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

Optimal. Leaf size=41 \[ \frac{1}{3} \log \left (x^2+x+1\right )-\frac{2}{3} \log (1-x)+\frac{4 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0272263, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1861, 31, 634, 618, 204, 628} \[ \frac{1}{3} \log \left (x^2+x+1\right )-\frac{2}{3} \log (1-x)+\frac{4 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(3 - x)/(1 - x^3),x]

[Out]

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

Rule 1861

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 3]], s = Denominato
r[Rt[-(a/b), 3]]}, Dist[(r*(B*r + A*s))/(3*a*s), Int[1/(r - s*x), x], x] - Dist[r/(3*a*s), Int[(r*(B*r - 2*A*s
) - s*(B*r + A*s)*x)/(r^2 + r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && Ne
gQ[a/b]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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{3-x}{1-x^3} \, dx &=-\left (\frac{1}{3} \int \frac{-7-2 x}{1+x+x^2} \, dx\right )+\frac{2}{3} \int \frac{1}{1-x} \, dx\\ &=-\frac{2}{3} \log (1-x)+\frac{1}{3} \int \frac{1+2 x}{1+x+x^2} \, dx+2 \int \frac{1}{1+x+x^2} \, dx\\ &=-\frac{2}{3} \log (1-x)+\frac{1}{3} \log \left (1+x+x^2\right )-4 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac{4 \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{2}{3} \log (1-x)+\frac{1}{3} \log \left (1+x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0075016, size = 41, normalized size = 1. \[ \frac{1}{3} \log \left (x^2+x+1\right )-\frac{2}{3} \log (1-x)+\frac{4 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 - x)/(1 - x^3),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 33, normalized size = 0.8 \begin{align*} -{\frac{2\,\ln \left ( -1+x \right ) }{3}}+{\frac{\ln \left ({x}^{2}+x+1 \right ) }{3}}+{\frac{4\,\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.4424, size = 43, normalized size = 1.05 \begin{align*} \frac{4}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac{2}{3} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.948494, size = 112, normalized size = 2.73 \begin{align*} \frac{4}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac{2}{3} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.135038, size = 44, normalized size = 1.07 \begin{align*} - \frac{2 \log{\left (x - 1 \right )}}{3} + \frac{\log{\left (x^{2} + x + 1 \right )}}{3} + \frac{4 \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((3-x)/(-x**3+1),x)

[Out]

-2*log(x - 1)/3 + log(x**2 + x + 1)/3 + 4*sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/3

________________________________________________________________________________________

Giac [A]  time = 1.0882, size = 45, normalized size = 1.1 \begin{align*} \frac{4}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac{2}{3} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/3*log(x^2 + x + 1) - 2/3*log(abs(x - 1))