3.2.40 \(\int \frac {x (9-9 x+2 x^2)}{\sqrt [3]{-3 x+x^2}} \, dx\)

Optimal. Leaf size=15 \[ \frac {3}{5} \left (x^2-3 x\right )^{5/3} \]

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1631, 629} \begin {gather*} \frac {3}{5} \left (x^2-3 x\right )^{5/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(9 - 9*x + 2*x^2))/(-3*x + x^2)^(1/3),x]

[Out]

(3*(-3*x + x^2)^(5/3))/5

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1631

Int[(Pq_)*((e_.)*(x_))^(m_.)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[e, Int[(e*x)^(m - 1)*Polynom
ialQuotient[Pq, b + c*x, x]*(b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{b, c, e, m, p}, x] && PolyQ[Pq, x] && EqQ[
PolynomialRemainder[Pq, b + c*x, x], 0]

Rubi steps

\begin {align*} \int \frac {x \left (9-9 x+2 x^2\right )}{\sqrt [3]{-3 x+x^2}} \, dx &=\int (-3+2 x) \left (-3 x+x^2\right )^{2/3} \, dx\\ &=\frac {3}{5} \left (-3 x+x^2\right )^{5/3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.87 \begin {gather*} \frac {3}{5} ((x-3) x)^{5/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(9 - 9*x + 2*x^2))/(-3*x + x^2)^(1/3),x]

[Out]

(3*((-3 + x)*x)^(5/3))/5

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IntegrateAlgebraic [A]  time = 0.02, size = 13, normalized size = 0.87 \begin {gather*} \frac {3}{5} ((x-3) x)^{5/3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(9 - 9*x + 2*x^2))/(-3*x + x^2)^(1/3),x]

[Out]

(3*((-3 + x)*x)^(5/3))/5

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fricas [A]  time = 0.99, size = 11, normalized size = 0.73 \begin {gather*} \frac {3}{5} \, {\left (x^{2} - 3 \, x\right )}^{\frac {5}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(2*x^2-9*x+9)/(x^2-3*x)^(1/3),x, algorithm="fricas")

[Out]

3/5*(x^2 - 3*x)^(5/3)

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giac [A]  time = 0.16, size = 11, normalized size = 0.73 \begin {gather*} \frac {3}{5} \, {\left (x^{2} - 3 \, x\right )}^{\frac {5}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(2*x^2-9*x+9)/(x^2-3*x)^(1/3),x, algorithm="giac")

[Out]

3/5*(x^2 - 3*x)^(5/3)

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maple [A]  time = 0.00, size = 20, normalized size = 1.33 \begin {gather*} \frac {3 \left (x -3\right )^{2} x^{2}}{5 \left (x^{2}-3 x \right )^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(2*x^2-9*x+9)/(x^2-3*x)^(1/3),x)

[Out]

3/5*(x-3)^2*x^2/(x^2-3*x)^(1/3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{2} - 9 \, x + 9\right )} x}{{\left (x^{2} - 3 \, x\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(2*x^2-9*x+9)/(x^2-3*x)^(1/3),x, algorithm="maxima")

[Out]

integrate((2*x^2 - 9*x + 9)*x/(x^2 - 3*x)^(1/3), x)

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mupad [B]  time = 3.68, size = 15, normalized size = 1.00 \begin {gather*} \frac {3\,x\,{\left (x^2-3\,x\right )}^{2/3}\,\left (x-3\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(2*x^2 - 9*x + 9))/(x^2 - 3*x)^(1/3),x)

[Out]

(3*x*(x^2 - 3*x)^(2/3)*(x - 3))/5

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (x - 3\right ) \left (2 x - 3\right )}{\sqrt [3]{x \left (x - 3\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(2*x**2-9*x+9)/(x**2-3*x)**(1/3),x)

[Out]

Integral(x*(x - 3)*(2*x - 3)/(x*(x - 3))**(1/3), x)

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