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]

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

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Rubi [A]  time = 0.04, antiderivative size = 113, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.56 \[ \frac {3 \left (a+b x^2\right )^{4/3} \left (-140 a^3+126 a^2 b x^2-108 a b^2 x^4+81 b^3 x^6\right )}{3640 a^4 c^9 x^8 (c x)^{2/3}} \]

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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fricas [A]  time = 1.21, size = 68, normalized size = 0.60 \[ \frac {3 \, {\left (81 \, b^{4} x^{8} - 27 \, a b^{3} x^{6} + 18 \, a^{2} b^{2} x^{4} - 14 \, a^{3} b x^{2} - 140 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}}}{3640 \, a^{4} c^{10} x^{9}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}}}{\left (c x\right )^{\frac {29}{3}}}\,{d x} \]

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)

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maple [A]  time = 0.01, size = 53, normalized size = 0.47 \[ -\frac {3 \left (b \,x^{2}+a \right )^{\frac {4}{3}} \left (-81 b^{3} x^{6}+108 a \,b^{2} x^{4}-126 a^{2} b \,x^{2}+140 a^{3}\right ) x}{3640 \left (c x \right )^{\frac {29}{3}} a^{4}} \]

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 = 1.47, size = 64, normalized size = 0.57 \[ \frac {3 \, {\left (81 \, b^{4} x^{9} - 27 \, a b^{3} x^{7} + 18 \, a^{2} b^{2} x^{5} - 14 \, a^{3} b x^{3} - 140 \, a^{4} x\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{3640 \, a^{4} c^{\frac {29}{3}} x^{\frac {29}{3}}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^{1/3}}{{\left (c\,x\right )}^{29/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/3)/(c*x)^(29/3),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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