3.336 \(\int \frac{1}{x^5 (1+x^4+x^8)} \, dx\)

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

[Out]

-1/(4*x^4) - 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 = 0.0512834, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {1357, 709, 800, 634, 618, 204, 628} \[ -\frac{1}{4 x^4}+\frac{1}{8} \log \left (x^8+x^4+1\right )-\frac{\tan ^{-1}\left (\frac{2 x^4+1}{\sqrt{3}}\right )}{4 \sqrt{3}}-\log (x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(1 + x^4 + x^8)),x]

[Out]

-1/(4*x^4) - 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 709

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

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{1}{x^5 \left (1+x^4+x^8\right )} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x+x^2\right )} \, dx,x,x^4\right )\\ &=-\frac{1}{4 x^4}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{-1-x}{x \left (1+x+x^2\right )} \, dx,x,x^4\right )\\ &=-\frac{1}{4 x^4}+\frac{1}{4} \operatorname{Subst}\left (\int \left (-\frac{1}{x}+\frac{x}{1+x+x^2}\right ) \, dx,x,x^4\right )\\ &=-\frac{1}{4 x^4}-\log (x)+\frac{1}{4} \operatorname{Subst}\left (\int \frac{x}{1+x+x^2} \, dx,x,x^4\right )\\ &=-\frac{1}{4 x^4}-\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 )\\ &=-\frac{1}{4 x^4}-\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{1}{4 x^4}-\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 = 0.104257, size = 141, normalized size = 2.94 \[ \frac{1}{24} \left (-\frac{6}{x^4}+\sqrt{3} \left (\sqrt{3}+i\right ) \log \left (x^2-\frac{i \sqrt{3}}{2}-\frac{1}{2}\right )+\sqrt{3} \left (\sqrt{3}-i\right ) \log \left (x^2+\frac{1}{2} i \left (\sqrt{3}+i\right )\right )+3 \log \left (x^2-x+1\right )+3 \log \left (x^2+x+1\right )-24 \log (x)+2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(1 + x^4 + x^8)),x]

[Out]

(-6/x^4 + 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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Maple [B]  time = 0.009, size = 94, normalized size = 2. \begin{align*}{\frac{\ln \left ({x}^{2}+x+1 \right ) }{8}}-{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }-{\frac{1}{4\,{x}^{4}}}-\ln \left ( x \right ) +{\frac{\ln \left ({x}^{2}-x+1 \right ) }{8}}+{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{\ln \left ({x}^{4}-{x}^{2}+1 \right ) }{8}}-{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 2\,{x}^{2}-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(x^8+x^4+1),x)

[Out]

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

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Maxima [A]  time = 1.47343, size = 55, normalized size = 1.15 \begin{align*} -\frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{4} + 1\right )}\right ) - \frac{1}{4 \, x^{4}} + \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^5/(x^8+x^4+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.45547, size = 143, normalized size = 2.98 \begin{align*} -\frac{2 \, \sqrt{3} x^{4} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{4} + 1\right )}\right ) - 3 \, x^{4} \log \left (x^{8} + x^{4} + 1\right ) + 24 \, x^{4} \log \left (x\right ) + 6}{24 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^8+x^4+1),x, algorithm="fricas")

[Out]

-1/24*(2*sqrt(3)*x^4*arctan(1/3*sqrt(3)*(2*x^4 + 1)) - 3*x^4*log(x^8 + x^4 + 1) + 24*x^4*log(x) + 6)/x^4

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Sympy [A]  time = 0.180634, size = 48, 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} - \frac{1}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(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 - 1/(4*x**4)

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Giac [A]  time = 1.1091, size = 62, normalized size = 1.29 \begin{align*} -\frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{4} + 1\right )}\right ) + \frac{x^{4} - 1}{4 \, x^{4}} + \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^5/(x^8+x^4+1),x, algorithm="giac")

[Out]

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