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]

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

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Rubi [A]  time = 0.14, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.69, size = 79, normalized size = 0.81 \[ \frac {1}{4} \, x^{4} + \frac {1}{3} \, x^{3} - \frac {3}{4} \, x^{2} - \frac {1}{72} \, \sqrt {5} \sqrt {3} \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {5}{4} \, x - \frac {13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) \]

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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giac [A]  time = 0.31, size = 73, normalized size = 0.75 \[ \frac {1}{4} \, x^{4} + \frac {1}{3} \, x^{3} - \frac {3}{4} \, x^{2} - \frac {1}{72} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {5}{4} \, x - \frac {13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) \]

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)

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maple [A]  time = 0.01, size = 74, normalized size = 0.76 \[ \frac {x^{4}}{4}+\frac {x^{3}}{3}-\frac {3 x^{2}}{4}+\frac {5 x}{4}-\frac {\sqrt {15}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {15}}{15}\right )}{72}-\frac {10 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {\ln \left (x^{2}+x +1\right )}{3}-\frac {13 \ln \left (2 x^{2}-x +2\right )}{48} \]

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

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maxima [A]  time = 1.52, size = 73, normalized size = 0.75 \[ \frac {1}{4} \, x^{4} + \frac {1}{3} \, x^{3} - \frac {3}{4} \, x^{2} - \frac {1}{72} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {5}{4} \, x - \frac {13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) \]

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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mupad [B]  time = 0.19, size = 97, normalized size = 1.00 \[ \frac {5\,x}{4}+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {13}{48}+\frac {\sqrt {15}\,1{}\mathrm {i}}{144}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {13}{48}+\frac {\sqrt {15}\,1{}\mathrm {i}}{144}\right )-\frac {3\,x^2}{4}+\frac {x^3}{3}+\frac {x^4}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*x)/4 + log(x - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*5i)/9 + 1/3) - log(x + (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*5i)/9
- 1/3) + log(x - (15^(1/2)*1i)/4 - 1/4)*((15^(1/2)*1i)/144 - 13/48) - log(x + (15^(1/2)*1i)/4 - 1/4)*((15^(1/
2)*1i)/144 + 13/48) - (3*x^2)/4 + x^3/3 + x^4/4

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sympy [A]  time = 0.25, size = 97, normalized size = 1.00 \[ \frac {x^{4}}{4} + \frac {x^{3}}{3} - \frac {3 x^{2}}{4} + \frac {5 x}{4} - \frac {13 \log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{48} + \frac {\log {\left (x^{2} + x + 1 \right )}}{3} - \frac {\sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{72} - \frac {10 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]

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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