3.275 \(\int \frac {1+2 x-x^2+8 x^3+x^4}{(x+x^2) (1+x^3)} \, dx\)

Optimal. Leaf size=44 \[ \log \left (x^2-x+1\right )-\frac {3}{x+1}+\log (x)-2 \log (x+1)-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

-3/(1+x)+ln(x)-2*ln(1+x)+ln(x^2-x+1)-2/3*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1593, 6725, 634, 618, 204, 628} \[ \log \left (x^2-x+1\right )-\frac {3}{x+1}+\log (x)-2 \log (x+1)-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x - x^2 + 8*x^3 + x^4)/((x + x^2)*(1 + x^3)),x]

[Out]

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

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx &=\int \frac {1+2 x-x^2+8 x^3+x^4}{x (1+x) \left (1+x^3\right )} \, dx\\ &=\int \left (\frac {1}{x}+\frac {3}{(1+x)^2}-\frac {2}{1+x}+\frac {2 x}{1-x+x^2}\right ) \, dx\\ &=-\frac {3}{1+x}+\log (x)-2 \log (1+x)+2 \int \frac {x}{1-x+x^2} \, dx\\ &=-\frac {3}{1+x}+\log (x)-2 \log (1+x)+\int \frac {1}{1-x+x^2} \, dx+\int \frac {-1+2 x}{1-x+x^2} \, dx\\ &=-\frac {3}{1+x}+\log (x)-2 \log (1+x)+\log \left (1-x+x^2\right )-2 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=-\frac {3}{1+x}-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-2 \log (1+x)+\log \left (1-x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 44, normalized size = 1.00 \[ \log \left (x^2-x+1\right )-\frac {3}{x+1}+\log (x)-2 \log (x+1)+\frac {2 \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x - x^2 + 8*x^3 + x^4)/((x + x^2)*(1 + x^3)),x]

[Out]

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

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fricas [A]  time = 0.88, size = 58, normalized size = 1.32 \[ \frac {2 \, \sqrt {3} {\left (x + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 3 \, {\left (x + 1\right )} \log \left (x^{2} - x + 1\right ) - 6 \, {\left (x + 1\right )} \log \left (x + 1\right ) + 3 \, {\left (x + 1\right )} \log \relax (x) - 9}{3 \, {\left (x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+8*x^3-x^2+2*x+1)/(x^2+x)/(x^3+1),x, algorithm="fricas")

[Out]

1/3*(2*sqrt(3)*(x + 1)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3*(x + 1)*log(x^2 - x + 1) - 6*(x + 1)*log(x + 1) + 3*(
x + 1)*log(x) - 9)/(x + 1)

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giac [A]  time = 0.36, size = 43, normalized size = 0.98 \[ \frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {3}{x + 1} + \log \left (x^{2} - x + 1\right ) - 2 \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+8*x^3-x^2+2*x+1)/(x^2+x)/(x^3+1),x, algorithm="giac")

[Out]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 3/(x + 1) + log(x^2 - x + 1) - 2*log(abs(x + 1)) + log(abs(x))

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maple [A]  time = 0.01, size = 42, normalized size = 0.95 \[ \frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}+\ln \relax (x )-2 \ln \left (x +1\right )+\ln \left (x^{2}-x +1\right )-\frac {3}{x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+8*x^3-x^2+2*x+1)/(x^2+x)/(x^3+1),x)

[Out]

ln(x^2-x+1)+2/3*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))-3/(x+1)-2*ln(x+1)+ln(x)

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maxima [A]  time = 2.25, size = 41, normalized size = 0.93 \[ \frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {3}{x + 1} + \log \left (x^{2} - x + 1\right ) - 2 \, \log \left (x + 1\right ) + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+8*x^3-x^2+2*x+1)/(x^2+x)/(x^3+1),x, algorithm="maxima")

[Out]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 3/(x + 1) + log(x^2 - x + 1) - 2*log(x + 1) + log(x)

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mupad [B]  time = 0.13, size = 55, normalized size = 1.25 \[ \ln \relax (x)-2\,\ln \left (x+1\right )-\frac {3}{x+1}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-1+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (1+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - x^2 + 8*x^3 + x^4 + 1)/((x^3 + 1)*(x + x^2)),x)

[Out]

log(x) - 2*log(x + 1) - 3/(x + 1) - log(x - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/3 - 1) + log(x + (3^(1/2)*1i)/
2 - 1/2)*((3^(1/2)*1i)/3 + 1)

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sympy [A]  time = 0.21, size = 49, normalized size = 1.11 \[ \log {\relax (x )} - 2 \log {\left (x + 1 \right )} + \log {\left (x^{2} - x + 1 \right )} + \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} - \frac {3}{x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+8*x**3-x**2+2*x+1)/(x**2+x)/(x**3+1),x)

[Out]

log(x) - 2*log(x + 1) + log(x**2 - x + 1) + 2*sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/3 - 3/(x + 1)

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