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

Optimal. Leaf size=90 \[ \frac{x^3}{3}+\frac{x^2}{2}+\frac{2}{3} \log \left (x^2+x+1\right )-\frac{1}{24} \log \left (2 x^2-x+2\right )-\frac{3 x}{2}+\frac{5}{12} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )+\frac{8 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

(-3*x)/2 + x^2/2 + x^3/3 + (5*Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/12 + (8*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt
[3]) + (2*Log[1 + x + x^2])/3 - Log[2 - x + 2*x^2]/24

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Rubi [A]  time = 0.122615, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2075, 634, 618, 204, 628} \[ \frac{x^3}{3}+\frac{x^2}{2}+\frac{2}{3} \log \left (x^2+x+1\right )-\frac{1}{24} \log \left (2 x^2-x+2\right )-\frac{3 x}{2}+\frac{5}{12} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )+\frac{8 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/2 + x^2/2 + x^3/3 + (5*Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/12 + (8*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt
[3]) + (2*Log[1 + x + x^2])/3 - Log[2 - x + 2*x^2]/24

Rule 2075

Int[(P_)^(p_)*(Qm_), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Qm, x], x] /; QuadraticProdu
ctQ[PP, x]] /; PolyQ[Qm, x] && PolyQ[P, x] && ILtQ[p, 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 \frac{x^3 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx &=\int \left (-\frac{3}{2}+x+x^2+\frac{2 (3+2 x)}{3 \left (1+x+x^2\right )}+\frac{-6-x}{6 \left (2-x+2 x^2\right )}\right ) \, dx\\ &=-\frac{3 x}{2}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{1}{6} \int \frac{-6-x}{2-x+2 x^2} \, dx+\frac{2}{3} \int \frac{3+2 x}{1+x+x^2} \, dx\\ &=-\frac{3 x}{2}+\frac{x^2}{2}+\frac{x^3}{3}-\frac{1}{24} \int \frac{-1+4 x}{2-x+2 x^2} \, dx+\frac{2}{3} \int \frac{1+2 x}{1+x+x^2} \, dx-\frac{25}{24} \int \frac{1}{2-x+2 x^2} \, dx+\frac{4}{3} \int \frac{1}{1+x+x^2} \, dx\\ &=-\frac{3 x}{2}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{2}{3} \log \left (1+x+x^2\right )-\frac{1}{24} \log \left (2-x+2 x^2\right )+\frac{25}{12} \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{3 x}{2}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{5}{12} \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}{24} \log \left (2-x+2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0229295, size = 78, normalized size = 0.87 \[ \frac{1}{72} \left (24 x^3+36 x^2+48 \log \left (x^2+x+1\right )-3 \log \left (2 x^2-x+2\right )-108 x+64 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )-10 \sqrt{15} \tan ^{-1}\left (\frac{4 x-1}{\sqrt{15}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-108*x + 36*x^2 + 24*x^3 + 64*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 10*Sqrt[15]*ArcTan[(-1 + 4*x)/Sqrt[15]] + 4
8*Log[1 + x + x^2] - 3*Log[2 - x + 2*x^2])/72

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Maple [A]  time = 0.006, size = 69, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{3}}+{\frac{{x}^{2}}{2}}-{\frac{3\,x}{2}}-{\frac{\ln \left ( 2\,{x}^{2}-x+2 \right ) }{24}}-{\frac{5\,\sqrt{15}}{36}\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(x^3*(2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x)

[Out]

1/3*x^3+1/2*x^2-3/2*x-1/24*ln(2*x^2-x+2)-5/36*15^(1/2)*arctan(1/15*(-1+4*x)*15^(1/2))+2/3*ln(x^2+x+1)+8/9*arct
an(1/3*(1+2*x)*3^(1/2))*3^(1/2)

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Maxima [A]  time = 1.55046, size = 92, normalized size = 1.02 \begin{align*} \frac{1}{3} \, x^{3} + \frac{1}{2} \, x^{2} - \frac{5}{36} \, \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{3}{2} \, x - \frac{1}{24} \, \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(x^3*(2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x, algorithm="maxima")

[Out]

1/3*x^3 + 1/2*x^2 - 5/36*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) + 8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1))
- 3/2*x - 1/24*log(2*x^2 - x + 2) + 2/3*log(x^2 + x + 1)

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Fricas [A]  time = 1.40235, size = 246, normalized size = 2.73 \begin{align*} \frac{1}{3} \, x^{3} + \frac{1}{2} \, x^{2} - \frac{5}{36} \, \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{3}{2} \, x - \frac{1}{24} \, \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(x^3*(2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x, algorithm="fricas")

[Out]

1/3*x^3 + 1/2*x^2 - 5/36*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)) - 3/2*x - 1/24*log(2*x^2 - x + 2) + 2/3*log(x^2 + x + 1)

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Sympy [A]  time = 0.209759, size = 92, normalized size = 1.02 \begin{align*} \frac{x^{3}}{3} + \frac{x^{2}}{2} - \frac{3 x}{2} - \frac{\log{\left (x^{2} - \frac{x}{2} + 1 \right )}}{24} + \frac{2 \log{\left (x^{2} + x + 1 \right )}}{3} - \frac{5 \sqrt{15} \operatorname{atan}{\left (\frac{4 \sqrt{15} x}{15} - \frac{\sqrt{15}}{15} \right )}}{36} + \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(x**3*(2*x**3+3*x**2+x+5)/(2*x**4+x**3+3*x**2+x+2),x)

[Out]

x**3/3 + x**2/2 - 3*x/2 - log(x**2 - x/2 + 1)/24 + 2*log(x**2 + x + 1)/3 - 5*sqrt(15)*atan(4*sqrt(15)*x/15 - s
qrt(15)/15)/36 + 8*sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/9

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Giac [A]  time = 1.21717, size = 92, normalized size = 1.02 \begin{align*} \frac{1}{3} \, x^{3} + \frac{1}{2} \, x^{2} - \frac{5}{36} \, \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{3}{2} \, x - \frac{1}{24} \, \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(x^3*(2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x, algorithm="giac")

[Out]

1/3*x^3 + 1/2*x^2 - 5/36*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) + 8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1))
- 3/2*x - 1/24*log(2*x^2 - x + 2) + 2/3*log(x^2 + x + 1)