3.1.23 \(\int x^2 (1+x^3)^{2/3} \, dx\) [23]

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

[Out]

1/5*(x^3+1)^(5/3)

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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 {1}{5} \left (x^3+1\right )^{5/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(1 + x^3)^(2/3),x]

[Out]

(1 + x^3)^(5/3)/5

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 \left (1+x^3\right )^{2/3} \, dx &=\frac {1}{5} \left (1+x^3\right )^{5/3}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

Integrate[x^2*(1 + x^3)^(2/3),x]

[Out]

(1 + x^3)^(5/3)/5

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Maple [A]
time = 0.26, size = 10, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {\left (x^{3}+1\right )^{\frac {5}{3}}}{5}\)	\(10\)
default	\(\frac {\left (x^{3}+1\right )^{\frac {5}{3}}}{5}\)	\(10\)
risch	\(\frac {\left (x^{3}+1\right )^{\frac {5}{3}}}{5}\)	\(10\)
trager	\(\left (\frac {1}{5}+\frac {x^{3}}{5}\right ) \left (x^{3}+1\right )^{\frac {2}{3}}\)	\(16\)
meijerg	\(\frac {x^{3} \hypergeom \left (\left [-\frac {2}{3}, 1\right ], \left [2\right ], -x^{3}\right )}{3}\)	\(17\)
gosper	\(\frac {\left (1+x \right ) \left (x^{2}-x +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{5}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^3+1)^(5/3)

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Maxima [A]
time = 0.26, size = 9, normalized size = 0.69 \begin {gather*} \frac {1}{5} \, {\left (x^{3} + 1\right )}^{\frac {5}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^3 + 1)^(5/3)

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Fricas [A]
time = 0.35, size = 9, normalized size = 0.69 \begin {gather*} \frac {1}{5} \, {\left (x^{3} + 1\right )}^{\frac {5}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^3 + 1)^(5/3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 22 vs. \(2 (8) = 16\).
time = 0.09, size = 22, normalized size = 1.69 \begin {gather*} \frac {x^{3} \left (x^{3} + 1\right )^{\frac {2}{3}}}{5} + \frac {\left (x^{3} + 1\right )^{\frac {2}{3}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(x**3+1)**(2/3),x)

[Out]

x**3*(x**3 + 1)**(2/3)/5 + (x**3 + 1)**(2/3)/5

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Giac [A]
time = 0.40, size = 9, normalized size = 0.69 \begin {gather*} \frac {1}{5} \, {\left (x^{3} + 1\right )}^{\frac {5}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^3 + 1)^(5/3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^3 + 1)^(2/3),x)

[Out]

(x^3 + 1)^(5/3)/5

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