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

Optimal. Leaf size=59 \[ \frac{3 a^2 (b x-a)^{2/3}}{2 b^3}+\frac{3 (b x-a)^{8/3}}{8 b^3}+\frac{6 a (b x-a)^{5/3}}{5 b^3} \]

[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)

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Rubi [A]  time = 0.0129849, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {43} \[ \frac{3 a^2 (b x-a)^{2/3}}{2 b^3}+\frac{3 (b x-a)^{8/3}}{8 b^3}+\frac{6 a (b x-a)^{5/3}}{5 b^3} \]

Antiderivative was successfully verified.

[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 43

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*} \int \frac{x^2}{\sqrt [3]{-a+b x}} \, dx &=\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]  time = 0.0579486, size = 37, normalized size = 0.63 \[ \frac{3 (b x-a)^{2/3} \left (9 a^2+6 a b x+5 b^2 x^2\right )}{40 b^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.004, size = 34, normalized size = 0.6 \begin{align*}{\frac{15\,{b}^{2}{x}^{2}+18\,abx+27\,{a}^{2}}{40\,{b}^{3}} \left ( bx-a \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02014, size = 63, normalized size = 1.07 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.71182, size = 76, normalized size = 1.29 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [C]  time = 2.55204, size = 1328, normalized size = 22.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/Abs(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 - 1
20*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)*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))

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Giac [A]  time = 1.26082, size = 58, normalized size = 0.98 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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