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

Optimal. Leaf size=97 \[ \frac{x^4}{4}+\frac{x^3}{3}-\frac{3 x^2}{4}+\frac{1}{3} \log \left (x^2+x+1\right )-\frac{13}{48} \log \left (2 x^2-x+2\right )+\frac{5 x}{4}+\frac{1}{24} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )-\frac{10 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

(5*x)/4 - (3*x^2)/4 + x^3/3 + x^4/4 + (Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/24 - (10*ArcTan[(1 + 2*x)/Sqrt[3]
])/(3*Sqrt[3]) + Log[1 + x + x^2]/3 - (13*Log[2 - x + 2*x^2])/48

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Rubi [A]  time = 0.137518, antiderivative size = 97, 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} \[ \frac{x^4}{4}+\frac{x^3}{3}-\frac{3 x^2}{4}+\frac{1}{3} \log \left (x^2+x+1\right )-\frac{13}{48} \log \left (2 x^2-x+2\right )+\frac{5 x}{4}+\frac{1}{24} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )-\frac{10 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/4 - (3*x^2)/4 + x^3/3 + x^4/4 + (Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/24 - (10*ArcTan[(1 + 2*x)/Sqrt[3]
])/(3*Sqrt[3]) + Log[1 + x + x^2]/3 - (13*Log[2 - x + 2*x^2])/48

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^4 \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx &=\int \left (\frac{5}{4}-\frac{3 x}{2}+x^2+x^3+\frac{2 (-2+x)}{3 \left (1+x+x^2\right )}+\frac{2-13 x}{12 \left (2-x+2 x^2\right )}\right ) \, dx\\ &=\frac{5 x}{4}-\frac{3 x^2}{4}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{1}{12} \int \frac{2-13 x}{2-x+2 x^2} \, dx+\frac{2}{3} \int \frac{-2+x}{1+x+x^2} \, dx\\ &=\frac{5 x}{4}-\frac{3 x^2}{4}+\frac{x^3}{3}+\frac{x^4}{4}-\frac{5}{48} \int \frac{1}{2-x+2 x^2} \, dx-\frac{13}{48} \int \frac{-1+4 x}{2-x+2 x^2} \, dx+\frac{1}{3} \int \frac{1+2 x}{1+x+x^2} \, dx-\frac{5}{3} \int \frac{1}{1+x+x^2} \, dx\\ &=\frac{5 x}{4}-\frac{3 x^2}{4}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{1}{3} \log \left (1+x+x^2\right )-\frac{13}{48} \log \left (2-x+2 x^2\right )+\frac{5}{24} \operatorname{Subst}\left (\int \frac{1}{-15-x^2} \, dx,x,-1+4 x\right )+\frac{10}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac{5 x}{4}-\frac{3 x^2}{4}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{1}{24} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )-\frac{10 \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}+\frac{1}{3} \log \left (1+x+x^2\right )-\frac{13}{48} \log \left (2-x+2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0368234, size = 83, normalized size = 0.86 \[ \frac{1}{144} \left (36 x^4+48 x^3-108 x^2+48 \log \left (x^2+x+1\right )-39 \log \left (2 x^2-x+2\right )+180 x-160 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )-2 \sqrt{15} \tan ^{-1}\left (\frac{4 x-1}{\sqrt{15}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(180*x - 108*x^2 + 48*x^3 + 36*x^4 - 160*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 2*Sqrt[15]*ArcTan[(-1 + 4*x)/Sqrt
[15]] + 48*Log[1 + x + x^2] - 39*Log[2 - x + 2*x^2])/144

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Maple [A]  time = 0.007, size = 74, normalized size = 0.8 \begin{align*}{\frac{{x}^{4}}{4}}+{\frac{{x}^{3}}{3}}-{\frac{3\,{x}^{2}}{4}}+{\frac{5\,x}{4}}-{\frac{13\,\ln \left ( 2\,{x}^{2}-x+2 \right ) }{48}}-{\frac{\sqrt{15}}{72}\arctan \left ({\frac{ \left ( -1+4\,x \right ) \sqrt{15}}{15}} \right ) }+{\frac{\ln \left ({x}^{2}+x+1 \right ) }{3}}-{\frac{10\,\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^4*(2*x^3+3*x^2+x+5)/(2*x^4+x^3+3*x^2+x+2),x)

[Out]

1/4*x^4+1/3*x^3-3/4*x^2+5/4*x-13/48*ln(2*x^2-x+2)-1/72*15^(1/2)*arctan(1/15*(-1+4*x)*15^(1/2))+1/3*ln(x^2+x+1)
-10/9*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)

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Maxima [A]  time = 1.50038, size = 99, normalized size = 1.02 \begin{align*} \frac{1}{4} \, x^{4} + \frac{1}{3} \, x^{3} - \frac{3}{4} \, x^{2} - \frac{1}{72} \, \sqrt{15} \arctan \left (\frac{1}{15} \, \sqrt{15}{\left (4 \, x - 1\right )}\right ) - \frac{10}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{5}{4} \, x - \frac{13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac{1}{3} \, \log \left (x^{2} + x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4 + 1/3*x^3 - 3/4*x^2 - 1/72*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) - 10/9*sqrt(3)*arctan(1/3*sqrt(3)*
(2*x + 1)) + 5/4*x - 13/48*log(2*x^2 - x + 2) + 1/3*log(x^2 + x + 1)

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Fricas [A]  time = 1.2355, size = 262, normalized size = 2.7 \begin{align*} \frac{1}{4} \, x^{4} + \frac{1}{3} \, x^{3} - \frac{3}{4} \, x^{2} - \frac{1}{72} \, \sqrt{5} \sqrt{3} \arctan \left (\frac{1}{15} \, \sqrt{5} \sqrt{3}{\left (4 \, x - 1\right )}\right ) - \frac{10}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{5}{4} \, x - \frac{13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac{1}{3} \, \log \left (x^{2} + x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4 + 1/3*x^3 - 3/4*x^2 - 1/72*sqrt(5)*sqrt(3)*arctan(1/15*sqrt(5)*sqrt(3)*(4*x - 1)) - 10/9*sqrt(3)*arcta
n(1/3*sqrt(3)*(2*x + 1)) + 5/4*x - 13/48*log(2*x^2 - x + 2) + 1/3*log(x^2 + x + 1)

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Sympy [A]  time = 0.212874, size = 97, normalized size = 1. \begin{align*} \frac{x^{4}}{4} + \frac{x^{3}}{3} - \frac{3 x^{2}}{4} + \frac{5 x}{4} - \frac{13 \log{\left (x^{2} - \frac{x}{2} + 1 \right )}}{48} + \frac{\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 )}}{72} - \frac{10 \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**4*(2*x**3+3*x**2+x+5)/(2*x**4+x**3+3*x**2+x+2),x)

[Out]

x**4/4 + x**3/3 - 3*x**2/4 + 5*x/4 - 13*log(x**2 - x/2 + 1)/48 + log(x**2 + x + 1)/3 - sqrt(15)*atan(4*sqrt(15
)*x/15 - sqrt(15)/15)/72 - 10*sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/9

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Giac [A]  time = 1.13766, size = 99, normalized size = 1.02 \begin{align*} \frac{1}{4} \, x^{4} + \frac{1}{3} \, x^{3} - \frac{3}{4} \, x^{2} - \frac{1}{72} \, \sqrt{15} \arctan \left (\frac{1}{15} \, \sqrt{15}{\left (4 \, x - 1\right )}\right ) - \frac{10}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{5}{4} \, x - \frac{13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac{1}{3} \, \log \left (x^{2} + x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4 + 1/3*x^3 - 3/4*x^2 - 1/72*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) - 10/9*sqrt(3)*arctan(1/3*sqrt(3)*
(2*x + 1)) + 5/4*x - 13/48*log(2*x^2 - x + 2) + 1/3*log(x^2 + x + 1)