3.8.68 \(\int (c x)^{2/3} (a+b x^2)^{4/3} \, dx\) [768]

Optimal. Leaf size=59 \[ \frac {3 a (c x)^{5/3} \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {4}{3},\frac {5}{6};\frac {11}{6};-\frac {b x^2}{a}\right )}{5 c \sqrt [3]{1+\frac {b x^2}{a}}} \]

[Out]

3/5*a*(c*x)^(5/3)*(b*x^2+a)^(1/3)*hypergeom([-4/3, 5/6],[11/6],-b*x^2/a)/c/(1+b*x^2/a)^(1/3)

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Rubi [A]
time = 0.01, antiderivative size = 59, 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 a (c x)^{5/3} \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {4}{3},\frac {5}{6};\frac {11}{6};-\frac {b x^2}{a}\right )}{5 c \sqrt [3]{\frac {b x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^(2/3)*(a + b*x^2)^(4/3),x]

[Out]

(3*a*(c*x)^(5/3)*(a + b*x^2)^(1/3)*Hypergeometric2F1[-4/3, 5/6, 11/6, -((b*x^2)/a)])/(5*c*(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 (c x)^{2/3} \left (a+b x^2\right )^{4/3} \, dx &=\frac {\left (a \sqrt [3]{a+b x^2}\right ) \int (c x)^{2/3} \left (1+\frac {b x^2}{a}\right )^{4/3} \, dx}{\sqrt [3]{1+\frac {b x^2}{a}}}\\ &=\frac {3 a (c x)^{5/3} \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {4}{3},\frac {5}{6};\frac {11}{6};-\frac {b x^2}{a}\right )}{5 c \sqrt [3]{1+\frac {b x^2}{a}}}\\ \end {align*}

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Mathematica [A]
time = 10.01, size = 57, normalized size = 0.97 \begin {gather*} \frac {3 a x (c x)^{2/3} \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {4}{3},\frac {5}{6};\frac {11}{6};-\frac {b x^2}{a}\right )}{5 \sqrt [3]{1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x*(c*x)^(2/3)*(a + b*x^2)^(1/3)*Hypergeometric2F1[-4/3, 5/6, 11/6, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^(1/3
))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 5.09, size = 46, normalized size = 0.78 \begin {gather*} \frac {a^{\frac {4}{3}} c^{\frac {2}{3}} x^{\frac {5}{3}} \Gamma \left (\frac {5}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {5}{6} \\ \frac {11}{6} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {11}{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(4/3)*c**(2/3)*x**(5/3)*gamma(5/6)*hyper((-4/3, 5/6), (11/6,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(11/6))

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(2/3)*(a + b*x^2)^(4/3),x)

[Out]

int((c*x)^(2/3)*(a + b*x^2)^(4/3), x)

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