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

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Rubi [A]  time = 0.16, antiderivative size = 91, 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} \[ -\frac {5}{4 x^2}+\frac {2}{3} \log \left (x^2+x+1\right )+\frac {13}{48} \log \left (2 x^2-x+2\right )+\frac {3}{4 x}-\frac {15 \log (x)}{8}+\frac {1}{24} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

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 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^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*}

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Mathematica [A]  time = 0.06, size = 82, normalized size = 0.90 \[ \frac {1}{144} \left (3 \left (-\frac {60}{x^2}+32 \log \left (x^2+x+1\right )+13 \log \left (2 x^2-x+2\right )+\frac {36}{x}-90 \log (x)\right )+128 \sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )-2 \sqrt {15} \tan ^{-1}\left (\frac {4 x-1}{\sqrt {15}}\right )\right ) \]

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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fricas [A]  time = 0.97, size = 89, normalized size = 0.98 \[ -\frac {2 \, \sqrt {5} \sqrt {3} x^{2} \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) - 128 \, \sqrt {3} x^{2} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 39 \, x^{2} \log \left (2 \, x^{2} - x + 2\right ) - 96 \, x^{2} \log \left (x^{2} + x + 1\right ) + 270 \, x^{2} \log \relax (x) - 108 \, x + 180}{144 \, x^{2}} \]

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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giac [A]  time = 0.29, size = 70, normalized size = 0.77 \[ -\frac {1}{72} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {3 \, x - 5}{4 \, x^{2}} + \frac {13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {15}{8} \, \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 70, normalized size = 0.77 \[ -\frac {\sqrt {15}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {15}}{15}\right )}{72}+\frac {8 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}-\frac {15 \ln \relax (x )}{8}+\frac {2 \ln \left (x^{2}+x +1\right )}{3}+\frac {13 \ln \left (2 x^{2}-x +2\right )}{48}+\frac {3}{4 x}-\frac {5}{4 x^{2}} \]

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]

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

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maxima [A]  time = 1.26, size = 69, normalized size = 0.76 \[ -\frac {1}{72} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {3 \, x - 5}{4 \, x^{2}} + \frac {13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {15}{8} \, \log \relax (x) \]

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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mupad [B]  time = 0.15, size = 92, normalized size = 1.01 \[ \frac {\frac {3\,x}{4}-\frac {5}{4}}{x^2}-\frac {15\,\ln \relax (x)}{8}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {2}{3}+\frac {\sqrt {3}\,4{}\mathrm {i}}{9}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {2}{3}+\frac {\sqrt {3}\,4{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {13}{48}+\frac {\sqrt {15}\,1{}\mathrm {i}}{144}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {13}{48}+\frac {\sqrt {15}\,1{}\mathrm {i}}{144}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*x)/4 - 5/4)/x^2 - (15*log(x))/8 - 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)/144 + 13/48) - log(x + (15
^(1/2)*1i)/4 - 1/4)*((15^(1/2)*1i)/144 - 13/48)

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sympy [A]  time = 0.32, size = 94, normalized size = 1.03 \[ - \frac {15 \log {\relax (x )}}{8} + \frac {13 \log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{48} + \frac {2 \log {\left (x^{2} + x + 1 \right )}}{3} - \frac {\sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{72} + \frac {8 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} + \frac {3 x - 5}{4 x^{2}} \]

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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