\(\int x^2 \sqrt {-b+a x^3} \, dx\) [171]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int x^2 \sqrt {-b+a x^3} \, dx=\frac {2 \left (-b+a x^3\right )^{3/2}}{9 a} \]

[Out]

2/9*(a*x^3-b)^(3/2)/a

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {267} \[ \int x^2 \sqrt {-b+a x^3} \, dx=\frac {2 \left (a x^3-b\right )^{3/2}}{9 a} \]

[In]

Int[x^2*Sqrt[-b + a*x^3],x]

[Out]

(2*(-b + a*x^3)^(3/2))/(9*a)

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 {2 \left (-b+a x^3\right )^{3/2}}{9 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int x^2 \sqrt {-b+a x^3} \, dx=\frac {2 \left (-b+a x^3\right )^{3/2}}{9 a} \]

[In]

Integrate[x^2*Sqrt[-b + a*x^3],x]

[Out]

(2*(-b + a*x^3)^(3/2))/(9*a)

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
gosper	\(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{9 a}\)	\(17\)
derivativedivides	\(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{9 a}\)	\(17\)
default	\(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{9 a}\)	\(17\)
trager	\(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{9 a}\)	\(17\)
risch	\(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{9 a}\)	\(17\)
pseudoelliptic	\(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{9 a}\)	\(17\)

[In]

int(x^2*(a*x^3-b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/9*(a*x^3-b)^(3/2)/a

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int x^2 \sqrt {-b+a x^3} \, dx=\frac {2 \, {\left (a x^{3} - b\right )}^{\frac {3}{2}}}{9 \, a} \]

[In]

integrate(x^2*(a*x^3-b)^(1/2),x, algorithm="fricas")

[Out]

2/9*(a*x^3 - b)^(3/2)/a

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (14) = 28\).

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.20 \[ \int x^2 \sqrt {-b+a x^3} \, dx=\begin {cases} \frac {2 x^{3} \sqrt {a x^{3} - b}}{9} - \frac {2 b \sqrt {a x^{3} - b}}{9 a} & \text {for}\: a \neq 0 \\\frac {x^{3} \sqrt {- b}}{3} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(a*x**3-b)**(1/2),x)

[Out]

Piecewise((2*x**3*sqrt(a*x**3 - b)/9 - 2*b*sqrt(a*x**3 - b)/(9*a), Ne(a, 0)), (x**3*sqrt(-b)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int x^2 \sqrt {-b+a x^3} \, dx=\frac {2 \, {\left (a x^{3} - b\right )}^{\frac {3}{2}}}{9 \, a} \]

[In]

integrate(x^2*(a*x^3-b)^(1/2),x, algorithm="maxima")

[Out]

2/9*(a*x^3 - b)^(3/2)/a

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int x^2 \sqrt {-b+a x^3} \, dx=\frac {2 \, {\left (a x^{3} - b\right )}^{\frac {3}{2}}}{9 \, a} \]

[In]

integrate(x^2*(a*x^3-b)^(1/2),x, algorithm="giac")

[Out]

2/9*(a*x^3 - b)^(3/2)/a

Mupad [B] (verification not implemented)

Time = 5.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int x^2 \sqrt {-b+a x^3} \, dx=\frac {2\,{\left (a\,x^3-b\right )}^{3/2}}{9\,a} \]

[In]

int(x^2*(a*x^3 - b)^(1/2),x)

[Out]

(2*(a*x^3 - b)^(3/2))/(9*a)