3.2.29 \(\int \frac {1}{(\frac {b}{x^2})^{2/3}} \, dx\) [129]

Optimal. Leaf size=14 \[ \frac {3 x}{7 \left (\frac {b}{x^2}\right )^{2/3}} \]

[Out]

3/7*x/(b/x^2)^(2/3)

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Rubi [A]
time = 0.00, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {15, 30} \begin {gather*} \frac {3 x}{7 \left (\frac {b}{x^2}\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b/x^2)^(-2/3),x]

[Out]

(3*x)/(7*(b/x^2)^(2/3))

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{\left (\frac {b}{x^2}\right )^{2/3}} \, dx &=\frac {\int x^{4/3} \, dx}{\left (\frac {b}{x^2}\right )^{2/3} x^{4/3}}\\ &=\frac {3 x}{7 \left (\frac {b}{x^2}\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} \frac {3 x}{7 \left (\frac {b}{x^2}\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b/x^2)^(-2/3),x]

[Out]

(3*x)/(7*(b/x^2)^(2/3))

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Maple [A]
time = 0.02, size = 11, normalized size = 0.79

method	result	size
gosper	\(\frac {3 x}{7 \left (\frac {b}{x^{2}}\right )^{\frac {2}{3}}}\)	\(11\)
risch	\(\frac {3 x}{7 \left (\frac {b}{x^{2}}\right )^{\frac {2}{3}}}\)	\(11\)
trager	\(\frac {3 \left (\frac {b}{x^{2}}\right )^{\frac {1}{3}} x^{3}}{7 b}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*x/(b/x^2)^(2/3)

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Maxima [A]
time = 0.30, size = 10, normalized size = 0.71 \begin {gather*} \frac {3 \, x}{7 \, \left (\frac {b}{x^{2}}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*x/(b/x^2)^(2/3)

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Fricas [A]
time = 0.34, size = 15, normalized size = 1.07 \begin {gather*} \frac {3 \, x^{3} \left (\frac {b}{x^{2}}\right )^{\frac {1}{3}}}{7 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*x^3*(b/x^2)^(1/3)/b

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Sympy [A]
time = 0.17, size = 12, normalized size = 0.86 \begin {gather*} \frac {3 x}{7 \left (\frac {b}{x^{2}}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x/(7*(b/x**2)**(2/3))

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Giac [A]
time = 1.63, size = 10, normalized size = 0.71 \begin {gather*} \frac {3 \, x}{7 \, \left (\frac {b}{x^{2}}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*x/(b/x^2)^(2/3)

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Mupad [B]
time = 0.92, size = 13, normalized size = 0.93 \begin {gather*} \frac {3\,x^3\,{\left (\frac {1}{x^2}\right )}^{1/3}}{7\,b^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/x^2)^(2/3),x)

[Out]

(3*x^3*(1/x^2)^(1/3))/(7*b^(2/3))

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