\(\int \frac {1}{x^2 (1+x^4+x^8)} \, dx\) [78]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [C] (verification not implemented)
Maxima [F]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 14, antiderivative size = 114 \[ \int \frac {1}{x^2 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{x}+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \arctan \left (\sqrt {3}+2 x\right )+\frac {1}{4} \text {arctanh}\left (\frac {x}{1+x^2}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{4 \sqrt {3}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^2 \left (1+x^4+x^8\right )} \, dx=\frac {1}{24} \left (-\frac {24}{x}+2 i \sqrt {-6+6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )-2 i \sqrt {-6-6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-3 \log \left (1-x+x^2\right )+3 \log \left (1+x+x^2\right )\right ) \] Input:

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

Output:

(-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]*Arc 
Tan[(-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
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.45, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {1704, 25, 1830, 1447, 1475, 1083, 217, 1478, 25, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1704

\(\displaystyle \int -\frac {x^2 \left (x^4+1\right )}{x^8+x^4+1}dx-\frac {1}{x}\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {x^2 \left (x^4+1\right )}{x^8+x^4+1}dx-\frac {1}{x}\)

\(\Big \downarrow \) 1830

\(\displaystyle -\frac {1}{2} \int \frac {x^2}{x^4-x^2+1}dx-\frac {1}{2} \int \frac {x^2}{x^4+x^2+1}dx-\frac {1}{x}\)

\(\Big \downarrow \) 1447

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx-\frac {1}{2} \int \frac {x^2+1}{x^4-x^2+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx-\frac {1}{2} \int \frac {x^2+1}{x^4+x^2+1}dx\right )-\frac {1}{x}\)

\(\Big \downarrow \) 1475

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-\sqrt {3} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {3} x+1}dx\right )+\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int \frac {1}{x^2+x+1}dx\right )+\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx\right )-\frac {1}{x}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \left (\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)+\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx+\frac {1}{2} \left (\int \frac {1}{-\left (2 x-\sqrt {3}\right )^2-1}d\left (2 x-\sqrt {3}\right )+\int \frac {1}{-\left (2 x+\sqrt {3}\right )^2-1}d\left (2 x+\sqrt {3}\right )\right )\right )-\frac {1}{x}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x\right )-\arctan \left (2 x+\sqrt {3}\right )\right )\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )-\frac {1}{x}\)

\(\Big \downarrow \) 1478

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx-\frac {1}{2} \int -\frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )+\frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int -\frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x\right )-\arctan \left (2 x+\sqrt {3}\right )\right )\right )-\frac {1}{x}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx+\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )+\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {\int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x\right )-\arctan \left (2 x+\sqrt {3}\right )\right )\right )-\frac {1}{x}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {1}{2} \log \left (x^2+x+1\right )-\frac {1}{2} \log \left (x^2-x+1\right )\right )\right )+\frac {1}{2} \left (\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x\right )-\arctan \left (2 x+\sqrt {3}\right )\right )+\frac {1}{2} \left (\frac {\log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}\right )\right )-\frac {1}{x}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1447
Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a/c, 2]}, Simp[1/2   Int[(q + x^2)/(a + b*x^2 + c*x^4), x], x] - Simp[1/2 
 Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && LtQ[b 
^2 - 4*a*c, 0] && PosQ[a*c]
 

rule 1475
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^ 
2, x], x], x] + Simp[e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F 
reeQ[{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] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] 
, 0]))
 

rule 1478
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[-2*(d/e) - b/c, 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e 
 + q*x - x^2, x], x], x] + Simp[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 1704
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] - Simp[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 1830
Int[(((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + 
(c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[a*c, 2]}, With[{r = Rt[2*c*q - 
 b*c, 2]}, Simp[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] + Simp[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]
 
Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.84

method result size
risch \(-\frac {1}{x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-6 \textit {\_R}^{3}-\textit {\_R} +x \right )\right )}{4}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (4 x^{2}-4 x +4\right )}{8}-\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}+\frac {\ln \left (4 x^{2}+4 x +4\right )}{8}\) \(96\)
default \(-\frac {\ln \left (x^{2}-x +1\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{2}+x +1\right )}{8}-\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}+\frac {\sqrt {3}\, \left (\frac {\ln \left (x^{2}+\sqrt {3}\, x +1\right )}{2}-\sqrt {3}\, \arctan \left (\sqrt {3}+2 x \right )\right )}{12}+\frac {\sqrt {3}\, \left (-\frac {\ln \left (x^{2}-\sqrt {3}\, x +1\right )}{2}-\sqrt {3}\, \arctan \left (-\sqrt {3}+2 x \right )\right )}{12}-\frac {1}{x}\) \(126\)

Input:

int(1/x^2/(x^8+x^4+1),x,method=_RETURNVERBOSE)
 

Output:

-1/x+1/4*sum(_R*ln(-6*_R^3-_R+x),_R=RootOf(9*_Z^4+3*_Z^2+1))-1/12*3^(1/2)* 
arctan(1/3*(2*x-1)*3^(1/2))-1/8*ln(4*x^2-4*x+4)-1/12*arctan(1/3*(2*x+1)*3^ 
(1/2))*3^(1/2)+1/8*ln(4*x^2+4*x+4)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^2 \left (1+x^4+x^8\right )} \, dx=-\frac {2 \, \sqrt {3} x \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + 2 \, \sqrt {3} x \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \sqrt {3} x \log \left (x^{2} + \sqrt {3} x + 1\right ) + \sqrt {3} x \log \left (x^{2} - \sqrt {3} x + 1\right ) + 6 \, x \arctan \left (2 \, x + \sqrt {3}\right ) - 6 \, x \arctan \left (-2 \, x + \sqrt {3}\right ) - 3 \, x \log \left (x^{2} + x + 1\right ) + 3 \, x \log \left (x^{2} - x + 1\right ) + 24}{24 \, x} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.91 \[ \int \frac {1}{x^2 \left (1+x^4+x^8\right )} \, dx=\left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 442368 \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{7} - 384 \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{3} \right )} + \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 384 \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{3} - 442368 \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{7} \right )} + \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 442368 \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{7} - 384 \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{3} \right )} + \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 384 \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{3} - 442368 \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{7} \right )} + \operatorname {RootSum} {\left (2304 t^{4} + 48 t^{2} + 1, \left ( t \mapsto t \log {\left (- 442368 t^{7} - 384 t^{3} + x \right )} \right )\right )} - \frac {1}{x} \] Input:

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

Output:

(-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 - 4 
42368*(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
 

Maxima [F]

\[ \int \frac {1}{x^2 \left (1+x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} + x^{4} + 1\right )} x^{2}} \,d x } \] Input:

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

Output:

-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)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^2 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) - \frac {1}{x} - \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) - \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 \left (1+x^4+x^8\right )} \, dx=\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}-\frac {1}{4}{}\mathrm {i}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}+\frac {1}{4}{}\mathrm {i}\right )-\frac {1}{x} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^2 \left (1+x^4+x^8\right )} \, dx=\frac {6 \mathit {atan} \left (\sqrt {3}-2 x \right ) x -6 \mathit {atan} \left (\sqrt {3}+2 x \right ) x -2 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) x -2 \sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) x -\sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right ) x +\sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right ) x -3 \,\mathrm {log}\left (x^{2}-x +1\right ) x +3 \,\mathrm {log}\left (x^{2}+x +1\right ) x -24}{24 x} \] Input:

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

Output:

(6*atan(sqrt(3) - 2*x)*x - 6*atan(sqrt(3) + 2*x)*x - 2*sqrt(3)*atan((2*x - 
 1)/sqrt(3))*x - 2*sqrt(3)*atan((2*x + 1)/sqrt(3))*x - sqrt(3)*log( - sqrt 
(3)*x + x**2 + 1)*x + sqrt(3)*log(sqrt(3)*x + x**2 + 1)*x - 3*log(x**2 - x 
 + 1)*x + 3*log(x**2 + x + 1)*x - 24)/(24*x)