\(\int x^4 (a+b x^3)^{2/3} \, dx\) [533]

Optimal result

Integrand size = 15, antiderivative size = 38 \[ \int x^4 \left (a+b x^3\right )^{2/3} \, dx=\frac {x^5 \left (a+b x^3\right )^{5/3} \operatorname {Hypergeometric2F1}\left (1,\frac {10}{3},\frac {8}{3},-\frac {b x^3}{a}\right )}{5 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {372, 371} \[ \int x^4 \left (a+b x^3\right )^{2/3} \, dx=\frac {x^5 \left (a+b x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )}{5 \left (\frac {b x^3}{a}+1\right )^{2/3}} \]

[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 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*} \text {integral}& = \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] (verified)

Time = 5.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.34 \[ \int x^4 \left (a+b x^3\right )^{2/3} \, dx=\frac {x^5 \left (a+b x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )}{5 \left (1+\frac {b x^3}{a}\right )^{2/3}} \]

[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))

Maple [F]

\[\int x^{4} \left (b \,x^{3}+a \right )^{\frac {2}{3}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x^4 \left (a+b x^3\right )^{2/3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{4} \,d x } \]

[In]

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

[Out]

integral((b*x^3 + a)^(2/3)*x^4, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int x^4 \left (a+b x^3\right )^{2/3} \, dx=\frac {a^{\frac {2}{3}} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} \]

[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))

Maxima [F]

\[ \int x^4 \left (a+b x^3\right )^{2/3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{4} \,d x } \]

[In]

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

[Out]

integrate((b*x^3 + a)^(2/3)*x^4, x)

Giac [F]

\[ \int x^4 \left (a+b x^3\right )^{2/3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{4} \,d x } \]

[In]

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

[Out]

integrate((b*x^3 + a)^(2/3)*x^4, x)

Mupad [F(-1)]

Timed out. \[ \int x^4 \left (a+b x^3\right )^{2/3} \, dx=\int x^4\,{\left (b\,x^3+a\right )}^{2/3} \,d x \]

[In]

int(x^4*(a + b*x^3)^(2/3),x)

[Out]

int(x^4*(a + b*x^3)^(2/3), x)