\(\int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx\) [2332]

Optimal result

Integrand size = 14, antiderivative size = 28 \[ \int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx=\frac {9 x^{2/3}}{2}-\frac {3 x^{5/3}}{5}+\frac {3 x^{8/3}}{8} \]

[Out]

9/2*x^(2/3)-3/5*x^(5/3)+3/8*x^(8/3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {14} \[ \int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx=\frac {3 x^{8/3}}{8}-\frac {3 x^{5/3}}{5}+\frac {9 x^{2/3}}{2} \]

[In]

Int[(3 - x + x^2)/x^(1/3),x]

[Out]

(9*x^(2/3))/2 - (3*x^(5/3))/5 + (3*x^(8/3))/8

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{\sqrt [3]{x}}-x^{2/3}+x^{5/3}\right ) \, dx \\ & = \frac {9 x^{2/3}}{2}-\frac {3 x^{5/3}}{5}+\frac {3 x^{8/3}}{8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx=\frac {3}{40} x^{2/3} \left (60-8 x+5 x^2\right ) \]

[In]

Integrate[(3 - x + x^2)/x^(1/3),x]

[Out]

(3*x^(2/3)*(60 - 8*x + 5*x^2))/40

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54

method	result	size
trager	\(\left (\frac {3}{8} x^{2}-\frac {3}{5} x +\frac {9}{2}\right ) x^{\frac {2}{3}}\)	\(15\)
gosper	\(\frac {3 x^{\frac {2}{3}} \left (5 x^{2}-8 x +60\right )}{40}\)	\(16\)
risch	\(\frac {3 x^{\frac {2}{3}} \left (5 x^{2}-8 x +60\right )}{40}\)	\(16\)
derivativedivides	\(\frac {9 x^{\frac {2}{3}}}{2}-\frac {3 x^{\frac {5}{3}}}{5}+\frac {3 x^{\frac {8}{3}}}{8}\)	\(17\)
default	\(\frac {9 x^{\frac {2}{3}}}{2}-\frac {3 x^{\frac {5}{3}}}{5}+\frac {3 x^{\frac {8}{3}}}{8}\)	\(17\)

[In]

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

[Out]

(3/8*x^2-3/5*x+9/2)*x^(2/3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx=\frac {3}{40} \, {\left (5 \, x^{2} - 8 \, x + 60\right )} x^{\frac {2}{3}} \]

[In]

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

[Out]

3/40*(5*x^2 - 8*x + 60)*x^(2/3)

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx=\frac {3 x^{\frac {8}{3}}}{8} - \frac {3 x^{\frac {5}{3}}}{5} + \frac {9 x^{\frac {2}{3}}}{2} \]

[In]

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

[Out]

3*x**(8/3)/8 - 3*x**(5/3)/5 + 9*x**(2/3)/2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.57 \[ \int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx=\frac {3}{8} \, x^{\frac {8}{3}} - \frac {3}{5} \, x^{\frac {5}{3}} + \frac {9}{2} \, x^{\frac {2}{3}} \]

[In]

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

[Out]

3/8*x^(8/3) - 3/5*x^(5/3) + 9/2*x^(2/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.57 \[ \int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx=\frac {3}{8} \, x^{\frac {8}{3}} - \frac {3}{5} \, x^{\frac {5}{3}} + \frac {9}{2} \, x^{\frac {2}{3}} \]

[In]

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

[Out]

3/8*x^(8/3) - 3/5*x^(5/3) + 9/2*x^(2/3)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {3-x+x^2}{\sqrt [3]{x}} \, dx=\frac {3\,x^{2/3}\,\left (5\,x^2-8\,x+60\right )}{40} \]

[In]

int((x^2 - x + 3)/x^(1/3),x)

[Out]

(3*x^(2/3)*(5*x^2 - 8*x + 60))/40