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

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

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Rubi [A]  time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.01, 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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IntegrateAlgebraic [A]  time = 4.88, size = 37, normalized size = 0.40 \[ \frac {3 (x-3) (x+1) \left (9 (x+1)^2-60 (x+1)+80\right )}{320 \left ((x-3)^2 (x+1)\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-3 + x)*(1 + x)*(80 - 60*(1 + x) + 9*(1 + x)^2))/(320*((-3 + x)^2*(1 + x))^(4/3))

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fricas [A]  time = 0.67, size = 44, normalized size = 0.48 \[ \frac {3 \, {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {2}{3}} {\left (9 \, x^{2} - 42 \, x + 29\right )}}{320 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {4}{3}}}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 29, normalized size = 0.32

method	result	size
risch	\(\frac {\frac {27}{320} x^{2}-\frac {63}{160} x +\frac {87}{320}}{\left (-3+x \right ) \left (\left (1+x \right ) \left (-3+x \right )^{2}\right )^{\frac {1}{3}}}\)	\(29\)
gosper	\(\frac {3 \left (1+x \right ) \left (-3+x \right ) \left (9 x^{2}-42 x +29\right )}{320 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {4}{3}}}\)	\(34\)
trager	\(\frac {3 \left (9 x^{2}-42 x +29\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}}{320 \left (-3+x \right )^{3} \left (1+x \right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {4}{3}}}\,{d x} \]

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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mupad [B]  time = 0.08, size = 37, normalized size = 0.40 \[ \frac {3\,\left (9\,x^2-42\,x+29\right )\,{\left (x^3-5\,x^2+3\,x+9\right )}^{2/3}}{320\,\left (x+1\right )\,{\left (x-3\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (x^{3} - 5 x^{2} + 3 x + 9\right )^{\frac {4}{3}}}\, dx \]

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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