\(\int \frac {1}{x^4 (1-3 x^4+x^8)} \, dx\) [134]

Optimal result

Integrand size = 16, antiderivative size = 182 \[ \int \frac {1}{x^4 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{3 x^3}-\frac {\sqrt [4]{\frac {1}{2} \left (843-377 \sqrt {5}\right )} \arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2 \sqrt {5}}+\frac {\left (3+\sqrt {5}\right )^{7/4} \arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{4\ 2^{3/4} \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (843-377 \sqrt {5}\right )} \text {arctanh}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2 \sqrt {5}}+\frac {\left (3+\sqrt {5}\right )^{7/4} \text {arctanh}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{4\ 2^{3/4} \sqrt {5}} \] Output:

-1/3/x^3-1/10*(843/2-377/2*5^(1/2))^(1/4)*arctan(2^(1/4)*(1/(3+5^(1/2)))^( 
1/4)*x)*5^(1/2)+1/40*(3+5^(1/2))^(7/4)*arctan((3/2+1/2*5^(1/2))^(1/4)*x)*2 
^(1/4)*5^(1/2)-1/10*(843/2-377/2*5^(1/2))^(1/4)*arctanh(2^(1/4)*(1/(3+5^(1 
/2)))^(1/4)*x)*5^(1/2)+1/40*(3+5^(1/2))^(7/4)*arctanh((3/2+1/2*5^(1/2))^(1 
/4)*x)*2^(1/4)*5^(1/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^4 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{3 x^3}+\frac {\left (2+\sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\left (-2+\sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {\left (2+\sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\left (-2+\sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}} \] Input:

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

Output:

-1/3*1/x^3 + ((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 = 184, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1704, 27, 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.

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

\(\Big \downarrow \) 1704

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1752

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/3*1/x^3 - ((5 - 3*Sqrt[5])*(-(ArcTan[(2/(3 + Sqrt[5]))^(1/4)*x]/(2^(1/4 
)*(3 + Sqrt[5])^(3/4))) - ArcTanh[(2/(3 + Sqrt[5]))^(1/4)*x]/(2^(1/4)*(3 + 
 Sqrt[5])^(3/4))))/10 - ((5 + 3*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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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 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 = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.40

method	result	size
risch	\(-\frac {1}{3 x^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+145 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (-35 \textit {\_R}^{3}-199 \textit {\_R} +13 x \right )\right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}-145 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (35 \textit {\_R}^{3}-199 \textit {\_R} +13 x \right )\right )}{4}\)	\(73\)
default	\(-\frac {1}{3 x^{3}}-\frac {\left (-2+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2+2 \sqrt {5}}}\right )}{5 \sqrt {2+2 \sqrt {5}}}+\frac {\left (2+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {-2+2 \sqrt {5}}}\right )}{5 \sqrt {-2+2 \sqrt {5}}}+\frac {\left (2+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {-2+2 \sqrt {5}}}\right )}{5 \sqrt {-2+2 \sqrt {5}}}-\frac {\left (-2+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2+2 \sqrt {5}}}\right )}{5 \sqrt {2+2 \sqrt {5}}}\)	\(135\)

Input:

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

Output:

-1/3/x^3+1/4*sum(_R*ln(-35*_R^3-199*_R+13*x),_R=RootOf(25*_Z^4+145*_Z^2-1) 
)+1/4*sum(_R*ln(35*_R^3-199*_R+13*x),_R=RootOf(25*_Z^4-145*_Z^2-1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^4 \left (1-3 x^4+x^8\right )} \, dx=-\frac {6 \, x^{3} \sqrt {\frac {13}{10} \, \sqrt {5} + \frac {29}{10}} \arctan \left ({\left (2 \, \sqrt {5} x - 5 \, x\right )} \sqrt {\frac {13}{10} \, \sqrt {5} + \frac {29}{10}}\right ) + 6 \, x^{3} \sqrt {\frac {13}{10} \, \sqrt {5} - \frac {29}{10}} \arctan \left ({\left (2 \, \sqrt {5} x + 5 \, x\right )} \sqrt {\frac {13}{10} \, \sqrt {5} - \frac {29}{10}}\right ) - 3 \, x^{3} \sqrt {\frac {13}{10} \, \sqrt {5} + \frac {29}{10}} \log \left ({\left (7 \, \sqrt {5} - 15\right )} \sqrt {\frac {13}{10} \, \sqrt {5} + \frac {29}{10}} + 2 \, x\right ) + 3 \, x^{3} \sqrt {\frac {13}{10} \, \sqrt {5} + \frac {29}{10}} \log \left (-{\left (7 \, \sqrt {5} - 15\right )} \sqrt {\frac {13}{10} \, \sqrt {5} + \frac {29}{10}} + 2 \, x\right ) + 3 \, x^{3} \sqrt {\frac {13}{10} \, \sqrt {5} - \frac {29}{10}} \log \left ({\left (7 \, \sqrt {5} + 15\right )} \sqrt {\frac {13}{10} \, \sqrt {5} - \frac {29}{10}} + 2 \, x\right ) - 3 \, x^{3} \sqrt {\frac {13}{10} \, \sqrt {5} - \frac {29}{10}} \log \left (-{\left (7 \, \sqrt {5} + 15\right )} \sqrt {\frac {13}{10} \, \sqrt {5} - \frac {29}{10}} + 2 \, x\right ) + 4}{12 \, x^{3}} \] Input:

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

Output:

-1/12*(6*x^3*sqrt(13/10*sqrt(5) + 29/10)*arctan((2*sqrt(5)*x - 5*x)*sqrt(1 
3/10*sqrt(5) + 29/10)) + 6*x^3*sqrt(13/10*sqrt(5) - 29/10)*arctan((2*sqrt( 
5)*x + 5*x)*sqrt(13/10*sqrt(5) - 29/10)) - 3*x^3*sqrt(13/10*sqrt(5) + 29/1 
0)*log((7*sqrt(5) - 15)*sqrt(13/10*sqrt(5) + 29/10) + 2*x) + 3*x^3*sqrt(13 
/10*sqrt(5) + 29/10)*log(-(7*sqrt(5) - 15)*sqrt(13/10*sqrt(5) + 29/10) + 2 
*x) + 3*x^3*sqrt(13/10*sqrt(5) - 29/10)*log((7*sqrt(5) + 15)*sqrt(13/10*sq 
rt(5) - 29/10) + 2*x) - 3*x^3*sqrt(13/10*sqrt(5) - 29/10)*log(-(7*sqrt(5) 
+ 15)*sqrt(13/10*sqrt(5) - 29/10) + 2*x) + 4)/x^3
 

Sympy [A] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^4 \left (1-3 x^4+x^8\right )} \, dx=\operatorname {RootSum} {\left (6400 t^{4} - 2320 t^{2} - 1, \left ( t \mapsto t \log {\left (\frac {179200 t^{5}}{377} - \frac {23112 t}{377} + x \right )} \right )\right )} + \operatorname {RootSum} {\left (6400 t^{4} + 2320 t^{2} - 1, \left ( t \mapsto t \log {\left (\frac {179200 t^{5}}{377} - \frac {23112 t}{377} + x \right )} \right )\right )} - \frac {1}{3 x^{3}} \] Input:

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

Output:

RootSum(6400*_t**4 - 2320*_t**2 - 1, Lambda(_t, _t*log(179200*_t**5/377 - 
23112*_t/377 + x))) + RootSum(6400*_t**4 + 2320*_t**2 - 1, Lambda(_t, _t*l 
og(179200*_t**5/377 - 23112*_t/377 + x))) - 1/(3*x**3)
 

Maxima [F]

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

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

Output:

-1/3/x^3 - 1/2*integrate((2*x^2 + 3)/(x^4 + x^2 - 1), x) + 1/2*integrate(( 
2*x^2 - 3)/(x^4 - x^2 - 1), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^4 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{20} \, \sqrt {130 \, \sqrt {5} - 290} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) + \frac {1}{20} \, \sqrt {130 \, \sqrt {5} + 290} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{40} \, \sqrt {130 \, \sqrt {5} - 290} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {130 \, \sqrt {5} - 290} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {130 \, \sqrt {5} + 290} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{40} \, \sqrt {130 \, \sqrt {5} + 290} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{3 \, x^{3}} \] Input:

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

Output:

-1/20*sqrt(130*sqrt(5) - 290)*arctan(x/sqrt(1/2*sqrt(5) + 1/2)) + 1/20*sqr 
t(130*sqrt(5) + 290)*arctan(x/sqrt(1/2*sqrt(5) - 1/2)) - 1/40*sqrt(130*sqr 
t(5) - 290)*log(abs(x + sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(130*sqrt(5) 
- 290)*log(abs(x - sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(130*sqrt(5) + 290 
)*log(abs(x + sqrt(1/2*sqrt(5) - 1/2))) - 1/40*sqrt(130*sqrt(5) + 290)*log 
(abs(x - sqrt(1/2*sqrt(5) - 1/2))) - 1/3/x^3
 

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.47 \[ \int \frac {1}{x^4 \left (1-3 x^4+x^8\right )} \, dx=\frac {\mathrm {atan}\left (\frac {x\,\sqrt {-130\,\sqrt {5}-290}\,20735{}\mathrm {i}}{2\,\left (87841\,\sqrt {5}+196417\right )}+\frac {\sqrt {5}\,x\,\sqrt {-130\,\sqrt {5}-290}\,46371{}\mathrm {i}}{10\,\left (87841\,\sqrt {5}+196417\right )}\right )\,\sqrt {-130\,\sqrt {5}-290}\,1{}\mathrm {i}}{20}+\frac {\mathrm {atan}\left (\frac {x\,\sqrt {290-130\,\sqrt {5}}\,20735{}\mathrm {i}}{2\,\left (87841\,\sqrt {5}-196417\right )}-\frac {\sqrt {5}\,x\,\sqrt {290-130\,\sqrt {5}}\,46371{}\mathrm {i}}{10\,\left (87841\,\sqrt {5}-196417\right )}\right )\,\sqrt {290-130\,\sqrt {5}}\,1{}\mathrm {i}}{20}-\frac {1}{3\,x^3}-\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {13\,\sqrt {5}-29}\,20735{}\mathrm {i}}{2\,\left (87841\,\sqrt {5}-196417\right )}-\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {13\,\sqrt {5}-29}\,46371{}\mathrm {i}}{10\,\left (87841\,\sqrt {5}-196417\right )}\right )\,\sqrt {13\,\sqrt {5}-29}\,1{}\mathrm {i}}{20}-\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {13\,\sqrt {5}+29}\,20735{}\mathrm {i}}{2\,\left (87841\,\sqrt {5}+196417\right )}+\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {13\,\sqrt {5}+29}\,46371{}\mathrm {i}}{10\,\left (87841\,\sqrt {5}+196417\right )}\right )\,\sqrt {13\,\sqrt {5}+29}\,1{}\mathrm {i}}{20} \] Input:

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

Output:

(atan((x*(- 130*5^(1/2) - 290)^(1/2)*20735i)/(2*(87841*5^(1/2) + 196417)) 
+ (5^(1/2)*x*(- 130*5^(1/2) - 290)^(1/2)*46371i)/(10*(87841*5^(1/2) + 1964 
17)))*(- 130*5^(1/2) - 290)^(1/2)*1i)/20 + (atan((x*(290 - 130*5^(1/2))^(1 
/2)*20735i)/(2*(87841*5^(1/2) - 196417)) - (5^(1/2)*x*(290 - 130*5^(1/2))^ 
(1/2)*46371i)/(10*(87841*5^(1/2) - 196417)))*(290 - 130*5^(1/2))^(1/2)*1i) 
/20 - 1/(3*x^3) - (10^(1/2)*atan((10^(1/2)*x*(13*5^(1/2) - 29)^(1/2)*20735 
i)/(2*(87841*5^(1/2) - 196417)) - (5^(1/2)*10^(1/2)*x*(13*5^(1/2) - 29)^(1 
/2)*46371i)/(10*(87841*5^(1/2) - 196417)))*(13*5^(1/2) - 29)^(1/2)*1i)/20 
- (10^(1/2)*atan((10^(1/2)*x*(13*5^(1/2) + 29)^(1/2)*20735i)/(2*(87841*5^( 
1/2) + 196417)) + (5^(1/2)*10^(1/2)*x*(13*5^(1/2) + 29)^(1/2)*46371i)/(10* 
(87841*5^(1/2) + 196417)))*(13*5^(1/2) + 29)^(1/2)*1i)/20
 

Reduce [F]

\[ \int \frac {1}{x^4 \left (1-3 x^4+x^8\right )} \, dx=\int \frac {1}{x^{12}-3 x^{8}+x^{4}}d x \] Input:

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

Output:

int(1/(x**12 - 3*x**8 + x**4),x)