3.1.24 \(\int \frac {1-x^3}{x^4 (1-x^3+x^6)} \, dx\)

Optimal. Leaf size=31 \[ \frac {2 \tan ^{-1}\left (\frac {1-2 x^3}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {1}{3 x^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1474, 800, 618, 204} \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {1-2 x^3}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {1}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^3)/(x^4*(1 - x^3 + x^6)),x]

[Out]

-1/(3*x^3) + (2*ArcTan[(1 - 2*x^3)/Sqrt[3]])/(3*Sqrt[3])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1474

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a,
b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1-x^3}{x^4 \left (1-x^3+x^6\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1-x}{x^2 \left (1-x+x^2\right )} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {1}{x^2}+\frac {1}{-1+x-x^2}\right ) \, dx,x,x^3\right )\\ &=-\frac {1}{3 x^3}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-1+x-x^2} \, dx,x,x^3\right )\\ &=-\frac {1}{3 x^3}-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 x^3\right )\\ &=-\frac {1}{3 x^3}+\frac {2 \tan ^{-1}\left (\frac {1-2 x^3}{\sqrt {3}}\right )}{3 \sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 45, normalized size = 1.45 \begin {gather*} -\frac {1}{3} \text {RootSum}\left [\text {$\#$1}^6-\text {$\#$1}^3+1\&,\frac {\log (x-\text {$\#$1})}{2 \text {$\#$1}^3-1}\&\right ]-\frac {1}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x^3)/(x^4*(1 - x^3 + x^6)),x]

[Out]

-1/3*1/x^3 - RootSum[1 - #1^3 + #1^6 & , Log[x - #1]/(-1 + 2*#1^3) & ]/3

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-x^3}{x^4 \left (1-x^3+x^6\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(1 - x^3)/(x^4*(1 - x^3 + x^6)),x]

[Out]

IntegrateAlgebraic[(1 - x^3)/(x^4*(1 - x^3 + x^6)), x]

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fricas [A]  time = 1.06, size = 28, normalized size = 0.90 \begin {gather*} -\frac {2 \, \sqrt {3} x^{3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) + 3}{9 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+1)/x^4/(x^6-x^3+1),x, algorithm="fricas")

[Out]

-1/9*(2*sqrt(3)*x^3*arctan(1/3*sqrt(3)*(2*x^3 - 1)) + 3)/x^3

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giac [A]  time = 0.45, size = 24, normalized size = 0.77 \begin {gather*} -\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) - \frac {1}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+1)/x^4/(x^6-x^3+1),x, algorithm="giac")

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^3 - 1)) - 1/3/x^3

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maple [A]  time = 0.01, size = 25, normalized size = 0.81 \begin {gather*} -\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x^{3}-1\right ) \sqrt {3}}{3}\right )}{9}-\frac {1}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^3+1)/x^4/(x^6-x^3+1),x)

[Out]

-2/9*3^(1/2)*arctan(1/3*(2*x^3-1)*3^(1/2))-1/3/x^3

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maxima [A]  time = 0.96, size = 24, normalized size = 0.77 \begin {gather*} -\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) - \frac {1}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+1)/x^4/(x^6-x^3+1),x, algorithm="maxima")

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^3 - 1)) - 1/3/x^3

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mupad [B]  time = 0.04, size = 26, normalized size = 0.84 \begin {gather*} \frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}}{3}-\frac {2\,\sqrt {3}\,x^3}{3}\right )}{9}-\frac {1}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3 - 1)/(x^4*(x^6 - x^3 + 1)),x)

[Out]

(2*3^(1/2)*atan(3^(1/2)/3 - (2*3^(1/2)*x^3)/3))/9 - 1/(3*x^3)

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sympy [A]  time = 0.14, size = 36, normalized size = 1.16 \begin {gather*} - \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{3}}{3} - \frac {\sqrt {3}}{3} \right )}}{9} - \frac {1}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**3+1)/x**4/(x**6-x**3+1),x)

[Out]

-2*sqrt(3)*atan(2*sqrt(3)*x**3/3 - sqrt(3)/3)/9 - 1/(3*x**3)

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