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

Optimal. Leaf size=84 \[ \frac {1}{3} \log \left (x^2+x+1\right )+\frac {1}{24} \log \left (2 x^2-x+2\right )-\frac {5}{2 x}-\frac {3 \log (x)}{4}+\frac {5}{12} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

[Out]

-5/2/x-3/4*ln(x)+1/3*ln(x^2+x+1)+1/24*ln(2*x^2-x+2)+5/36*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 = 0.15, antiderivative size = 84, 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} \[ \frac {1}{3} \log \left (x^2+x+1\right )+\frac {1}{24} \log \left (2 x^2-x+2\right )-\frac {5}{2 x}-\frac {3 \log (x)}{4}+\frac {5}{12} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-5/(2*x) + (5*Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/12 - (10*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) - (3*Log[x
])/4 + Log[1 + x + x^2]/3 + Log[2 - x + 2*x^2]/24

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^2 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx &=-\left (\frac {1}{3} \int \frac {-6+4 x}{x^2 \left (4-2 x+4 x^2\right )} \, dx\right )+\frac {1}{3} \int \frac {24+16 x}{x^2 \left (4+4 x+4 x^2\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {6}{x^2}-\frac {2}{x}+\frac {2 (-2+x)}{1+x+x^2}\right ) \, dx-\frac {1}{3} \int \left (-\frac {3}{2 x^2}+\frac {1}{4 x}+\frac {13-2 x}{4 \left (2-x+2 x^2\right )}\right ) \, dx\\ &=-\frac {5}{2 x}-\frac {3 \log (x)}{4}-\frac {1}{12} \int \frac {13-2 x}{2-x+2 x^2} \, dx+\frac {2}{3} \int \frac {-2+x}{1+x+x^2} \, dx\\ &=-\frac {5}{2 x}-\frac {3 \log (x)}{4}+\frac {1}{24} \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 {25}{24} \int \frac {1}{2-x+2 x^2} \, dx-\frac {5}{3} \int \frac {1}{1+x+x^2} \, dx\\ &=-\frac {5}{2 x}-\frac {3 \log (x)}{4}+\frac {1}{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 {10}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-\frac {5}{2 x}+\frac {5}{12} \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 {3 \log (x)}{4}+\frac {1}{3} \log \left (1+x+x^2\right )+\frac {1}{24} \log \left (2-x+2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 78, normalized size = 0.93 \[ -\frac {-24 x \log \left (x^2+x+1\right )-3 x \log \left (2 x^2-x+2\right )+54 x \log (x)+80 \sqrt {3} x \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )+10 \sqrt {15} x \tan ^{-1}\left (\frac {4 x-1}{\sqrt {15}}\right )+180}{72 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/72*(180 + 80*Sqrt[3]*x*ArcTan[(1 + 2*x)/Sqrt[3]] + 10*Sqrt[15]*x*ArcTan[(-1 + 4*x)/Sqrt[15]] + 54*x*Log[x]
- 24*x*Log[1 + x + x^2] - 3*x*Log[2 - x + 2*x^2])/x

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fricas [A]  time = 0.89, size = 76, normalized size = 0.90 \[ -\frac {10 \, \sqrt {5} \sqrt {3} x \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) + 80 \, \sqrt {3} x \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 3 \, x \log \left (2 \, x^{2} - x + 2\right ) - 24 \, x \log \left (x^{2} + x + 1\right ) + 54 \, x \log \relax (x) + 180}{72 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/72*(10*sqrt(5)*sqrt(3)*x*arctan(1/15*sqrt(5)*sqrt(3)*(4*x - 1)) + 80*sqrt(3)*x*arctan(1/3*sqrt(3)*(2*x + 1)
) - 3*x*log(2*x^2 - x + 2) - 24*x*log(x^2 + x + 1) + 54*x*log(x) + 180)/x

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giac [A]  time = 0.26, size = 65, normalized size = 0.77 \[ -\frac {5}{36} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {5}{2 \, x} + \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {3}{4} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/36*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) - 10/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 5/2/x + 1/24*log
(2*x^2 - x + 2) + 1/3*log(x^2 + x + 1) - 3/4*log(abs(x))

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maple [A]  time = 0.01, size = 65, normalized size = 0.77 \[ -\frac {5 \sqrt {15}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {15}}{15}\right )}{36}-\frac {10 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}-\frac {3 \ln \relax (x )}{4}+\frac {\ln \left (x^{2}+x +1\right )}{3}+\frac {\ln \left (2 x^{2}-x +2\right )}{24}-\frac {5}{2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*ln(2*x^2-x+2)-5/36*15^(1/2)*arctan(1/15*(4*x-1)*15^(1/2))-5/2/x-3/4*ln(x)+1/3*ln(x^2+x+1)-10/9*3^(1/2)*ar
ctan(1/3*(2*x+1)*3^(1/2))

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maxima [A]  time = 1.44, size = 64, normalized size = 0.76 \[ -\frac {5}{36} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {5}{2 \, x} + \frac {1}{24} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {3}{4} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/36*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) - 10/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 5/2/x + 1/24*log
(2*x^2 - x + 2) + 1/3*log(x^2 + x + 1) - 3/4*log(x)

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mupad [B]  time = 2.28, size = 88, normalized size = 1.05 \[ -\frac {3\,\ln \relax (x)}{4}+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{24}+\frac {\sqrt {15}\,5{}\mathrm {i}}{72}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {15}\,5{}\mathrm {i}}{72}\right )-\frac {5}{2\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*5i)/9 + 1/3) - (3*log(x))/4 - 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)*5i)/72 + 1/24) - log(x + (15^(1/2)*1i)/4 - 1/4)*((15^
(1/2)*5i)/72 - 1/24) - 5/(2*x)

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sympy [A]  time = 0.31, size = 87, normalized size = 1.04 \[ - \frac {3 \log {\relax (x )}}{4} + \frac {\log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{24} + \frac {\log {\left (x^{2} + x + 1 \right )}}{3} - \frac {5 \sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{36} - \frac {10 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} - \frac {5}{2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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