\(\int \frac {1}{1-x^8} \, dx\) [1485]

Optimal result

Integrand size = 9, antiderivative size = 97 \[ \int \frac {1}{1-x^8} \, dx=\frac {\arctan (x)}{4}-\frac {\arctan \left (1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\text {arctanh}(x)}{4}-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.111, Rules used = {220, 218, 212, 209, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{1-x^8} \, dx=\frac {\arctan (x)}{4}-\frac {\arctan \left (1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\arctan \left (\sqrt {2} x+1\right )}{4 \sqrt {2}}+\frac {\text {arctanh}(x)}{4}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}} \]

[In]

Int[(1 - x^8)^(-1),x]

[Out]

ArcTan[x]/4 - ArcTan[1 - Sqrt[2]*x]/(4*Sqrt[2]) + ArcTan[1 + Sqrt[2]*x]/(4*Sqrt[2]) + ArcTanh[x]/4 - Log[1 - S
qrt[2]*x + x^2]/(8*Sqrt[2]) + Log[1 + Sqrt[2]*x + x^2]/(8*Sqrt[2])

Rule 209

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

Rule 210

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 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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 218

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

Rule 220

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]
}, Dist[r/(2*a), Int[1/(r - s*x^(n/2)), x], x] + Dist[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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {1}{1-x^4} \, dx+\frac {1}{2} \int \frac {1}{1+x^4} \, dx \\ & = \frac {1}{4} \int \frac {1}{1-x^2} \, dx+\frac {1}{4} \int \frac {1}{1+x^2} \, dx+\frac {1}{4} \int \frac {1-x^2}{1+x^4} \, dx+\frac {1}{4} \int \frac {1+x^2}{1+x^4} \, dx \\ & = \frac {1}{4} \tan ^{-1}(x)+\frac {1}{4} \tanh ^{-1}(x)+\frac {1}{8} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {1}{8} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx-\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{8 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{8 \sqrt {2}} \\ & = \frac {1}{4} \tan ^{-1}(x)+\frac {1}{4} \tanh ^{-1}(x)-\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}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{4 \sqrt {2}} \\ & = \frac {1}{4} \tan ^{-1}(x)-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {1}{4} \tanh ^{-1}(x)-\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}} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[(1 - x^8)^(-1),x]

[Out]

(4*ArcTan[x] - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] + 2*Sqrt[2]*ArcTan[1 + Sqrt[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

Maple [C] (verified)

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

Time = 3.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.36

method	result	size
risch	\(\frac {\ln \left (1+x \right )}{8}-\frac {\ln \left (-1+x \right )}{8}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R} +x \right )\right )}{8}+\frac {\arctan \left (x \right )}{4}\)	\(35\)
default	\(\frac {\operatorname {arctanh}\left (x \right )}{4}+\frac {\arctan \left (x \right )}{4}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+\sqrt {2}\, x}{1+x^{2}-\sqrt {2}\, x}\right )+2 \arctan \left (\sqrt {2}\, x +1\right )+2 \arctan \left (\sqrt {2}\, x -1\right )\right )}{16}\)	\(61\)
meijerg	\(-\frac {x \left (\ln \left (1-\left (x^{8}\right )^{\frac {1}{8}}\right )-\ln \left (1+\left (x^{8}\right )^{\frac {1}{8}}\right )+\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{2}-\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}{2-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}\right )-2 \arctan \left (\left (x^{8}\right )^{\frac {1}{8}}\right )-\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{2}-\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}{2+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}\right )\right )}{8 \left (x^{8}\right )^{\frac {1}{8}}}\)	\(143\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.79 \[ \int \frac {1}{1-x^8} \, dx=\left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) + \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) - \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) + \frac {1}{4} \, \arctan \left (x\right ) + \frac {1}{8} \, \log \left (x + 1\right ) - \frac {1}{8} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

(1/16*I + 1/16)*sqrt(2)*log(2*x + (I + 1)*sqrt(2)) - (1/16*I - 1/16)*sqrt(2)*log(2*x - (I - 1)*sqrt(2)) + (1/1
6*I - 1/16)*sqrt(2)*log(2*x + (I - 1)*sqrt(2)) - (1/16*I + 1/16)*sqrt(2)*log(2*x - (I + 1)*sqrt(2)) + 1/4*arct
an(x) + 1/8*log(x + 1) - 1/8*log(x - 1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 132.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.45 \[ \int \frac {1}{1-x^8} \, dx=- \frac {\log {\left (x - 1 \right )}}{8} + \frac {\log {\left (x + 1 \right )}}{8} - \frac {i \log {\left (x - i \right )}}{8} + \frac {i \log {\left (x + i \right )}}{8} - \operatorname {RootSum} {\left (4096 t^{4} + 1, \left ( t \mapsto t \log {\left (- 8 t + x \right )} \right )\right )} \]

[In]

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

[Out]

-log(x - 1)/8 + log(x + 1)/8 - I*log(x - I)/8 + I*log(x + I)/8 - RootSum(4096*_t**4 + 1, Lambda(_t, _t*log(-8*
_t + x)))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \frac {1}{1-x^8} \, dx=\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {1}{4} \, \arctan \left (x\right ) + \frac {1}{8} \, \log \left (x + 1\right ) - \frac {1}{8} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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*sqrt(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)

none

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93 \[ \int \frac {1}{1-x^8} \, dx=\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {1}{4} \, \arctan \left (x\right ) + \frac {1}{8} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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*sqrt(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 = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.46 \[ \int \frac {1}{1-x^8} \, dx=\frac {\mathrm {atan}\left (x\right )}{4}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \]

[In]

int(-1/(x^8 - 1),x)

[Out]

atan(x)/4 - (atan(x*1i)*1i)/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)