3.312 \(\int \frac{x (1+2 x)}{1-x^3} \, dx\)

Optimal. Leaf size=39 \[ -\frac{1}{2} \log \left (x^2+x+1\right )-\log (1-x)-\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0395694, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {1875, 31, 634, 618, 204, 628} \[ -\frac{1}{2} \log \left (x^2+x+1\right )-\log (1-x)-\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(1 + 2*x))/(1 - x^3),x]

[Out]

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

Rule 1875

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2], q = (-(a/b))^(1/3)}, Dist[(q*(A + B*q + C*q^2))/(3*a), Int[1/(q - x), x], x] + Dist[q/(3*a), Int[(q*(2*A
- B*q - C*q^2) + (A + B*q - 2*C*q^2)*x)/(q^2 + q*x + x^2), x], x] /; NeQ[a*B^3 - b*A^3, 0] && NeQ[A + B*q + C*
q^2, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2] && LtQ[a/b, 0]

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

Mathematica [A]  time = 0.010297, size = 53, normalized size = 1.36 \[ \frac{1}{6} \log \left (x^2+x+1\right )-\frac{2}{3} \log \left (1-x^3\right )-\frac{1}{3} \log (1-x)-\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(1 + 2*x))/(1 - x^3),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 33, normalized size = 0.9 \begin{align*} -\ln \left ( -1+x \right ) -{\frac{\ln \left ({x}^{2}+x+1 \right ) }{2}}-{\frac{\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(x*(1+2*x)/(-x^3+1),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.40839, size = 43, normalized size = 1.1 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{2} \, \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1+2*x)/(-x^3+1),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.50109, size = 108, normalized size = 2.77 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{2} \, \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.12768, size = 41, normalized size = 1.05 \begin{align*} - \log{\left (x - 1 \right )} - \frac{\log{\left (x^{2} + x + 1 \right )}}{2} - \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(x*(1+2*x)/(-x**3+1),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.05645, size = 45, normalized size = 1.15 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{2} \, \log \left (x^{2} + x + 1\right ) - \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1+2*x)/(-x^3+1),x, algorithm="giac")

[Out]

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