3.2.24 \(\int \frac {1}{(b x^n)^{2/3}} \, dx\) [124]

Optimal. Leaf size=19 \[ \frac {3 x}{(3-2 n) \left (b x^n\right )^{2/3}} \]

[Out]

3*x/(3-2*n)/(b*x^n)^(2/3)

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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 = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {15, 30} \begin {gather*} \frac {3 x}{(3-2 n) \left (b x^n\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x)/((3 - 2*n)*(b*x^n)^(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 (b x^n\right )^{2/3}} \, dx &=\frac {x^{2 n/3} \int x^{-2 n/3} \, dx}{\left (b x^n\right )^{2/3}}\\ &=\frac {3 x}{(3-2 n) \left (b x^n\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 20, normalized size = 1.05 \begin {gather*} \frac {x}{\left (1-\frac {2 n}{3}\right ) \left (b x^n\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/((1 - (2*n)/3)*(b*x^n)^(2/3))

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Maple [A]
time = 0.01, size = 18, normalized size = 0.95

method	result	size
gosper	\(-\frac {3 x}{\left (2 n -3\right ) \left (b \,x^{n}\right )^{\frac {2}{3}}}\)	\(18\)
risch	\(-\frac {3 x}{\left (2 n -3\right ) \left (b \,x^{n}\right )^{\frac {2}{3}}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x/(2*n-3)/(b*x^n)^(2/3)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-(2*n)/3>0)', see `assume?` fo
r more detai

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} - \frac {3 x}{2 n \left (b x^{n}\right )^{\frac {2}{3}} - 3 \left (b x^{n}\right )^{\frac {2}{3}}} & \text {for}\: n \neq \frac {3}{2} \\\int \frac {1}{\left (b x^{\frac {3}{2}}\right )^{\frac {2}{3}}}\, dx & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*x/(2*n*(b*x**n)**(2/3) - 3*(b*x**n)**(2/3)), Ne(n, 3/2)), (Integral((b*x**(3/2))**(-2/3), x), Tr
ue))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^n)^(-2/3), x)

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Mupad [B]
time = 0.98, size = 26, normalized size = 1.37 \begin {gather*} -\frac {3\,x^{1-n}\,{\left (b\,x^n\right )}^{1/3}}{b\,\left (2\,n-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*x^(1 - n)*(b*x^n)^(1/3))/(b*(2*n - 3))

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