3.560 \(\int \frac{1}{x^{12} \sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=92 \[ \frac{81 b^3 \left (a+b x^3\right )^{2/3}}{440 a^4 x^2}-\frac{27 b^2 \left (a+b x^3\right )^{2/3}}{220 a^3 x^5}+\frac{9 b \left (a+b x^3\right )^{2/3}}{88 a^2 x^8}-\frac{\left (a+b x^3\right )^{2/3}}{11 a x^{11}} \]

[Out]

-(a + b*x^3)^(2/3)/(11*a*x^11) + (9*b*(a + b*x^3)^(2/3))/(88*a^2*x^8) - (27*b^2*(a + b*x^3)^(2/3))/(220*a^3*x^
5) + (81*b^3*(a + b*x^3)^(2/3))/(440*a^4*x^2)

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Rubi [A]  time = 0.0289558, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {271, 264} \[ \frac{81 b^3 \left (a+b x^3\right )^{2/3}}{440 a^4 x^2}-\frac{27 b^2 \left (a+b x^3\right )^{2/3}}{220 a^3 x^5}+\frac{9 b \left (a+b x^3\right )^{2/3}}{88 a^2 x^8}-\frac{\left (a+b x^3\right )^{2/3}}{11 a x^{11}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^12*(a + b*x^3)^(1/3)),x]

[Out]

-(a + b*x^3)^(2/3)/(11*a*x^11) + (9*b*(a + b*x^3)^(2/3))/(88*a^2*x^8) - (27*b^2*(a + b*x^3)^(2/3))/(220*a^3*x^
5) + (81*b^3*(a + b*x^3)^(2/3))/(440*a^4*x^2)

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -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}{x^{12} \sqrt [3]{a+b x^3}} \, dx &=-\frac{\left (a+b x^3\right )^{2/3}}{11 a x^{11}}-\frac{(9 b) \int \frac{1}{x^9 \sqrt [3]{a+b x^3}} \, dx}{11 a}\\ &=-\frac{\left (a+b x^3\right )^{2/3}}{11 a x^{11}}+\frac{9 b \left (a+b x^3\right )^{2/3}}{88 a^2 x^8}+\frac{\left (27 b^2\right ) \int \frac{1}{x^6 \sqrt [3]{a+b x^3}} \, dx}{44 a^2}\\ &=-\frac{\left (a+b x^3\right )^{2/3}}{11 a x^{11}}+\frac{9 b \left (a+b x^3\right )^{2/3}}{88 a^2 x^8}-\frac{27 b^2 \left (a+b x^3\right )^{2/3}}{220 a^3 x^5}-\frac{\left (81 b^3\right ) \int \frac{1}{x^3 \sqrt [3]{a+b x^3}} \, dx}{220 a^3}\\ &=-\frac{\left (a+b x^3\right )^{2/3}}{11 a x^{11}}+\frac{9 b \left (a+b x^3\right )^{2/3}}{88 a^2 x^8}-\frac{27 b^2 \left (a+b x^3\right )^{2/3}}{220 a^3 x^5}+\frac{81 b^3 \left (a+b x^3\right )^{2/3}}{440 a^4 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0215756, size = 53, normalized size = 0.58 \[ \frac{\left (a+b x^3\right )^{2/3} \left (45 a^2 b x^3-40 a^3-54 a b^2 x^6+81 b^3 x^9\right )}{440 a^4 x^{11}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^12*(a + b*x^3)^(1/3)),x]

[Out]

((a + b*x^3)^(2/3)*(-40*a^3 + 45*a^2*b*x^3 - 54*a*b^2*x^6 + 81*b^3*x^9))/(440*a^4*x^11)

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Maple [A]  time = 0.006, size = 50, normalized size = 0.5 \begin{align*} -{\frac{-81\,{b}^{3}{x}^{9}+54\,a{b}^{2}{x}^{6}-45\,{a}^{2}b{x}^{3}+40\,{a}^{3}}{440\,{x}^{11}{a}^{4}} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^12/(b*x^3+a)^(1/3),x)

[Out]

-1/440*(b*x^3+a)^(2/3)*(-81*b^3*x^9+54*a*b^2*x^6-45*a^2*b*x^3+40*a^3)/x^11/a^4

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Maxima [A]  time = 1.06923, size = 93, normalized size = 1.01 \begin{align*} \frac{\frac{220 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} b^{3}}{x^{2}} - \frac{264 \,{\left (b x^{3} + a\right )}^{\frac{5}{3}} b^{2}}{x^{5}} + \frac{165 \,{\left (b x^{3} + a\right )}^{\frac{8}{3}} b}{x^{8}} - \frac{40 \,{\left (b x^{3} + a\right )}^{\frac{11}{3}}}{x^{11}}}{440 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^12/(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

1/440*(220*(b*x^3 + a)^(2/3)*b^3/x^2 - 264*(b*x^3 + a)^(5/3)*b^2/x^5 + 165*(b*x^3 + a)^(8/3)*b/x^8 - 40*(b*x^3
+ a)^(11/3)/x^11)/a^4

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Fricas [A]  time = 1.46566, size = 119, normalized size = 1.29 \begin{align*} \frac{{\left (81 \, b^{3} x^{9} - 54 \, a b^{2} x^{6} + 45 \, a^{2} b x^{3} - 40 \, a^{3}\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{440 \, a^{4} x^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^12/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

1/440*(81*b^3*x^9 - 54*a*b^2*x^6 + 45*a^2*b*x^3 - 40*a^3)*(b*x^3 + a)^(2/3)/(a^4*x^11)

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Sympy [B]  time = 3.32519, size = 692, normalized size = 7.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**12/(b*x**3+a)**(1/3),x)

[Out]

-80*a**6*b**(29/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**10*x**12*g
amma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1/3)) - 150*a**5*b**(32/3)*x**3*(a/(b*
x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a**5*b**1
1*x**15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1/3)) - 78*a**4*b**(35/3)*x**6*(a/(b*x**3) + 1)**(2/3)*gamma(-1
1/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 81*a*
*4*b**12*x**18*gamma(1/3)) + 28*a**3*b**(38/3)*x**9*(a/(b*x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*ga
mma(1/3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1/3))
+ 252*a**2*b**(41/3)*x**12*(a/(b*x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**1
0*x**12*gamma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1/3)) + 378*a*b**(44/3)*x**15
*(a/(b*x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a*
*5*b**11*x**15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1/3)) + 162*b**(47/3)*x**18*(a/(b*x**3) + 1)**(2/3)*gamm
a(-11/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 8
1*a**4*b**12*x**18*gamma(1/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{\frac{1}{3}} x^{12}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^12/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(1/3)*x^12), x)