3.246 \(\int \frac{5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx\)

Optimal. Leaf size=71 \[ \frac{2}{3} \log \left (x^2+x+1\right )-\frac{1}{6} \log \left (2 x^2-x+2\right )-\frac{1}{3} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )+\frac{8 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

-(Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/3 + (8*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) + (2*Log[1 + x + x^2])/3
- Log[2 - x + 2*x^2]/6

________________________________________________________________________________________

Rubi [A]  time = 0.0782529, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {2074, 634, 618, 204, 628} \[ \frac{2}{3} \log \left (x^2+x+1\right )-\frac{1}{6} \log \left (2 x^2-x+2\right )-\frac{1}{3} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )+\frac{8 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(5 + x + 3*x^2 + 2*x^3)/(2 + x + 3*x^2 + x^3 + 2*x^4),x]

[Out]

-(Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/3 + (8*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) + (2*Log[1 + x + x^2])/3
- Log[2 - x + 2*x^2]/6

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]

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 \frac{5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx &=\int \left (\frac{2 (3+2 x)}{3 \left (1+x+x^2\right )}+\frac{3-2 x}{3 \left (2-x+2 x^2\right )}\right ) \, dx\\ &=\frac{1}{3} \int \frac{3-2 x}{2-x+2 x^2} \, dx+\frac{2}{3} \int \frac{3+2 x}{1+x+x^2} \, dx\\ &=-\left (\frac{1}{6} \int \frac{-1+4 x}{2-x+2 x^2} \, dx\right )+\frac{2}{3} \int \frac{1+2 x}{1+x+x^2} \, dx+\frac{5}{6} \int \frac{1}{2-x+2 x^2} \, dx+\frac{4}{3} \int \frac{1}{1+x+x^2} \, dx\\ &=\frac{2}{3} \log \left (1+x+x^2\right )-\frac{1}{6} \log \left (2-x+2 x^2\right )-\frac{5}{3} \operatorname{Subst}\left (\int \frac{1}{-15-x^2} \, dx,x,-1+4 x\right )-\frac{8}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-\frac{1}{3} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )+\frac{8 \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}+\frac{2}{3} \log \left (1+x+x^2\right )-\frac{1}{6} \log \left (2-x+2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.017241, size = 65, normalized size = 0.92 \[ \frac{1}{18} \left (12 \log \left (x^2+x+1\right )-3 \log \left (2 x^2-x+2\right )+16 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )+2 \sqrt{15} \tan ^{-1}\left (\frac{4 x-1}{\sqrt{15}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + x + 3*x^2 + 2*x^3)/(2 + x + 3*x^2 + x^3 + 2*x^4),x]

[Out]

(16*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] + 2*Sqrt[15]*ArcTan[(-1 + 4*x)/Sqrt[15]] + 12*Log[1 + x + x^2] - 3*Log[2
- x + 2*x^2])/18

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 56, normalized size = 0.8 \begin{align*} -{\frac{\ln \left ( 2\,{x}^{2}-x+2 \right ) }{6}}+{\frac{\sqrt{15}}{9}\arctan \left ({\frac{ \left ( -1+4\,x \right ) \sqrt{15}}{15}} \right ) }+{\frac{2\,\ln \left ({x}^{2}+x+1 \right ) }{3}}+{\frac{8\,\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x)

[Out]

-1/6*ln(2*x^2-x+2)+1/9*15^(1/2)*arctan(1/15*(-1+4*x)*15^(1/2))+2/3*ln(x^2+x+1)+8/9*arctan(1/3*(1+2*x)*3^(1/2))
*3^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 1.47321, size = 74, normalized size = 1.04 \begin{align*} \frac{1}{9} \, \sqrt{15} \arctan \left (\frac{1}{15} \, \sqrt{15}{\left (4 \, x - 1\right )}\right ) + \frac{8}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac{2}{3} \, \log \left (x^{2} + x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x, algorithm="maxima")

[Out]

1/9*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) + 8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/6*log(2*x^2 - x +
2) + 2/3*log(x^2 + x + 1)

________________________________________________________________________________________

Fricas [A]  time = 1.43082, size = 205, normalized size = 2.89 \begin{align*} \frac{1}{9} \, \sqrt{5} \sqrt{3} \arctan \left (\frac{1}{15} \, \sqrt{5} \sqrt{3}{\left (4 \, x - 1\right )}\right ) + \frac{8}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac{2}{3} \, \log \left (x^{2} + x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x, algorithm="fricas")

[Out]

1/9*sqrt(5)*sqrt(3)*arctan(1/15*sqrt(5)*sqrt(3)*(4*x - 1)) + 8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/6*l
og(2*x^2 - x + 2) + 2/3*log(x^2 + x + 1)

________________________________________________________________________________________

Sympy [A]  time = 0.19463, size = 75, normalized size = 1.06 \begin{align*} - \frac{\log{\left (x^{2} - \frac{x}{2} + 1 \right )}}{6} + \frac{2 \log{\left (x^{2} + x + 1 \right )}}{3} + \frac{\sqrt{15} \operatorname{atan}{\left (\frac{4 \sqrt{15} x}{15} - \frac{\sqrt{15}}{15} \right )}}{9} + \frac{8 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} + \frac{\sqrt{3}}{3} \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3+3*x**2+x+5)/(2*x**4+x**3+3*x**2+x+2),x)

[Out]

-log(x**2 - x/2 + 1)/6 + 2*log(x**2 + x + 1)/3 + sqrt(15)*atan(4*sqrt(15)*x/15 - sqrt(15)/15)/9 + 8*sqrt(3)*at
an(2*sqrt(3)*x/3 + sqrt(3)/3)/9

________________________________________________________________________________________

Giac [A]  time = 1.19205, size = 74, normalized size = 1.04 \begin{align*} \frac{1}{9} \, \sqrt{15} \arctan \left (\frac{1}{15} \, \sqrt{15}{\left (4 \, x - 1\right )}\right ) + \frac{8}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac{2}{3} \, \log \left (x^{2} + x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x, algorithm="giac")

[Out]

1/9*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) + 8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/6*log(2*x^2 - x +
2) + 2/3*log(x^2 + x + 1)