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

Optimal. Leaf size=58 \[ \frac{3 (c x)^{2/3} \left (\frac{b x^2}{a}+1\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^2}{a}\right )}{2 c \left (a+b x^2\right )^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.0189458, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {365, 364} \[ \frac{3 (c x)^{2/3} \left (\frac{b x^2}{a}+1\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^2}{a}\right )}{2 c \left (a+b x^2\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0104509, size = 56, normalized size = 0.97 \[ \frac{3 x \left (\frac{b x^2}{a}+1\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^2}{a}\right )}{2 \sqrt [3]{c x} \left (a+b x^2\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt [3]{cx}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.51933, size = 46, normalized size = 0.79 \begin{align*} \frac{\Gamma \left (- \frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{2}{3} \\ \frac{4}{3} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac{2}{3}} \sqrt [3]{c} x^{\frac{2}{3}} \Gamma \left (\frac{2}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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