3.1.16 \(\int x^2 \sqrt [4]{-1+x^3} \, dx\) [16]

Optimal. Leaf size=13 \[ \frac {4}{15} \left (-1+x^3\right )^{5/4} \]

[Out]

4/15*(x^3-1)^(5/4)

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Rubi [A]
time = 0.00, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \begin {gather*} \frac {4}{15} \left (x^3-1\right )^{5/4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(-1 + x^3)^(1/4),x]

[Out]

(4*(-1 + x^3)^(5/4))/15

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x^2 \sqrt [4]{-1+x^3} \, dx &=\frac {4}{15} \left (-1+x^3\right )^{5/4}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} \frac {4}{15} \left (-1+x^3\right )^{5/4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(-1 + x^3)^(1/4),x]

[Out]

(4*(-1 + x^3)^(5/4))/15

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Maple [A]
time = 0.24, size = 10, normalized size = 0.77

method result size
derivativedivides \(\frac {4 \left (x^{3}-1\right )^{\frac {5}{4}}}{15}\) \(10\)
default \(\frac {4 \left (x^{3}-1\right )^{\frac {5}{4}}}{15}\) \(10\)
risch \(\frac {4 \left (x^{3}-1\right )^{\frac {5}{4}}}{15}\) \(10\)
trager \(\left (\frac {4 x^{3}}{15}-\frac {4}{15}\right ) \left (x^{3}-1\right )^{\frac {1}{4}}\) \(16\)
gosper \(\frac {4 \left (-1+x \right ) \left (x^{2}+x +1\right ) \left (x^{3}-1\right )^{\frac {1}{4}}}{15}\) \(19\)
meijerg \(\frac {\mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{3} \hypergeom \left (\left [-\frac {1}{4}, 1\right ], \left [2\right ], x^{3}\right )}{3 \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{4}}}\) \(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^3-1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/15*(x^3-1)^(5/4)

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Maxima [A]
time = 0.27, size = 9, normalized size = 0.69 \begin {gather*} \frac {4}{15} \, {\left (x^{3} - 1\right )}^{\frac {5}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^3-1)^(1/4),x, algorithm="maxima")

[Out]

4/15*(x^3 - 1)^(5/4)

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Fricas [A]
time = 0.33, size = 9, normalized size = 0.69 \begin {gather*} \frac {4}{15} \, {\left (x^{3} - 1\right )}^{\frac {5}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^3-1)^(1/4),x, algorithm="fricas")

[Out]

4/15*(x^3 - 1)^(5/4)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).
time = 0.08, size = 26, normalized size = 2.00 \begin {gather*} \frac {4 x^{3} \sqrt [4]{x^{3} - 1}}{15} - \frac {4 \sqrt [4]{x^{3} - 1}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(x**3-1)**(1/4),x)

[Out]

4*x**3*(x**3 - 1)**(1/4)/15 - 4*(x**3 - 1)**(1/4)/15

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Giac [A]
time = 0.39, size = 9, normalized size = 0.69 \begin {gather*} \frac {4}{15} \, {\left (x^{3} - 1\right )}^{\frac {5}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^3-1)^(1/4),x, algorithm="giac")

[Out]

4/15*(x^3 - 1)^(5/4)

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Mupad [B]
time = 0.13, size = 9, normalized size = 0.69 \begin {gather*} \frac {4\,{\left (x^3-1\right )}^{5/4}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^3 - 1)^(1/4),x)

[Out]

(4*(x^3 - 1)^(5/4))/15

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