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]

-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])

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Rubi [A]  time = 0.0520014, antiderivative size = 97, normalized size of antiderivative = 1., 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 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 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 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 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 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.0257536, 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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Maple [A]  time = 0., size = 66, normalized size = 0.7 \begin{align*} -{\frac{{\it Artanh} \left ( x \right ) }{4}}-{\frac{\arctan \left ( x \right ) }{4}}-{\frac{\sqrt{2}}{16}\ln \left ({\frac{1+{x}^{2}+x\sqrt{2}}{1+{x}^{2}-x\sqrt{2}}} \right ) }-{\frac{\arctan \left ( -1+x\sqrt{2} \right ) \sqrt{2}}{8}}-{\frac{\arctan \left ( 1+x\sqrt{2} \right ) \sqrt{2}}{8}} \end{align*}

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((1+x^2+x*2^(1/2))/(1+x^2-x*2^(1/2)))-1/8*arctan(-1+x*2^(1/2))*2^
(1/2)-1/8*arctan(1+x*2^(1/2))*2^(1/2)

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Maxima [A]  time = 1.43254, size = 119, normalized size = 1.23 \begin{align*} -\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{align*}

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 = 2.11882, size = 371, normalized size = 3.82 \begin{align*} \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 \left (x\right ) - \frac{1}{8} \, \log \left (x + 1\right ) + \frac{1}{8} \, \log \left (x - 1\right ) \end{align*}

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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Sympy [C]  time = 128.354, size = 44, normalized size = 0.45 \begin{align*} \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{align*}

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 = 1.08627, size = 122, normalized size = 1.26 \begin{align*} -\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{align*}

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))