3.24.97 \(\int (a+\frac {b}{\sqrt [3]{x}}) x \, dx\) [2397]

Optimal. Leaf size=19 \[ \frac {3}{5} b x^{5/3}+\frac {a x^2}{2} \]

[Out]

3/5*b*x^(5/3)+1/2*a*x^2

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Rubi [A]
time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14} \begin {gather*} \frac {a x^2}{2}+\frac {3}{5} b x^{5/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^(1/3))*x,x]

[Out]

(3*b*x^(5/3))/5 + (a*x^2)/2

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*} \int \left (a+\frac {b}{\sqrt [3]{x}}\right ) x \, dx &=\int \left (b x^{2/3}+a x\right ) \, dx\\ &=\frac {3}{5} b x^{5/3}+\frac {a x^2}{2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 19, normalized size = 1.00 \begin {gather*} \frac {3}{5} b x^{5/3}+\frac {a x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^(1/3))*x,x]

[Out]

(3*b*x^(5/3))/5 + (a*x^2)/2

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Maple [A]
time = 0.02, size = 14, normalized size = 0.74

method	result	size
derivativedivides	\(\frac {3 b \,x^{\frac {5}{3}}}{5}+\frac {a \,x^{2}}{2}\)	\(14\)
default	\(\frac {3 b \,x^{\frac {5}{3}}}{5}+\frac {a \,x^{2}}{2}\)	\(14\)
trager	\(\frac {\left (x -1\right ) a \left (x +1\right )}{2}+\frac {3 b \,x^{\frac {5}{3}}}{5}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/5*b*x^(5/3)+1/2*a*x^2

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Maxima [A]
time = 0.30, size = 15, normalized size = 0.79 \begin {gather*} \frac {1}{10} \, {\left (5 \, a + \frac {6 \, b}{x^{\frac {1}{3}}}\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(5*a + 6*b/x^(1/3))*x^2

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Fricas [A]
time = 0.37, size = 13, normalized size = 0.68 \begin {gather*} \frac {1}{2} \, a x^{2} + \frac {3}{5} \, b x^{\frac {5}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 3/5*b*x^(5/3)

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Sympy [A]
time = 0.07, size = 15, normalized size = 0.79 \begin {gather*} \frac {a x^{2}}{2} + \frac {3 b x^{\frac {5}{3}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**(1/3))*x,x)

[Out]

a*x**2/2 + 3*b*x**(5/3)/5

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Giac [A]
time = 0.56, size = 13, normalized size = 0.68 \begin {gather*} \frac {1}{2} \, a x^{2} + \frac {3}{5} \, b x^{\frac {5}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 3/5*b*x^(5/3)

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Mupad [B]
time = 0.03, size = 13, normalized size = 0.68 \begin {gather*} \frac {a\,x^2}{2}+\frac {3\,b\,x^{5/3}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/x^(1/3)),x)

[Out]

(a*x^2)/2 + (3*b*x^(5/3))/5

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