\(\int \frac {1}{-1+x^8} \, dx\) [50]

Optimal result

Integrand size = 7, antiderivative size = 97 \[ \int \frac {1}{-1+x^8} \, dx=-\frac {\arctan (x)}{4}+\frac {\arctan \left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\text {arctanh}(x)}{4}+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}} \] Output:

-1/4*arctan(x)-1/4*arctanh(x)-1/8*arctan(-1+x*2^(1/2))*2^(1/2)-1/8*arctan( 
1+x*2^(1/2))*2^(1/2)+1/16*ln(1+x^2-x*2^(1/2))*2^(1/2)-1/16*ln(1+x^2+x*2^(1 
/2))*2^(1/2)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01 \[ \int \frac {1}{-1+x^8} \, dx=\frac {1}{16} \left (-4 \arctan (x)+2 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+2 \log (1-x)-2 \log (1+x)+\sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-\sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \] Input:

Integrate[(-1 + x^8)^(-1),x]
 

Output:

(-4*ArcTan[x] + 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 2*Sqrt[2]*ArcTan[1 + Sqr 
t[2]*x] + 2*Log[1 - x] - 2*Log[1 + x] + Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] - 
 Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/16
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.714, Rules used = {758, 755, 756, 216, 219, 1476, 1082, 217, 1479, 25, 27, 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.

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

\(\Big \downarrow \) 758

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx-\frac {1}{2} \int \frac {x^2+1}{x^4+1}dx\right )+\frac {1}{2} \left (-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2}\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-\sqrt {2} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {2} x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx\right )+\frac {1}{2} \left (-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\left (\sqrt {2} x+1\right )^2-1}d\left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\left (1-\sqrt {2} x\right )^2-1}d\left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx\right )+\frac {1}{2} \left (-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx\right )+\frac {1}{2} \left (-\frac {\arctan (x)}{2}-\frac {\text {arctanh}(x)}{2}\right )\)

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(-1 + x^8)^(-1),x]
 

Output:

(-1/2*ArcTan[x] - ArcTanh[x]/2)/2 + ((ArcTan[1 - Sqrt[2]*x]/Sqrt[2] - ArcT 
an[1 + Sqrt[2]*x]/Sqrt[2])/2 + (Log[1 - Sqrt[2]*x + x^2]/(2*Sqrt[2]) - Log 
[1 + Sqrt[2]*x + x^2]/(2*Sqrt[2]))/2)/2
 

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 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 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 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 755

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

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 758

Int[((a_) + (b_.)*(x_)^(n_))^(-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^(n/2)), 
 x], x] + Simp[r/(2*a)   Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a, b}, x] 
 && IGtQ[n/4, 1] &&  !GtQ[a/b, 0]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{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 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 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]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 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]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [C] (verified)

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

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.38

method	result	size
risch	\(-\frac {\ln \left (1+x \right )}{8}+\frac {\ln \left (-1+x \right )}{8}-\frac {\arctan \left (x \right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (x -\textit {\_R} \right )\right )}{8}\)	\(37\)
default	\(-\frac {\arctan \left (x \right )}{4}-\frac {\operatorname {arctanh}\left (x \right )}{4}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+x \sqrt {2}}{1+x^{2}-x \sqrt {2}}\right )+2 \arctan \left (1+x \sqrt {2}\right )+2 \arctan \left (-1+x \sqrt {2}\right )\right )}{16}\)	\(61\)
meijerg	\(\frac {x \left (\ln \left (1-\left (x^{8}\right )^{\frac {1}{8}}\right )-\ln \left (1+\left (x^{8}\right )^{\frac {1}{8}}\right )+\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{2}-\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}{2-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}\right )-2 \arctan \left (\left (x^{8}\right )^{\frac {1}{8}}\right )-\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{2}-\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}{2+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}\right )\right )}{8 \left (x^{8}\right )^{\frac {1}{8}}}\)	\(143\)

Input:

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

Output:

-1/8*ln(1+x)+1/8*ln(-1+x)-1/4*arctan(x)+1/8*sum(_R*ln(x-_R),_R=RootOf(_Z^4 
+1))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \frac {1}{-1+x^8} \, dx=-\frac {1}{8} \, \sqrt {2} \arctan \left (\sqrt {2} x + 1\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\sqrt {2} x - 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{4} \, \arctan \left (x\right ) - \frac {1}{8} \, \log \left (x + 1\right ) + \frac {1}{8} \, \log \left (x - 1\right ) \] Input:

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

Output:

-1/8*sqrt(2)*arctan(sqrt(2)*x + 1) - 1/8*sqrt(2)*arctan(sqrt(2)*x - 1) - 1 
/16*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 1/16*sqrt(2)*log(x^2 - sqrt(2)*x + 
1) - 1/4*arctan(x) - 1/8*log(x + 1) + 1/8*log(x - 1)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 118.67 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.45 \[ \int \frac {1}{-1+x^8} \, dx=\frac {\log {\left (x - 1 \right )}}{8} - \frac {\log {\left (x + 1 \right )}}{8} + \frac {i \log {\left (x - i \right )}}{8} - \frac {i \log {\left (x + i \right )}}{8} + \operatorname {RootSum} {\left (4096 t^{4} + 1, \left ( t \mapsto t \log {\left (- 8 t + x \right )} \right )\right )} \] Input:

integrate(1/(x**8-1),x)
 

Output:

log(x - 1)/8 - log(x + 1)/8 + I*log(x - I)/8 - I*log(x + I)/8 + RootSum(40 
96*_t**4 + 1, Lambda(_t, _t*log(-8*_t + x)))
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \frac {1}{-1+x^8} \, dx=-\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{4} \, \arctan \left (x\right ) - \frac {1}{8} \, \log \left (x + 1\right ) + \frac {1}{8} \, \log \left (x - 1\right ) \] Input:

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

Output:

-1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 1/8*sqrt(2)*arctan(1/2* 
sqrt(2)*(2*x - sqrt(2))) - 1/16*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 1/16*sq 
rt(2)*log(x^2 - sqrt(2)*x + 1) - 1/4*arctan(x) - 1/8*log(x + 1) + 1/8*log( 
x - 1)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93 \[ \int \frac {1}{-1+x^8} \, dx=-\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{4} \, \arctan \left (x\right ) - \frac {1}{8} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{8} \, \log \left ({\left | x - 1 \right |}\right ) \] Input:

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

Output:

-1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 1/8*sqrt(2)*arctan(1/2* 
sqrt(2)*(2*x - sqrt(2))) - 1/16*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 1/16*sq 
rt(2)*log(x^2 - sqrt(2)*x + 1) - 1/4*arctan(x) - 1/8*log(abs(x + 1)) + 1/8 
*log(abs(x - 1))
 

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.46 \[ \int \frac {1}{-1+x^8} \, dx=\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}-\frac {\mathrm {atan}\left (x\right )}{4}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right ) \] Input:

int(1/(x^8 - 1),x)
 

Output:

(atan(x*1i)*1i)/4 - atan(x)/4 - 2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(1/8 
+ 1i/8) - 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2))*(1/8 - 1i/8)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \frac {1}{-1+x^8} \, dx=\frac {\sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right )}{8}-\frac {\sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right )}{8}-\frac {\mathit {atan} \left (x \right )}{4}+\frac {\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right )}{16}-\frac {\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right )}{16}+\frac {\mathrm {log}\left (x -1\right )}{8}-\frac {\mathrm {log}\left (x +1\right )}{8} \] Input:

int(1/(x^8-1),x)
 

Output:

(2*sqrt(2)*atan((sqrt(2) - 2*x)/sqrt(2)) - 2*sqrt(2)*atan((sqrt(2) + 2*x)/ 
sqrt(2)) - 4*atan(x) + sqrt(2)*log( - sqrt(2)*x + x**2 + 1) - sqrt(2)*log( 
sqrt(2)*x + x**2 + 1) + 2*log(x - 1) - 2*log(x + 1))/16