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) + (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

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Rubi [A]  time = 0.151957, antiderivative size = 84, normalized size of antiderivative = 1., 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 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]

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 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{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*}

Mathematica [A]  time = 0.0332324, 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]

-(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])/(72*x)

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Maple [A]  time = 0.007, size = 65, normalized size = 0.8 \begin{align*} -{\frac{5}{2\,x}}-{\frac{3\,\ln \left ( x \right ) }{4}}+{\frac{\ln \left ( 2\,{x}^{2}-x+2 \right ) }{24}}-{\frac{5\,\sqrt{15}}{36}\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((2*x^3+3*x^2+x+5)/x^2/(2*x^4+x^3+3*x^2+x+2),x)

[Out]

-5/2/x-3/4*ln(x)+1/24*ln(2*x^2-x+2)-5/36*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 = 1.51129, size = 86, normalized size = 1.02 \begin{align*} -\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 (x\right ) \end{align*}

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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Fricas [A]  time = 1.40958, size = 250, normalized size = 2.98 \begin{align*} -\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 \left (x\right ) + 180}{72 \, x} \end{align*}

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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Sympy [A]  time = 0.275504, size = 87, normalized size = 1.04 \begin{align*} - \frac{3 \log{\left (x \right )}}{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} \end{align*}

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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Giac [A]  time = 1.19833, size = 88, normalized size = 1.05 \begin{align*} -\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 ) \end{align*}

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))