3.759 \(\int \frac{(a+b x^2)^{4/3}}{(c x)^{17/3}} \, dx\)

Optimal. Leaf size=28 \[ -\frac{3 \left (a+b x^2\right )^{7/3}}{14 a c (c x)^{14/3}} \]

[Out]

(-3*(a + b*x^2)^(7/3))/(14*a*c*(c*x)^(14/3))

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Rubi [A]  time = 0.0060914, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {264} \[ -\frac{3 \left (a+b x^2\right )^{7/3}}{14 a c (c x)^{14/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/(c*x)^(17/3),x]

[Out]

(-3*(a + b*x^2)^(7/3))/(14*a*c*(c*x)^(14/3))

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{4/3}}{(c x)^{17/3}} \, dx &=-\frac{3 \left (a+b x^2\right )^{7/3}}{14 a c (c x)^{14/3}}\\ \end{align*}

Mathematica [A]  time = 0.0101296, size = 26, normalized size = 0.93 \[ -\frac{3 x \left (a+b x^2\right )^{7/3}}{14 a (c x)^{17/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/(c*x)^(17/3),x]

[Out]

(-3*x*(a + b*x^2)^(7/3))/(14*a*(c*x)^(17/3))

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Maple [A]  time = 0.004, size = 21, normalized size = 0.8 \begin{align*} -{\frac{3\,x}{14\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{3}}} \left ( cx \right ) ^{-{\frac{17}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(4/3)/(c*x)^(17/3),x)

[Out]

-3/14*x*(b*x^2+a)^(7/3)/a/(c*x)^(17/3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{4}{3}}}{\left (c x\right )^{\frac{17}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(17/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(17/3), x)

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Fricas [A]  time = 2.1771, size = 104, normalized size = 3.71 \begin{align*} -\frac{3 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{3}} \left (c x\right )^{\frac{1}{3}}}{14 \, a c^{6} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(17/3),x, algorithm="fricas")

[Out]

-3/14*(b^2*x^4 + 2*a*b*x^2 + a^2)*(b*x^2 + a)^(1/3)*(c*x)^(1/3)/(a*c^6*x^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(4/3)/(c*x)**(17/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{4}{3}}}{\left (c x\right )^{\frac{17}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(17/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(17/3), x)