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

Optimal. Leaf size=91 \[ -\frac{5}{4 x^2}+\frac{2}{3} \log \left (x^2+x+1\right )+\frac{13}{48} \log \left (2 x^2-x+2\right )+\frac{3}{4 x}-\frac{15 \log (x)}{8}+\frac{1}{24} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )+\frac{8 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

-5/(4*x^2) + 3/(4*x) + (Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/24 + (8*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) -
(15*Log[x])/8 + (2*Log[1 + x + x^2])/3 + (13*Log[2 - x + 2*x^2])/48

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Rubi [A]  time = 0.156817, antiderivative size = 91, 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} \[ -\frac{5}{4 x^2}+\frac{2}{3} \log \left (x^2+x+1\right )+\frac{13}{48} \log \left (2 x^2-x+2\right )+\frac{3}{4 x}-\frac{15 \log (x)}{8}+\frac{1}{24} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{1-4 x}{\sqrt{15}}\right )+\frac{8 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-5/(4*x^2) + 3/(4*x) + (Sqrt[5/3]*ArcTan[(1 - 4*x)/Sqrt[15]])/24 + (8*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) -
(15*Log[x])/8 + (2*Log[1 + x + x^2])/3 + (13*Log[2 - x + 2*x^2])/48

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^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx &=-\left (\frac{1}{3} \int \frac{-6+4 x}{x^3 \left (4-2 x+4 x^2\right )} \, dx\right )+\frac{1}{3} \int \frac{24+16 x}{x^3 \left (4+4 x+4 x^2\right )} \, dx\\ &=\frac{1}{3} \int \left (\frac{6}{x^3}-\frac{2}{x^2}-\frac{4}{x}+\frac{2 (3+2 x)}{1+x+x^2}\right ) \, dx-\frac{1}{3} \int \left (-\frac{3}{2 x^3}+\frac{1}{4 x^2}+\frac{13}{8 x}+\frac{9-26 x}{8 \left (2-x+2 x^2\right )}\right ) \, dx\\ &=-\frac{5}{4 x^2}+\frac{3}{4 x}-\frac{15 \log (x)}{8}-\frac{1}{24} \int \frac{9-26 x}{2-x+2 x^2} \, dx+\frac{2}{3} \int \frac{3+2 x}{1+x+x^2} \, dx\\ &=-\frac{5}{4 x^2}+\frac{3}{4 x}-\frac{15 \log (x)}{8}-\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{2}{3} \int \frac{1+2 x}{1+x+x^2} \, dx+\frac{4}{3} \int \frac{1}{1+x+x^2} \, dx\\ &=-\frac{5}{4 x^2}+\frac{3}{4 x}-\frac{15 \log (x)}{8}+\frac{2}{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{8}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-\frac{5}{4 x^2}+\frac{3}{4 x}+\frac{1}{24} \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{15 \log (x)}{8}+\frac{2}{3} \log \left (1+x+x^2\right )+\frac{13}{48} \log \left (2-x+2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0547384, size = 82, normalized size = 0.9 \[ \frac{1}{144} \left (3 \left (-\frac{60}{x^2}+32 \log \left (x^2+x+1\right )+13 \log \left (2 x^2-x+2\right )+\frac{36}{x}-90 \log (x)\right )+128 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )-2 \sqrt{15} \tan ^{-1}\left (\frac{4 x-1}{\sqrt{15}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(128*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 2*Sqrt[15]*ArcTan[(-1 + 4*x)/Sqrt[15]] + 3*(-60/x^2 + 36/x - 90*Log[x
] + 32*Log[1 + x + x^2] + 13*Log[2 - x + 2*x^2]))/144

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Maple [A]  time = 0.008, size = 70, normalized size = 0.8 \begin{align*} -{\frac{5}{4\,{x}^{2}}}+{\frac{3}{4\,x}}-{\frac{15\,\ln \left ( x \right ) }{8}}+{\frac{13\,\ln \left ( 2\,{x}^{2}-x+2 \right ) }{48}}-{\frac{\sqrt{15}}{72}\arctan \left ({\frac{ \left ( -1+4\,x \right ) \sqrt{15}}{15}} \right ) }+{\frac{2\,\ln \left ({x}^{2}+x+1 \right ) }{3}}+{\frac{8\,\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^3/(2*x^4+x^3+3*x^2+x+2),x)

[Out]

-5/4/x^2+3/4/x-15/8*ln(x)+13/48*ln(2*x^2-x+2)-1/72*15^(1/2)*arctan(1/15*(-1+4*x)*15^(1/2))+2/3*ln(x^2+x+1)+8/9
*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)

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Maxima [A]  time = 1.4885, size = 93, normalized size = 1.02 \begin{align*} -\frac{1}{72} \, \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{3 \, x - 5}{4 \, x^{2}} + \frac{13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac{2}{3} \, \log \left (x^{2} + x + 1\right ) - \frac{15}{8} \, \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^3/(2*x^4+x^3+3*x^2+x+2),x, algorithm="maxima")

[Out]

-1/72*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) + 8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/4*(3*x - 5)/x^2
+ 13/48*log(2*x^2 - x + 2) + 2/3*log(x^2 + x + 1) - 15/8*log(x)

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Fricas [A]  time = 1.30494, size = 281, normalized size = 3.09 \begin{align*} -\frac{2 \, \sqrt{5} \sqrt{3} x^{2} \arctan \left (\frac{1}{15} \, \sqrt{5} \sqrt{3}{\left (4 \, x - 1\right )}\right ) - 128 \, \sqrt{3} x^{2} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - 39 \, x^{2} \log \left (2 \, x^{2} - x + 2\right ) - 96 \, x^{2} \log \left (x^{2} + x + 1\right ) + 270 \, x^{2} \log \left (x\right ) - 108 \, x + 180}{144 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/144*(2*sqrt(5)*sqrt(3)*x^2*arctan(1/15*sqrt(5)*sqrt(3)*(4*x - 1)) - 128*sqrt(3)*x^2*arctan(1/3*sqrt(3)*(2*x
+ 1)) - 39*x^2*log(2*x^2 - x + 2) - 96*x^2*log(x^2 + x + 1) + 270*x^2*log(x) - 108*x + 180)/x^2

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Sympy [A]  time = 0.28945, size = 94, normalized size = 1.03 \begin{align*} - \frac{15 \log{\left (x \right )}}{8} + \frac{13 \log{\left (x^{2} - \frac{x}{2} + 1 \right )}}{48} + \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 )}}{72} + \frac{8 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} + \frac{\sqrt{3}}{3} \right )}}{9} + \frac{3 x - 5}{4 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*log(x)/8 + 13*log(x**2 - x/2 + 1)/48 + 2*log(x**2 + x + 1)/3 - sqrt(15)*atan(4*sqrt(15)*x/15 - sqrt(15)/15
)/72 + 8*sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/9 + (3*x - 5)/(4*x**2)

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Giac [A]  time = 1.13334, size = 95, normalized size = 1.04 \begin{align*} -\frac{1}{72} \, \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{3 \, x - 5}{4 \, x^{2}} + \frac{13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac{2}{3} \, \log \left (x^{2} + x + 1\right ) - \frac{15}{8} \, \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^3/(2*x^4+x^3+3*x^2+x+2),x, algorithm="giac")

[Out]

-1/72*sqrt(15)*arctan(1/15*sqrt(15)*(4*x - 1)) + 8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/4*(3*x - 5)/x^2
+ 13/48*log(2*x^2 - x + 2) + 2/3*log(x^2 + x + 1) - 15/8*log(abs(x))