3.81 \(\int \log (1+x+x^2) \, dx\)

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

[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)

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Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {2523, 773, 634, 618, 204, 628} \[ x \log \left (x^2+x+1\right )+\frac {1}{2} \log \left (x^2+x+1\right )-2 x+\sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[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 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 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 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]

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 773

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 2523

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*} \int \log \left (1+x+x^2\right ) \, dx &=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 \operatorname {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*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.83 \[ \left (x+\frac {1}{2}\right ) \log \left (x^2+x+1\right )-2 x+\sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.44, size = 33, normalized size = 0.79 \[ \frac {1}{2} \, {\left (2 \, x + 1\right )} \log \left (x^{2} + x + 1\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.17, size = 37, normalized size = 0.88 \[ x \log \left (x^{2} + x + 1\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x + \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.07, size = 38, normalized size = 0.90 \[ x \ln \left (x^{2}+x +1\right )-2 x +\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )+\frac {\ln \left (x^{2}+x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x^2+x+1),x)

[Out]

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

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maxima [A]  time = 1.34, size = 37, normalized size = 0.88 \[ x \log \left (x^{2} + x + 1\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x + \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.06, size = 39, normalized size = 0.93 \[ \frac {\ln \left (x^2+x+1\right )}{2}-2\,x+\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}+\frac {\sqrt {3}}{3}\right )+x\,\ln \left (x^2+x+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.15, size = 46, normalized size = 1.10 \[ x \log {\left (x^{2} + x + 1 \right )} - 2 x + \frac {\log {\left (x^{2} + x + 1 \right )}}{2} + \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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