Integrand size = 14, antiderivative size = 109 \[ \int \frac {x^4}{1+x^4+x^8} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \arctan \left (\sqrt {3}+2 x\right )+\frac {1}{4} \text {arctanh}\left (\frac {x}{1+x^2}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{4 \sqrt {3}} \] Output:
1/12*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/4*arctan(-3^(1/2)+2*x)-1/12*arc tan(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)
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.24 \[ \int \frac {x^4}{1+x^4+x^8} \, dx=\frac {1}{24} \left (-2 i \sqrt {-6+6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )+2 i \sqrt {-6-6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-3 \log \left (1-x+x^2\right )+3 \log \left (1+x+x^2\right )\right ) \] Input:
Integrate[x^4/(1 + x^4 + x^8),x]
Output:
((-2*I)*Sqrt[-6 + (6*I)*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x)/2] + (2*I)*Sqr t[-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
Time = 0.39 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.47, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {1709, 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.
| \(\displaystyle \int \frac {x^4}{x^8+x^4+1} \, dx\) |
| \(\Big \downarrow \) 1709 |
| \(\displaystyle \frac {1}{2} \int \frac {x^2}{x^4-x^2+1}dx-\frac {1}{2} \int \frac {x^2}{x^4+x^2+1}dx\) |
| \(\Big \downarrow \) 1447 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {x^2+1}{x^4-x^2+1}dx-\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx-\frac {1}{2} \int \frac {x^2+1}{x^4+x^2+1}dx\right )\) |
| \(\Big \downarrow \) 1475 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {1}{x^2+\sqrt {3} x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int \frac {1}{x^2+x+1}dx\right )+\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \left (\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)+\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )\right )+\frac {1}{2} \left (\frac {1}{2} \left (-\int \frac {1}{-\left (2 x-\sqrt {3}\right )^2-1}d\left (2 x-\sqrt {3}\right )-\int \frac {1}{-\left (2 x+\sqrt {3}\right )^2-1}d\left (2 x+\sqrt {3}\right )\right )-\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\arctan \left (2 x+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x\right )\right )-\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx+\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )\) |
| \(\Big \downarrow \) 1478 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx-\frac {1}{2} \int -\frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )+\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {\int -\frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (2 x+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x\right )\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx+\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )+\frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (2 x+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x\right )\right )\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {1}{2} \log \left (x^2+x+1\right )-\frac {1}{2} \log \left (x^2-x+1\right )\right )\right )+\frac {1}{2} \left (\frac {1}{2} \left (\arctan \left (2 x+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x\right )\right )+\frac {1}{2} \left (\frac {\log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}\right )\right )\) |
Input:
Int[x^4/(1 + x^4 + x^8),x]
Output:
((-(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 + ((-ArcTan[Sqrt[ 3] - 2*x] + ArcTan[Sqrt[3] + 2*x])/2 + (Log[1 - Sqrt[3]*x + x^2]/(2*Sqrt[3 ]) - Log[1 + Sqrt[3]*x + x^2]/(2*Sqrt[3]))/2)/2
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])
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]
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]
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]
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]))
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]
Int[(x_)^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> W ith[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*r) Int[x^( m - n/2)/(q - r*x^(n/2) + x^n), x], x] - Simp[1/(2*c*r) Int[x^(m - n/2)/( q + r*x^(n/2) + x^n), x], x]]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && Ne Q[b^2 - 4*a*c, 0] && IGtQ[n/2, 0] && IGtQ[m, 0] && GeQ[m, n/2] && LtQ[m, 3* (n/2)] && NegQ[b^2 - 4*a*c]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.82
| method | result | size |
| risch | \(-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (4 x^{2}-4 x +4\right )}{8}-\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}+\frac {\ln \left (4 x^{2}+4 x +4\right )}{8}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (6 \textit {\_R}^{3}+\textit {\_R} +x \right )\right )}{4}\) | \(89\) |
| default | \(-\frac {\ln \left (x^{2}-x +1\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{2}+x +1\right )}{8}-\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\sqrt {3}\, \left (\frac {\ln \left (x^{2}+\sqrt {3}\, x +1\right )}{2}-\sqrt {3}\, \arctan \left (\sqrt {3}+2 x \right )\right )}{12}-\frac {\sqrt {3}\, \left (-\frac {\ln \left (x^{2}-\sqrt {3}\, x +1\right )}{2}-\sqrt {3}\, \arctan \left (-\sqrt {3}+2 x \right )\right )}{12}\) | \(121\) |
Input:
int(x^4/(x^8+x^4+1),x,method=_RETURNVERBOSE)
Output:
-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)+1/4*sum(_R*ln(6*_R^3+_R+x ),_R=RootOf(9*_Z^4+3*_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97 \[ \int \frac {x^4}{1+x^4+x^8} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) - \frac {1}{4} \, \arctan \left (-2 \, x + \sqrt {3}\right ) + \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^4/(x^8+x^4+1),x, algorithm="fricas")
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)
Result contains complex when optimal does not.
Time = 0.41 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.81 \[ \int \frac {x^4}{1+x^4+x^8} \, dx=\left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x - \frac {1}{2} + \frac {\sqrt {3} i}{6} - 18432 \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x - \frac {1}{2} - 18432 \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} - \frac {\sqrt {3} i}{6} \right )} + \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x + \frac {1}{2} + \frac {\sqrt {3} i}{6} - 18432 \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x + \frac {1}{2} - 18432 \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} - \frac {\sqrt {3} i}{6} \right )} + \operatorname {RootSum} {\left (2304 t^{4} + 48 t^{2} + 1, \left ( t \mapsto t \log {\left (- 18432 t^{5} - 4 t + x \right )} \right )\right )} \] Input:
integrate(x**4/(x**8+x**4+1),x)
Output:
(1/8 - sqrt(3)*I/24)*log(x - 1/2 + sqrt(3)*I/6 - 18432*(1/8 - sqrt(3)*I/24 )**5) + (1/8 + sqrt(3)*I/24)*log(x - 1/2 - 18432*(1/8 + sqrt(3)*I/24)**5 - sqrt(3)*I/6) + (-1/8 - sqrt(3)*I/24)*log(x + 1/2 + sqrt(3)*I/6 - 18432*(- 1/8 - sqrt(3)*I/24)**5) + (-1/8 + sqrt(3)*I/24)*log(x + 1/2 - 18432*(-1/8 + sqrt(3)*I/24)**5 - sqrt(3)*I/6) + RootSum(2304*_t**4 + 48*_t**2 + 1, Lam bda(_t, _t*log(-18432*_t**5 - 4*_t + x)))
\[ \int \frac {x^4}{1+x^4+x^8} \, dx=\int { \frac {x^{4}}{x^{8} + x^{4} + 1} \,d x } \] Input:
integrate(x^4/(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(x^2/(x^4 - x^2 + 1), x) + 1/8*log(x^2 + x + 1) - 1/8*log(x^2 - x + 1)
Time = 0.14 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.99 \[ \int \frac {x^4}{1+x^4+x^8} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^4/(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)
Time = 18.92 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int \frac {x^4}{1+x^4+x^8} \, dx=-\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}-\frac {1}{4}{}\mathrm {i}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}+\frac {1}{4}{}\mathrm {i}\right ) \] Input:
int(x^4/(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)
Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {x^4}{1+x^4+x^8} \, dx=-\frac {\mathit {atan} \left (\sqrt {3}-2 x \right )}{4}+\frac {\mathit {atan} \left (\sqrt {3}+2 x \right )}{4}-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{12}-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )}{12}+\frac {\sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )}{24}-\frac {\sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )}{24}-\frac {\mathrm {log}\left (x^{2}-x +1\right )}{8}+\frac {\mathrm {log}\left (x^{2}+x +1\right )}{8} \] Input:
int(x^4/(x^8+x^4+1),x)
Output:
( - 6*atan(sqrt(3) - 2*x) + 6*atan(sqrt(3) + 2*x) - 2*sqrt(3)*atan((2*x - 1)/sqrt(3)) - 2*sqrt(3)*atan((2*x + 1)/sqrt(3)) + sqrt(3)*log( - sqrt(3)*x + x**2 + 1) - sqrt(3)*log(sqrt(3)*x + x**2 + 1) - 3*log(x**2 - x + 1) + 3 *log(x**2 + x + 1))/24