3.8.53 \(\int \frac {\sqrt [3]{a+b x^2}}{(c x)^{4/3}} \, dx\) [753]

Optimal. Leaf size=56 \[ -\frac {3 \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {1}{3},-\frac {1}{6};\frac {5}{6};-\frac {b x^2}{a}\right )}{c \sqrt [3]{c x} \sqrt [3]{1+\frac {b x^2}{a}}} \]

[Out]

-3*(b*x^2+a)^(1/3)*hypergeom([-1/3, -1/6],[5/6],-b*x^2/a)/c/(c*x)^(1/3)/(1+b*x^2/a)^(1/3)

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Rubi [A]
time = 0.01, antiderivative size = 56, normalized size of antiderivative = 1.00, 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 = {372, 371} \begin {gather*} -\frac {3 \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {1}{3},-\frac {1}{6};\frac {5}{6};-\frac {b x^2}{a}\right )}{c \sqrt [3]{c x} \sqrt [3]{\frac {b x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/3)/(c*x)^(4/3),x]

[Out]

(-3*(a + b*x^2)^(1/3)*Hypergeometric2F1[-1/3, -1/6, 5/6, -((b*x^2)/a)])/(c*(c*x)^(1/3)*(1 + (b*x^2)/a)^(1/3))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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])

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{4/3}} \, dx &=\frac {\sqrt [3]{a+b x^2} \int \frac {\sqrt [3]{1+\frac {b x^2}{a}}}{(c x)^{4/3}} \, dx}{\sqrt [3]{1+\frac {b x^2}{a}}}\\ &=-\frac {3 \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {1}{3},-\frac {1}{6};\frac {5}{6};-\frac {b x^2}{a}\right )}{c \sqrt [3]{c x} \sqrt [3]{1+\frac {b x^2}{a}}}\\ \end {align*}

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Mathematica [A]
time = 10.02, size = 54, normalized size = 0.96 \begin {gather*} -\frac {3 x \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {1}{3},-\frac {1}{6};\frac {5}{6};-\frac {b x^2}{a}\right )}{(c x)^{4/3} \sqrt [3]{1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-1/3, -1/6, 5/6, -((b*x^2)/a)])/((c*x)^(4/3)*(1 + (b*x^2)/a)^(1/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{\left (c x \right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/3)/(c*x)^(4/3),x)

[Out]

int((b*x^2+a)^(1/3)/(c*x)^(4/3),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(4/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(4/3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(4/3),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.02, size = 49, normalized size = 0.88 \begin {gather*} \frac {\sqrt [3]{a} \Gamma \left (- \frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{6} \\ \frac {5}{6} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {4}{3}} \sqrt [3]{x} \Gamma \left (\frac {5}{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(1/3)/(c*x)**(4/3),x)

[Out]

a**(1/3)*gamma(-1/6)*hyper((-1/3, -1/6), (5/6,), b*x**2*exp_polar(I*pi)/a)/(2*c**(4/3)*x**(1/3)*gamma(5/6))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(4/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(4/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{1/3}}{{\left (c\,x\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/3)/(c*x)^(4/3),x)

[Out]

int((a + b*x^2)^(1/3)/(c*x)^(4/3), x)

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