\(\int \log (1+x+x^2) \, dx\) [81]

Optimal result

Integrand size = 7, antiderivative size = 42 \[ \int \log \left (1+x+x^2\right ) \, dx=-2 x+\sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+\frac {1}{2} \log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {2603, 787, 648, 632, 210, 642} \[ \int \log \left (1+x+x^2\right ) \, dx=\sqrt {3} \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )+x \log \left (x^2+x+1\right )+\frac {1}{2} \log \left (x^2+x+1\right )-2 x \]

[In]

Int[Log[1 + x + x^2],x]

[Out]

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

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 787

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*g*(x/c
), x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*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]

Rule 2603

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[x*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83 \[ \int \log \left (1+x+x^2\right ) \, dx=-2 x+\sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+\left (\frac {1}{2}+x\right ) \log \left (1+x+x^2\right ) \]

[In]

Integrate[Log[1 + x + x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90

method	result	size
default	\(-2 x +\frac {\ln \left (x^{2}+x +1\right )}{2}+x \ln \left (x^{2}+x +1\right )+\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}\)	\(38\)
parts	\(-2 x +\frac {\ln \left (x^{2}+x +1\right )}{2}+x \ln \left (x^{2}+x +1\right )+\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}\)	\(38\)
risch	\(x \ln \left (x^{2}+x +1\right )-2 x +\frac {\ln \left (4 x^{2}+4 x +4\right )}{2}+\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79 \[ \int \log \left (1+x+x^2\right ) \, dx=\frac {1}{2} \, {\left (2 \, x + 1\right )} \log \left (x^{2} + x + 1\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10 \[ \int \log \left (1+x+x^2\right ) \, dx=x \log {\left (x^{2} + x + 1 \right )} - 2 x + \frac {\log {\left (x^{2} + x + 1 \right )}}{2} + \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )} \]

[In]

integrate(ln(x**2+x+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \log \left (1+x+x^2\right ) \, dx=x \log \left (x^{2} + x + 1\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x + \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \log \left (1+x+x^2\right ) \, dx=x \log \left (x^{2} + x + 1\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x + \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93 \[ \int \log \left (1+x+x^2\right ) \, dx=\frac {\ln \left (x^2+x+1\right )}{2}-2\,x+\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}+\frac {\sqrt {3}}{3}\right )+x\,\ln \left (x^2+x+1\right ) \]

[In]

int(log(x + x^2 + 1),x)

[Out]

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