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A quotient in x^2 over a power of a linear in x^2 and a biquadratic is split in x^2 #1809

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Rafael-SOWNet merged 1 commit into from
Oct 6, 2026
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A quotient in x^2 over a power of a linear in x^2 and a biquadratic is split in x^2 #1809

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17 changes: 17 additions & 0 deletions BREAKING-CHANGES.md
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Expand Up @@ -336,6 +336,23 @@ quotient of two such linears the sum is `(b - d t^2)^2 + (c t^2 - a)^2`. Rubi's
| `"(c + d*tan(x))^(3/2)/(a + b*tan(x))^3".ToEntity().Integrate("x")` | `integral(...)` | the same |
| `"1/((a + b*tan(x))^(3/2)*(c + d*tan(x))^(3/2))".ToEntity().Integrate("x")` | `integral(...)`; past a minute on the unreleased master | in the root of the quotient of the two, in a second |

### A quotient in `x^2` over a power of a linear in `x^2` and a biquadratic is split in `x^2`

**Shorter answers, sooner.** `sqrt(c + d tan(x)) (A + B tan(x) + C tan(x)^2)/(a + b tan(x))^3` was
declined in 2.5.0 and answered in 3.3 million characters since; it is answered in seven thousand.
Under `u = tan(x)` and `t = sqrt(c + d u)` it is a polynomial in `t^2` over
`(a d + b (t^2 - c))^3 ((t^2 - c)^2 + d^2)`, which was split in `t`, the sum of two squares over its
conjugates. Such a quotient, one linear in `x^2` to a power beside one biquadratic, is split in
`s = x^2 - r` now, `r` the linear's root: in powers of `s` it is the split over a power of `x^2` and a
biquadratic that the rule above makes, and the terms in `1/(x^2 - r)^j` go by their reduction to
`atan(x/sqrt(-r))/sqrt(-r)` ([#718](https://github.com/asc-community/AngouriMath/issues/718)).

| Input | Was (2.5.0) | Now |
|---|---|---|
| `"sqrt(c + d*tan(x))*(A + B*tan(x) + C*tan(x)^2)/(a + b*tan(x))^3".ToEntity().Integrate("x")` | `integral(...)`; 3,299,915 characters after 24 seconds on the unreleased master | 6,912 characters |
| `"sqrt(c + d*tan(x))/(a + b*tan(x))^2".ToEntity().Integrate("x")` | `integral(...)`; 40,540 characters on the unreleased master | 1,465 characters |
| `"(c + d*tan(x))^(3/2)/(a + b*tan(x))^3".ToEntity().Integrate("x")` | `integral(...)`; 71,293 characters on the unreleased master | 2,786 characters |

### A quotient in `x^2` over a power of `x` and a biquadratic is split in `x^2`

**Shorter answers, sooner.** `cot(x)^(13/2) (a + b tan(x))^(5/2) (A + B tan(x))` was declined in 2.5.0
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217 changes: 186 additions & 31 deletions Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs
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Expand Up @@ -2485,59 +2485,214 @@ over is Entity.Powf(var @base, var power) ?
if (Functions.PartialFractions.IsZeroAsAValue(discriminant))
return null;

// The numerator in w = x^2, divided down by w^k (c w^2 + b w + a) to a remainder of a
// degree below k + 2.
// The numerator in w = x^2 over w^k (c w^2 + b w + a).
var degree = above.Keys.Max() / 2;
var inW = new Entity[System.Math.Max(degree, k + 1) + 1];
for (var j = 0; j < inW.Length; j++)
var inW = new Entity[degree + 1];
for (var j = 0; j <= degree; j++)
inW[j] = above.TryGetValue(2 * j, out var at) ? at : Number.Integer.Zero;
var (polynomial, pole, d, e) = SplitOverAPowerAndAQuadratic(inW, k, a, b, c);
Entity polynomialPart = Number.Integer.Zero;
for (var j = polynomial.Length - 1; j >= 0; j--)
if (polynomial[j] != Number.Integer.Zero)
polynomialPart += polynomial[j] * MathS.Pow(x, 2 * j);

Entity total = Number.Integer.Zero;
if (polynomialPart != Number.Integer.Zero)
{
if (Integration.ComputeIndefiniteIntegral(polynomialPart, x, integrateByParts) is not { } whole)
return null;
total += whole;
}
// r_m x^(2m - 2k), whose exponent is odd and so never minus one.
for (var m = 0; m < k; m++)
{
if (pole[m] == Number.Integer.Zero)
continue;
var exponent = 2 * (m - k) + 1;
total += pole[m] * MathS.Pow(x, exponent) / exponent;
}
return ArctangentsOverABiquadratic(total, d, e, a, b, c, discriminant, x);
}

/// <summary>
/// A polynomial in <c>x^2</c> over a power of a linear in <c>x^2</c> and a biquadratic, with a
/// symbol in it: <c>N(x^2)/(L(x^2)^k Q(x^2))</c>, <c>L(w) = l (w - r)</c>, split in
/// <c>s = x^2 - r</c>. Written in powers of <c>s</c>, the quotient is the one
/// <see cref="SolveAnEvenPolynomialOverASymbolicBiquadratic"/> splits over <c>w^k Q</c>, and the
/// terms in <c>s^(-j)</c> go by the reduction
/// <c>int dx/(x^2 + p)^j = x/(2 p (j - 1) (x^2 + p)^(j - 1)) + (2 j - 3)/(2 p (j - 1)) int dx/(x^2 + p)^(j - 1)</c>,
/// <c>p = -r</c>, down to <c>atan(x/sqrt(p))/sqrt(p)</c>, which holds for every complex
/// <c>p</c> but zero, as the biquadratic's own terms do.
/// </summary>
/// <remarks>
/// Under <c>t = sqrt(c + d u)</c> Rubi's <c>sqrt(c + d u)/((a + b u)^n (1 + u^2))</c>, which the
/// tangent substitution makes of <c>sqrt(c + d tan(x))/(a + b tan(x))^n</c>, is
/// <c>t^2/((a d + b (t^2 - c))^n ((t^2 - c)^2 + d^2))</c> up to a constant. Split in <c>t</c>, the
/// sum of two squares went over its conjugates, and the answer to
/// <c>sqrt(c + d tan(x)) (A + B tan(x) + C tan(x)^2)/(a + b tan(x))^3</c> ran to 3.5 million
/// characters. Only one linear in <c>x^2</c>, to a power, beside one biquadratic: with
/// <c>r</c> zero it is <see cref="SolveAnEvenPolynomialOverASymbolicBiquadratic"/>'s, and a root
/// <c>r</c> shared with the biquadratic declines.
/// https://github.com/asc-community/AngouriMath/issues/718
/// </remarks>
internal static Entity? SolveAnEvenQuotientOverAPowerOfALinearInTheSquareAndABiquadratic(Entity expr, Entity.Variable x, bool integrateByParts)
{
if (!TryReadAsQuotient(expr, out var numerator, out var denominator) || !denominator.Vars.Any(v => v != x))
return null;
Entity constant = Number.Integer.One;
(Entity Zeroth, Entity First, int Power)? linear = null;
(Entity A, Entity B, Entity C)? quadratic = null;
foreach (var factor in Mulf.LinearChildren(denominator))
{
var (@base, power) = factor is Powf(var raised, Number.Integer { EInteger: var n }) && n.Sign > 0 && n.CanFitInInt32()
? (raised, n.ToInt32Unchecked()) : (factor, 1);
if (!@base.ContainsNode(x))
{
constant *= factor;
continue;
}
if (!TreeAnalyzer.TryGetPolynomial(@base, x, out var read) || read.Count == 0
|| read.Values.Any(coefficient => coefficient.ContainsNode(x))
|| read.Keys.Any(p => p.Sign < 0 || !p.IsEven || !p.CanFitInInt32()))
return null;
Entity At(int p) => read.TryGetValue(EInteger.FromInt32(p), out var value) ? value : Number.Integer.Zero;
var top = read.Keys.Max()!.ToInt32Unchecked();
if (top == 2 && linear is null)
linear = (At(0), At(2), power);
else if (top == 4 && quadratic is null && power == 1)
quadratic = (At(0), At(2), At(4));
else
return null;
}
if (linear is not { } l || quadratic is not { } quartic)
return null;
if (!TreeAnalyzer.TryGetPolynomial(numerator, x, out var above) || above.Count == 0
|| above.Values.Any(coefficient => coefficient.ContainsNode(x))
|| above.Keys.Any(p => p.Sign < 0 || !p.IsEven || !p.CanFitInInt32()))
return null;
var (a, b, c) = quartic;
if (Functions.PartialFractions.IsZeroAsAValue(a) || Functions.PartialFractions.IsZeroAsAValue(c)
|| Functions.PartialFractions.IsZeroAsAValue(l.First))
return null;
var r = Functions.PartialFractions.InLowestTermsOverTheSymbols(-l.Zeroth / l.First);
if (Functions.PartialFractions.IsZeroAsAValue(r))
return null;
var discriminant = Functions.PartialFractions.Bare((b * b - 4 * a * c).Simplify());
if (Functions.PartialFractions.IsZeroAsAValue(discriminant))
return null;
// The biquadratic in s: Q(r + s) = Q(r) + Q'(r) s + c s^2, and Q(r) is not zero, or the
// linear's root is one of the biquadratic's.
var atRoot = Functions.PartialFractions.InLowestTermsOverTheSymbols(a + b * r + c * r * r);
if (Functions.PartialFractions.IsZeroAsAValue(atRoot))
return null;
var slopeAtRoot = Functions.PartialFractions.InLowestTermsOverTheSymbols(b + 2 * c * r);

// N(r + s) in powers of s.
var degree = above.Keys.Max()!.ToInt32Unchecked() / 2;
var inW = new Entity[degree + 1];
for (var j = 0; j <= degree; j++)
inW[j] = above.TryGetValue(EInteger.FromInt32(2 * j), out var at) ? at : Number.Integer.Zero;
var inS = new Entity[degree + 1];
for (var m = 0; m <= degree; m++)
{
Entity sum = Number.Integer.Zero;
EInteger binomial = EInteger.One;
for (var j = m; j <= degree; j++)
{
if (j > m)
binomial = binomial * j / (j - m);
if (inW[j] != Number.Integer.Zero)
sum += inW[j] * Number.Integer.Create(binomial) * MathS.Pow(r, Number.Integer.Create(j - m));
}
inS[m] = Functions.PartialFractions.InLowestTermsOverTheSymbols(sum);
}
var k = l.Power;
var (polynomial, pole, dInS, eInS) = SplitOverAPowerAndAQuadratic(inS, k, atRoot, slopeAtRoot, c);
var square = MathS.Pow(x, 2) - r;

Entity total = Number.Integer.Zero;
Entity polynomialPart = Number.Integer.Zero;
for (var j = 0; j < polynomial.Length; j++)
if (polynomial[j] != Number.Integer.Zero)
polynomialPart += polynomial[j] * MathS.Pow(square, j);
if (polynomialPart != Number.Integer.Zero)
{
if (Integration.ComputeIndefiniteIntegral(polynomialPart.Expand(), x, integrateByParts) is not { } whole)
return null;
total += whole;
}
// pole[m] s^(m - k) is a term in 1/(x^2 - r)^j with j = k - m, by the reduction.
var p = -r;
var root = MathS.Sqrt(p);
Entity reduced = MathS.Arctan(x / root) / root;
var byPower = new Entity[k + 1];
byPower[1] = reduced;
for (var j = 2; j <= k; j++)
byPower[j] = x / (2 * p * (j - 1) * MathS.Pow(square, j - 1)) + Number.Integer.Create(2 * j - 3) / (2 * p * (j - 1)) * byPower[j - 1];
for (var m = 0; m < k; m++)
if (pole[m] != Number.Integer.Zero)
total += pole[m] * byPower[k - m];
// (d + e s)/Q in w = x^2 is (d - e r + e w)/Q(w).
var d = Functions.PartialFractions.InLowestTermsOverTheSymbols(dInS - eInS * r);
total = ArctangentsOverABiquadratic(total, d, eInS, a, b, c, discriminant, x);
return total / (constant * MathS.Pow(l.First, k));
}

/// <summary>
/// <c>N(s)/(s^k (a + b s + c s^2))</c>, the coefficients of <c>N</c> in <paramref name="above"/>,
/// as a polynomial in <c>s</c>, the terms in <c>s^(m - k)</c> for <c>m</c> below <c>k</c>, and
/// <c>(d + e s)/(a + b s + c s^2)</c>: <c>N</c> divided down by <c>s^k</c> times the quadratic first,
/// then the remainder <c>R</c> over it expanded at <c>s = 0</c> by
/// <c>r_m = (R_m - b r_(m-1) - c r_(m-2))/a</c>, and <c>d + e s</c> what
/// <c>(R - Q sum r_m s^m)/s^k</c> leaves, every lower coefficient cancelling by the recurrence.
/// </summary>
private static (Entity[] Polynomial, Entity[] Pole, Entity D, Entity E) SplitOverAPowerAndAQuadratic(
Entity[] above, int k, Entity a, Entity b, Entity c)
{
var degree = above.Length - 1;
var inS = new Entity[System.Math.Max(degree, k + 1) + 1];
for (var j = 0; j < inS.Length; j++)
inS[j] = j <= degree ? above[j] : Number.Integer.Zero;
var polynomial = new Entity[System.Math.Max(degree - k - 1, 0)];
for (var j = 0; j < polynomial.Length; j++)
polynomial[j] = Number.Integer.Zero;
for (var j = degree; j >= k + 2; j--)
{
var lead = Functions.PartialFractions.InLowestTermsOverTheSymbols(inW[j] / c);
var lead = Functions.PartialFractions.InLowestTermsOverTheSymbols(inS[j] / c);
if (lead == Number.Integer.Zero)
continue;
polynomialPart += lead * MathS.Pow(x, 2 * (j - k - 2));
inW[j - 1] = Functions.PartialFractions.InLowestTermsOverTheSymbols(inW[j - 1] - lead * b);
inW[j - 2] = Functions.PartialFractions.InLowestTermsOverTheSymbols(inW[j - 2] - lead * a);
polynomial[j - k - 2] = lead;
inS[j - 1] = Functions.PartialFractions.InLowestTermsOverTheSymbols(inS[j - 1] - lead * b);
inS[j - 2] = Functions.PartialFractions.InLowestTermsOverTheSymbols(inS[j - 2] - lead * a);
}
// The remainder R over w^k Q is sum r_m w^(m - k) over m < k, the expansion of R/Q at
// w = 0, plus (d + e w)/Q: r_m = (R_m - b r_(m-1) - c r_(m-2))/a, and d + e w is what
// (R - Q sum r_m w^m)/w^k leaves, every lower coefficient cancelling by the recurrence.
var pole = new Entity[k];
for (var m = 0; m < k; m++)
{
var term = inW[m];
var term = inS[m];
if (m >= 1) term -= b * pole[m - 1];
if (m >= 2) term -= c * pole[m - 2];
pole[m] = Functions.PartialFractions.InLowestTermsOverTheSymbols(term / a);
}
var dTerm = inW[k];
if (k == 0)
return (polynomial, pole, inS[0], inS[1]);
var dTerm = inS[k];
if (k >= 1) dTerm -= b * pole[k - 1];
if (k >= 2) dTerm -= c * pole[k - 2];
var eTerm = inW[k + 1];
var eTerm = inS[k + 1];
if (k >= 1) eTerm -= c * pole[k - 1];
var d = k == 0 ? inW[0] : Functions.PartialFractions.InLowestTermsOverTheSymbols(dTerm);
var e = k == 0 ? (degree >= 1 ? inW[1] : Number.Integer.Zero) : Functions.PartialFractions.InLowestTermsOverTheSymbols(eTerm);
return (polynomial, pole, Functions.PartialFractions.InLowestTermsOverTheSymbols(dTerm),
Functions.PartialFractions.InLowestTermsOverTheSymbols(eTerm));
}

/// <summary>
/// <paramref name="total"/> plus the antiderivative of <c>(d + e x^2)/(a + b x^2 + c x^4)</c> by
/// the two roots in <c>x^2</c>, <paramref name="discriminant"/> being <c>b^2 - 4 a c</c> and not zero.
/// </summary>
private static Entity ArctangentsOverABiquadratic(Entity total, Entity d, Entity e, Entity a, Entity b, Entity c, Entity discriminant, Entity.Variable x)
{
var q = MathS.Sqrt(discriminant);
var firstRoot = (-b + q) / (2 * c);
var secondRoot = (-b - q) / (2 * c);
Entity total = Number.Integer.Zero;
if (polynomialPart != Number.Integer.Zero)
{
if (Integration.ComputeIndefiniteIntegral(polynomialPart, x, integrateByParts) is not { } whole)
return null;
total += whole;
}
// r_m x^(2m - 2k), whose exponent is odd and so never minus one.
for (var m = 0; m < k; m++)
{
if (pole[m] == Number.Integer.Zero)
continue;
var exponent = 2 * (m - k) + 1;
total += pole[m] * MathS.Pow(x, exponent) / exponent;
}
// `1/(x^2 - r)` is `atan(x/s)/s` with `s = sqrt(-r)` for every complex `r` but zero:
// `d/dx atan(x/s)/s` is `1/(s^2 + x^2)` whatever `s` is, and for a positive `r` the
// arctangent of an imaginary argument is the hyperbolic one, `-atanh(x/sqrt(r))/sqrt(r)`,
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2 changes: 2 additions & 0 deletions Sources/AngouriMath/Functions/Continuous/Integration/Integration.Definition.cs
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Expand Up @@ -1017,6 +1017,8 @@ private static Entity Normalized(Entity expr, Entity.Variable x) =>
// in x^2 written with the root of the discriminant: partial fractions read a
// written factor, and `a + b x^2 + c x^4` is written as one.
if ((answer = IndefiniteIntegralSolver.SolveAnEvenPolynomialOverASymbolicBiquadratic(expr, x, integrateByParts)) is { }) return answer;
// And over a power of a linear in x^2 beside the biquadratic, split in x^2 minus its root.
if ((answer = IndefiniteIntegralSolver.SolveAnEvenQuotientOverAPowerOfALinearInTheSquareAndABiquadratic(expr, x, integrateByParts)) is { }) return answer;
if ((answer = IndefiniteIntegralSolver.SolveByPartialFractions(expr, x, integrateByParts)) is { }) return answer;
// A whole negative power of a polynomial of several terms among the factors,
// written below the bar and asked again: the gathering on the way in writes
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