Integrand size = 14, antiderivative size = 110 \[ \int \frac {x^8}{1+x^4+x^8} \, dx=x+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \arctan \left (\sqrt {3}+2 x\right )-\frac {1}{4} \text {arctanh}\left (\frac {x}{1+x^2}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{4 \sqrt {3}} \] Output:
x+1/12*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)-1/4*arctan(-3^(1/2)+2*x)-1/12*a rctan(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 = 0.17 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.26 \[ \int \frac {x^8}{1+x^4+x^8} \, dx=-\frac {i \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )}{\sqrt {-6+6 i \sqrt {3}}}+\frac {i \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )}{\sqrt {-6-6 i \sqrt {3}}}+\frac {1}{24} \left (24 x-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+3 \log \left (1-x+x^2\right )-3 \log \left (1+x+x^2\right )\right ) \] Input:
Integrate[x^8/(1 + x^4 + x^8),x]
Output:
((-I)*ArcTan[((1 - I*Sqrt[3])*x)/2])/Sqrt[-6 + (6*I)*Sqrt[3]] + (I*ArcTan[ ((1 + I*Sqrt[3])*x)/2])/Sqrt[-6 - (6*I)*Sqrt[3]] + (24*x - 2*Sqrt[3]*ArcTa n[(-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 = 0.38 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.56, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {1703, 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.
| \(\displaystyle \int \frac {x^8}{x^8+x^4+1} \, dx\) |
| \(\Big \downarrow \) 1703 |
| \(\displaystyle x-\int \frac {x^4+1}{x^8+x^4+1}dx\) |
| \(\Big \downarrow \) 1749 |
| \(\displaystyle -\frac {1}{2} \int \frac {1}{x^4-x^2+1}dx-\frac {1}{2} \int \frac {1}{x^4+x^2+1}dx+x\) |
| \(\Big \downarrow \) 1407 |
| \(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1-x}{x^2-x+1}dx-\frac {1}{2} \int \frac {x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (-\frac {\int \frac {\sqrt {3}-x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int \frac {x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+x\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx-\frac {1}{2} \int \frac {1}{x^2-x+1}dx\right )+\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2+x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )\right )+\frac {1}{2} \left (-\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2-\sqrt {3} x+1}dx-\frac {1}{2} \int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2+\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+x\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2+x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )\right )+\frac {1}{2} \left (-\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2-\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2+\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+x\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )\right )+\frac {1}{2} \left (-\frac {\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx-\sqrt {3} \int \frac {1}{-\left (2 x-\sqrt {3}\right )^2-1}d\left (2 x-\sqrt {3}\right )}{2 \sqrt {3}}-\frac {\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx-\sqrt {3} \int \frac {1}{-\left (2 x+\sqrt {3}\right )^2-1}d\left (2 x+\sqrt {3}\right )}{2 \sqrt {3}}\right )+x\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )+\frac {1}{2} \left (-\frac {\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx-\sqrt {3} \arctan \left (\sqrt {3}-2 x\right )}{2 \sqrt {3}}-\frac {\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx+\sqrt {3} \arctan \left (2 x+\sqrt {3}\right )}{2 \sqrt {3}}\right )+x\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \log \left (x^2-x+1\right )-\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^2+x+1\right )\right )\right )+\frac {1}{2} \left (-\frac {-\sqrt {3} \arctan \left (\sqrt {3}-2 x\right )-\frac {1}{2} \log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\sqrt {3} \arctan \left (2 x+\sqrt {3}\right )+\frac {1}{2} \log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}\right )+x\) |
Input:
Int[x^8/(1 + x^4 + x^8),x]
Output:
x + ((-(ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3]) + Log[1 - x + x^2]/2)/2 + (-(A rcTan[(1 + 2*x)/Sqrt[3]]/Sqrt[3]) - Log[1 + x + x^2]/2)/2)/2 + (-1/2*(-(Sq rt[3]*ArcTan[Sqrt[3] - 2*x]) - Log[1 - Sqrt[3]*x + x^2]/2)/Sqrt[3] - (Sqrt [3]*ArcTan[Sqrt[3] + 2*x] + Log[1 + Sqrt[3]*x + x^2]/2)/(2*Sqrt[3]))/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[((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]
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]
Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x _Symbol] :> Simp[d^(2*n - 1)*(d*x)^(m - 2*n + 1)*((a + b*x^n + c*x^(2*n))^( p + 1)/(c*(m + 2*n*p + 1))), x] - Simp[d^(2*n)/(c*(m + 2*n*p + 1)) Int[(d *x)^(m - 2*n)*Simp[a*(m - 2*n + 1) + b*(m + n*(p - 1) + 1)*x^n, x]*(a + b*x ^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[n2, 2*n] && N eQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1] && NeQ[m + 2*n*p + 1, 0 ] && IntegerQ[p]
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]]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84
| method | result | size |
| risch | \(x -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (4 x^{2}-4 x +4\right )}{8}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (3 \textit {\_R}^{3}-\textit {\_R} +x \right )\right )}{4}-\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\ln \left (4 x^{2}+4 x +4\right )}{8}\) | \(92\) |
| default | \(x +\frac {\ln \left (x^{2}-x +1\right )}{8}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x^{2}+x +1\right )}{8}-\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{24}-\frac {\arctan \left (\sqrt {3}+2 x \right )}{4}+\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{24}-\frac {\arctan \left (-\sqrt {3}+2 x \right )}{4}\) | \(110\) |
Input:
int(x^8/(x^8+x^4+1),x,method=_RETURNVERBOSE)
Output:
x-1/12*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))+1/8*ln(4*x^2-4*x+4)+1/4*sum(_R* ln(3*_R^3-_R+x),_R=RootOf(9*_Z^4+3*_Z^2+1))-1/12*arctan(1/3*(2*x+1)*3^(1/2 ))*3^(1/2)-1/8*ln(4*x^2+4*x+4)
Time = 0.07 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.97 \[ \int \frac {x^8}{1+x^4+x^8} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + x - \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (-2 \, x + \sqrt {3}\right ) - \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^8/(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) + x - 1/4*arctan(2*x + sqrt(3)) + 1/4*arctan(-2*x + s qrt(3)) - 1/8*log(x^2 + x + 1) + 1/8*log(x^2 - x + 1)
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.75 \[ \int \frac {x^8}{1+x^4+x^8} \, dx=x + \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 1 - \frac {\sqrt {3} i}{3} - 9216 \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 1 - 9216 \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} + \frac {\sqrt {3} i}{3} \right )} + \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x + 1 - \frac {\sqrt {3} i}{3} - 9216 \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x + 1 - 9216 \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} + \frac {\sqrt {3} i}{3} \right )} + \operatorname {RootSum} {\left (2304 t^{4} + 48 t^{2} + 1, \left ( t \mapsto t \log {\left (- 9216 t^{5} - 8 t + x \right )} \right )\right )} \] Input:
integrate(x**8/(x**8+x**4+1),x)
Output:
x + (1/8 + sqrt(3)*I/24)*log(x - 1 - sqrt(3)*I/3 - 9216*(1/8 + sqrt(3)*I/2 4)**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 + sqrt(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*log(-9216*_t**5 - 8*_t + x)))
\[ \int \frac {x^8}{1+x^4+x^8} \, dx=\int { \frac {x^{8}}{x^{8} + x^{4} + 1} \,d x } \] Input:
integrate(x^8/(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)) + x - 1/2*integrate(1/(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 = 109, normalized size of antiderivative = 0.99 \[ \int \frac {x^8}{1+x^4+x^8} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + x - \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) - \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) - \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^8/(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) + x - 1/4*arctan(2*x + sqrt(3)) - 1/4*arctan(2*x - sq rt(3)) - 1/8*log(x^2 + x + 1) + 1/8*log(x^2 - x + 1)
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.91 \[ \int \frac {x^8}{1+x^4+x^8} \, dx=x-\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}+\frac {1}{4}{}\mathrm {i}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}-\frac {1}{4}{}\mathrm {i}\right ) \] Input:
int(x^8/(x^4 + x^8 + 1),x)
Output:
x - 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 = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.90 \[ \int \frac {x^8}{1+x^4+x^8} \, dx=\frac {\mathit {atan} \left (\sqrt {3}-2 x \right )}{4}-\frac {\mathit {atan} \left (\sqrt {3}+2 x \right )}{4}-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{12}-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )}{12}+\frac {\sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )}{24}-\frac {\sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )}{24}+\frac {\mathrm {log}\left (x^{2}-x +1\right )}{8}-\frac {\mathrm {log}\left (x^{2}+x +1\right )}{8}+x \] Input:
int(x^8/(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*lo g(x**2 + x + 1) + 24*x)/24