\(\int \frac {1}{x^6 (1-2 x^4+x^8)} \, dx\) [307]

Optimal result

Integrand size = 16, antiderivative size = 43 \[ \int \frac {1}{x^6 \left (1-2 x^4+x^8\right )} \, dx=-\frac {9}{20 x^5}-\frac {9}{4 x}+\frac {1}{4 x^5 \left (1-x^4\right )}-\frac {9 \arctan (x)}{8}+\frac {9 \text {arctanh}(x)}{8} \]

[Out]

-9/20/x^5-9/4/x+1/4/x^5/(-x^4+1)-9/8*arctan(x)+9/8*arctanh(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {28, 296, 331, 304, 209, 212} \[ \int \frac {1}{x^6 \left (1-2 x^4+x^8\right )} \, dx=-\frac {9 \arctan (x)}{8}+\frac {9 \text {arctanh}(x)}{8}-\frac {9}{20 x^5}+\frac {1}{4 x^5 \left (1-x^4\right )}-\frac {9}{4 x} \]

[In]

Int[1/(x^6*(1 - 2*x^4 + x^8)),x]

[Out]

-9/(20*x^5) - 9/(4*x) + 1/(4*x^5*(1 - x^4)) - (9*ArcTan[x])/8 + (9*ArcTanh[x])/8

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 209

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

Rule 212

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

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 304

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 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^6 \left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{4 x^5 \left (1-x^4\right )}-\frac {9}{4} \int \frac {1}{x^6 \left (-1+x^4\right )} \, dx \\ & = -\frac {9}{20 x^5}+\frac {1}{4 x^5 \left (1-x^4\right )}-\frac {9}{4} \int \frac {1}{x^2 \left (-1+x^4\right )} \, dx \\ & = -\frac {9}{20 x^5}-\frac {9}{4 x}+\frac {1}{4 x^5 \left (1-x^4\right )}-\frac {9}{4} \int \frac {x^2}{-1+x^4} \, dx \\ & = -\frac {9}{20 x^5}-\frac {9}{4 x}+\frac {1}{4 x^5 \left (1-x^4\right )}+\frac {9}{8} \int \frac {1}{1-x^2} \, dx-\frac {9}{8} \int \frac {1}{1+x^2} \, dx \\ & = -\frac {9}{20 x^5}-\frac {9}{4 x}+\frac {1}{4 x^5 \left (1-x^4\right )}-\frac {9}{8} \tan ^{-1}(x)+\frac {9}{8} \tanh ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^6 \left (1-2 x^4+x^8\right )} \, dx=-\frac {1}{5 x^5}-\frac {2}{x}-\frac {x^3}{4 \left (-1+x^4\right )}-\frac {9 \arctan (x)}{8}-\frac {9}{16} \log (1-x)+\frac {9}{16} \log (1+x) \]

[In]

Integrate[1/(x^6*(1 - 2*x^4 + x^8)),x]

[Out]

-1/5*1/x^5 - 2/x - x^3/(4*(-1 + x^4)) - (9*ArcTan[x])/8 - (9*Log[1 - x])/16 + (9*Log[1 + x])/16

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {-\frac {9}{4} x^{8}+\frac {9}{5} x^{4}+\frac {1}{5}}{x^{5} \left (x^{4}-1\right )}-\frac {9 \ln \left (x -1\right )}{16}+\frac {9 \ln \left (x +1\right )}{16}-\frac {9 \arctan \left (x \right )}{8}\)	\(41\)
default	\(-\frac {1}{5 x^{5}}-\frac {2}{x}-\frac {1}{16 \left (x +1\right )}+\frac {9 \ln \left (x +1\right )}{16}-\frac {x}{8 \left (x^{2}+1\right )}-\frac {9 \arctan \left (x \right )}{8}-\frac {1}{16 \left (x -1\right )}-\frac {9 \ln \left (x -1\right )}{16}\)	\(52\)
parallelrisch	\(-\frac {-45 i \ln \left (x -i\right ) x^{9}+45 i \ln \left (x +i\right ) x^{9}+45 \ln \left (x -1\right ) x^{9}-45 \ln \left (x +1\right ) x^{9}-16+180 x^{8}+45 i \ln \left (x -i\right ) x^{5}-45 i \ln \left (x +i\right ) x^{5}-45 \ln \left (x -1\right ) x^{5}+45 \ln \left (x +1\right ) x^{5}-144 x^{4}}{80 x^{5} \left (x^{4}-1\right )}\)	\(105\)

[In]

int(1/x^6/(x^8-2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

(-9/4*x^8+9/5*x^4+1/5)/x^5/(x^4-1)-9/16*ln(x-1)+9/16*ln(x+1)-9/8*arctan(x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (31) = 62\).

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.58 \[ \int \frac {1}{x^6 \left (1-2 x^4+x^8\right )} \, dx=-\frac {180 \, x^{8} - 144 \, x^{4} + 90 \, {\left (x^{9} - x^{5}\right )} \arctan \left (x\right ) - 45 \, {\left (x^{9} - x^{5}\right )} \log \left (x + 1\right ) + 45 \, {\left (x^{9} - x^{5}\right )} \log \left (x - 1\right ) - 16}{80 \, {\left (x^{9} - x^{5}\right )}} \]

[In]

integrate(1/x^6/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

-1/80*(180*x^8 - 144*x^4 + 90*(x^9 - x^5)*arctan(x) - 45*(x^9 - x^5)*log(x + 1) + 45*(x^9 - x^5)*log(x - 1) -
16)/(x^9 - x^5)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^6 \left (1-2 x^4+x^8\right )} \, dx=- \frac {9 \log {\left (x - 1 \right )}}{16} + \frac {9 \log {\left (x + 1 \right )}}{16} - \frac {9 \operatorname {atan}{\left (x \right )}}{8} + \frac {- 45 x^{8} + 36 x^{4} + 4}{20 x^{9} - 20 x^{5}} \]

[In]

integrate(1/x**6/(x**8-2*x**4+1),x)

[Out]

-9*log(x - 1)/16 + 9*log(x + 1)/16 - 9*atan(x)/8 + (-45*x**8 + 36*x**4 + 4)/(20*x**9 - 20*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^6 \left (1-2 x^4+x^8\right )} \, dx=-\frac {45 \, x^{8} - 36 \, x^{4} - 4}{20 \, {\left (x^{9} - x^{5}\right )}} - \frac {9}{8} \, \arctan \left (x\right ) + \frac {9}{16} \, \log \left (x + 1\right ) - \frac {9}{16} \, \log \left (x - 1\right ) \]

[In]

integrate(1/x^6/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

-1/20*(45*x^8 - 36*x^4 - 4)/(x^9 - x^5) - 9/8*arctan(x) + 9/16*log(x + 1) - 9/16*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^6 \left (1-2 x^4+x^8\right )} \, dx=-\frac {x^{3}}{4 \, {\left (x^{4} - 1\right )}} - \frac {10 \, x^{4} + 1}{5 \, x^{5}} - \frac {9}{8} \, \arctan \left (x\right ) + \frac {9}{16} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {9}{16} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate(1/x^6/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

-1/4*x^3/(x^4 - 1) - 1/5*(10*x^4 + 1)/x^5 - 9/8*arctan(x) + 9/16*log(abs(x + 1)) - 9/16*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^6 \left (1-2 x^4+x^8\right )} \, dx=\frac {9\,\mathrm {atanh}\left (x\right )}{8}-\frac {9\,\mathrm {atan}\left (x\right )}{8}-\frac {-\frac {9\,x^8}{4}+\frac {9\,x^4}{5}+\frac {1}{5}}{x^5-x^9} \]

[In]

int(1/(x^6*(x^8 - 2*x^4 + 1)),x)

[Out]

(9*atanh(x))/8 - (9*atan(x))/8 - ((9*x^4)/5 - (9*x^8)/4 + 1/5)/(x^5 - x^9)