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

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

[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 = 0.05, 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 = {214, 212, 206, 203, 211, 1165, 628, 1162, 617, 204} \[ \frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{4 \sqrt {2}}-\frac {1}{4} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[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 -
Sqrt[2]*x + x^2]/(8*Sqrt[2]) - Log[1 + Sqrt[2]*x + x^2]/(8*Sqrt[2])

Rule 203

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

Rule 204

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

Rule 206

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

Rule 211

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 212

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] &&
!GtQ[a/b, 0]

Rule 214

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 617

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 628

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 1162

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 1165

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 {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\operatorname {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 = 0.02, size = 98, normalized size = 1.01 \[ \frac {1}{16} \left (\sqrt {2} \log \left (x^2-\sqrt {2} x+1\right )-\sqrt {2} \log \left (x^2+\sqrt {2} x+1\right )+2 \log (1-x)-2 \log (x+1)-4 \tan ^{-1}(x)+2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} x\right )-2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} x+1\right )\right ) \]

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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fricas [A]  time = 0.44, size = 111, normalized size = 1.14 \[ \frac {1}{4} \, \sqrt {2} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{4} \, \arctan \relax (x) - \frac {1}{8} \, \log \left (x + 1\right ) + \frac {1}{8} \, \log \left (x - 1\right ) \]

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(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)

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giac [A]  time = 0.96, size = 90, normalized size = 0.93 \[ -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{4} \, \arctan \relax (x) - \frac {1}{8} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{8} \, \log \left ({\left | x - 1 \right |}\right ) \]

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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maple [A]  time = 0.00, size = 66, normalized size = 0.68 \[ -\frac {\arctanh \relax (x )}{4}-\frac {\arctan \relax (x )}{4}-\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x -1\right )}{8}-\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x +1\right )}{8}-\frac {\sqrt {2}\, \ln \left (\frac {x^{2}+\sqrt {2}\, x +1}{x^{2}-\sqrt {2}\, x +1}\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.95, size = 88, normalized size = 0.91 \[ -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{4} \, \arctan \relax (x) - \frac {1}{8} \, \log \left (x + 1\right ) + \frac {1}{8} \, \log \left (x - 1\right ) \]

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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mupad [B]  time = 0.16, size = 45, normalized size = 0.46 \[ \frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}-\frac {\mathrm {atan}\relax (x)}{4}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right ) \]

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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sympy [C]  time = 161.87, size = 44, normalized size = 0.45 \[ \frac {\log {\left (x - 1 \right )}}{8} - \frac {\log {\left (x + 1 \right )}}{8} + \frac {i \log {\left (x - i \right )}}{8} - \frac {i \log {\left (x + i \right )}}{8} + \operatorname {RootSum} {\left (4096 t^{4} + 1, \left (t \mapsto t \log {\left (- 8 t + x \right )} \right )\right )} \]

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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