\(\int x^2 (a+b x^3)^{2/3} \, dx\) [530]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int x^2 \left (a+b x^3\right )^{2/3} \, dx=\frac {\left (a+b x^3\right )^{5/3}}{5 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {267} \[ \int x^2 \left (a+b x^3\right )^{2/3} \, dx=\frac {\left (a+b x^3\right )^{5/3}}{5 b} \]

[In]

Int[x^2*(a + b*x^3)^(2/3),x]

[Out]

(a + b*x^3)^(5/3)/(5*b)

Rule 267

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x^3\right )^{5/3}}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b x^3\right )^{2/3} \, dx=\frac {\left (a+b x^3\right )^{5/3}}{5 b} \]

[In]

Integrate[x^2*(a + b*x^3)^(2/3),x]

[Out]

(a + b*x^3)^(5/3)/(5*b)

Maple [A] (verified)

Time = 3.79 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 b}\)	\(15\)
derivativedivides	\(\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 b}\)	\(15\)
default	\(\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 b}\)	\(15\)
trager	\(\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 b}\)	\(15\)
risch	\(\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 b}\)	\(15\)
pseudoelliptic	\(\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 b}\)	\(15\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int x^2 \left (a+b x^3\right )^{2/3} \, dx=\frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{5 \, b} \]

[In]

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

[Out]

1/5*(b*x^3 + a)^(5/3)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (12) = 24\).

Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.17 \[ \int x^2 \left (a+b x^3\right )^{2/3} \, dx=\begin {cases} \frac {a \left (a + b x^{3}\right )^{\frac {2}{3}}}{5 b} + \frac {x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{5} & \text {for}\: b \neq 0 \\\frac {a^{\frac {2}{3}} x^{3}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*(a + b*x**3)**(2/3)/(5*b) + x**3*(a + b*x**3)**(2/3)/5, Ne(b, 0)), (a**(2/3)*x**3/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int x^2 \left (a+b x^3\right )^{2/3} \, dx=\frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{5 \, b} \]

[In]

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

[Out]

1/5*(b*x^3 + a)^(5/3)/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int x^2 \left (a+b x^3\right )^{2/3} \, dx=\frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{5 \, b} \]

[In]

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

[Out]

1/5*(b*x^3 + a)^(5/3)/b

Mupad [B] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int x^2 \left (a+b x^3\right )^{2/3} \, dx=\frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,b} \]

[In]

int(x^2*(a + b*x^3)^(2/3),x)

[Out]

(a + b*x^3)^(5/3)/(5*b)