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

Optimal. Leaf size=31 \[ \log \left (x^2+x+1\right )+\log (x+1)-\frac {2 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2074, 634, 618, 204, 628} \[ \log \left (x^2+x+1\right )+\log (x+1)-\frac {2 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 3*x + 3*x^2)/(1 + 2*x + 2*x^2 + x^3),x]

[Out]

(-2*ArcTan[(1 + 2*x)/Sqrt[3]])/Sqrt[3] + Log[1 + 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 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.00 \[ \log \left (x^2+x+1\right )+\log (x+1)-\frac {2 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 3*x + 3*x^2)/(1 + 2*x + 2*x^2 + x^3),x]

[Out]

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

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fricas [A]  time = 0.89, size = 28, normalized size = 0.90 \[ -\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \log \left (x^{2} + x + 1\right ) + \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 29, normalized size = 0.94 \[ -\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \log \left (x^{2} + x + 1\right ) + \log \left ({\left | x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 29, normalized size = 0.94 \[ -\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}+\ln \left (x +1\right )+\ln \left (x^{2}+x +1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+3*x+1)/(x^3+2*x^2+2*x+1),x)

[Out]

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

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maxima [A]  time = 2.36, size = 28, normalized size = 0.90 \[ -\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \log \left (x^{2} + x + 1\right ) + \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 57, normalized size = 1.84 \[ \ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )+\ln \left (x+1\right )+\frac {\sqrt {3}\,\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{3}-\frac {\sqrt {3}\,\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 3*x^2 + 1)/(2*x + 2*x^2 + x^3 + 1),x)

[Out]

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

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sympy [A]  time = 0.13, size = 3, normalized size = 0.10 \[ \log {\left (x + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+3*x+1)/(x**3+2*x**2+2*x+1),x)

[Out]

log(x + 1)

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