\(\int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx\) [246]

Optimal result

Integrand size = 32, antiderivative size = 71 \[ \int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx=-\frac {1}{3} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {2}{3} \log \left (1+x+x^2\right )-\frac {1}{6} \log \left (2-x+2 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2099, 648, 632, 210, 642} \[ \int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx=-\frac {1}{3} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {2}{3} \log \left (x^2+x+1\right )-\frac {1}{6} \log \left (2 x^2-x+2\right ) \]

[In]

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

[Out]

-1/3*(Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]]) + (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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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 642

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 648

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 2099

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]

Rubi steps \begin{align*} \text {integral}& = \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} \text {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,-1+4 x\right )-\frac {8}{3} \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {1}{18} \left (16 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+2 \sqrt {15} \arctan \left (\frac {-1+4 x}{\sqrt {15}}\right )+12 \log \left (1+x+x^2\right )-3 \log \left (2-x+2 x^2\right )\right ) \]

[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] (verified)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {2 \ln \left (x^{2}+x +1\right )}{3}+\frac {8 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}-\frac {\ln \left (2 x^{2}-x +2\right )}{6}+\frac {\sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{9}\)	\(56\)
risch	\(\frac {2 \ln \left (4 x^{2}+4 x +4\right )}{3}+\frac {8 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}-\frac {\ln \left (16 x^{2}-8 x +16\right )}{6}+\frac {\sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{9}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {1}{9} \, \sqrt {5} \sqrt {3} \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) \]

[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] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx=- \frac {\log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{6} + \frac {2 \log {\left (x^{2} + x + 1 \right )}}{3} + \frac {\sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{9} + \frac {8 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {1}{9} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx=\frac {1}{9} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.11 \[ \int \frac {5+x+3 x^2+2 x^3}{2+x+3 x^2+x^3+2 x^4} \, dx=-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {2}{3}+\frac {\sqrt {3}\,4{}\mathrm {i}}{9}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {2}{3}+\frac {\sqrt {3}\,4{}\mathrm {i}}{9}\right )-\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{6}+\frac {\sqrt {15}\,1{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {15}\,1{}\mathrm {i}}{18}\right ) \]

[In]

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

[Out]

log(x + (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*4i)/9 + 2/3) - log(x - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*4i)/9 - 2/3) -
log(x - (15^(1/2)*1i)/4 - 1/4)*((15^(1/2)*1i)/18 + 1/6) + log(x + (15^(1/2)*1i)/4 - 1/4)*((15^(1/2)*1i)/18 - 1
/6)