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

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int x^2 \sqrt [4]{-1+x^3} \, dx=\frac {4}{15} \left (-1+x^3\right )^{5/4} \]

[Out]

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

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

[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*} \text {integral}& = \frac {4}{15} \left (-1+x^3\right )^{5/4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int x^2 \sqrt [4]{-1+x^3} \, dx=\frac {4}{15} \left (-1+x^3\right )^{5/4} \]

[In]

Integrate[x^2*(-1 + x^3)^(1/4),x]

[Out]

(4*(-1 + x^3)^(5/4))/15

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 10, normalized size of antiderivative = 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	\(\frac {4 \left (x^{3}-1\right )^{\frac {5}{4}}}{15}\)	\(10\)
pseudoelliptic	\(\frac {4 \left (x^{3}-1\right )^{\frac {5}{4}}}{15}\)	\(10\)
trager	\(\left (\frac {4 x^{3}}{15}-\frac {4}{15}\right ) \left (x^{3}-1\right )^{\frac {1}{4}}\)	\(16\)
gosper	\(\frac {4 \left (x -1\right ) \left (x^{2}+x +1\right ) \left (x^{3}-1\right )^{\frac {1}{4}}}{15}\)	\(19\)
meijerg	\(\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, 1\right ], \left [2\right ], x^{3}\right )}{3 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}}}\)	\(33\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt [4]{-1+x^3} \, dx=\frac {4}{15} \, {\left (x^{3} - 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

4/15*(x^3 - 1)^(5/4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.00 \[ \int x^2 \sqrt [4]{-1+x^3} \, dx=\frac {4 x^{3} \sqrt [4]{x^{3} - 1}}{15} - \frac {4 \sqrt [4]{x^{3} - 1}}{15} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt [4]{-1+x^3} \, dx=\frac {4}{15} \, {\left (x^{3} - 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

4/15*(x^3 - 1)^(5/4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt [4]{-1+x^3} \, dx=\frac {4}{15} \, {\left (x^{3} - 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

4/15*(x^3 - 1)^(5/4)

Mupad [B] (verification not implemented)

Time = 5.35 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt [4]{-1+x^3} \, dx=\frac {4\,{\left (x^3-1\right )}^{5/4}}{15} \]

[In]

int(x^2*(x^3 - 1)^(1/4),x)

[Out]

(4*(x^3 - 1)^(5/4))/15