3.196 \(\int x^2 (a (b x^m)^n)^{-\frac {1}{m n}} \, dx\)

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 = 0.05, 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 = {6679, 30} \[ \frac {1}{2} x^3 \left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}} \]

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 6679

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

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 \[ \frac {1}{2} x^3 \left (a \left (b x^m\right )^n\right )^{-\frac {1}{m n}} \]

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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fricas [A]  time = 0.70, size = 21, normalized size = 0.84 \[ \frac {1}{2} \, x^{2} e^{\left (-\frac {n \log \relax (b) + \log \relax (a)}{m n}\right )} \]

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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giac [A]  time = 0.24, size = 21, normalized size = 0.84 \[ \frac {1}{2} \, x^{2} e^{\left (-\frac {n \log \relax (b) + \log \relax (a)}{m n}\right )} \]

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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maple [A]  time = 0.00, size = 25, normalized size = 1.00 \[ \frac {x^{3} \left (a \left (b \,x^{m}\right )^{n}\right )^{-\frac {1}{m n}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.05, size = 38, normalized size = 1.52 \[ \frac {x^{3}}{2 \, a^{\frac {1}{m n}} b^{\left (\frac {1}{m}\right )} {\left ({\left (x^{m}\right )}^{n}\right )}^{\frac {1}{m n}}} \]

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

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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sympy [A]  time = 20.01, size = 248, normalized size = 9.92 \[ \begin {cases} - \frac {x^{3}}{0^{m n} \tilde {\infty }^{m n} \left (0^{m n}\right )^{\frac {1}{m n}} \left (\left (x^{m}\right )^{n}\right )^{\frac {1}{m n}} \left (\left (\left (0^{m n}\right )^{\frac {1}{n}}\right )^{n}\right )^{\frac {1}{m n}} - 3 \left (0^{m n}\right )^{\frac {1}{m n}} \left (\left (x^{m}\right )^{n}\right )^{\frac {1}{m n}} \left (\left (\left (0^{m n}\right )^{\frac {1}{n}}\right )^{n}\right )^{\frac {1}{m n}}} & \text {for}\: a = 0^{m n} \wedge b = \left (0^{m n}\right )^{\frac {1}{n}} \\\frac {a^{- \frac {1}{m n}} x^{3} \left (\left (x^{m}\right )^{n}\right )^{- \frac {1}{m n}} \left (\left (\left (0^{m n}\right )^{\frac {1}{n}}\right )^{n}\right )^{- \frac {1}{m n}}}{2} & \text {for}\: b = \left (0^{m n}\right )^{\frac {1}{n}} \\- \frac {x^{3}}{0^{m n} \tilde {\infty }^{m n} \left (0^{m n}\right )^{\frac {1}{m n}} \left (b^{n}\right )^{\frac {1}{m n}} \left (\left (x^{m}\right )^{n}\right )^{\frac {1}{m n}} - 3 \left (0^{m n}\right )^{\frac {1}{m n}} \left (b^{n}\right )^{\frac {1}{m n}} \left (\left (x^{m}\right )^{n}\right )^{\frac {1}{m n}}} & \text {for}\: a = 0^{m n} \\\frac {a^{- \frac {1}{m n}} x^{3} \left (b^{n}\right )^{- \frac {1}{m n}} \left (\left (x^{m}\right )^{n}\right )^{- \frac {1}{m n}}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x**3/(0**(m*n)*zoo**(m*n)*(0**(m*n))**(1/(m*n))*((x**m)**n)**(1/(m*n))*(((0**(m*n))**(1/n))**n)**(
1/(m*n)) - 3*(0**(m*n))**(1/(m*n))*((x**m)**n)**(1/(m*n))*(((0**(m*n))**(1/n))**n)**(1/(m*n))), Eq(a, 0**(m*n)
) & Eq(b, (0**(m*n))**(1/n))), (a**(-1/(m*n))*x**3*((x**m)**n)**(-1/(m*n))*(((0**(m*n))**(1/n))**n)**(-1/(m*n)
)/2, Eq(b, (0**(m*n))**(1/n))), (-x**3/(0**(m*n)*zoo**(m*n)*(0**(m*n))**(1/(m*n))*(b**n)**(1/(m*n))*((x**m)**n
)**(1/(m*n)) - 3*(0**(m*n))**(1/(m*n))*(b**n)**(1/(m*n))*((x**m)**n)**(1/(m*n))), Eq(a, 0**(m*n))), (a**(-1/(m
*n))*x**3*(b**n)**(-1/(m*n))*((x**m)**n)**(-1/(m*n))/2, True))

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