3.1478 \(\int \frac{1}{x^5 (1-x^8)} \, dx\)

Optimal. Leaf size=16 \[ \frac{1}{4} \tanh ^{-1}\left (x^4\right )-\frac{1}{4 x^4} \]

[Out]

-1/(4*x^4) + ArcTanh[x^4]/4

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Rubi [A]  time = 0.0071306, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {275, 325, 206} \[ \frac{1}{4} \tanh ^{-1}\left (x^4\right )-\frac{1}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(1 - x^8)),x]

[Out]

-1/(4*x^4) + ArcTanh[x^4]/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 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^5 \left (1-x^8\right )} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-x^2\right )} \, dx,x,x^4\right )\\ &=-\frac{1}{4 x^4}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,x^4\right )\\ &=-\frac{1}{4 x^4}+\frac{1}{4} \tanh ^{-1}\left (x^4\right )\\ \end{align*}

Mathematica [A]  time = 0.0038907, size = 30, normalized size = 1.88 \[ -\frac{1}{4 x^4}-\frac{1}{8} \log \left (1-x^4\right )+\frac{1}{8} \log \left (x^4+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(1 - x^8)),x]

[Out]

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

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Maple [B]  time = 0.011, size = 35, normalized size = 2.2 \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{1}{4\,{x}^{4}}}+{\frac{\ln \left ({x}^{4}+1 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(-x^8+1),x)

[Out]

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

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Maxima [A]  time = 0.975933, size = 30, normalized size = 1.88 \begin{align*} -\frac{1}{4 \, x^{4}} + \frac{1}{8} \, \log \left (x^{4} + 1\right ) - \frac{1}{8} \, \log \left (x^{4} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^8+1),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.34249, size = 69, normalized size = 4.31 \begin{align*} \frac{x^{4} \log \left (x^{4} + 1\right ) - x^{4} \log \left (x^{4} - 1\right ) - 2}{8 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^8+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.138007, size = 22, normalized size = 1.38 \begin{align*} - \frac{\log{\left (x^{4} - 1 \right )}}{8} + \frac{\log{\left (x^{4} + 1 \right )}}{8} - \frac{1}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(-x**8+1),x)

[Out]

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

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Giac [A]  time = 1.13938, size = 31, normalized size = 1.94 \begin{align*} -\frac{1}{4 \, x^{4}} + \frac{1}{8} \, \log \left (x^{4} + 1\right ) - \frac{1}{8} \, \log \left ({\left | x^{4} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^8+1),x, algorithm="giac")

[Out]

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