3.1486 \(\int \frac {1}{x^2 (1-x^8)} \, dx\)

Optimal. Leaf size=102 \[ -\frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {1}{x}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{4 \sqrt {2}}+\frac {1}{4} \tanh ^{-1}(x) \]

[Out]

-1/x-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 = 0.06, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {325, 301, 297, 1162, 617, 204, 1165, 628, 298, 203, 206} \[ -\frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {1}{x}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{4 \sqrt {2}}+\frac {1}{4} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(1 - x^8)),x]

[Out]

-x^(-1) - 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 297

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

Rule 298

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

Rule 301

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(
a/b), 2]]}, Dist[s/(2*b), Int[x^(m - n/2)/(r + s*x^(n/2)), x], x] - Dist[s/(2*b), Int[x^(m - n/2)/(r - s*x^(n/
2)), x], x]] /; FreeQ[{a, b}, x] && IGtQ[n/4, 0] && IGtQ[m, 0] && LeQ[n/2, m] && LtQ[m, n] &&  !GtQ[a/b, 0]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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}{x^2 \left (1-x^8\right )} \, dx &=-\frac {1}{x}+\int \frac {x^6}{1-x^8} \, dx\\ &=-\frac {1}{x}+\frac {1}{2} \int \frac {x^2}{1-x^4} \, dx-\frac {1}{2} \int \frac {x^2}{1+x^4} \, dx\\ &=-\frac {1}{x}+\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}{x}-\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}{x}-\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}{x}-\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 = 0.04, size = 109, normalized size = 1.07 \[ -\frac {\sqrt {2} x \log \left (x^2-\sqrt {2} x+1\right )-\sqrt {2} x \log \left (x^2+\sqrt {2} x+1\right )+2 x \log (1-x)-2 x \log (x+1)+4 x \tan ^{-1}(x)-2 \sqrt {2} x \tan ^{-1}\left (1-\sqrt {2} x\right )+2 \sqrt {2} x \tan ^{-1}\left (\sqrt {2} x+1\right )+16}{16 x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(1 - x^8)),x]

[Out]

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

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fricas [A]  time = 0.61, size = 123, normalized size = 1.21 \[ \frac {4 \, \sqrt {2} x \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) + 4 \, \sqrt {2} x \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) + \sqrt {2} x \log \left (x^{2} + \sqrt {2} x + 1\right ) - \sqrt {2} x \log \left (x^{2} - \sqrt {2} x + 1\right ) - 4 \, x \arctan \relax (x) + 2 \, x \log \left (x + 1\right ) - 2 \, x \log \left (x - 1\right ) - 16}{16 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-x^8+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 95, normalized size = 0.93 \[ -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{x} - \frac {1}{4} \, \arctan \relax (x) + \frac {1}{8} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-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/x - 1/4*arctan(x) + 1/8*log(abs(x + 1
)) - 1/8*log(abs(x - 1))

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maple [A]  time = 0.01, size = 79, normalized size = 0.77 \[ -\frac {\arctan \relax (x )}{4}-\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x -1\right )}{8}-\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x +1\right )}{8}-\frac {\sqrt {2}\, \ln \left (\frac {x^{2}-\sqrt {2}\, x +1}{x^{2}+\sqrt {2}\, x +1}\right )}{16}-\frac {\ln \left (x -1\right )}{8}+\frac {\ln \left (x +1\right )}{8}-\frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(-x^8+1),x)

[Out]

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

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maxima [A]  time = 2.35, size = 93, normalized size = 0.91 \[ -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{x} - \frac {1}{4} \, \arctan \relax (x) + \frac {1}{8} \, \log \left (x + 1\right ) - \frac {1}{8} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.04, size = 50, normalized size = 0.49 \[ -\frac {\mathrm {atan}\relax (x)}{4}-\frac {1}{x}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^2*(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) - 1/x

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(-x**8+1),x)

[Out]

Timed out

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