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

Optimal. Leaf size=145 \[ -\frac{1}{8} \log \left (x^2-x+1\right )+\frac{1}{8} \log \left (x^2+x+1\right )-\frac{\log \left (x^2-\sqrt{3} x+1\right )}{8 \sqrt{3}}+\frac{\log \left (x^2+\sqrt{3} x+1\right )}{8 \sqrt{3}}-\frac{1}{x}+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{1}{4} \tan ^{-1}\left (\sqrt{3}-2 x\right )-\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{4 \sqrt{3}}-\frac{1}{4} \tan ^{-1}\left (2 x+\sqrt{3}\right ) \]

[Out]

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

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Rubi [A]  time = 0.111633, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {1368, 1506, 1127, 1161, 618, 204, 1164, 628} \[ -\frac{1}{8} \log \left (x^2-x+1\right )+\frac{1}{8} \log \left (x^2+x+1\right )-\frac{\log \left (x^2-\sqrt{3} x+1\right )}{8 \sqrt{3}}+\frac{\log \left (x^2+\sqrt{3} x+1\right )}{8 \sqrt{3}}-\frac{1}{x}+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{1}{4} \tan ^{-1}\left (\sqrt{3}-2 x\right )-\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{4 \sqrt{3}}-\frac{1}{4} \tan ^{-1}\left (2 x+\sqrt{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(1 + x^4 + x^8)),x]

[Out]

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

Rule 1368

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

Rule 1506

Int[(((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Wit
h[{q = Rt[a*c, 2]}, With[{r = Rt[2*c*q - b*c, 2]}, Dist[c/(2*q*r), Int[((f*x)^m*Simp[d*r - (c*d - e*q)*x^(n/2)
, x])/(q - r*x^(n/2) + c*x^n), x], x] + Dist[c/(2*q*r), Int[((f*x)^m*Simp[d*r + (c*d - e*q)*x^(n/2), x])/(q +
r*x^(n/2) + c*x^n), x], x]] /;  !LtQ[2*c*q - b*c, 0]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[n2, 2*n] && LtQ[b
^2 - 4*a*c, 0] && IntegersQ[m, n/2] && LtQ[0, m, n] && PosQ[a*c]

Rule 1127

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 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 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 1164

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

Rubi steps

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

Mathematica [C]  time = 0.217175, size = 140, normalized size = 0.97 \[ \frac{1}{24} \left (-3 \log \left (x^2-x+1\right )+3 \log \left (x^2+x+1\right )-\frac{24}{x}+2 i \sqrt{-6+6 i \sqrt{3}} \tan ^{-1}\left (\frac{1}{2} \left (1-i \sqrt{3}\right ) x\right )-2 i \sqrt{-6-6 i \sqrt{3}} \tan ^{-1}\left (\frac{1}{2} \left (1+i \sqrt{3}\right ) x\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^2*(1 + x^4 + x^8)),x]

[Out]

(-24/x + (2*I)*Sqrt[-6 + (6*I)*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x)/2] - (2*I)*Sqrt[-6 - (6*I)*Sqrt[3]]*ArcTan[
((1 + I*Sqrt[3])*x)/2] - 2*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] - 2*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 3*Log[1
- x + x^2] + 3*Log[1 + x + x^2])/24

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Maple [A]  time = 0.01, size = 114, normalized size = 0.8 \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{\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 ( 1+{x}^{2}+x\sqrt{3} \right ) \sqrt{3}}{24}}-{\frac{\arctan \left ( 2\,x+\sqrt{3} \right ) }{4}}-{\frac{\ln \left ( 1+{x}^{2}-x\sqrt{3} \right ) \sqrt{3}}{24}}-{\frac{\arctan \left ( 2\,x-\sqrt{3} \right ) }{4}}-{x}^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(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)-1/8*ln(x^2-x+1)-1/12*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/
2))+1/24*ln(1+x^2+x*3^(1/2))*3^(1/2)-1/4*arctan(2*x+3^(1/2))-1/24*ln(1+x^2-x*3^(1/2))*3^(1/2)-1/4*arctan(2*x-3
^(1/2))-1/x

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{x} - \frac{1}{2} \, \int \frac{x^{2}}{x^{4} - x^{2} + 1}\,{d x} + \frac{1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac{1}{8} \, \log \left (x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.68442, size = 720, normalized size = 4.97 \begin{align*} \frac{4 \, \sqrt{6} \sqrt{3} \sqrt{2} x \arctan \left (-\frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{2} x + \frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{\sqrt{6} \sqrt{2} x + 2 \, x^{2} + 2} - \sqrt{3}\right ) + 4 \, \sqrt{6} \sqrt{3} \sqrt{2} x \arctan \left (-\frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{2} x + \frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{-\sqrt{6} \sqrt{2} x + 2 \, x^{2} + 2} + \sqrt{3}\right ) + \sqrt{6} \sqrt{2} x \log \left (\sqrt{6} \sqrt{2} x + 2 \, x^{2} + 2\right ) - \sqrt{6} \sqrt{2} x \log \left (-\sqrt{6} \sqrt{2} x + 2 \, x^{2} + 2\right ) - 4 \, \sqrt{3} x \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - 4 \, \sqrt{3} x \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 6 \, x \log \left (x^{2} + x + 1\right ) - 6 \, x \log \left (x^{2} - x + 1\right ) - 48}{48 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(4*sqrt(6)*sqrt(3)*sqrt(2)*x*arctan(-1/3*sqrt(6)*sqrt(3)*sqrt(2)*x + 1/3*sqrt(6)*sqrt(3)*sqrt(sqrt(6)*sqr
t(2)*x + 2*x^2 + 2) - sqrt(3)) + 4*sqrt(6)*sqrt(3)*sqrt(2)*x*arctan(-1/3*sqrt(6)*sqrt(3)*sqrt(2)*x + 1/3*sqrt(
6)*sqrt(3)*sqrt(-sqrt(6)*sqrt(2)*x + 2*x^2 + 2) + sqrt(3)) + sqrt(6)*sqrt(2)*x*log(sqrt(6)*sqrt(2)*x + 2*x^2 +
2) - sqrt(6)*sqrt(2)*x*log(-sqrt(6)*sqrt(2)*x + 2*x^2 + 2) - 4*sqrt(3)*x*arctan(1/3*sqrt(3)*(2*x + 1)) - 4*sq
rt(3)*x*arctan(1/3*sqrt(3)*(2*x - 1)) + 6*x*log(x^2 + x + 1) - 6*x*log(x^2 - x + 1) - 48)/x

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Sympy [C]  time = 0.722479, size = 218, normalized size = 1.5 \begin{align*} \left (- \frac{1}{8} - \frac{\sqrt{3} i}{24}\right ) \log{\left (x - 442368 \left (- \frac{1}{8} - \frac{\sqrt{3} i}{24}\right )^{7} - 384 \left (- \frac{1}{8} - \frac{\sqrt{3} i}{24}\right )^{3} \right )} + \left (- \frac{1}{8} + \frac{\sqrt{3} i}{24}\right ) \log{\left (x - 384 \left (- \frac{1}{8} + \frac{\sqrt{3} i}{24}\right )^{3} - 442368 \left (- \frac{1}{8} + \frac{\sqrt{3} i}{24}\right )^{7} \right )} + \left (\frac{1}{8} - \frac{\sqrt{3} i}{24}\right ) \log{\left (x - 442368 \left (\frac{1}{8} - \frac{\sqrt{3} i}{24}\right )^{7} - 384 \left (\frac{1}{8} - \frac{\sqrt{3} i}{24}\right )^{3} \right )} + \left (\frac{1}{8} + \frac{\sqrt{3} i}{24}\right ) \log{\left (x - 384 \left (\frac{1}{8} + \frac{\sqrt{3} i}{24}\right )^{3} - 442368 \left (\frac{1}{8} + \frac{\sqrt{3} i}{24}\right )^{7} \right )} + \operatorname{RootSum}{\left (2304 t^{4} + 48 t^{2} + 1, \left ( t \mapsto t \log{\left (- 442368 t^{7} - 384 t^{3} + x \right )} \right )\right )} - \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/8 - sqrt(3)*I/24)*log(x - 442368*(-1/8 - sqrt(3)*I/24)**7 - 384*(-1/8 - sqrt(3)*I/24)**3) + (-1/8 + sqrt(3
)*I/24)*log(x - 384*(-1/8 + sqrt(3)*I/24)**3 - 442368*(-1/8 + sqrt(3)*I/24)**7) + (1/8 - sqrt(3)*I/24)*log(x -
442368*(1/8 - sqrt(3)*I/24)**7 - 384*(1/8 - sqrt(3)*I/24)**3) + (1/8 + sqrt(3)*I/24)*log(x - 384*(1/8 + sqrt(
3)*I/24)**3 - 442368*(1/8 + sqrt(3)*I/24)**7) + RootSum(2304*_t**4 + 48*_t**2 + 1, Lambda(_t, _t*log(-442368*_
t**7 - 384*_t**3 + x))) - 1/x

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{8} + x^{4} + 1\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^8 + x^4 + 1)*x^2), x)