\(\int \frac {1}{(-1+x) (1+x+x^2)} \, dx\) [2234]

Optimal result

Integrand size = 14, antiderivative size = 41 \[ \int \frac {1}{(-1+x) \left (1+x+x^2\right )} \, dx=-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (1+x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {719, 31, 648, 632, 210, 642} \[ \int \frac {1}{(-1+x) \left (1+x+x^2\right )} \, dx=-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{6} \log \left (x^2+x+1\right )+\frac {1}{3} \log (1-x) \]

[In]

Int[1/((-1 + x)*(1 + x + x^2)),x]

[Out]

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

Rule 31

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

Rule 210

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

Rule 632

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 642

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]

Rule 648

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 719

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {1}{-1+x} \, dx+\frac {1}{3} \int \frac {-2-x}{1+x+x^2} \, dx \\ & = \frac {1}{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 \\ & = \frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (1+x+x^2\right )+\text {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}}+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (1+x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(-1+x) \left (1+x+x^2\right )} \, dx=-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (1+x+x^2\right ) \]

[In]

Integrate[1/((-1 + x)*(1 + x + x^2)),x]

[Out]

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

Maple [A] (verified)

Time = 22.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {\ln \left (-1+x \right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x +\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}\)	\(31\)
default	\(\frac {\ln \left (-1+x \right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{6}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(33\)

[In]

int(1/(-1+x)/(x^2+x+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(-1+x) \left (1+x+x^2\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(-1+x) \left (1+x+x^2\right )} \, dx=\frac {\log {\left (x - 1 \right )}}{3} - \frac {\log {\left (x^{2} + x + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

integrate(1/(-1+x)/(x**2+x+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(-1+x) \left (1+x+x^2\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(-1+x) \left (1+x+x^2\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(-1+x) \left (1+x+x^2\right )} \, dx=\frac {\ln \left (x-1\right )}{3}+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]

[In]

int(1/((x - 1)*(x + x^2 + 1)),x)

[Out]

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