3.2.100 \(\int \frac {(a (b x^m)^n)^{-\frac {1}{m n}}}{x^2} \, dx\) [200]

Optimal. Leaf size=25 \[ -\frac {\left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}}}{2 x} \]

[Out]

-1/2/x/((a*(b*x^m)^n)^(1/m/n))

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1971, 30} \begin {gather*} -\frac {\left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}}}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a*(b*x^m)^n)^(1/(m*n))),x]

[Out]

-1/2*1/(x*(a*(b*x^m)^n)^(1/(m*n)))

Rule 30

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

Rule 1971

Int[(u_.)*((c_.)*((d_.)*((a_.) + (b_.)*(x_))^(n_))^(q_))^(p_), x_Symbol] :> Dist[(c*(d*(a + b*x)^n)^q)^p/(a +
b*x)^(n*p*q), Int[u*(a + b*x)^(n*p*q), x], x] /; FreeQ[{a, b, c, d, n, q, p}, x] &&  !IntegerQ[q] &&  !Integer
Q[p]

Rubi steps

\begin {align*} \int \frac {\left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}}}{x^2} \, dx &=\left (x \left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}}\right ) \int \frac {1}{x^3} \, dx\\ &=-\frac {\left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}}}{2 x}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 25, normalized size = 1.00 \begin {gather*} -\frac {\left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}}}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a*(b*x^m)^n)^(1/(m*n))),x]

[Out]

-1/2*1/(x*(a*(b*x^m)^n)^(1/(m*n)))

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Maple [A]
time = 0.05, size = 25, normalized size = 1.00

method	result	size
gosper	\(-\frac {\left (a \left (b \,x^{m}\right )^{n}\right )^{-\frac {1}{m n}}}{2 x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x/((a*(b*x^m)^n)^(1/m/n))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(((b*x^m)^n*a)^(1/(m*n))*x^2), x)

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Fricas [A]
time = 0.34, size = 21, normalized size = 0.84 \begin {gather*} -\frac {e^{\left (-\frac {n \log \left (b\right ) + \log \left (a\right )}{m n}\right )}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e^(-(n*log(b) + log(a))/(m*n))/x^2

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Sympy [A]
time = 1.65, size = 19, normalized size = 0.76 \begin {gather*} - \frac {\left (a \left (b x^{m}\right )^{n}\right )^{- \frac {1}{m n}}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/((a*(b*x**m)**n)**(1/m/n)),x)

[Out]

-1/(2*x*(a*(b*x**m)**n)**(1/(m*n)))

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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/x^2/((a*(b*x^m)^n)^(1/m/n)),x, algorithm="giac")

[Out]

integrate(1/(((b*x^m)^n*a)^(1/(m*n))*x^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{x^2\,{\left (a\,{\left (b\,x^m\right )}^n\right )}^{\frac {1}{m\,n}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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