3.554 \(\int \frac{1}{x^5 \sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=38 \[ -\frac{\left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{2}{3},1;-\frac{1}{3};-\frac{b x^3}{a}\right )}{4 a x^4} \]

[Out]

-((a + b*x^3)^(2/3)*Hypergeometric2F1[-2/3, 1, -1/3, -((b*x^3)/a)])/(4*a*x^4)

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Rubi [A]  time = 0.0150739, antiderivative size = 51, normalized size of antiderivative = 1.34, 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 = {365, 364} \[ -\frac{\sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (-\frac{4}{3},\frac{1}{3};-\frac{1}{3};-\frac{b x^3}{a}\right )}{4 x^4 \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^3)^(1/3)),x]

[Out]

-((1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[-4/3, 1/3, -1/3, -((b*x^3)/a)])/(4*x^4*(a + b*x^3)^(1/3))

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^5 \sqrt [3]{a+b x^3}} \, dx &=\frac{\sqrt [3]{1+\frac{b x^3}{a}} \int \frac{1}{x^5 \sqrt [3]{1+\frac{b x^3}{a}}} \, dx}{\sqrt [3]{a+b x^3}}\\ &=-\frac{\sqrt [3]{1+\frac{b x^3}{a}} \, _2F_1\left (-\frac{4}{3},\frac{1}{3};-\frac{1}{3};-\frac{b x^3}{a}\right )}{4 x^4 \sqrt [3]{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0090192, size = 51, normalized size = 1.34 \[ -\frac{\sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (-\frac{4}{3},\frac{1}{3};-\frac{1}{3};-\frac{b x^3}{a}\right )}{4 x^4 \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[-4/3, 1/3, -1/3, -((b*x^3)/a)])/(4*x^4*(a + b*x^3)^(1/3))

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Maple [F]  time = 0.031, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5}}{\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{\frac{1}{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)^(1/3),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{b x^{8} + a x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3 + a)^(2/3)/(b*x^8 + a*x^5), x)

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Sympy [C]  time = 1.27512, size = 44, normalized size = 1.16 \begin{align*} \frac{\Gamma \left (- \frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{4}{3}, \frac{1}{3} \\ - \frac{1}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} x^{4} \Gamma \left (- \frac{1}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-4/3)*hyper((-4/3, 1/3), (-1/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*x**4*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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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