3.2.16 \(\int \frac {1}{\sqrt [3]{b x^n}} \, dx\) [116]

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

[Out]

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

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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-n) \sqrt [3]{b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^n)^(-1/3),x]

[Out]

(3*x)/((3 - n)*(b*x^n)^(1/3))

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}{\sqrt [3]{b x^n}} \, dx &=\frac {x^{n/3} \int x^{-n/3} \, dx}{\sqrt [3]{b x^n}}\\ &=\frac {3 x}{(3-n) \sqrt [3]{b x^n}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 17, normalized size = 0.89 \begin {gather*} -\frac {3 x}{(-3+n) \sqrt [3]{b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^n)^(-1/3),x]

[Out]

(-3*x)/((-3 + n)*(b*x^n)^(1/3))

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Maple [A]
time = 0.01, size = 16, normalized size = 0.84

method	result	size
gosper	\(-\frac {3 x}{\left (-3+n \right ) \left (b \,x^{n}\right )^{\frac {1}{3}}}\)	\(16\)
risch	\(-\frac {3 x}{\left (-3+n \right ) \left (b \,x^{n}\right )^{\frac {1}{3}}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/3),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(-n/3>0)', see `assume?` for mo
re details)I

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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)^(1/3),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}{n \sqrt [3]{b x^{n}} - 3 \sqrt [3]{b x^{n}}} & \text {for}\: n \neq 3 \\\int \frac {1}{\sqrt [3]{b x^{3}}}\, dx & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*x/(n*(b*x**n)**(1/3) - 3*(b*x**n)**(1/3)), Ne(n, 3)), (Integral((b*x**3)**(-1/3), x), True))

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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)^(1/3),x, algorithm="giac")

[Out]

integrate((b*x^n)^(-1/3), x)

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Mupad [B]
time = 0.98, size = 24, normalized size = 1.26 \begin {gather*} -\frac {3\,x^{1-n}\,{\left (b\,x^n\right )}^{2/3}}{b\,\left (n-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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