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

Optimal. Leaf size=38 \[ -\frac{\left (a+b x^3\right )^{5/3} \, _2F_1\left (\frac{1}{3},1;-\frac{1}{3};-\frac{b x^3}{a}\right )}{4 a x^4} \]

[Out]

-((a + b*x^3)^(5/3)*Hypergeometric2F1[1/3, 1, -1/3, -((b*x^3)/a)])/(4*a*x^4)

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Rubi [A]  time = 0.0159126, 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} \[ -\frac{\left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{4}{3},-\frac{2}{3};-\frac{1}{3};-\frac{b x^3}{a}\right )}{4 x^4 \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(2/3)/x^5,x]

[Out]

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

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{\left (a+b x^3\right )^{2/3}}{x^5} \, dx &=\frac{\left (a+b x^3\right )^{2/3} \int \frac{\left (1+\frac{b x^3}{a}\right )^{2/3}}{x^5} \, dx}{\left (1+\frac{b x^3}{a}\right )^{2/3}}\\ &=-\frac{\left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{4}{3},-\frac{2}{3};-\frac{1}{3};-\frac{b x^3}{a}\right )}{4 x^4 \left (1+\frac{b x^3}{a}\right )^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0090865, size = 51, normalized size = 1.34 \[ -\frac{\left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{4}{3},-\frac{2}{3};-\frac{1}{3};-\frac{b x^3}{a}\right )}{4 x^4 \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(2/3)/x^5,x]

[Out]

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

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5}} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((b*x^3+a)^(2/3)/x^5,x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3 + a)^(2/3)/x^5, x)

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Sympy [C]  time = 1.32472, size = 46, normalized size = 1.21 \begin{align*} \frac{a^{\frac{2}{3}} \Gamma \left (- \frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{4}{3}, - \frac{2}{3} \\ - \frac{1}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{4} \Gamma \left (- \frac{1}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(2/3)*gamma(-4/3)*hyper((-4/3, -2/3), (-1/3,), b*x**3*exp_polar(I*pi)/a)/(3*x**4*gamma(-1/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((b*x^3+a)^(2/3)/x^5,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(2/3)/x^5, x)