -0.7, 0.1, 0.57, 1.4, 2.9 })
{
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}");
// And the antiderivative itself is real there: the two conjugate pieces add to one.
var value = antiderivative.Substitute("x", at).EvalNumerical();
Assert.True(Math.Abs((double)value.ImaginaryPart) < 1e-9 * Math.Max(1, Math.Abs((double)value.RealPart)),
$"the antiderivative of {integrand} is {value} at x = {at}");
}
}
}
}
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