3.233 \(\int \frac{1}{(9+3 x-5 x^2+x^3)^{2/3}} \, dx\)

Optimal. Leaf size=29 \[ \frac{3 (3-x) (x+1)}{4 \left (x^3-5 x^2+3 x+9\right )^{2/3}} \]

[Out]

(3*(3 - x)*(1 + x))/(4*(9 + 3*x - 5*x^2 + x^3)^(2/3))

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Rubi [A]  time = 0.0329589, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2067, 2064, 37} \[ \frac{3 (3-x) (x+1)}{4 \left (x^3-5 x^2+3 x+9\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(9 + 3*x - 5*x^2 + x^3)^(-2/3),x]

[Out]

(3*(3 - x)*(1 + x))/(4*(9 + 3*x - 5*x^2 + x^3)^(2/3))

Rule 2067

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rule 2064

Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*
x)^(2*p)), Int[(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0]
&&  !IntegerQ[p]

Rule 37

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

Rubi steps

\begin{align*} \int \frac{1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{27}-\frac{16 x}{3}+x^3\right )^{2/3}} \, dx,x,-\frac{5}{3}+x\right )\\ &=\frac{\left (512 \sqrt [3]{2} (3-x)^{4/3} (1+x)^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right )^{4/3} \left (\frac{128}{9}+\frac{16 x}{3}\right )^{2/3}} \, dx,x,-\frac{5}{3}+x\right )}{9 \left (9+3 x-5 x^2+x^3\right )^{2/3}}\\ &=\frac{3 (3-x) (1+x)}{4 \left (9+3 x-5 x^2+x^3\right )^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0077999, size = 23, normalized size = 0.79 \[ -\frac{3 (x-3) (x+1)}{4 \left ((x-3)^2 (x+1)\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(9 + 3*x - 5*x^2 + x^3)^(-2/3),x]

[Out]

(-3*(-3 + x)*(1 + x))/(4*((-3 + x)^2*(1 + x))^(2/3))

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Maple [A]  time = 0.002, size = 24, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 3+3\,x \right ) \left ( -3+x \right ) }{4} \left ({x}^{3}-5\,{x}^{2}+3\,x+9 \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-5*x^2+3*x+9)^(2/3),x)

[Out]

-3/4*(1+x)*(-3+x)/(x^3-5*x^2+3*x+9)^(2/3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(2/3),x, algorithm="maxima")

[Out]

integrate((x^3 - 5*x^2 + 3*x + 9)^(-2/3), x)

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Fricas [A]  time = 1.74807, size = 59, normalized size = 2.03 \begin{align*} -\frac{3 \,{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac{1}{3}}}{4 \,{\left (x - 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(2/3),x, algorithm="fricas")

[Out]

-3/4*(x^3 - 5*x^2 + 3*x + 9)^(1/3)/(x - 3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (x^{3} - 5 x^{2} + 3 x + 9\right )^{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3-5*x**2+3*x+9)**(2/3),x)

[Out]

Integral((x**3 - 5*x**2 + 3*x + 9)**(-2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(2/3),x, algorithm="giac")

[Out]

integrate((x^3 - 5*x^2 + 3*x + 9)^(-2/3), x)