3.776 \(\int \frac{1}{(c x)^{11/3} (a+b x^2)^{2/3}} \, dx\)

Optimal. Leaf size=57 \[ \frac{9 \left (a+b x^2\right )^{4/3}}{8 a^2 c (c x)^{8/3}}-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{8/3}} \]

[Out]

(-3*(a + b*x^2)^(1/3))/(2*a*c*(c*x)^(8/3)) + (9*(a + b*x^2)^(4/3))/(8*a^2*c*(c*x)^(8/3))

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Rubi [A]  time = 0.0153678, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {273, 264} \[ \frac{9 \left (a+b x^2\right )^{4/3}}{8 a^2 c (c x)^{8/3}}-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{8/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(11/3)*(a + b*x^2)^(2/3)),x]

[Out]

(-3*(a + b*x^2)^(1/3))/(2*a*c*(c*x)^(8/3)) + (9*(a + b*x^2)^(4/3))/(8*a^2*c*(c*x)^(8/3))

Rule 273

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

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{1}{(c x)^{11/3} \left (a+b x^2\right )^{2/3}} \, dx &=-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{8/3}}-\frac{3 \int \frac{\sqrt [3]{a+b x^2}}{(c x)^{11/3}} \, dx}{a}\\ &=-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{8/3}}+\frac{9 \left (a+b x^2\right )^{4/3}}{8 a^2 c (c x)^{8/3}}\\ \end{align*}

Mathematica [A]  time = 0.0180092, size = 34, normalized size = 0.6 \[ -\frac{3 x \left (a-3 b x^2\right ) \sqrt [3]{a+b x^2}}{8 a^2 (c x)^{11/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(11/3)*(a + b*x^2)^(2/3)),x]

[Out]

(-3*x*(a - 3*b*x^2)*(a + b*x^2)^(1/3))/(8*a^2*(c*x)^(11/3))

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Maple [A]  time = 0.003, size = 29, normalized size = 0.5 \begin{align*} -{\frac{3\,x \left ( -3\,b{x}^{2}+a \right ) }{8\,{a}^{2}}\sqrt [3]{b{x}^{2}+a} \left ( cx \right ) ^{-{\frac{11}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(11/3)/(b*x^2+a)^(2/3),x)

[Out]

-3/8*x*(b*x^2+a)^(1/3)*(-3*b*x^2+a)/a^2/(c*x)^(11/3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{2}{3}} \left (c x\right )^{\frac{11}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/3)/(b*x^2+a)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(2/3)*(c*x)^(11/3)), x)

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Fricas [A]  time = 2.04231, size = 85, normalized size = 1.49 \begin{align*} \frac{3 \,{\left (3 \, b x^{2} - a\right )}{\left (b x^{2} + a\right )}^{\frac{1}{3}} \left (c x\right )^{\frac{1}{3}}}{8 \, a^{2} c^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/3)/(b*x^2+a)^(2/3),x, algorithm="fricas")

[Out]

3/8*(3*b*x^2 - a)*(b*x^2 + a)^(1/3)*(c*x)^(1/3)/(a^2*c^4*x^3)

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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(1/(c*x)**(11/3)/(b*x**2+a)**(2/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{2}{3}} \left (c x\right )^{\frac{11}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/3)/(b*x^2+a)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(2/3)*(c*x)^(11/3)), x)