\(\int \frac {1}{x^6 (1-x^4+x^8)} \, dx\) [99]

Optimal result
Mathematica [C] (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 [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 231 \[ \int \frac {1}{x^6 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{5 x^5}-\frac {1}{x}+\frac {\arctan \left (\frac {\sqrt {2-\sqrt {3}}-2 x}{\sqrt {2+\sqrt {3}}}\right )}{2 \sqrt {6}}+\frac {\arctan \left (\frac {\sqrt {2+\sqrt {3}}-2 x}{\sqrt {2-\sqrt {3}}}\right )}{2 \sqrt {6}}-\frac {\arctan \left (\frac {\sqrt {2-\sqrt {3}}+2 x}{\sqrt {2+\sqrt {3}}}\right )}{2 \sqrt {6}}-\frac {\arctan \left (\frac {\sqrt {2+\sqrt {3}}+2 x}{\sqrt {2-\sqrt {3}}}\right )}{2 \sqrt {6}}+\frac {\text {arctanh}\left (\frac {\sqrt {2-\sqrt {3}} x}{1+x^2}\right )}{2 \sqrt {6}}+\frac {\text {arctanh}\left (\frac {\sqrt {2+\sqrt {3}} x}{1+x^2}\right )}{2 \sqrt {6}} \] Output:

-1/5/x^5-1/x+1/12*arctan((1/2*6^(1/2)-1/2*2^(1/2)-2*x)/(1/2*6^(1/2)+1/2*2^ 
(1/2)))*6^(1/2)+1/12*arctan((1/2*6^(1/2)+1/2*2^(1/2)-2*x)/(1/2*6^(1/2)-1/2 
*2^(1/2)))*6^(1/2)-1/12*arctan((1/2*6^(1/2)-1/2*2^(1/2)+2*x)/(1/2*6^(1/2)+ 
1/2*2^(1/2)))*6^(1/2)-1/12*arctan((1/2*6^(1/2)+1/2*2^(1/2)+2*x)/(1/2*6^(1/ 
2)-1/2*2^(1/2)))*6^(1/2)+1/12*arctanh((1/2*6^(1/2)-1/2*2^(1/2))*x/(x^2+1)) 
*6^(1/2)+1/12*arctanh((1/2*6^(1/2)+1/2*2^(1/2))*x/(x^2+1))*6^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.23 \[ \int \frac {1}{x^6 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{5 x^5}-\frac {1}{x}-\frac {1}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x-\text {$\#$1}) \text {$\#$1}^3}{-1+2 \text {$\#$1}^4}\&\right ] \] Input:

Integrate[1/(x^6*(1 - x^4 + x^8)),x]
 

Output:

-1/5*1/x^5 - x^(-1) - RootSum[1 - #1^4 + #1^8 & , (Log[x - #1]*#1^3)/(-1 + 
 2*#1^4) & ]/4
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.84, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {1704, 27, 1828, 1708, 1483, 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.

\(\displaystyle \int \frac {1}{x^6 \left (x^8-x^4+1\right )} \, dx\)

\(\Big \downarrow \) 1704

\(\displaystyle \frac {1}{5} \int \frac {5 \left (1-x^4\right )}{x^2 \left (x^8-x^4+1\right )}dx-\frac {1}{5 x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {1-x^4}{x^2 \left (x^8-x^4+1\right )}dx-\frac {1}{5 x^5}\)

\(\Big \downarrow \) 1828

\(\displaystyle -\int \frac {x^6}{x^8-x^4+1}dx-\frac {1}{5 x^5}-\frac {1}{x}\)

\(\Big \downarrow \) 1708

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

\(\Big \downarrow \) 1483

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {\frac {\sqrt {\frac {2}{2+\sqrt {3}}} \arctan \left (\frac {2 x-\sqrt {2-\sqrt {3}}}{\sqrt {2+\sqrt {3}}}\right )-\frac {1}{2} \left (1-\sqrt {3}\right ) \log \left (x^2-\sqrt {2-\sqrt {3}} x+1\right )}{2 \sqrt {2-\sqrt {3}}}+\frac {\sqrt {\frac {2}{2+\sqrt {3}}} \arctan \left (\frac {2 x+\sqrt {2-\sqrt {3}}}{\sqrt {2+\sqrt {3}}}\right )+\frac {1}{2} \left (1-\sqrt {3}\right ) \log \left (x^2+\sqrt {2-\sqrt {3}} x+1\right )}{2 \sqrt {2-\sqrt {3}}}}{2 \sqrt {3}}+\frac {\frac {-\sqrt {\frac {2}{2-\sqrt {3}}} \arctan \left (\frac {2 x-\sqrt {2+\sqrt {3}}}{\sqrt {2-\sqrt {3}}}\right )-\frac {1}{2} \left (1+\sqrt {3}\right ) \log \left (x^2-\sqrt {2+\sqrt {3}} x+1\right )}{2 \sqrt {2+\sqrt {3}}}+\frac {\frac {1}{2} \left (1+\sqrt {3}\right ) \log \left (x^2+\sqrt {2+\sqrt {3}} x+1\right )-\sqrt {\frac {2}{2-\sqrt {3}}} \arctan \left (\frac {2 x+\sqrt {2+\sqrt {3}}}{\sqrt {2-\sqrt {3}}}\right )}{2 \sqrt {2+\sqrt {3}}}}{2 \sqrt {3}}-\frac {1}{5 x^5}-\frac {1}{x}\)

Input:

Int[1/(x^6*(1 - x^4 + x^8)),x]
 

Output:

-1/5*1/x^5 - x^(-1) - ((Sqrt[2/(2 + Sqrt[3])]*ArcTan[(-Sqrt[2 - Sqrt[3]] + 
 2*x)/Sqrt[2 + Sqrt[3]]] - ((1 - Sqrt[3])*Log[1 - Sqrt[2 - Sqrt[3]]*x + x^ 
2])/2)/(2*Sqrt[2 - Sqrt[3]]) + (Sqrt[2/(2 + Sqrt[3])]*ArcTan[(Sqrt[2 - Sqr 
t[3]] + 2*x)/Sqrt[2 + Sqrt[3]]] + ((1 - Sqrt[3])*Log[1 + Sqrt[2 - Sqrt[3]] 
*x + x^2])/2)/(2*Sqrt[2 - Sqrt[3]]))/(2*Sqrt[3]) + ((-(Sqrt[2/(2 - Sqrt[3] 
)]*ArcTan[(-Sqrt[2 + Sqrt[3]] + 2*x)/Sqrt[2 - Sqrt[3]]]) - ((1 + Sqrt[3])* 
Log[1 - Sqrt[2 + Sqrt[3]]*x + x^2])/2)/(2*Sqrt[2 + Sqrt[3]]) + (-(Sqrt[2/( 
2 - Sqrt[3])]*ArcTan[(Sqrt[2 + Sqrt[3]] + 2*x)/Sqrt[2 - Sqrt[3]]]) + ((1 + 
 Sqrt[3])*Log[1 + Sqrt[2 + Sqrt[3]]*x + x^2])/2)/(2*Sqrt[2 + Sqrt[3]]))/(2 
*Sqrt[3])
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 1483
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r)   In 
t[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Simp[1/(2*c*q*r)   Int[(d*r 
 + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && N 
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]
 

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 1708
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 - 3*(n/2))*((q - r*x^(n/2))/(q - r*x^(n/2) + x^n)), x], x] + Simp[1/(2*c 
*r)   Int[x^(m - 3*(n/2))*((q + r*x^(n/2))/(q + r*x^(n/2) + x^n)), x], x]]] 
 /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n/2, 
 0] && IGtQ[m, 0] && GeQ[m, 3*(n/2)] && LtQ[m, 2*n] && NegQ[b^2 - 4*a*c]
 

rule 1828
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + ( 
c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^n + c*x^ 
(2*n))^(p + 1)/(a*f*(m + 1))), x] + Simp[1/(a*f^n*(m + 1))   Int[(f*x)^(m + 
 n)*(a + b*x^n + c*x^(2*n))^p*Simp[a*e*(m + 1) - b*d*(m + n*(p + 1) + 1) - 
c*d*(m + 2*n*(p + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x 
] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -1] && Int 
egerQ[p]
 
Maple [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.19

method result size
default \(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (9 x \,\textit {\_R}^{3}-3 \textit {\_R}^{2}+x^{2}\right )\right )}{4}-\frac {1}{5 x^{5}}-\frac {1}{x}\) \(43\)
risch \(\frac {-x^{4}-\frac {1}{5}}{x^{5}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (-9 x \,\textit {\_R}^{3}-3 \textit {\_R}^{2}+x^{2}\right )\right )}{4}\) \(44\)

Input:

int(1/x^6/(x^8-x^4+1),x,method=_RETURNVERBOSE)
 

Output:

-1/4*sum(_R*ln(9*_R^3*x-3*_R^2+x^2),_R=RootOf(9*_Z^4+1))-1/5/x^5-1/x
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^6 \left (1-x^4+x^8\right )} \, dx=-\frac {10 \, \sqrt {\frac {2}{3}} x^{5} \arctan \left (4 \, x^{2} + 3 \, \sqrt {\frac {2}{3}} {\left (x^{3} + 2 \, x\right )} + 3\right ) + 10 \, \sqrt {\frac {2}{3}} x^{5} \arctan \left (-4 \, x^{2} + 3 \, \sqrt {\frac {2}{3}} {\left (x^{3} + 2 \, x\right )} - 3\right ) + 10 \, \sqrt {\frac {2}{3}} x^{5} \arctan \left (\sqrt {\frac {2}{3}} x + \frac {1}{3}\right ) + 10 \, \sqrt {\frac {2}{3}} x^{5} \arctan \left (\sqrt {\frac {2}{3}} x - \frac {1}{3}\right ) - 5 \, \sqrt {\frac {2}{3}} x^{5} \log \left (x^{4} + 3 \, x^{2} + 3 \, \sqrt {\frac {2}{3}} {\left (x^{3} + x\right )} + 1\right ) + 5 \, \sqrt {\frac {2}{3}} x^{5} \log \left (x^{4} + 3 \, x^{2} - 3 \, \sqrt {\frac {2}{3}} {\left (x^{3} + x\right )} + 1\right ) + 40 \, x^{4} + 8}{40 \, x^{5}} \] Input:

integrate(1/x^6/(x^8-x^4+1),x, algorithm="fricas")
 

Output:

-1/40*(10*sqrt(2/3)*x^5*arctan(4*x^2 + 3*sqrt(2/3)*(x^3 + 2*x) + 3) + 10*s 
qrt(2/3)*x^5*arctan(-4*x^2 + 3*sqrt(2/3)*(x^3 + 2*x) - 3) + 10*sqrt(2/3)*x 
^5*arctan(sqrt(2/3)*x + 1/3) + 10*sqrt(2/3)*x^5*arctan(sqrt(2/3)*x - 1/3) 
- 5*sqrt(2/3)*x^5*log(x^4 + 3*x^2 + 3*sqrt(2/3)*(x^3 + x) + 1) + 5*sqrt(2/ 
3)*x^5*log(x^4 + 3*x^2 - 3*sqrt(2/3)*(x^3 + x) + 1) + 40*x^4 + 8)/x^5
 

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^6 \left (1-x^4+x^8\right )} \, dx=\frac {\sqrt {6} \left (- 2 \operatorname {atan}{\left (\frac {\sqrt {6} x}{3} - \frac {1}{3} \right )} - 2 \operatorname {atan}{\left (\sqrt {6} x^{3} - 4 x^{2} + 2 \sqrt {6} x - 3 \right )}\right )}{24} + \frac {\sqrt {6} \left (- 2 \operatorname {atan}{\left (\frac {\sqrt {6} x}{3} + \frac {1}{3} \right )} - 2 \operatorname {atan}{\left (\sqrt {6} x^{3} + 4 x^{2} + 2 \sqrt {6} x + 3 \right )}\right )}{24} - \frac {\sqrt {6} \log {\left (x^{4} - \sqrt {6} x^{3} + 3 x^{2} - \sqrt {6} x + 1 \right )}}{24} + \frac {\sqrt {6} \log {\left (x^{4} + \sqrt {6} x^{3} + 3 x^{2} + \sqrt {6} x + 1 \right )}}{24} + \frac {- 5 x^{4} - 1}{5 x^{5}} \] Input:

integrate(1/x**6/(x**8-x**4+1),x)
 

Output:

sqrt(6)*(-2*atan(sqrt(6)*x/3 - 1/3) - 2*atan(sqrt(6)*x**3 - 4*x**2 + 2*sqr 
t(6)*x - 3))/24 + sqrt(6)*(-2*atan(sqrt(6)*x/3 + 1/3) - 2*atan(sqrt(6)*x** 
3 + 4*x**2 + 2*sqrt(6)*x + 3))/24 - sqrt(6)*log(x**4 - sqrt(6)*x**3 + 3*x* 
*2 - sqrt(6)*x + 1)/24 + sqrt(6)*log(x**4 + sqrt(6)*x**3 + 3*x**2 + sqrt(6 
)*x + 1)/24 + (-5*x**4 - 1)/(5*x**5)
 

Maxima [F]

\[ \int \frac {1}{x^6 \left (1-x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} - x^{4} + 1\right )} x^{6}} \,d x } \] Input:

integrate(1/x^6/(x^8-x^4+1),x, algorithm="maxima")
 

Output:

-1/5*(5*x^4 + 1)/x^5 - integrate(x^6/(x^8 - x^4 + 1), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^6 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{12} \, \sqrt {6} \arctan \left (\frac {4 \, x + \sqrt {6} - \sqrt {2}}{\sqrt {6} + \sqrt {2}}\right ) - \frac {1}{12} \, \sqrt {6} \arctan \left (\frac {4 \, x - \sqrt {6} + \sqrt {2}}{\sqrt {6} + \sqrt {2}}\right ) - \frac {1}{12} \, \sqrt {6} \arctan \left (\frac {4 \, x + \sqrt {6} + \sqrt {2}}{\sqrt {6} - \sqrt {2}}\right ) - \frac {1}{12} \, \sqrt {6} \arctan \left (\frac {4 \, x - \sqrt {6} - \sqrt {2}}{\sqrt {6} - \sqrt {2}}\right ) + \frac {1}{24} \, \sqrt {6} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {6} + \sqrt {2}\right )} + 1\right ) - \frac {1}{24} \, \sqrt {6} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {6} + \sqrt {2}\right )} + 1\right ) + \frac {1}{24} \, \sqrt {6} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {6} - \sqrt {2}\right )} + 1\right ) - \frac {1}{24} \, \sqrt {6} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {6} - \sqrt {2}\right )} + 1\right ) - \frac {5 \, x^{4} + 1}{5 \, x^{5}} \] Input:

integrate(1/x^6/(x^8-x^4+1),x, algorithm="giac")
 

Output:

-1/12*sqrt(6)*arctan((4*x + sqrt(6) - sqrt(2))/(sqrt(6) + sqrt(2))) - 1/12 
*sqrt(6)*arctan((4*x - sqrt(6) + sqrt(2))/(sqrt(6) + sqrt(2))) - 1/12*sqrt 
(6)*arctan((4*x + sqrt(6) + sqrt(2))/(sqrt(6) - sqrt(2))) - 1/12*sqrt(6)*a 
rctan((4*x - sqrt(6) - sqrt(2))/(sqrt(6) - sqrt(2))) + 1/24*sqrt(6)*log(x^ 
2 + 1/2*x*(sqrt(6) + sqrt(2)) + 1) - 1/24*sqrt(6)*log(x^2 - 1/2*x*(sqrt(6) 
 + sqrt(2)) + 1) + 1/24*sqrt(6)*log(x^2 + 1/2*x*(sqrt(6) - sqrt(2)) + 1) - 
 1/24*sqrt(6)*log(x^2 - 1/2*x*(sqrt(6) - sqrt(2)) + 1) - 1/5*(5*x^4 + 1)/x 
^5
 

Mupad [B] (verification not implemented)

Time = 19.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.27 \[ \int \frac {1}{x^6 \left (1-x^4+x^8\right )} \, dx=-\frac {x^4+\frac {1}{5}}{x^5}+\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x\,\left (\frac {1}{3}+\frac {1}{3}{}\mathrm {i}\right )}{\frac {2\,x^2}{3}-\frac {2}{3}{}\mathrm {i}}\right )\,\left (\frac {1}{12}-\frac {1}{12}{}\mathrm {i}\right )+\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x\,\left (\frac {1}{3}-\frac {1}{3}{}\mathrm {i}\right )}{\frac {2\,x^2}{3}+\frac {2}{3}{}\mathrm {i}}\right )\,\left (\frac {1}{12}+\frac {1}{12}{}\mathrm {i}\right ) \] Input:

int(1/(x^6*(x^8 - x^4 + 1)),x)
 

Output:

6^(1/2)*atan((6^(1/2)*x*(1/3 + 1i/3))/((2*x^2)/3 - 2i/3))*(1/12 - 1i/12) + 
 6^(1/2)*atan((6^(1/2)*x*(1/3 - 1i/3))/((2*x^2)/3 + 2i/3))*(1/12 + 1i/12) 
- (x^4 + 1/5)/x^5
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.57 \[ \int \frac {1}{x^6 \left (1-x^4+x^8\right )} \, dx=\frac {10 \sqrt {-\sqrt {3}+2}\, \sqrt {3}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}-4 x}{2 \sqrt {-\sqrt {3}+2}}\right ) x^{5}+30 \sqrt {-\sqrt {3}+2}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}-4 x}{2 \sqrt {-\sqrt {3}+2}}\right ) x^{5}-10 \sqrt {-\sqrt {3}+2}\, \sqrt {3}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}+4 x}{2 \sqrt {-\sqrt {3}+2}}\right ) x^{5}-30 \sqrt {-\sqrt {3}+2}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}+4 x}{2 \sqrt {-\sqrt {3}+2}}\right ) x^{5}+10 \sqrt {6}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}-4 x}{\sqrt {6}+\sqrt {2}}\right ) x^{5}-10 \sqrt {6}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}+4 x}{\sqrt {6}+\sqrt {2}}\right ) x^{5}-5 \sqrt {-\sqrt {3}+2}\, \sqrt {3}\, \mathrm {log}\left (-\sqrt {-\sqrt {3}+2}\, x +x^{2}+1\right ) x^{5}+5 \sqrt {-\sqrt {3}+2}\, \sqrt {3}\, \mathrm {log}\left (\sqrt {-\sqrt {3}+2}\, x +x^{2}+1\right ) x^{5}-15 \sqrt {-\sqrt {3}+2}\, \mathrm {log}\left (-\sqrt {-\sqrt {3}+2}\, x +x^{2}+1\right ) x^{5}+15 \sqrt {-\sqrt {3}+2}\, \mathrm {log}\left (\sqrt {-\sqrt {3}+2}\, x +x^{2}+1\right ) x^{5}-5 \sqrt {6}\, \mathrm {log}\left (-\frac {\sqrt {6}\, x}{2}-\frac {\sqrt {2}\, x}{2}+x^{2}+1\right ) x^{5}+5 \sqrt {6}\, \mathrm {log}\left (\frac {\sqrt {6}\, x}{2}+\frac {\sqrt {2}\, x}{2}+x^{2}+1\right ) x^{5}-120 x^{4}-24}{120 x^{5}} \] Input:

int(1/x^6/(x^8-x^4+1),x)
 

Output:

(10*sqrt( - sqrt(3) + 2)*sqrt(3)*atan((sqrt(6) + sqrt(2) - 4*x)/(2*sqrt( - 
 sqrt(3) + 2)))*x**5 + 30*sqrt( - sqrt(3) + 2)*atan((sqrt(6) + sqrt(2) - 4 
*x)/(2*sqrt( - sqrt(3) + 2)))*x**5 - 10*sqrt( - sqrt(3) + 2)*sqrt(3)*atan( 
(sqrt(6) + sqrt(2) + 4*x)/(2*sqrt( - sqrt(3) + 2)))*x**5 - 30*sqrt( - sqrt 
(3) + 2)*atan((sqrt(6) + sqrt(2) + 4*x)/(2*sqrt( - sqrt(3) + 2)))*x**5 + 1 
0*sqrt(6)*atan((2*sqrt( - sqrt(3) + 2) - 4*x)/(sqrt(6) + sqrt(2)))*x**5 - 
10*sqrt(6)*atan((2*sqrt( - sqrt(3) + 2) + 4*x)/(sqrt(6) + sqrt(2)))*x**5 - 
 5*sqrt( - sqrt(3) + 2)*sqrt(3)*log( - sqrt( - sqrt(3) + 2)*x + x**2 + 1)* 
x**5 + 5*sqrt( - sqrt(3) + 2)*sqrt(3)*log(sqrt( - sqrt(3) + 2)*x + x**2 + 
1)*x**5 - 15*sqrt( - sqrt(3) + 2)*log( - sqrt( - sqrt(3) + 2)*x + x**2 + 1 
)*x**5 + 15*sqrt( - sqrt(3) + 2)*log(sqrt( - sqrt(3) + 2)*x + x**2 + 1)*x* 
*5 - 5*sqrt(6)*log(( - sqrt(6)*x - sqrt(2)*x + 2*x**2 + 2)/2)*x**5 + 5*sqr 
t(6)*log((sqrt(6)*x + sqrt(2)*x + 2*x**2 + 2)/2)*x**5 - 120*x**4 - 24)/(12 
0*x**5)