\(\int x^5 (a+b x^3)^{2/3} \, dx\) [529]

Optimal result

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

[Out]

-1/5*a*(b*x^3+a)^(5/3)/b^2+1/8*(b*x^3+a)^(8/3)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int x^5 \left (a+b x^3\right )^{2/3} \, dx=\frac {\left (a+b x^3\right )^{8/3}}{8 b^2}-\frac {a \left (a+b x^3\right )^{5/3}}{5 b^2} \]

[In]

Int[x^5*(a + b*x^3)^(2/3),x]

[Out]

-1/5*(a*(a + b*x^3)^(5/3))/b^2 + (a + b*x^3)^(8/3)/(8*b^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int x (a+b x)^{2/3} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (-\frac {a (a+b x)^{2/3}}{b}+\frac {(a+b x)^{5/3}}{b}\right ) \, dx,x,x^3\right ) \\ & = -\frac {a \left (a+b x^3\right )^{5/3}}{5 b^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int x^5 \left (a+b x^3\right )^{2/3} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-3 a^2+2 a b x^3+5 b^2 x^6\right )}{40 b^2} \]

[In]

Integrate[x^5*(a + b*x^3)^(2/3),x]

[Out]

((a + b*x^3)^(2/3)*(-3*a^2 + 2*a*b*x^3 + 5*b^2*x^6))/(40*b^2)

Maple [A] (verified)

Time = 3.76 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66

method	result	size
gosper	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-5 b \,x^{3}+3 a \right )}{40 b^{2}}\)	\(25\)
pseudoelliptic	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-5 b \,x^{3}+3 a \right )}{40 b^{2}}\)	\(25\)
trager	\(-\frac {\left (-5 b^{2} x^{6}-2 a b \,x^{3}+3 a^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{40 b^{2}}\)	\(36\)
risch	\(-\frac {\left (-5 b^{2} x^{6}-2 a b \,x^{3}+3 a^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{40 b^{2}}\)	\(36\)

[In]

int(x^5*(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)

[Out]

-1/40*(b*x^3+a)^(5/3)*(-5*b*x^3+3*a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int x^5 \left (a+b x^3\right )^{2/3} \, dx=\frac {{\left (5 \, b^{2} x^{6} + 2 \, a b x^{3} - 3 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{40 \, b^{2}} \]

[In]

integrate(x^5*(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

1/40*(5*b^2*x^6 + 2*a*b*x^3 - 3*a^2)*(b*x^3 + a)^(2/3)/b^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (31) = 62\).

Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.66 \[ \int x^5 \left (a+b x^3\right )^{2/3} \, dx=\begin {cases} - \frac {3 a^{2} \left (a + b x^{3}\right )^{\frac {2}{3}}}{40 b^{2}} + \frac {a x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{20 b} + \frac {x^{6} \left (a + b x^{3}\right )^{\frac {2}{3}}}{8} & \text {for}\: b \neq 0 \\\frac {a^{\frac {2}{3}} x^{6}}{6} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5*(b*x**3+a)**(2/3),x)

[Out]

Piecewise((-3*a**2*(a + b*x**3)**(2/3)/(40*b**2) + a*x**3*(a + b*x**3)**(2/3)/(20*b) + x**6*(a + b*x**3)**(2/3
)/8, Ne(b, 0)), (a**(2/3)*x**6/6, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int x^5 \left (a+b x^3\right )^{2/3} \, dx=\frac {{\left (b x^{3} + a\right )}^{\frac {8}{3}}}{8 \, b^{2}} - \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}} a}{5 \, b^{2}} \]

[In]

integrate(x^5*(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

1/8*(b*x^3 + a)^(8/3)/b^2 - 1/5*(b*x^3 + a)^(5/3)*a/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int x^5 \left (a+b x^3\right )^{2/3} \, dx=\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} - 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a}{40 \, b^{2}} \]

[In]

integrate(x^5*(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

1/40*(5*(b*x^3 + a)^(8/3) - 8*(b*x^3 + a)^(5/3)*a)/b^2

Mupad [B] (verification not implemented)

Time = 5.78 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int x^5 \left (a+b x^3\right )^{2/3} \, dx={\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {x^6}{8}-\frac {3\,a^2}{40\,b^2}+\frac {a\,x^3}{20\,b}\right ) \]

[In]

int(x^5*(a + b*x^3)^(2/3),x)

[Out]

(a + b*x^3)^(2/3)*(x^6/8 - (3*a^2)/(40*b^2) + (a*x^3)/(20*b))