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

Optimal. Leaf size=36 \[ -\frac{\left (a+b x^3\right )^{5/3} \, _2F_1\left (1,\frac{4}{3};\frac{2}{3};-\frac{b x^3}{a}\right )}{a x} \]

[Out]

-(((a + b*x^3)^(5/3)*Hypergeometric2F1[1, 4/3, 2/3, -((b*x^3)/a)])/(a*x))

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Rubi [A]  time = 0.0161101, 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} \[ -\frac{\left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{2}{3},-\frac{1}{3};\frac{2}{3};-\frac{b x^3}{a}\right )}{x \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(2/3)/x^2,x]

[Out]

-(((a + b*x^3)^(2/3)*Hypergeometric2F1[-2/3, -1/3, 2/3, -((b*x^3)/a)])/(x*(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^2} \, dx &=\frac{\left (a+b x^3\right )^{2/3} \int \frac{\left (1+\frac{b x^3}{a}\right )^{2/3}}{x^2} \, dx}{\left (1+\frac{b x^3}{a}\right )^{2/3}}\\ &=-\frac{\left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{2}{3},-\frac{1}{3};\frac{2}{3};-\frac{b x^3}{a}\right )}{x \left (1+\frac{b x^3}{a}\right )^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.009383, size = 49, normalized size = 1.36 \[ -\frac{\left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{2}{3},-\frac{1}{3};\frac{2}{3};-\frac{b x^3}{a}\right )}{x \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(2/3)/x^2,x]

[Out]

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

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \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^2,x)

[Out]

int((b*x^3+a)^(2/3)/x^2,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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3 + a)^(2/3)/x^2, x)

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Sympy [C]  time = 0.999123, size = 41, normalized size = 1.14 \begin{align*} \frac{a^{\frac{2}{3}} \Gamma \left (- \frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{2}{3}, - \frac{1}{3} \\ \frac{2}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 x \Gamma \left (\frac{2}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(2/3)/x^2, x)