\(\int \frac {x^8}{1-3 x^4+x^8} \, dx\) [128]

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

Optimal result

Integrand size = 16, antiderivative size = 178 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=x-\frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2 \sqrt {5}}+\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \text {arctanh}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2 \sqrt {5}}+\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \text {arctanh}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2 \sqrt {5}} \] Output:

x-1/10*(123/2+55/2*5^(1/2))^(1/4)*arctan(2^(1/4)*(1/(3+5^(1/2)))^(1/4)*x)* 
5^(1/2)+1/10*(123/2-55/2*5^(1/2))^(1/4)*arctan((3/2+1/2*5^(1/2))^(1/4)*x)* 
5^(1/2)-1/10*(123/2+55/2*5^(1/2))^(1/4)*arctanh(2^(1/4)*(1/(3+5^(1/2)))^(1 
/4)*x)*5^(1/2)+1/10*(123/2-55/2*5^(1/2))^(1/4)*arctanh((3/2+1/2*5^(1/2))^( 
1/4)*x)*5^(1/2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.90 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=x+\frac {\left (-2+\sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\left (2+\sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {\left (-2+\sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\left (2+\sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}} \] Input:

Integrate[x^8/(1 - 3*x^4 + x^8),x]
 

Output:

x + ((-2 + Sqrt[5])*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x])/Sqrt[10*(-1 + Sqrt[5 
])] - ((2 + Sqrt[5])*ArcTan[Sqrt[2/(1 + Sqrt[5])]*x])/Sqrt[10*(1 + Sqrt[5] 
)] + ((-2 + Sqrt[5])*ArcTanh[Sqrt[2/(-1 + Sqrt[5])]*x])/Sqrt[10*(-1 + Sqrt 
[5])] - ((2 + Sqrt[5])*ArcTanh[Sqrt[2/(1 + Sqrt[5])]*x])/Sqrt[10*(1 + Sqrt 
[5])]
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1703, 1752, 756, 216, 219}

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.

\(\displaystyle \int \frac {x^8}{x^8-3 x^4+1} \, dx\)

\(\Big \downarrow \) 1703

\(\displaystyle x-\int \frac {1-3 x^4}{x^8-3 x^4+1}dx\)

\(\Big \downarrow \) 1752

\(\displaystyle \frac {1}{10} \left (15+7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (-3-\sqrt {5}\right )}dx+\frac {1}{10} \left (15-7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (-3+\sqrt {5}\right )}dx+x\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {1}{10} \left (15-7 \sqrt {5}\right ) \left (-\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2}dx}{\sqrt {3-\sqrt {5}}}-\frac {\int \frac {1}{\sqrt {2} x^2+\sqrt {3-\sqrt {5}}}dx}{\sqrt {3-\sqrt {5}}}\right )+\frac {1}{10} \left (15+7 \sqrt {5}\right ) \left (-\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2}dx}{\sqrt {3+\sqrt {5}}}-\frac {\int \frac {1}{\sqrt {2} x^2+\sqrt {3+\sqrt {5}}}dx}{\sqrt {3+\sqrt {5}}}\right )+x\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{10} \left (15-7 \sqrt {5}\right ) \left (-\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2}dx}{\sqrt {3-\sqrt {5}}}-\frac {\arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{\sqrt [4]{2} \left (3-\sqrt {5}\right )^{3/4}}\right )+\frac {1}{10} \left (15+7 \sqrt {5}\right ) \left (-\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2}dx}{\sqrt {3+\sqrt {5}}}-\frac {\arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{\sqrt [4]{2} \left (3+\sqrt {5}\right )^{3/4}}\right )+x\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{10} \left (15+7 \sqrt {5}\right ) \left (-\frac {\arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{\sqrt [4]{2} \left (3+\sqrt {5}\right )^{3/4}}-\frac {\text {arctanh}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{\sqrt [4]{2} \left (3+\sqrt {5}\right )^{3/4}}\right )+\frac {1}{10} \left (15-7 \sqrt {5}\right ) \left (-\frac {\arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{\sqrt [4]{2} \left (3-\sqrt {5}\right )^{3/4}}-\frac {\text {arctanh}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{\sqrt [4]{2} \left (3-\sqrt {5}\right )^{3/4}}\right )+x\)

Input:

Int[x^8/(1 - 3*x^4 + x^8),x]
 

Output:

x + ((15 + 7*Sqrt[5])*(-(ArcTan[(2/(3 + Sqrt[5]))^(1/4)*x]/(2^(1/4)*(3 + S 
qrt[5])^(3/4))) - ArcTanh[(2/(3 + Sqrt[5]))^(1/4)*x]/(2^(1/4)*(3 + Sqrt[5] 
)^(3/4))))/10 + ((15 - 7*Sqrt[5])*(-(ArcTan[((3 + Sqrt[5])/2)^(1/4)*x]/(2^ 
(1/4)*(3 - Sqrt[5])^(3/4))) - ArcTanh[((3 + Sqrt[5])/2)^(1/4)*x]/(2^(1/4)* 
(3 - Sqrt[5])^(3/4))))/10
 

Defintions of rubi rules used

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 756
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 1703
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]
 

rule 1752
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x 
_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) 
   Int[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   I 
nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 
, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 
 - 4*a*c] ||  !IGtQ[n/2, 0])
 
Maple [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.39

method result size
risch \(x +\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+55 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (15 \textit {\_R}^{3}+29 \textit {\_R} +5 x \right )\right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}-55 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (-15 \textit {\_R}^{3}+29 \textit {\_R} +5 x \right )\right )}{4}\) \(69\)
default \(x -\frac {\left (2+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2+2 \sqrt {5}}}\right )}{5 \sqrt {2+2 \sqrt {5}}}+\frac {\left (-2+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {-2+2 \sqrt {5}}}\right )}{5 \sqrt {-2+2 \sqrt {5}}}+\frac {\left (-2+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {-2+2 \sqrt {5}}}\right )}{5 \sqrt {-2+2 \sqrt {5}}}-\frac {\left (2+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2+2 \sqrt {5}}}\right )}{5 \sqrt {2+2 \sqrt {5}}}\) \(131\)

Input:

int(x^8/(x^8-3*x^4+1),x,method=_RETURNVERBOSE)
 

Output:

x+1/4*sum(_R*ln(15*_R^3+29*_R+5*x),_R=RootOf(25*_Z^4+55*_Z^2-1))+1/4*sum(_ 
R*ln(-15*_R^3+29*_R+5*x),_R=RootOf(25*_Z^4-55*_Z^2-1))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.12 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=\frac {1}{2} \, \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} \arctan \left ({\left (2 \, \sqrt {5} x - 5 \, x\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}}\right ) + \frac {1}{2} \, \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} \arctan \left ({\left (2 \, \sqrt {5} x + 5 \, x\right )} \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}}\right ) - \frac {1}{4} \, \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} \log \left ({\left (3 \, \sqrt {5} - 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} + 2 \, x\right ) + \frac {1}{4} \, \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} \log \left (-{\left (3 \, \sqrt {5} - 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} + 2 \, x\right ) + \frac {1}{4} \, \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} \log \left ({\left (3 \, \sqrt {5} + 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} + 2 \, x\right ) - \frac {1}{4} \, \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} \log \left (-{\left (3 \, \sqrt {5} + 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} + 2 \, x\right ) + x \] Input:

integrate(x^8/(x^8-3*x^4+1),x, algorithm="fricas")
 

Output:

1/2*sqrt(1/2*sqrt(5) + 11/10)*arctan((2*sqrt(5)*x - 5*x)*sqrt(1/2*sqrt(5) 
+ 11/10)) + 1/2*sqrt(1/2*sqrt(5) - 11/10)*arctan((2*sqrt(5)*x + 5*x)*sqrt( 
1/2*sqrt(5) - 11/10)) - 1/4*sqrt(1/2*sqrt(5) + 11/10)*log((3*sqrt(5) - 5)* 
sqrt(1/2*sqrt(5) + 11/10) + 2*x) + 1/4*sqrt(1/2*sqrt(5) + 11/10)*log(-(3*s 
qrt(5) - 5)*sqrt(1/2*sqrt(5) + 11/10) + 2*x) + 1/4*sqrt(1/2*sqrt(5) - 11/1 
0)*log((3*sqrt(5) + 5)*sqrt(1/2*sqrt(5) - 11/10) + 2*x) - 1/4*sqrt(1/2*sqr 
t(5) - 11/10)*log(-(3*sqrt(5) + 5)*sqrt(1/2*sqrt(5) - 11/10) + 2*x) + x
 

Sympy [A] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.33 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=x + \operatorname {RootSum} {\left (6400 t^{4} - 880 t^{2} - 1, \left ( t \mapsto t \log {\left (- \frac {15360 t^{5}}{11} + \frac {1288 t}{55} + x \right )} \right )\right )} + \operatorname {RootSum} {\left (6400 t^{4} + 880 t^{2} - 1, \left ( t \mapsto t \log {\left (- \frac {15360 t^{5}}{11} + \frac {1288 t}{55} + x \right )} \right )\right )} \] Input:

integrate(x**8/(x**8-3*x**4+1),x)
 

Output:

x + RootSum(6400*_t**4 - 880*_t**2 - 1, Lambda(_t, _t*log(-15360*_t**5/11 
+ 1288*_t/55 + x))) + RootSum(6400*_t**4 + 880*_t**2 - 1, Lambda(_t, _t*lo 
g(-15360*_t**5/11 + 1288*_t/55 + x)))
 

Maxima [F]

\[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=\int { \frac {x^{8}}{x^{8} - 3 \, x^{4} + 1} \,d x } \] Input:

integrate(x^8/(x^8-3*x^4+1),x, algorithm="maxima")
 

Output:

x + 1/2*integrate((2*x^2 + 1)/(x^4 - x^2 - 1), x) - 1/2*integrate((2*x^2 - 
 1)/(x^4 + x^2 - 1), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.83 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=-\frac {1}{20} \, \sqrt {50 \, \sqrt {5} + 110} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) + \frac {1}{20} \, \sqrt {50 \, \sqrt {5} - 110} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{40} \, \sqrt {50 \, \sqrt {5} + 110} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {50 \, \sqrt {5} + 110} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {50 \, \sqrt {5} - 110} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{40} \, \sqrt {50 \, \sqrt {5} - 110} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) + x \] Input:

integrate(x^8/(x^8-3*x^4+1),x, algorithm="giac")
 

Output:

-1/20*sqrt(50*sqrt(5) + 110)*arctan(x/sqrt(1/2*sqrt(5) + 1/2)) + 1/20*sqrt 
(50*sqrt(5) - 110)*arctan(x/sqrt(1/2*sqrt(5) - 1/2)) - 1/40*sqrt(50*sqrt(5 
) + 110)*log(abs(x + sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(50*sqrt(5) + 11 
0)*log(abs(x - sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(50*sqrt(5) - 110)*log 
(abs(x + sqrt(1/2*sqrt(5) - 1/2))) - 1/40*sqrt(50*sqrt(5) - 110)*log(abs(x 
 - sqrt(1/2*sqrt(5) - 1/2))) + x
 

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.38 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=x-\frac {\mathrm {atan}\left (\frac {x\,\sqrt {-50\,\sqrt {5}-110}\,55{}\mathrm {i}}{2\,\left (275\,\sqrt {5}+605\right )}+\frac {\sqrt {5}\,x\,\sqrt {-50\,\sqrt {5}-110}\,33{}\mathrm {i}}{2\,\left (275\,\sqrt {5}+605\right )}\right )\,\sqrt {-50\,\sqrt {5}-110}\,1{}\mathrm {i}}{20}-\frac {\mathrm {atan}\left (\frac {x\,\sqrt {110-50\,\sqrt {5}}\,55{}\mathrm {i}}{2\,\left (275\,\sqrt {5}-605\right )}-\frac {\sqrt {5}\,x\,\sqrt {110-50\,\sqrt {5}}\,33{}\mathrm {i}}{2\,\left (275\,\sqrt {5}-605\right )}\right )\,\sqrt {110-50\,\sqrt {5}}\,1{}\mathrm {i}}{20}+\frac {\mathrm {atan}\left (\frac {x\,\sqrt {50\,\sqrt {5}-110}\,55{}\mathrm {i}}{2\,\left (275\,\sqrt {5}-605\right )}-\frac {\sqrt {5}\,x\,\sqrt {50\,\sqrt {5}-110}\,33{}\mathrm {i}}{2\,\left (275\,\sqrt {5}-605\right )}\right )\,\sqrt {50\,\sqrt {5}-110}\,1{}\mathrm {i}}{20}+\frac {\mathrm {atan}\left (\frac {x\,\sqrt {50\,\sqrt {5}+110}\,55{}\mathrm {i}}{2\,\left (275\,\sqrt {5}+605\right )}+\frac {\sqrt {5}\,x\,\sqrt {50\,\sqrt {5}+110}\,33{}\mathrm {i}}{2\,\left (275\,\sqrt {5}+605\right )}\right )\,\sqrt {50\,\sqrt {5}+110}\,1{}\mathrm {i}}{20} \] Input:

int(x^8/(x^8 - 3*x^4 + 1),x)
 

Output:

x - (atan((x*(- 50*5^(1/2) - 110)^(1/2)*55i)/(2*(275*5^(1/2) + 605)) + (5^ 
(1/2)*x*(- 50*5^(1/2) - 110)^(1/2)*33i)/(2*(275*5^(1/2) + 605)))*(- 50*5^( 
1/2) - 110)^(1/2)*1i)/20 - (atan((x*(110 - 50*5^(1/2))^(1/2)*55i)/(2*(275* 
5^(1/2) - 605)) - (5^(1/2)*x*(110 - 50*5^(1/2))^(1/2)*33i)/(2*(275*5^(1/2) 
 - 605)))*(110 - 50*5^(1/2))^(1/2)*1i)/20 + (atan((x*(50*5^(1/2) - 110)^(1 
/2)*55i)/(2*(275*5^(1/2) - 605)) - (5^(1/2)*x*(50*5^(1/2) - 110)^(1/2)*33i 
)/(2*(275*5^(1/2) - 605)))*(50*5^(1/2) - 110)^(1/2)*1i)/20 + (atan((x*(50* 
5^(1/2) + 110)^(1/2)*55i)/(2*(275*5^(1/2) + 605)) + (5^(1/2)*x*(50*5^(1/2) 
 + 110)^(1/2)*33i)/(2*(275*5^(1/2) + 605)))*(50*5^(1/2) + 110)^(1/2)*1i)/2 
0
 

Reduce [F]

\[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=\frac {3 \sqrt {\sqrt {5}+1}\, \sqrt {10}\, \mathit {atan} \left (\frac {2 x}{\sqrt {\sqrt {5}+1}\, \sqrt {2}}\right )}{20}-\frac {3 \sqrt {\sqrt {5}+1}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 x}{\sqrt {\sqrt {5}+1}\, \sqrt {2}}\right )}{4}+\frac {3 \sqrt {\sqrt {5}-1}\, \sqrt {10}\, \mathrm {log}\left (-\sqrt {\sqrt {5}-1}+\sqrt {2}\, x \right )}{40}-\frac {3 \sqrt {\sqrt {5}-1}\, \sqrt {10}\, \mathrm {log}\left (\sqrt {\sqrt {5}-1}+\sqrt {2}\, x \right )}{40}+\frac {3 \sqrt {\sqrt {5}-1}\, \sqrt {2}\, \mathrm {log}\left (-\sqrt {\sqrt {5}-1}+\sqrt {2}\, x \right )}{8}-\frac {3 \sqrt {\sqrt {5}-1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {\sqrt {5}-1}+\sqrt {2}\, x \right )}{8}+3 \left (\int \frac {x^{2}}{x^{8}-3 x^{4}+1}d x \right )+2 \left (\int \frac {1}{x^{8}-3 x^{4}+1}d x \right )+x \] Input:

int(x^8/(x^8-3*x^4+1),x)
 

Output:

(6*sqrt(sqrt(5) + 1)*sqrt(10)*atan((2*x)/(sqrt(sqrt(5) + 1)*sqrt(2))) - 30 
*sqrt(sqrt(5) + 1)*sqrt(2)*atan((2*x)/(sqrt(sqrt(5) + 1)*sqrt(2))) + 3*sqr 
t(sqrt(5) - 1)*sqrt(10)*log( - sqrt(sqrt(5) - 1) + sqrt(2)*x) - 3*sqrt(sqr 
t(5) - 1)*sqrt(10)*log(sqrt(sqrt(5) - 1) + sqrt(2)*x) + 15*sqrt(sqrt(5) - 
1)*sqrt(2)*log( - sqrt(sqrt(5) - 1) + sqrt(2)*x) - 15*sqrt(sqrt(5) - 1)*sq 
rt(2)*log(sqrt(sqrt(5) - 1) + sqrt(2)*x) + 120*int(x**2/(x**8 - 3*x**4 + 1 
),x) + 80*int(1/(x**8 - 3*x**4 + 1),x) + 40*x)/40