3.2.96 \(\int x^2 (a (b x^m)^n)^{-\frac {1}{m n}} \, dx\) [196]

Optimal. Leaf size=25 \[ \frac {1}{2} x^3 \left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}} \]

[Out]

1/2*x^3/((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 {1}{2} x^3 \left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{2} x^3 \left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^3/(2*(a*(b*x^m)^n)^(1/(m*n)))

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Maple [A]
time = 0.10, size = 25, normalized size = 1.00

method	result	size
gosper	\(\frac {x^{3} \left (a \left (b \,x^{m}\right )^{n}\right )^{-\frac {1}{m n}}}{2}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^3/((a*(b*x^m)^n)^(1/m/n))

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Maxima [A]
time = 0.39, size = 38, normalized size = 1.52 \begin {gather*} \frac {x^{3}}{2 \, a^{\frac {1}{m n}} b^{\left (\frac {1}{m}\right )} {\left ({\left (x^{m}\right )}^{n}\right )}^{\frac {1}{m n}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^3/(a^(1/(m*n))*b^(1/m)*((x^m)^n)^(1/(m*n)))

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Fricas [A]
time = 0.36, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (-\frac {n \log \left (b\right ) + \log \left (a\right )}{m n}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.87, size = 17, normalized size = 0.68 \begin {gather*} \frac {x^{3} \left (a \left (b x^{m}\right )^{n}\right )^{- \frac {1}{m n}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3/(2*(a*(b*x**m)**n)**(1/(m*n)))

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Giac [A]
time = 1.44, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (-\frac {n \log \left (b\right ) + \log \left (a\right )}{m n}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.01, size = 24, normalized size = 0.96 \begin {gather*} \frac {x^3}{2\,{\left (a\,{\left (b\,x^m\right )}^n\right )}^{\frac {1}{m\,n}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3/(2*(a*(b*x^m)^n)^(1/(m*n)))

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