3.553 \(\int \frac{1}{x^2 \sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=36 \[ -\frac{\left (a+b x^3\right )^{2/3} \, _2F_1\left (\frac{1}{3},1;\frac{2}{3};-\frac{b x^3}{a}\right )}{a x} \]

[Out]

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

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Rubi [A]  time = 0.0158883, antiderivative size = 49, normalized size of antiderivative = 1.36, 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} \[ -\frac{\sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{2}{3};-\frac{b x^3}{a}\right )}{x \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^3)^(1/3)),x]

[Out]

-(((1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[-1/3, 1/3, 2/3, -((b*x^3)/a)])/(x*(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^2 \sqrt [3]{a+b x^3}} \, dx &=\frac{\sqrt [3]{1+\frac{b x^3}{a}} \int \frac{1}{x^2 \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{1}{3},\frac{1}{3};\frac{2}{3};-\frac{b x^3}{a}\right )}{x \sqrt [3]{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0085581, size = 49, normalized size = 1.36 \[ -\frac{\sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{2}{3};-\frac{b x^3}{a}\right )}{x \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}{\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^2/(b*x^3+a)^(1/3),x)

[Out]

int(1/x^2/(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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^3 + a)^(1/3)*x^2), 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^{5} + a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.713758, size = 39, normalized size = 1.08 \begin{align*} \frac{\Gamma \left (- \frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{3}, \frac{1}{3} \\ \frac{2}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} x \Gamma \left (\frac{2}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-1/3)*hyper((-1/3, 1/3), (2/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*x*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{1}{3}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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