3.308 \(\int \frac{1-x}{x (1+x^3)} \, dx\)

Optimal. Leaf size=42 \[ -\frac{1}{6} \log \left (x^2-x+1\right )+\log (x)-\frac{2}{3} \log (x+1)+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

ArcTan[(1 - 2*x)/Sqrt[3]]/Sqrt[3] + Log[x] - (2*Log[1 + x])/3 - Log[1 - x + x^2]/6

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Rubi [A]  time = 0.0493821, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {1834, 634, 618, 204, 628} \[ -\frac{1}{6} \log \left (x^2-x+1\right )+\log (x)-\frac{2}{3} \log (x+1)+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)/(x*(1 + x^3)),x]

[Out]

ArcTan[(1 - 2*x)/Sqrt[3]]/Sqrt[3] + Log[x] - (2*Log[1 + x])/3 - Log[1 - x + x^2]/6

Rule 1834

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[((c*x)^m*Pq)/(a + b*
x^n), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]

Rule 634

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

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 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 628

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

Rubi steps

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

Mathematica [A]  time = 0.0076564, size = 53, normalized size = 1.26 \[ \frac{1}{6} \log \left (x^2-x+1\right )-\frac{1}{3} \log \left (x^3+1\right )+\log (x)-\frac{1}{3} \log (x+1)-\frac{\tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)/(x*(1 + x^3)),x]

[Out]

-(ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3]) + Log[x] - Log[1 + x]/3 + Log[1 - x + x^2]/6 - Log[1 + x^3]/3

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Maple [A]  time = 0.006, size = 37, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ({x}^{2}-x+1 \right ) }{6}}-{\frac{\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+\ln \left ( x \right ) -{\frac{2\,\ln \left ( 1+x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*ln(x^2-x+1)-1/3*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))+ln(x)-2/3*ln(1+x)

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Maxima [A]  time = 1.42337, size = 49, normalized size = 1.17 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac{2}{3} \, \log \left (x + 1\right ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*log(x^2 - x + 1) - 2/3*log(x + 1) + log(x)

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Fricas [A]  time = 1.4933, size = 126, normalized size = 3. \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac{2}{3} \, \log \left (x + 1\right ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*log(x^2 - x + 1) - 2/3*log(x + 1) + log(x)

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Sympy [A]  time = 0.175426, size = 46, normalized size = 1.1 \begin{align*} \log{\left (x \right )} - \frac{2 \log{\left (x + 1 \right )}}{3} - \frac{\log{\left (x^{2} - x + 1 \right )}}{6} - \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - 2*log(x + 1)/3 - log(x**2 - x + 1)/6 - sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/3

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Giac [A]  time = 1.08784, size = 51, normalized size = 1.21 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac{2}{3} \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*log(x^2 - x + 1) - 2/3*log(abs(x + 1)) + log(abs(x))