3.292 \(\int \frac {5+2 x-x^2+x^3}{1+x+x^2} \, dx\)

Optimal. Leaf size=41 \[ \frac {x^2}{2}+\frac {3}{2} \log \left (x^2+x+1\right )-2 x+\frac {11 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

-2*x+1/2*x^2+3/2*ln(x^2+x+1)+11/3*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1657, 634, 618, 204, 628} \[ \frac {x^2}{2}+\frac {3}{2} \log \left (x^2+x+1\right )-2 x+\frac {11 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 2*x - x^2 + x^3)/(1 + x + x^2),x]

[Out]

-2*x + x^2/2 + (11*ArcTan[(1 + 2*x)/Sqrt[3]])/Sqrt[3] + (3*Log[1 + x + x^2])/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 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {5+2 x-x^2+x^3}{1+x+x^2} \, dx &=\int \left (-2+x+\frac {7+3 x}{1+x+x^2}\right ) \, dx\\ &=-2 x+\frac {x^2}{2}+\int \frac {7+3 x}{1+x+x^2} \, dx\\ &=-2 x+\frac {x^2}{2}+\frac {3}{2} \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {11}{2} \int \frac {1}{1+x+x^2} \, dx\\ &=-2 x+\frac {x^2}{2}+\frac {3}{2} \log \left (1+x+x^2\right )-11 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-2 x+\frac {x^2}{2}+\frac {11 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {3}{2} \log \left (1+x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 1.00 \[ \frac {x^2}{2}+\frac {3}{2} \log \left (x^2+x+1\right )-2 x+\frac {11 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + 2*x - x^2 + x^3)/(1 + x + x^2),x]

[Out]

-2*x + x^2/2 + (11*ArcTan[(1 + 2*x)/Sqrt[3]])/Sqrt[3] + (3*Log[1 + x + x^2])/2

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fricas [A]  time = 0.92, size = 34, normalized size = 0.83 \[ \frac {1}{2} \, x^{2} + \frac {11}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x + \frac {3}{2} \, \log \left (x^{2} + x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-x^2+2*x+5)/(x^2+x+1),x, algorithm="fricas")

[Out]

1/2*x^2 + 11/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 2*x + 3/2*log(x^2 + x + 1)

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giac [A]  time = 0.29, size = 34, normalized size = 0.83 \[ \frac {1}{2} \, x^{2} + \frac {11}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x + \frac {3}{2} \, \log \left (x^{2} + x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-x^2+2*x+5)/(x^2+x+1),x, algorithm="giac")

[Out]

1/2*x^2 + 11/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 2*x + 3/2*log(x^2 + x + 1)

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maple [A]  time = 0.00, size = 35, normalized size = 0.85 \[ \frac {x^{2}}{2}-2 x +\frac {11 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}+\frac {3 \ln \left (x^{2}+x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-x^2+2*x+5)/(x^2+x+1),x)

[Out]

-2*x+1/2*x^2+3/2*ln(x^2+x+1)+11/3*3^(1/2)*arctan(1/3*(2*x+1)*3^(1/2))

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maxima [A]  time = 2.18, size = 34, normalized size = 0.83 \[ \frac {1}{2} \, x^{2} + \frac {11}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, x + \frac {3}{2} \, \log \left (x^{2} + x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-x^2+2*x+5)/(x^2+x+1),x, algorithm="maxima")

[Out]

1/2*x^2 + 11/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 2*x + 3/2*log(x^2 + x + 1)

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mupad [B]  time = 0.04, size = 36, normalized size = 0.88 \[ \frac {3\,\ln \left (x^2+x+1\right )}{2}-2\,x+\frac {11\,\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}+\frac {\sqrt {3}}{3}\right )}{3}+\frac {x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - x^2 + x^3 + 5)/(x + x^2 + 1),x)

[Out]

(3*log(x + x^2 + 1))/2 - 2*x + (11*3^(1/2)*atan((2*3^(1/2)*x)/3 + 3^(1/2)/3))/3 + x^2/2

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sympy [A]  time = 0.12, size = 46, normalized size = 1.12 \[ \frac {x^{2}}{2} - 2 x + \frac {3 \log {\left (x^{2} + x + 1 \right )}}{2} + \frac {11 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-x**2+2*x+5)/(x**2+x+1),x)

[Out]

x**2/2 - 2*x + 3*log(x**2 + x + 1)/2 + 11*sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/3

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