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

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Rubi [A]  time = 0.0263058, antiderivative size = 42, normalized size of antiderivative = 1., 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 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]

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 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 \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*}

Mathematica [A]  time = 0.012705, 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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Maple [A]  time = 0.007, size = 38, normalized size = 0.9 \begin{align*} -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} \end{align*}

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

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Maxima [A]  time = 1.77586, size = 50, normalized size = 1.19 \begin{align*} 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 ) \end{align*}

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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Fricas [A]  time = 2.03413, size = 105, normalized size = 2.5 \begin{align*} \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 \end{align*}

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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Sympy [A]  time = 0.140535, size = 46, normalized size = 1.1 \begin{align*} 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 )} \end{align*}

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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Giac [A]  time = 1.16446, size = 50, normalized size = 1.19 \begin{align*} 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 ) \end{align*}

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)