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

Optimal. Leaf size=97 \[ -\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {1}{4} \tanh ^{-1}(x)+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}} \]

[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)

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Rubi [A]
time = 0.04, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.429, Rules used = {220, 218, 212, 209, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {\text {ArcTan}(x)}{4}+\frac {\text {ArcTan}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\text {ArcTan}\left (\sqrt {2} x+1\right )}{4 \sqrt {2}}+\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}}-\frac {1}{4} \tanh ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^8)^(-1),x]

[Out]

-1/4*ArcTan[x] + ArcTan[1 - Sqrt[2]*x]/(4*Sqrt[2]) - ArcTan[1 + Sqrt[2]*x]/(4*Sqrt[2]) - ArcTanh[x]/4 + Log[1
- Sqrt[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*} \int \frac {1}{-1+x^8} \, dx &=-\left (\frac {1}{2} \int \frac {1}{1-x^4} \, dx\right )-\frac {1}{2} \int \frac {1}{1+x^4} \, dx\\ &=-\left (\frac {1}{4} \int \frac {1}{1-x^2} \, dx\right )-\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*}

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Mathematica [A]
time = 0.01, size = 98, normalized size = 1.01 \begin {gather*} \frac {1}{16} \left (-4 \tan ^{-1}(x)+2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} x\right )-2 \sqrt {2} \tan ^{-1}\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 ) \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 0.04, size = 61, normalized size = 0.63

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.89, size = 88, normalized size = 0.91 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*lo
g(x - 1)

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Fricas [A]
time = 1.06, size = 116, normalized size = 1.20 \begin {gather*} \frac {1}{4} \, \sqrt {2} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (4 \, x^{2} + 4 \, \sqrt {2} x + 4\right ) + \frac {1}{16} \, \sqrt {2} \log \left (4 \, x^{2} - 4 \, \sqrt {2} x + 4\right ) - \frac {1}{4} \, \arctan \left (x\right ) - \frac {1}{8} \, \log \left (x + 1\right ) + \frac {1}{8} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 82.25, size = 44, normalized size = 0.45 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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Giac [A]
time = 0.47, size = 90, normalized size = 0.93 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Mupad [B]
time = 0.16, size = 45, normalized size = 0.46 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8 - 1),x)

[Out]

(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)

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