\(\int \frac {1}{x^7 (1-2 x^4+x^8)} \, dx\) [299]

Optimal result

Integrand size = 16, antiderivative size = 39 \[ \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx=-\frac {5}{12 x^6}-\frac {5}{4 x^2}+\frac {1}{4 x^6 \left (1-x^4\right )}+\frac {5 \text {arctanh}\left (x^2\right )}{4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {28, 281, 296, 331, 213} \[ \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx=\frac {5 \text {arctanh}\left (x^2\right )}{4}-\frac {5}{12 x^6}-\frac {5}{4 x^2}+\frac {1}{4 x^6 \left (1-x^4\right )} \]

[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 213

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

Rule 281

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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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*} \text {integral}& = \int \frac {1}{x^7 \left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {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} \text {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} \text {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} \text {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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx=-\frac {1}{6 x^6}-\frac {1}{x^2}-\frac {x^2}{4 \left (-1+x^4\right )}-\frac {5}{8} \log \left (1-x^2\right )+\frac {5}{8} \log \left (1+x^2\right ) \]

[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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05

method	result	size
risch	\(\frac {\frac {1}{6}+\frac {5}{6} x^{4}-\frac {5}{4} x^{8}}{x^{6} \left (x^{4}-1\right )}+\frac {5 \ln \left (x^{2}+1\right )}{8}-\frac {5 \ln \left (x^{2}-1\right )}{8}\)	\(41\)
norman	\(\frac {\frac {1}{6}+\frac {5}{6} x^{4}-\frac {5}{4} x^{8}}{x^{6} \left (x^{4}-1\right )}-\frac {5 \ln \left (x -1\right )}{8}-\frac {5 \ln \left (x +1\right )}{8}+\frac {5 \ln \left (x^{2}+1\right )}{8}\)	\(45\)
default	\(-\frac {1}{6 x^{6}}-\frac {1}{x^{2}}+\frac {1}{16 x +16}-\frac {5 \ln \left (x +1\right )}{8}+\frac {5 \ln \left (x^{2}+1\right )}{8}-\frac {1}{8 \left (x^{2}+1\right )}-\frac {1}{16 \left (x -1\right )}-\frac {5 \ln \left (x -1\right )}{8}\)	\(55\)
parallelrisch	\(-\frac {15 \ln \left (x -1\right ) x^{10}+15 \ln \left (x +1\right ) x^{10}-15 \ln \left (x^{2}+1\right ) x^{10}-4+30 x^{8}-15 \ln \left (x -1\right ) x^{6}-15 \ln \left (x +1\right ) x^{6}+15 \ln \left (x^{2}+1\right ) x^{6}-20 x^{4}}{24 x^{6} \left (x^{4}-1\right )}\)	\(83\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (29) = 58\).

Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.51 \[ \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx=-\frac {30 \, x^{8} - 20 \, x^{4} - 15 \, {\left (x^{10} - x^{6}\right )} \log \left (x^{2} + 1\right ) + 15 \, {\left (x^{10} - x^{6}\right )} \log \left (x^{2} - 1\right ) - 4}{24 \, {\left (x^{10} - x^{6}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx=- \frac {5 \log {\left (x^{2} - 1 \right )}}{8} + \frac {5 \log {\left (x^{2} + 1 \right )}}{8} + \frac {- 15 x^{8} + 10 x^{4} + 2}{12 x^{10} - 12 x^{6}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx=-\frac {15 \, x^{8} - 10 \, x^{4} - 2}{12 \, {\left (x^{10} - x^{6}\right )}} + \frac {5}{8} \, \log \left (x^{2} + 1\right ) - \frac {5}{8} \, \log \left (x^{2} - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx=-\frac {x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {6 \, x^{4} + 1}{6 \, x^{6}} + \frac {5}{8} \, \log \left (x^{2} + 1\right ) - \frac {5}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^7 \left (1-2 x^4+x^8\right )} \, dx=\frac {5\,\mathrm {atanh}\left (x^2\right )}{4}-\frac {-\frac {5\,x^8}{4}+\frac {5\,x^4}{6}+\frac {1}{6}}{x^6-x^{10}} \]

[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)