\(\int x^2 \sqrt [3]{1+x^3} \, dx\) [21]

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int x^2 \sqrt [3]{1+x^3} \, dx=\frac {1}{4} \left (1+x^3\right )^{4/3} \]

[Out]

1/4*(x^3+1)^(4/3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \[ \int x^2 \sqrt [3]{1+x^3} \, dx=\frac {1}{4} \left (x^3+1\right )^{4/3} \]

[In]

Int[x^2*(1 + x^3)^(1/3),x]

[Out]

(1 + x^3)^(4/3)/4

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*} \text {integral}& = \frac {1}{4} \left (1+x^3\right )^{4/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int x^2 \sqrt [3]{1+x^3} \, dx=\frac {1}{4} \left (1+x^3\right )^{4/3} \]

[In]

Integrate[x^2*(1 + x^3)^(1/3),x]

[Out]

(1 + x^3)^(4/3)/4

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {\left (x^{3}+1\right )^{\frac {4}{3}}}{4}\)	\(10\)
default	\(\frac {\left (x^{3}+1\right )^{\frac {4}{3}}}{4}\)	\(10\)
risch	\(\frac {\left (x^{3}+1\right )^{\frac {4}{3}}}{4}\)	\(10\)
pseudoelliptic	\(\frac {\left (x^{3}+1\right )^{\frac {4}{3}}}{4}\)	\(10\)
trager	\(\left (\frac {x^{3}}{4}+\frac {1}{4}\right ) \left (x^{3}+1\right )^{\frac {1}{3}}\)	\(16\)
meijerg	\(\frac {x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, 1\right ], \left [2\right ], -x^{3}\right )}{3}\)	\(17\)
gosper	\(\frac {\left (1+x \right ) \left (x^{2}-x +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}}{4}\)	\(21\)

[In]

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

[Out]

1/4*(x^3+1)^(4/3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt [3]{1+x^3} \, dx=\frac {1}{4} \, {\left (x^{3} + 1\right )}^{\frac {4}{3}} \]

[In]

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

[Out]

1/4*(x^3 + 1)^(4/3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 22 vs. \(2 (8) = 16\).

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.69 \[ \int x^2 \sqrt [3]{1+x^3} \, dx=\frac {x^{3} \sqrt [3]{x^{3} + 1}}{4} + \frac {\sqrt [3]{x^{3} + 1}}{4} \]

[In]

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

[Out]

x**3*(x**3 + 1)**(1/3)/4 + (x**3 + 1)**(1/3)/4

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt [3]{1+x^3} \, dx=\frac {1}{4} \, {\left (x^{3} + 1\right )}^{\frac {4}{3}} \]

[In]

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

[Out]

1/4*(x^3 + 1)^(4/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt [3]{1+x^3} \, dx=\frac {1}{4} \, {\left (x^{3} + 1\right )}^{\frac {4}{3}} \]

[In]

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

[Out]

1/4*(x^3 + 1)^(4/3)

Mupad [B] (verification not implemented)

Time = 5.10 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt [3]{1+x^3} \, dx=\frac {{\left (x^3+1\right )}^{4/3}}{4} \]

[In]

int(x^2*(x^3 + 1)^(1/3),x)

[Out]

(x^3 + 1)^(4/3)/4