3.2283 \(\int \frac{1}{x^2 (a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=40 \[ -\frac{\sqrt [3]{a+b x^{3/2}} \, _2F_1\left (-\frac{1}{3},1;\frac{1}{3};-\frac{b x^{3/2}}{a}\right )}{a x} \]

[Out]

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

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Rubi [A]  time = 0.0330628, antiderivative size = 55, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {341, 365, 364} \[ -\frac{\left (\frac{b x^{3/2}}{a}+1\right )^{2/3} \, _2F_1\left (-\frac{2}{3},\frac{2}{3};\frac{1}{3};-\frac{b x^{3/2}}{a}\right )}{x \left (a+b x^{3/2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^(3/2))^(2/3)),x]

[Out]

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

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

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

Mathematica [A]  time = 0.0109728, size = 55, normalized size = 1.38 \[ -\frac{\left (\frac{b x^{3/2}}{a}+1\right )^{2/3} \, _2F_1\left (-\frac{2}{3},\frac{2}{3};\frac{1}{3};-\frac{b x^{3/2}}{a}\right )}{x \left (a+b x^{3/2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \left ( a+b{x}^{{\frac{3}{2}}} \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 3.4647, size = 42, normalized size = 1.05 \begin{align*} \frac{2 \Gamma \left (- \frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{2}{3}, \frac{2}{3} \\ \frac{1}{3} \end{matrix}\middle |{\frac{b x^{\frac{3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac{2}{3}} x \Gamma \left (\frac{1}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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