3.2282 \(\int \frac{x}{(a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=42 \[ \frac{x^2 \sqrt [3]{a+b x^{3/2}} \, _2F_1\left (1,\frac{5}{3};\frac{7}{3};-\frac{b x^{3/2}}{a}\right )}{2 a} \]

[Out]

(x^2*(a + b*x^(3/2))^(1/3)*Hypergeometric2F1[1, 5/3, 7/3, -((b*x^(3/2))/a)])/(2*a)

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Rubi [A]  time = 0.0299314, antiderivative size = 57, normalized size of antiderivative = 1.36, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {341, 365, 364} \[ \frac{x^2 \left (\frac{b x^{3/2}}{a}+1\right )^{2/3} \, _2F_1\left (\frac{2}{3},\frac{4}{3};\frac{7}{3};-\frac{b x^{3/2}}{a}\right )}{2 \left (a+b x^{3/2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x^(3/2))^(2/3),x]

[Out]

(x^2*(1 + (b*x^(3/2))/a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^(3/2))/a)])/(2*(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{x}{\left (a+b x^{3/2}\right )^{2/3}} \, dx &=2 \operatorname{Subst}\left (\int \frac{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{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{x^2 \left (1+\frac{b x^{3/2}}{a}\right )^{2/3} \, _2F_1\left (\frac{2}{3},\frac{4}{3};\frac{7}{3};-\frac{b x^{3/2}}{a}\right )}{2 \left (a+b x^{3/2}\right )^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.010949, size = 57, normalized size = 1.36 \[ \frac{x^2 \left (\frac{b x^{3/2}}{a}+1\right )^{2/3} \, _2F_1\left (\frac{2}{3},\frac{4}{3};\frac{7}{3};-\frac{b x^{3/2}}{a}\right )}{2 \left (a+b x^{3/2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*x^(3/2))^(2/3),x]

[Out]

(x^2*(1 + (b*x^(3/2))/a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^(3/2))/a)])/(2*(a + b*x^(3/2))^(2/3))

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Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int{x \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(x/(a+b*x^(3/2))^(2/3),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*x^(3/2) + a)^(2/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{\frac{5}{2}} - a x\right )}{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}}}{b^{2} x^{3} - a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.754297, size = 41, normalized size = 0.98 \begin{align*} \frac{2 x^{2} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{2}{3}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{b x^{\frac{3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac{2}{3}} \Gamma \left (\frac{7}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**2*gamma(4/3)*hyper((2/3, 4/3), (7/3,), b*x**(3/2)*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(7/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*x^(3/2) + a)^(2/3), x)