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

Optimal. Leaf size=75 \[ -\log \left (x^2+x+1\right )-\frac {1}{4} \log \left (2 x^2-x+2\right )+\frac {5 \log (x)}{2}+\frac {1}{6} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {2 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

[Out]

5/2*ln(x)-ln(x^2+x+1)-1/4*ln(2*x^2-x+2)+1/18*arctan(1/15*(1-4*x)*15^(1/2))*15^(1/2)+2/9*arctan(1/3*(1+2*x)*3^(
1/2))*3^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2087, 800, 634, 618, 204, 628} \[ -\log \left (x^2+x+1\right )-\frac {1}{4} \log \left (2 x^2-x+2\right )+\frac {5 \log (x)}{2}+\frac {1}{6} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {2 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2087

Int[((P3_)*(x_)^(m_.))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2 + (d_.)*(x_)^3 + (e_.)*(x_)^4), x_Symbol] :> With[{q
= Sqrt[8*a^2 + b^2 - 4*a*c], A = Coeff[P3, x, 0], B = Coeff[P3, x, 1], C = Coeff[P3, x, 2], D = Coeff[P3, x, 3
]}, Dist[1/q, Int[(x^m*(b*A - 2*a*B + 2*a*D + A*q + (2*a*A - 2*a*C + b*D + D*q)*x))/(2*a + (b + q)*x + 2*a*x^2
), x], x] - Dist[1/q, Int[(x^m*(b*A - 2*a*B + 2*a*D - A*q + (2*a*A - 2*a*C + b*D - D*q)*x))/(2*a + (b - q)*x +
2*a*x^2), x], x]] /; FreeQ[{a, b, c, m}, x] && PolyQ[P3, x, 3] && EqQ[a, e] && EqQ[b, d]

Rubi steps

\begin {align*} \int \frac {5+x+3 x^2+2 x^3}{x \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx &=-\left (\frac {1}{3} \int \frac {-6+4 x}{x \left (4-2 x+4 x^2\right )} \, dx\right )+\frac {1}{3} \int \frac {24+16 x}{x \left (4+4 x+4 x^2\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {6}{x}-\frac {2 (1+3 x)}{1+x+x^2}\right ) \, dx-\frac {1}{3} \int \left (-\frac {3}{2 x}+\frac {1+6 x}{2 \left (2-x+2 x^2\right )}\right ) \, dx\\ &=\frac {5 \log (x)}{2}-\frac {1}{6} \int \frac {1+6 x}{2-x+2 x^2} \, dx-\frac {2}{3} \int \frac {1+3 x}{1+x+x^2} \, dx\\ &=\frac {5 \log (x)}{2}-\frac {1}{4} \int \frac {-1+4 x}{2-x+2 x^2} \, dx+\frac {1}{3} \int \frac {1}{1+x+x^2} \, dx-\frac {5}{12} \int \frac {1}{2-x+2 x^2} \, dx-\int \frac {1+2 x}{1+x+x^2} \, dx\\ &=\frac {5 \log (x)}{2}-\log \left (1+x+x^2\right )-\frac {1}{4} \log \left (2-x+2 x^2\right )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )+\frac {5}{6} \operatorname {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,-1+4 x\right )\\ &=\frac {1}{6} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {2 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {5 \log (x)}{2}-\log \left (1+x+x^2\right )-\frac {1}{4} \log \left (2-x+2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 69, normalized size = 0.92 \[ \frac {1}{36} \left (-36 \log \left (x^2+x+1\right )-9 \log \left (2 x^2-x+2\right )+90 \log (x)+8 \sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )-2 \sqrt {15} \tan ^{-1}\left (\frac {4 x-1}{\sqrt {15}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 2*Sqrt[15]*ArcTan[(-1 + 4*x)/Sqrt[15]] + 90*Log[x] - 36*Log[1 + x + x^2
] - 9*Log[2 - x + 2*x^2])/36

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fricas [A]  time = 0.96, size = 65, normalized size = 0.87 \[ -\frac {1}{18} \, \sqrt {5} \sqrt {3} \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) + \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - x + 2\right ) - \log \left (x^{2} + x + 1\right ) + \frac {5}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*sqrt(5)*sqrt(3)*arctan(1/15*sqrt(5)*sqrt(3)*(4*x - 1)) + 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/4
*log(2*x^2 - x + 2) - log(x^2 + x + 1) + 5/2*log(x)

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giac [A]  time = 0.31, size = 60, normalized size = 0.80 \[ -\frac {1}{18} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - x + 2\right ) - \log \left (x^{2} + x + 1\right ) + \frac {5}{2} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) + 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/4*log(2*x^2 - x
+ 2) - log(x^2 + x + 1) + 5/2*log(abs(x))

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maple [A]  time = 0.01, size = 60, normalized size = 0.80 \[ -\frac {\sqrt {15}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {15}}{15}\right )}{18}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {5 \ln \relax (x )}{2}-\ln \left (x^{2}+x +1\right )-\frac {\ln \left (2 x^{2}-x +2\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*ln(2*x^2-x+2)-1/18*15^(1/2)*arctan(1/15*(4*x-1)*15^(1/2))+5/2*ln(x)-ln(x^2+x+1)+2/9*3^(1/2)*arctan(1/3*(2
*x+1)*3^(1/2))

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maxima [A]  time = 1.57, size = 59, normalized size = 0.79 \[ -\frac {1}{18} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - x + 2\right ) - \log \left (x^{2} + x + 1\right ) + \frac {5}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) + 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/4*log(2*x^2 - x
+ 2) - log(x^2 + x + 1) + 5/2*log(x)

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mupad [B]  time = 0.15, size = 83, normalized size = 1.11 \[ \frac {5\,\ln \relax (x)}{2}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (1+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-1+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{36}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{36}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*log(x))/2 - log(x - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*1i)/9 + 1) + log(x + (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*1i
)/9 - 1) + log(x - (15^(1/2)*1i)/4 - 1/4)*((15^(1/2)*1i)/36 - 1/4) - log(x + (15^(1/2)*1i)/4 - 1/4)*((15^(1/2)
*1i)/36 + 1/4)

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sympy [A]  time = 0.29, size = 78, normalized size = 1.04 \[ \frac {5 \log {\relax (x )}}{2} - \frac {\log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{4} - \log {\left (x^{2} + x + 1 \right )} - \frac {\sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{18} + \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*log(x)/2 - log(x**2 - x/2 + 1)/4 - log(x**2 + x + 1) - sqrt(15)*atan(4*sqrt(15)*x/15 - sqrt(15)/15)/18 + 2*s
qrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/9

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