3.299 \(\int \frac {1}{x^7 (1-2 x^4+x^8)} \, dx\)

Optimal. Leaf size=39 \[ -\frac {5}{12 x^6}-\frac {5}{4 x^2}+\frac {5}{4} \tanh ^{-1}\left (x^2\right )+\frac {1}{4 x^6 \left (1-x^4\right )} \]

[Out]

-5/12/x^6-5/4/x^2+1/4/x^6/(-x^4+1)+5/4*arctanh(x^2)

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {28, 275, 290, 325, 207} \[ \frac {1}{4 x^6 \left (1-x^4\right )}-\frac {5}{4 x^2}-\frac {5}{12 x^6}+\frac {5}{4} \tanh ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(1 - 2*x^4 + x^8)),x]

[Out]

-5/(12*x^6) - 5/(4*x^2) + 1/(4*x^6*(1 - x^4)) + (5*ArcTanh[x^2])/4

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 207

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

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 290

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

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]

Rubi steps

\begin {align*} \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx &=\int \frac {1}{x^7 \left (-1+x^4\right )^2} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 \left (-1+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac {1}{4 x^6 \left (1-x^4\right )}-\frac {5}{4} \operatorname {Subst}\left (\int \frac {1}{x^4 \left (-1+x^2\right )} \, dx,x,x^2\right )\\ &=-\frac {5}{12 x^6}+\frac {1}{4 x^6 \left (1-x^4\right )}-\frac {5}{4} \operatorname {Subst}\left (\int \frac {1}{x^2 \left (-1+x^2\right )} \, dx,x,x^2\right )\\ &=-\frac {5}{12 x^6}-\frac {5}{4 x^2}+\frac {1}{4 x^6 \left (1-x^4\right )}-\frac {5}{4} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,x^2\right )\\ &=-\frac {5}{12 x^6}-\frac {5}{4 x^2}+\frac {1}{4 x^6 \left (1-x^4\right )}+\frac {5}{4} \tanh ^{-1}\left (x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.26 \[ -\frac {1}{6 x^6}-\frac {1}{x^2}-\frac {5}{8} \log \left (1-x^2\right )+\frac {5}{8} \log \left (x^2+1\right )-\frac {x^2}{4 \left (x^4-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(1 - 2*x^4 + x^8)),x]

[Out]

-1/6*1/x^6 - x^(-2) - x^2/(4*(-1 + x^4)) - (5*Log[1 - x^2])/8 + (5*Log[1 + x^2])/8

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fricas [B]  time = 0.91, size = 59, normalized size = 1.51 \[ -\frac {30 \, x^{8} - 20 \, x^{4} - 15 \, {\left (x^{10} - x^{6}\right )} \log \left (x^{2} + 1\right ) + 15 \, {\left (x^{10} - x^{6}\right )} \log \left (x^{2} - 1\right ) - 4}{24 \, {\left (x^{10} - x^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

-1/24*(30*x^8 - 20*x^4 - 15*(x^10 - x^6)*log(x^2 + 1) + 15*(x^10 - x^6)*log(x^2 - 1) - 4)/(x^10 - x^6)

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giac [A]  time = 0.33, size = 42, normalized size = 1.08 \[ -\frac {x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {6 \, x^{4} + 1}{6 \, x^{6}} + \frac {5}{8} \, \log \left (x^{2} + 1\right ) - \frac {5}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

-1/4*x^2/(x^4 - 1) - 1/6*(6*x^4 + 1)/x^6 + 5/8*log(x^2 + 1) - 5/8*log(abs(x^2 - 1))

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maple [A]  time = 0.02, size = 55, normalized size = 1.41 \[ -\frac {5 \ln \left (x -1\right )}{8}-\frac {5 \ln \left (x +1\right )}{8}+\frac {5 \ln \left (x^{2}+1\right )}{8}-\frac {1}{x^{2}}-\frac {1}{6 x^{6}}+\frac {1}{16 x +16}-\frac {1}{8 \left (x^{2}+1\right )}-\frac {1}{16 \left (x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^8-2*x^4+1),x)

[Out]

-1/6/x^6-1/x^2+1/16/(x+1)-5/8*ln(x+1)+5/8*ln(x^2+1)-1/8/(x^2+1)-1/16/(x-1)-5/8*ln(x-1)

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maxima [A]  time = 0.94, size = 42, normalized size = 1.08 \[ -\frac {15 \, x^{8} - 10 \, x^{4} - 2}{12 \, {\left (x^{10} - x^{6}\right )}} + \frac {5}{8} \, \log \left (x^{2} + 1\right ) - \frac {5}{8} \, \log \left (x^{2} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

-1/12*(15*x^8 - 10*x^4 - 2)/(x^10 - x^6) + 5/8*log(x^2 + 1) - 5/8*log(x^2 - 1)

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mupad [B]  time = 0.05, size = 32, normalized size = 0.82 \[ \frac {5\,\mathrm {atanh}\left (x^2\right )}{4}-\frac {-\frac {5\,x^8}{4}+\frac {5\,x^4}{6}+\frac {1}{6}}{x^6-x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(x^8 - 2*x^4 + 1)),x)

[Out]

(5*atanh(x^2))/4 - ((5*x^4)/6 - (5*x^8)/4 + 1/6)/(x^6 - x^10)

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sympy [A]  time = 0.18, size = 41, normalized size = 1.05 \[ - \frac {5 \log {\left (x^{2} - 1 \right )}}{8} + \frac {5 \log {\left (x^{2} + 1 \right )}}{8} + \frac {- 15 x^{8} + 10 x^{4} + 2}{12 x^{10} - 12 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(x**8-2*x**4+1),x)

[Out]

-5*log(x**2 - 1)/8 + 5*log(x**2 + 1)/8 + (-15*x**8 + 10*x**4 + 2)/(12*x**10 - 12*x**6)

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