\(\int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx\) [400]

Optimal result

Integrand size = 15, antiderivative size = 59 \[ \int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx=\frac {3 a^2 (-a+b x)^{2/3}}{2 b^3}+\frac {6 a (-a+b x)^{5/3}}{5 b^3}+\frac {3 (-a+b x)^{8/3}}{8 b^3} \]

[Out]

3/2*a^2*(b*x-a)^(2/3)/b^3+6/5*a*(b*x-a)^(5/3)/b^3+3/8*(b*x-a)^(8/3)/b^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx=\frac {3 a^2 (b x-a)^{2/3}}{2 b^3}+\frac {3 (b x-a)^{8/3}}{8 b^3}+\frac {6 a (b x-a)^{5/3}}{5 b^3} \]

[In]

Int[x^2/(-a + b*x)^(1/3),x]

[Out]

(3*a^2*(-a + b*x)^(2/3))/(2*b^3) + (6*a*(-a + b*x)^(5/3))/(5*b^3) + (3*(-a + b*x)^(8/3))/(8*b^3)

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])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^2}{b^2 \sqrt [3]{-a+b x}}+\frac {2 a (-a+b x)^{2/3}}{b^2}+\frac {(-a+b x)^{5/3}}{b^2}\right ) \, dx \\ & = \frac {3 a^2 (-a+b x)^{2/3}}{2 b^3}+\frac {6 a (-a+b x)^{5/3}}{5 b^3}+\frac {3 (-a+b x)^{8/3}}{8 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.63 \[ \int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx=\frac {3 (-a+b x)^{2/3} \left (9 a^2+6 a b x+5 b^2 x^2\right )}{40 b^3} \]

[In]

Integrate[x^2/(-a + b*x)^(1/3),x]

[Out]

(3*(-a + b*x)^(2/3)*(9*a^2 + 6*a*b*x + 5*b^2*x^2))/(40*b^3)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.58

method	result	size
gosper	\(\frac {3 \left (b x -a \right )^{\frac {2}{3}} \left (5 b^{2} x^{2}+6 a b x +9 a^{2}\right )}{40 b^{3}}\)	\(34\)
trager	\(\frac {3 \left (b x -a \right )^{\frac {2}{3}} \left (5 b^{2} x^{2}+6 a b x +9 a^{2}\right )}{40 b^{3}}\)	\(34\)
pseudoelliptic	\(\frac {3 \left (b x -a \right )^{\frac {2}{3}} \left (5 b^{2} x^{2}+6 a b x +9 a^{2}\right )}{40 b^{3}}\)	\(34\)
risch	\(-\frac {3 \left (-b x +a \right ) \left (5 b^{2} x^{2}+6 a b x +9 a^{2}\right )}{40 b^{3} \left (b x -a \right )^{\frac {1}{3}}}\)	\(40\)
derivativedivides	\(\frac {\frac {3 \left (b x -a \right )^{\frac {8}{3}}}{8}+\frac {6 a \left (b x -a \right )^{\frac {5}{3}}}{5}+\frac {3 a^{2} \left (b x -a \right )^{\frac {2}{3}}}{2}}{b^{3}}\)	\(44\)
default	\(\frac {\frac {3 \left (b x -a \right )^{\frac {8}{3}}}{8}+\frac {6 a \left (b x -a \right )^{\frac {5}{3}}}{5}+\frac {3 a^{2} \left (b x -a \right )^{\frac {2}{3}}}{2}}{b^{3}}\)	\(44\)

[In]

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

[Out]

3/40/b^3*(b*x-a)^(2/3)*(5*b^2*x^2+6*a*b*x+9*a^2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.56 \[ \int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx=\frac {3 \, {\left (5 \, b^{2} x^{2} + 6 \, a b x + 9 \, a^{2}\right )} {\left (b x - a\right )}^{\frac {2}{3}}}{40 \, b^{3}} \]

[In]

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

[Out]

3/40*(5*b^2*x^2 + 6*a*b*x + 9*a^2)*(b*x - a)^(2/3)/b^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.24 (sec) , antiderivative size = 1326, normalized size of antiderivative = 22.47 \[ \int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-27*a**(32/3)*(-1 + b*x/a)**(2/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b
**6*x**3) + 27*a**(32/3)*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x*
*3) + 63*a**(29/3)*b*x*(-1 + b*x/a)**(2/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**
6*x**3) - 81*a**(29/3)*b*x*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*
x**3) - 42*a**(26/3)*b**2*x**2*(-1 + b*x/a)**(2/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*
a**5*b**6*x**3) + 81*a**(26/3)*b**2*x**2*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 +
40*a**5*b**6*x**3) + 18*a**(23/3)*b**3*x**3*(-1 + b*x/a)**(2/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b
**5*x**2 + 40*a**5*b**6*x**3) - 27*a**(23/3)*b**3*x**3*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a*
*6*b**5*x**2 + 40*a**5*b**6*x**3) - 27*a**(20/3)*b**4*x**4*(-1 + b*x/a)**(2/3)/(-40*a**8*b**3 + 120*a**7*b**4*
x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 15*a**(17/3)*b**5*x**5*(-1 + b*x/a)**(2/3)/(-40*a**8*b**3 + 120*
a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3), Abs(b*x/a) > 1), (-27*a**(32/3)*(1 - b*x/a)**(2/3)*exp(
2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 27*a**(32/3)*exp(2*I*pi
/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 63*a**(29/3)*b*x*(1 - b*x/a)*
*(2/3)*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) - 81*a**(29/3)
*b*x*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) - 42*a**(26/3)*b
**2*x**2*(1 - b*x/a)**(2/3)*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6
*x**3) + 81*a**(26/3)*b**2*x**2*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*
b**6*x**3) + 18*a**(23/3)*b**3*x**3*(1 - b*x/a)**(2/3)*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a*
*6*b**5*x**2 + 40*a**5*b**6*x**3) - 27*a**(23/3)*b**3*x**3*exp(2*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 12
0*a**6*b**5*x**2 + 40*a**5*b**6*x**3) - 27*a**(20/3)*b**4*x**4*(1 - b*x/a)**(2/3)*exp(2*I*pi/3)/(-40*a**8*b**3
+ 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 15*a**(17/3)*b**5*x**5*(1 - b*x/a)**(2/3)*exp(2
*I*pi/3)/(-40*a**8*b**3 + 120*a**7*b**4*x - 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx=\frac {3 \, {\left (b x - a\right )}^{\frac {8}{3}}}{8 \, b^{3}} + \frac {6 \, {\left (b x - a\right )}^{\frac {5}{3}} a}{5 \, b^{3}} + \frac {3 \, {\left (b x - a\right )}^{\frac {2}{3}} a^{2}}{2 \, b^{3}} \]

[In]

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

[Out]

3/8*(b*x - a)^(8/3)/b^3 + 6/5*(b*x - a)^(5/3)*a/b^3 + 3/2*(b*x - a)^(2/3)*a^2/b^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx=\frac {3 \, {\left (5 \, {\left (b x - a\right )}^{\frac {8}{3}} + 16 \, {\left (b x - a\right )}^{\frac {5}{3}} a + 20 \, {\left (b x - a\right )}^{\frac {2}{3}} a^{2}\right )}}{40 \, b^{3}} \]

[In]

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

[Out]

3/40*(5*(b*x - a)^(8/3) + 16*(b*x - a)^(5/3)*a + 20*(b*x - a)^(2/3)*a^2)/b^3

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int \frac {x^2}{\sqrt [3]{-a+b x}} \, dx=\frac {48\,a\,{\left (b\,x-a\right )}^{5/3}+15\,{\left (b\,x-a\right )}^{8/3}+60\,a^2\,{\left (b\,x-a\right )}^{2/3}}{40\,b^3} \]

[In]

int(x^2/(b*x - a)^(1/3),x)

[Out]

(48*a*(b*x - a)^(5/3) + 15*(b*x - a)^(8/3) + 60*a^2*(b*x - a)^(2/3))/(40*b^3)