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

Optimal. Leaf size=41 \[ -\frac{1}{8} \log \left (x^8-x^4+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^4}{\sqrt{3}}\right )}{4 \sqrt{3}}+\log (x) \]

[Out]

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

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Rubi [A]  time = 0.0378776, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {1357, 705, 29, 634, 618, 204, 628} \[ -\frac{1}{8} \log \left (x^8-x^4+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^4}{\sqrt{3}}\right )}{4 \sqrt{3}}+\log (x) \]

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 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]]

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 29

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

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{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*}

Mathematica [C]  time = 0.0130928, size = 55, normalized size = 1.34 \[ \log (x)-\frac{1}{4} \text{RootSum}\left [\text{$\#$1}^8-\text{$\#$1}^4+1\& ,\frac{\text{$\#$1}^4 \log (x-\text{$\#$1})-\log (x-\text{$\#$1})}{2 \text{$\#$1}^4-1}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 - x^4 + x^8)),x]

[Out]

Log[x] - RootSum[1 - #1^4 + #1^8 & , (-Log[x - #1] + Log[x - #1]*#1^4)/(-1 + 2*#1^4) & ]/4

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Maple [A]  time = 0.007, size = 35, normalized size = 0.9 \begin{align*} \ln \left ( x \right ) -{\frac{\ln \left ({x}^{8}-{x}^{4}+1 \right ) }{8}}+{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 2\,{x}^{4}-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

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

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Maxima [A]  time = 1.48284, size = 51, normalized size = 1.24 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 1.4634, size = 108, normalized size = 2.63 \begin{align*} \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 \left (x\right ) \end{align*}

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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Sympy [A]  time = 0.150035, size = 41, normalized size = 1. \begin{align*} \log{\left (x \right )} - \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} \end{align*}

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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Giac [A]  time = 1.13535, size = 51, normalized size = 1.24 \begin{align*} \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 ) \end{align*}

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)