3.334 \(\int \frac {1}{x (1+x^4+x^8)} \, dx\)

Optimal. Leaf size=39 \[ -\frac {\tan ^{-1}\left (\frac {2 x^4+1}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{8} \log \left (x^8+x^4+1\right )+\log (x) \]

[Out]

ln(x)-1/8*ln(x^8+x^4+1)-1/12*arctan(1/3*(2*x^4+1)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1357, 705, 29, 634, 618, 204, 628} \[ -\frac {1}{8} \log \left (x^8+x^4+1\right )-\frac {\tan ^{-1}\left (\frac {2 x^4+1}{\sqrt {3}}\right )}{4 \sqrt {3}}+\log (x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + x^4 + x^8)),x]

[Out]

-ArcTan[(1 + 2*x^4)/Sqrt[3]]/(4*Sqrt[3]) + Log[x] - Log[1 + x^4 + x^8]/8

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [C]  time = 0.08, size = 138, normalized size = 3.54 \[ \frac {1}{24} \left (-\sqrt {3} \left (\sqrt {3}-i\right ) \log \left (x^2-\frac {i \sqrt {3}}{2}-\frac {1}{2}\right )-\sqrt {3} \left (\sqrt {3}+i\right ) \log \left (x^2+\frac {1}{2} i \left (\sqrt {3}+i\right )\right )-3 \log \left (x^2-x+1\right )-3 \log \left (x^2+x+1\right )+24 \log (x)+2 \sqrt {3} \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x^4 + x^8)),x]

[Out]

(2*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] - 2*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] + 24*Log[x] - Sqrt[3]*(-I + Sqrt[3
])*Log[-1/2 - (I/2)*Sqrt[3] + x^2] - Sqrt[3]*(I + Sqrt[3])*Log[(I/2)*(I + Sqrt[3]) + x^2] - 3*Log[1 - x + x^2]
- 3*Log[1 + x + x^2])/24

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fricas [A]  time = 0.79, size = 32, normalized size = 0.82 \[ -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} + x^{4} + 1\right ) + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(x^8+x^4+1),x, algorithm="fricas")

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1)) - 1/8*log(x^8 + x^4 + 1) + log(x)

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giac [A]  time = 0.38, size = 36, normalized size = 0.92 \[ -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} + x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(x^8+x^4+1),x, algorithm="giac")

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1)) - 1/8*log(x^8 + x^4 + 1) + 1/4*log(x^4)

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maple [B]  time = 0.01, size = 87, normalized size = 2.23 \[ -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}+\ln \relax (x )-\frac {\ln \left (x^{2}-x +1\right )}{8}-\frac {\ln \left (x^{2}+x +1\right )}{8}-\frac {\ln \left (x^{4}-x^{2}+1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(x^8+x^4+1),x)

[Out]

ln(x)-1/8*ln(x^2+x+1)-1/12*3^(1/2)*arctan(1/3*(2*x+1)*3^(1/2))-1/8*ln(x^4-x^2+1)-1/12*3^(1/2)*arctan(1/3*(2*x^
2-1)*3^(1/2))-1/8*ln(x^2-x+1)+1/12*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))

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maxima [A]  time = 3.05, size = 36, normalized size = 0.92 \[ -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} + x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(x^8+x^4+1),x, algorithm="maxima")

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1)) - 1/8*log(x^8 + x^4 + 1) + 1/4*log(x^4)

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mupad [B]  time = 1.30, size = 34, normalized size = 0.87 \[ \ln \relax (x)-\frac {\ln \left (x^8+x^4+1\right )}{8}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x^4}{3}+\frac {\sqrt {3}}{3}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x^4 + x^8 + 1)),x)

[Out]

log(x) - log(x^4 + x^8 + 1)/8 - (3^(1/2)*atan(3^(1/2)/3 + (2*3^(1/2)*x^4)/3))/12

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sympy [A]  time = 0.15, size = 41, normalized size = 1.05 \[ \log {\relax (x )} - \frac {\log {\left (x^{8} + x^{4} + 1 \right )}}{8} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{4}}{3} + \frac {\sqrt {3}}{3} \right )}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(x**8+x**4+1),x)

[Out]

log(x) - log(x**8 + x**4 + 1)/8 - sqrt(3)*atan(2*sqrt(3)*x**4/3 + sqrt(3)/3)/12

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