3.234 \(\int \frac{1}{(9+3 x-5 x^2+x^3)^{4/3}} \, dx\)

Optimal. Leaf size=92 \[ -\frac{27 (x+1) (3-x)^3}{320 \left (x^3-5 x^2+3 x+9\right )^{4/3}}+\frac{9 (x+1) (3-x)^2}{80 \left (x^3-5 x^2+3 x+9\right )^{4/3}}+\frac{3 (x+1) (3-x)}{20 \left (x^3-5 x^2+3 x+9\right )^{4/3}} \]

[Out]

(3*(3 - x)*(1 + x))/(20*(9 + 3*x - 5*x^2 + x^3)^(4/3)) + (9*(3 - x)^2*(1 + x))/(80*(9 + 3*x - 5*x^2 + x^3)^(4/
3)) - (27*(3 - x)^3*(1 + x))/(320*(9 + 3*x - 5*x^2 + x^3)^(4/3))

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Rubi [A]  time = 0.0850329, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2067, 2064, 45, 37} \[ -\frac{27 (x+1) (3-x)^3}{320 \left (x^3-5 x^2+3 x+9\right )^{4/3}}+\frac{9 (x+1) (3-x)^2}{80 \left (x^3-5 x^2+3 x+9\right )^{4/3}}+\frac{3 (x+1) (3-x)}{20 \left (x^3-5 x^2+3 x+9\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[(9 + 3*x - 5*x^2 + x^3)^(-4/3),x]

[Out]

(3*(3 - x)*(1 + x))/(20*(9 + 3*x - 5*x^2 + x^3)^(4/3)) + (9*(3 - x)^2*(1 + x))/(80*(9 + 3*x - 5*x^2 + x^3)^(4/
3)) - (27*(3 - x)^3*(1 + x))/(320*(9 + 3*x - 5*x^2 + x^3)^(4/3))

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 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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 )^{4/3}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{27}-\frac{16 x}{3}+x^3\right )^{4/3}} \, dx,x,-\frac{5}{3}+x\right )\\ &=\frac{\left (262144\ 2^{2/3} (3-x)^{8/3} (1+x)^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right )^{8/3} \left (\frac{128}{9}+\frac{16 x}{3}\right )^{4/3}} \, dx,x,-\frac{5}{3}+x\right )}{81 \left (9+3 x-5 x^2+x^3\right )^{4/3}}\\ &=\frac{3 (3-x) (1+x)}{20 \left (9+3 x-5 x^2+x^3\right )^{4/3}}+\frac{\left (4096\ 2^{2/3} (3-x)^{8/3} (1+x)^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right )^{5/3} \left (\frac{128}{9}+\frac{16 x}{3}\right )^{4/3}} \, dx,x,-\frac{5}{3}+x\right )}{45 \left (9+3 x-5 x^2+x^3\right )^{4/3}}\\ &=\frac{3 (3-x) (1+x)}{20 \left (9+3 x-5 x^2+x^3\right )^{4/3}}+\frac{9 (3-x)^2 (1+x)}{80 \left (9+3 x-5 x^2+x^3\right )^{4/3}}+\frac{\left (16\ 2^{2/3} (3-x)^{8/3} (1+x)^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right )^{2/3} \left (\frac{128}{9}+\frac{16 x}{3}\right )^{4/3}} \, dx,x,-\frac{5}{3}+x\right )}{5 \left (9+3 x-5 x^2+x^3\right )^{4/3}}\\ &=\frac{3 (3-x) (1+x)}{20 \left (9+3 x-5 x^2+x^3\right )^{4/3}}+\frac{9 (3-x)^2 (1+x)}{80 \left (9+3 x-5 x^2+x^3\right )^{4/3}}-\frac{27 (3-x)^3 (1+x)}{320 \left (9+3 x-5 x^2+x^3\right )^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.0080243, size = 32, normalized size = 0.35 \[ \frac{3 \left (9 x^2-42 x+29\right )}{320 (x-3) \sqrt [3]{(x-3)^2 (x+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(9 + 3*x - 5*x^2 + x^3)^(-4/3),x]

[Out]

(3*(29 - 42*x + 9*x^2))/(320*(-3 + x)*((-3 + x)^2*(1 + x))^(1/3))

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Maple [A]  time = 0.003, size = 34, normalized size = 0.4 \begin{align*}{\frac{ \left ( 3+3\,x \right ) \left ( -3+x \right ) \left ( 9\,{x}^{2}-42\,x+29 \right ) }{320} \left ({x}^{3}-5\,{x}^{2}+3\,x+9 \right ) ^{-{\frac{4}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/320*(1+x)*(-3+x)*(9*x^2-42*x+29)/(x^3-5*x^2+3*x+9)^(4/3)

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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{4}{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)^(4/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.62835, size = 115, normalized size = 1.25 \begin{align*} \frac{3 \,{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac{2}{3}}{\left (9 \, x^{2} - 42 \, x + 29\right )}}{320 \,{\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(4/3),x, algorithm="fricas")

[Out]

3/320*(x^3 - 5*x^2 + 3*x + 9)^(2/3)*(9*x^2 - 42*x + 29)/(x^4 - 8*x^3 + 18*x^2 - 27)

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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{4}{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)**(4/3),x)

[Out]

Integral((x**3 - 5*x**2 + 3*x + 9)**(-4/3), 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{4}{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)^(4/3),x, algorithm="giac")

[Out]

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