3.1.12 \(\int \frac {1+x^4}{1+x^4+x^8} \, dx\) [12]

3.1.12.1 Optimal result
3.1.12.2 Mathematica [C] (verified)
3.1.12.3 Rubi [A] (verified)
3.1.12.4 Maple [C] (verified)
3.1.12.5 Fricas [C] (verification not implemented)
3.1.12.6 Sympy [C] (verification not implemented)
3.1.12.7 Maxima [F]
3.1.12.8 Giac [A] (verification not implemented)
3.1.12.9 Mupad [B] (verification not implemented)

3.1.12.1 Optimal result

Integrand size = 16, antiderivative size = 140 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \arctan \left (\sqrt {3}+2 x\right )-\frac {1}{8} \log \left (1-x+x^2\right )+\frac {1}{8} \log \left (1+x+x^2\right )-\frac {\log \left (1-\sqrt {3} x+x^2\right )}{8 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x+x^2\right )}{8 \sqrt {3}} \]

output
1/4*arctan(2*x-3^(1/2))+1/4*arctan(2*x+3^(1/2))-1/8*ln(x^2-x+1)+1/8*ln(x^2 
+x+1)-1/12*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/12*arctan(1/3*(1+2*x)*3^( 
1/2))*3^(1/2)-1/24*ln(1+x^2-x*3^(1/2))*3^(1/2)+1/24*ln(1+x^2+x*3^(1/2))*3^ 
(1/2)
 
3.1.12.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\frac {1}{48} \left (4 i \sqrt {-6-6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )-4 i \sqrt {-6+6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )+4 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+4 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-6 \log \left (1-x+x^2\right )+6 \log \left (1+x+x^2\right )\right ) \]

input
Integrate[(1 + x^4)/(1 + x^4 + x^8),x]
 
output
((4*I)*Sqrt[-6 - (6*I)*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x)/2] - (4*I)*Sqrt 
[-6 + (6*I)*Sqrt[3]]*ArcTan[((1 + I*Sqrt[3])*x)/2] + 4*Sqrt[3]*ArcTan[(-1 
+ 2*x)/Sqrt[3]] + 4*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 6*Log[1 - x + x^2] 
 + 6*Log[1 + x + x^2])/48
 
3.1.12.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1749, 1407, 1142, 25, 1083, 217, 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.

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

\(\Big \downarrow \) 1749

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

\(\Big \downarrow \) 1407

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

input
Int[(1 + x^4)/(1 + x^4 + x^8),x]
 
output
((ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3] - Log[1 - x + x^2]/2)/2 + (ArcTan[(1 
+ 2*x)/Sqrt[3]]/Sqrt[3] + Log[1 + x + x^2]/2)/2)/2 + ((-(Sqrt[3]*ArcTan[Sq 
rt[3] - 2*x]) - Log[1 - Sqrt[3]*x + x^2]/2)/(2*Sqrt[3]) + (Sqrt[3]*ArcTan[ 
Sqrt[3] + 2*x] + Log[1 + Sqrt[3]*x + x^2]/2)/(2*Sqrt[3]))/2
 

3.1.12.3.1 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 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1407
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/ 
c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r)   Int[(r - x)/(q - r* 
x + x^2), x], x] + Simp[1/(2*c*q*r)   Int[(r + x)/(q + r*x + x^2), x], x]]] 
 /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]
 

rule 1749
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x 
_Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c)   Int[1/Simp[d/e 
+ q*x^(n/2) + x^n, x], x], x] + Simp[e/(2*c)   Int[1/Simp[d/e - q*x^(n/2) + 
 x^n, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 
 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && IGtQ[n/2, 0] && (GtQ[2*(d/e) - b/c, 
0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d, e*Rt[a/c, 2]]))
 
3.1.12.4 Maple [C] (verified)

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

Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.64

method result size
risch \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-3 \textit {\_R}^{3}+\textit {\_R} +x \right )\right )}{4}-\frac {\ln \left (4 x^{2}-4 x +4\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (4 x^{2}+4 x +4\right )}{8}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}\) \(89\)
default \(-\frac {\ln \left (x^{2}-x +1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{2}+x +1\right )}{8}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\ln \left (1+x^{2}-x \sqrt {3}\right ) \sqrt {3}}{24}+\frac {\arctan \left (2 x -\sqrt {3}\right )}{4}+\frac {\ln \left (1+x^{2}+x \sqrt {3}\right ) \sqrt {3}}{24}+\frac {\arctan \left (2 x +\sqrt {3}\right )}{4}\) \(109\)

input
int((x^4+1)/(x^8+x^4+1),x,method=_RETURNVERBOSE)
 
output
1/4*sum(_R*ln(-3*_R^3+_R+x),_R=RootOf(9*_Z^4+3*_Z^2+1))-1/8*ln(4*x^2-4*x+4 
)+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*(1+2*x)*3^(1/2))*3^(1/2)
 
3.1.12.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.49 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=-\frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (i \, \sqrt {3} - 3\right )} + 12 \, x\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} + 3\right )} + 12 \, x\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (\sqrt {6} {\left (i \, \sqrt {3} + 3\right )} \sqrt {-i \, \sqrt {3} - 1} + 12 \, x\right ) - \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {-i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} - 3\right )} + 12 \, x\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \]

input
integrate((x^4+1)/(x^8+x^4+1),x, algorithm="fricas")
 
output
-1/24*sqrt(6)*sqrt(I*sqrt(3) - 1)*log(sqrt(6)*sqrt(I*sqrt(3) - 1)*(I*sqrt( 
3) - 3) + 12*x) + 1/24*sqrt(6)*sqrt(I*sqrt(3) - 1)*log(sqrt(6)*sqrt(I*sqrt 
(3) - 1)*(-I*sqrt(3) + 3) + 12*x) + 1/24*sqrt(6)*sqrt(-I*sqrt(3) - 1)*log( 
sqrt(6)*(I*sqrt(3) + 3)*sqrt(-I*sqrt(3) - 1) + 12*x) - 1/24*sqrt(6)*sqrt(- 
I*sqrt(3) - 1)*log(sqrt(6)*sqrt(-I*sqrt(3) - 1)*(-I*sqrt(3) - 3) + 12*x) + 
 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt 
(3)*(2*x - 1)) + 1/8*log(x^2 + x + 1) - 1/8*log(x^2 - x + 1)
 
3.1.12.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.36 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 1 - \frac {\sqrt {3} i}{3} + 9216 \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 1 + 9216 \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} + \frac {\sqrt {3} i}{3} \right )} + \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x + 1 - \frac {\sqrt {3} i}{3} + 9216 \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x + 1 + 9216 \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} + \frac {\sqrt {3} i}{3} \right )} + \operatorname {RootSum} {\left (2304 t^{4} + 48 t^{2} + 1, \left ( t \mapsto t \log {\left (9216 t^{5} + 8 t + x \right )} \right )\right )} \]

input
integrate((x**4+1)/(x**8+x**4+1),x)
 
output
(-1/8 - sqrt(3)*I/24)*log(x - 1 - sqrt(3)*I/3 + 9216*(-1/8 - sqrt(3)*I/24) 
**5) + (-1/8 + sqrt(3)*I/24)*log(x - 1 + 9216*(-1/8 + sqrt(3)*I/24)**5 + s 
qrt(3)*I/3) + (1/8 - sqrt(3)*I/24)*log(x + 1 - sqrt(3)*I/3 + 9216*(1/8 - s 
qrt(3)*I/24)**5) + (1/8 + sqrt(3)*I/24)*log(x + 1 + 9216*(1/8 + sqrt(3)*I/ 
24)**5 + sqrt(3)*I/3) + RootSum(2304*_t**4 + 48*_t**2 + 1, Lambda(_t, _t*l 
og(9216*_t**5 + 8*_t + x)))
 
3.1.12.7 Maxima [F]

\[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\int { \frac {x^{4} + 1}{x^{8} + x^{4} + 1} \,d x } \]

input
integrate((x^4+1)/(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)) + 1/2*integrate(1/(x^4 - x^2 + 1), x) + 1/8*log(x^2 + x + 1) 
 - 1/8*log(x^2 - x + 1)
 
3.1.12.8 Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \]

input
integrate((x^4+1)/(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) + 1/4*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)
 
3.1.12.9 Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.68 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}+\frac {1}{4}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}-\frac {1}{4}{}\mathrm {i}\right ) \]

input
int((x^4 + 1)/(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)