3.1477 \(\int \frac{1}{x^3 (1-x^8)} \, dx\)

Optimal. Leaf size=24 \[ -\frac{1}{2 x^2}-\frac{1}{4} \tan ^{-1}\left (x^2\right )+\frac{1}{4} \tanh ^{-1}\left (x^2\right ) \]

[Out]

-1/(2*x^2) - ArcTan[x^2]/4 + ArcTanh[x^2]/4

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Rubi [A]  time = 0.0110137, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {275, 325, 298, 203, 206} \[ -\frac{1}{2 x^2}-\frac{1}{4} \tan ^{-1}\left (x^2\right )+\frac{1}{4} \tanh ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 - x^8)),x]

[Out]

-1/(2*x^2) - ArcTan[x^2]/4 + ArcTanh[x^2]/4

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (1-x^8\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-x^4\right )} \, dx,x,x^2\right )\\ &=-\frac{1}{2 x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,x^2\right )\\ &=-\frac{1}{2 x^2}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{2 x^2}-\frac{1}{4} \tan ^{-1}\left (x^2\right )+\frac{1}{4} \tanh ^{-1}\left (x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0052489, size = 38, normalized size = 1.58 \[ -\frac{1}{2 x^2}-\frac{1}{8} \log \left (1-x^2\right )+\frac{1}{8} \log \left (x^2+1\right )+\frac{1}{4} \tan ^{-1}\left (\frac{1}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 - x^8)),x]

[Out]

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

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Maple [A]  time = 0.011, size = 33, normalized size = 1.4 \begin{align*}{\frac{\ln \left ({x}^{2}+1 \right ) }{8}}-{\frac{\ln \left ( 1+x \right ) }{8}}-{\frac{\ln \left ( -1+x \right ) }{8}}-{\frac{\arctan \left ({x}^{2} \right ) }{4}}-{\frac{1}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(-x^8+1),x)

[Out]

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

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Maxima [A]  time = 1.468, size = 38, normalized size = 1.58 \begin{align*} -\frac{1}{2 \, x^{2}} - \frac{1}{4} \, \arctan \left (x^{2}\right ) + \frac{1}{8} \, \log \left (x^{2} + 1\right ) - \frac{1}{8} \, \log \left (x^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-x^8+1),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.2555, size = 97, normalized size = 4.04 \begin{align*} -\frac{2 \, x^{2} \arctan \left (x^{2}\right ) - x^{2} \log \left (x^{2} + 1\right ) + x^{2} \log \left (x^{2} - 1\right ) + 4}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-x^8+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.152678, size = 29, normalized size = 1.21 \begin{align*} - \frac{\log{\left (x^{2} - 1 \right )}}{8} + \frac{\log{\left (x^{2} + 1 \right )}}{8} - \frac{\operatorname{atan}{\left (x^{2} \right )}}{4} - \frac{1}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(-x**8+1),x)

[Out]

-log(x**2 - 1)/8 + log(x**2 + 1)/8 - atan(x**2)/4 - 1/(2*x**2)

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Giac [A]  time = 1.17224, size = 39, normalized size = 1.62 \begin{align*} -\frac{1}{2 \, x^{2}} - \frac{1}{4} \, \arctan \left (x^{2}\right ) + \frac{1}{8} \, \log \left (x^{2} + 1\right ) - \frac{1}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-x^8+1),x, algorithm="giac")

[Out]

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