3.533 \(\int x^4 (a+b x^3)^{2/3} \, dx\)

Optimal. Leaf size=38 \[ \frac{x^5 \left (a+b x^3\right )^{5/3} \, _2F_1\left (1,\frac{10}{3};\frac{8}{3};-\frac{b x^3}{a}\right )}{5 a} \]

[Out]

(x^5*(a + b*x^3)^(5/3)*Hypergeometric2F1[1, 10/3, 8/3, -((b*x^3)/a)])/(5*a)

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Rubi [A]  time = 0.0153125, antiderivative size = 51, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {365, 364} \[ \frac{x^5 \left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{2}{3},\frac{5}{3};\frac{8}{3};-\frac{b x^3}{a}\right )}{5 \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^3)^(2/3),x]

[Out]

(x^5*(a + b*x^3)^(2/3)*Hypergeometric2F1[-2/3, 5/3, 8/3, -((b*x^3)/a)])/(5*(1 + (b*x^3)/a)^(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 x^4 \left (a+b x^3\right )^{2/3} \, dx &=\frac{\left (a+b x^3\right )^{2/3} \int x^4 \left (1+\frac{b x^3}{a}\right )^{2/3} \, dx}{\left (1+\frac{b x^3}{a}\right )^{2/3}}\\ &=\frac{x^5 \left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{2}{3},\frac{5}{3};\frac{8}{3};-\frac{b x^3}{a}\right )}{5 \left (1+\frac{b x^3}{a}\right )^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0123162, size = 51, normalized size = 1.34 \[ \frac{x^5 \left (a+b x^3\right )^{2/3} \, _2F_1\left (-\frac{2}{3},\frac{5}{3};\frac{8}{3};-\frac{b x^3}{a}\right )}{5 \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^3)^(2/3),x]

[Out]

(x^5*(a + b*x^3)^(2/3)*Hypergeometric2F1[-2/3, 5/3, 8/3, -((b*x^3)/a)])/(5*(1 + (b*x^3)/a)^(2/3))

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{x}^{4} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{2}{3}} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(2/3)*x^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} x^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3 + a)^(2/3)*x^4, x)

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Sympy [C]  time = 1.29292, size = 39, normalized size = 1.03 \begin{align*} \frac{a^{\frac{2}{3}} x^{5} \Gamma \left (\frac{5}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{2}{3}, \frac{5}{3} \\ \frac{8}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{8}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(2/3)*x**5*gamma(5/3)*hyper((-2/3, 5/3), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(8/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{2}{3}} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(2/3)*x^4, x)