3.571 \(\int \frac{1}{x^5 (a+b x^3)^{2/3}} \, dx\)

Optimal. Leaf size=44 \[ \frac{3 b \sqrt [3]{a+b x^3}}{4 a^2 x}-\frac{\sqrt [3]{a+b x^3}}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.0111215, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {271, 264} \[ \frac{3 b \sqrt [3]{a+b x^3}}{4 a^2 x}-\frac{\sqrt [3]{a+b x^3}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^3)^(2/3)),x]

[Out]

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

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^5 \left (a+b x^3\right )^{2/3}} \, dx &=-\frac{\sqrt [3]{a+b x^3}}{4 a x^4}-\frac{(3 b) \int \frac{1}{x^2 \left (a+b x^3\right )^{2/3}} \, dx}{4 a}\\ &=-\frac{\sqrt [3]{a+b x^3}}{4 a x^4}+\frac{3 b \sqrt [3]{a+b x^3}}{4 a^2 x}\\ \end{align*}

Mathematica [A]  time = 0.0096342, size = 29, normalized size = 0.66 \[ -\frac{\left (a-3 b x^3\right ) \sqrt [3]{a+b x^3}}{4 a^2 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^3)^(2/3)),x]

[Out]

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

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Maple [A]  time = 0.004, size = 26, normalized size = 0.6 \begin{align*} -{\frac{-3\,b{x}^{3}+a}{4\,{a}^{2}{x}^{4}}\sqrt [3]{b{x}^{3}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^3+a)^(2/3),x)

[Out]

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

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Maxima [A]  time = 0.991176, size = 47, normalized size = 1.07 \begin{align*} \frac{\frac{4 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} b}{x} - \frac{{\left (b x^{3} + a\right )}^{\frac{4}{3}}}{x^{4}}}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51496, size = 63, normalized size = 1.43 \begin{align*} \frac{{\left (3 \, b x^{3} - a\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{4 \, a^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.985059, size = 68, normalized size = 1.55 \begin{align*} - \frac{\sqrt [3]{b} \sqrt [3]{\frac{a}{b x^{3}} + 1} \Gamma \left (- \frac{4}{3}\right )}{9 a x^{3} \Gamma \left (\frac{2}{3}\right )} + \frac{b^{\frac{4}{3}} \sqrt [3]{\frac{a}{b x^{3}} + 1} \Gamma \left (- \frac{4}{3}\right )}{3 a^{2} \Gamma \left (\frac{2}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**3+a)**(2/3),x)

[Out]

-b**(1/3)*(a/(b*x**3) + 1)**(1/3)*gamma(-4/3)/(9*a*x**3*gamma(2/3)) + b**(4/3)*(a/(b*x**3) + 1)**(1/3)*gamma(-
4/3)/(3*a**2*gamma(2/3))

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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{2}{3}} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(2/3)*x^5), x)