3.745 \(\int \frac{\sqrt [3]{a+b x^2}}{(c x)^{29/3}} \, dx\)

Optimal. Leaf size=113 \[ \frac{243 \left (a+b x^2\right )^{13/3}}{3640 a^4 c (c x)^{26/3}}-\frac{81 \left (a+b x^2\right )^{10/3}}{280 a^3 c (c x)^{26/3}}+\frac{27 \left (a+b x^2\right )^{7/3}}{56 a^2 c (c x)^{26/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{26/3}} \]

[Out]

(-3*(a + b*x^2)^(4/3))/(8*a*c*(c*x)^(26/3)) + (27*(a + b*x^2)^(7/3))/(56*a^2*c*(c*x)^(26/3)) - (81*(a + b*x^2)
^(10/3))/(280*a^3*c*(c*x)^(26/3)) + (243*(a + b*x^2)^(13/3))/(3640*a^4*c*(c*x)^(26/3))

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Rubi [A]  time = 0.0394777, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {273, 264} \[ \frac{243 \left (a+b x^2\right )^{13/3}}{3640 a^4 c (c x)^{26/3}}-\frac{81 \left (a+b x^2\right )^{10/3}}{280 a^3 c (c x)^{26/3}}+\frac{27 \left (a+b x^2\right )^{7/3}}{56 a^2 c (c x)^{26/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{26/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/3)/(c*x)^(29/3),x]

[Out]

(-3*(a + b*x^2)^(4/3))/(8*a*c*(c*x)^(26/3)) + (27*(a + b*x^2)^(7/3))/(56*a^2*c*(c*x)^(26/3)) - (81*(a + b*x^2)
^(10/3))/(280*a^3*c*(c*x)^(26/3)) + (243*(a + b*x^2)^(13/3))/(3640*a^4*c*(c*x)^(26/3))

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{\sqrt [3]{a+b x^2}}{(c x)^{29/3}} \, dx &=-\frac{3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{26/3}}-\frac{9 \int \frac{\left (a+b x^2\right )^{4/3}}{(c x)^{29/3}} \, dx}{4 a}\\ &=-\frac{3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{26/3}}+\frac{27 \left (a+b x^2\right )^{7/3}}{56 a^2 c (c x)^{26/3}}+\frac{27 \int \frac{\left (a+b x^2\right )^{7/3}}{(c x)^{29/3}} \, dx}{14 a^2}\\ &=-\frac{3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{26/3}}+\frac{27 \left (a+b x^2\right )^{7/3}}{56 a^2 c (c x)^{26/3}}-\frac{81 \left (a+b x^2\right )^{10/3}}{280 a^3 c (c x)^{26/3}}-\frac{81 \int \frac{\left (a+b x^2\right )^{10/3}}{(c x)^{29/3}} \, dx}{140 a^3}\\ &=-\frac{3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{26/3}}+\frac{27 \left (a+b x^2\right )^{7/3}}{56 a^2 c (c x)^{26/3}}-\frac{81 \left (a+b x^2\right )^{10/3}}{280 a^3 c (c x)^{26/3}}+\frac{243 \left (a+b x^2\right )^{13/3}}{3640 a^4 c (c x)^{26/3}}\\ \end{align*}

Mathematica [A]  time = 0.0170239, size = 63, normalized size = 0.56 \[ \frac{3 \left (a+b x^2\right )^{4/3} \left (126 a^2 b x^2-140 a^3-108 a b^2 x^4+81 b^3 x^6\right )}{3640 a^4 c^9 x^8 (c x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/3)/(c*x)^(29/3),x]

[Out]

(3*(a + b*x^2)^(4/3)*(-140*a^3 + 126*a^2*b*x^2 - 108*a*b^2*x^4 + 81*b^3*x^6))/(3640*a^4*c^9*x^8*(c*x)^(2/3))

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Maple [A]  time = 0.006, size = 53, normalized size = 0.5 \begin{align*} -{\frac{3\,x \left ( -81\,{b}^{3}{x}^{6}+108\,a{b}^{2}{x}^{4}-126\,{a}^{2}b{x}^{2}+140\,{a}^{3} \right ) }{3640\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{4}{3}}} \left ( cx \right ) ^{-{\frac{29}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/3)/(c*x)^(29/3),x)

[Out]

-3/3640*x*(b*x^2+a)^(4/3)*(-81*b^3*x^6+108*a*b^2*x^4-126*a^2*b*x^2+140*a^3)/a^4/(c*x)^(29/3)

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Maxima [A]  time = 1.38082, size = 86, normalized size = 0.76 \begin{align*} \frac{3 \,{\left (81 \, b^{4} x^{9} - 27 \, a b^{3} x^{7} + 18 \, a^{2} b^{2} x^{5} - 14 \, a^{3} b x^{3} - 140 \, a^{4} x\right )}{\left (b x^{2} + a\right )}^{\frac{1}{3}}}{3640 \, a^{4} c^{\frac{29}{3}} x^{\frac{29}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(29/3),x, algorithm="maxima")

[Out]

3/3640*(81*b^4*x^9 - 27*a*b^3*x^7 + 18*a^2*b^2*x^5 - 14*a^3*b*x^3 - 140*a^4*x)*(b*x^2 + a)^(1/3)/(a^4*c^(29/3)
*x^(29/3))

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Fricas [A]  time = 1.69862, size = 166, normalized size = 1.47 \begin{align*} \frac{3 \,{\left (81 \, b^{4} x^{8} - 27 \, a b^{3} x^{6} + 18 \, a^{2} b^{2} x^{4} - 14 \, a^{3} b x^{2} - 140 \, a^{4}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{3}} \left (c x\right )^{\frac{1}{3}}}{3640 \, a^{4} c^{10} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(29/3),x, algorithm="fricas")

[Out]

3/3640*(81*b^4*x^8 - 27*a*b^3*x^6 + 18*a^2*b^2*x^4 - 14*a^3*b*x^2 - 140*a^4)*(b*x^2 + a)^(1/3)*(c*x)^(1/3)/(a^
4*c^10*x^9)

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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)**(1/3)/(c*x)**(29/3),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{1}{3}}}{\left (c x\right )^{\frac{29}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(29/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(29/3), x)