\(\int \frac {1}{x^2 (1-2 x^4+x^8)} \, dx\) [305]

Optimal result

Integrand size = 16, antiderivative size = 36 \[ \int \frac {1}{x^2 \left (1-2 x^4+x^8\right )} \, dx=-\frac {5}{4 x}+\frac {1}{4 x \left (1-x^4\right )}-\frac {5 \arctan (x)}{8}+\frac {5 \text {arctanh}(x)}{8} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {28, 296, 331, 304, 209, 212} \[ \int \frac {1}{x^2 \left (1-2 x^4+x^8\right )} \, dx=-\frac {5 \arctan (x)}{8}+\frac {5 \text {arctanh}(x)}{8}+\frac {1}{4 x \left (1-x^4\right )}-\frac {5}{4 x} \]

[In]

Int[1/(x^2*(1 - 2*x^4 + x^8)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^2 \left (1-2 x^4+x^8\right )} \, dx=\frac {1}{16} \left (-\frac {16}{x}-\frac {4 x^3}{-1+x^4}-10 \arctan (x)-5 \log (1-x)+5 \log (1+x)\right ) \]

[In]

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

[Out]

(-16/x - (4*x^3)/(-1 + x^4) - 10*ArcTan[x] - 5*Log[1 - x] + 5*Log[1 + x])/16

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {-\frac {5 x^{4}}{4}+1}{\left (x^{4}-1\right ) x}+\frac {5 \ln \left (x +1\right )}{16}-\frac {5 \ln \left (x -1\right )}{16}-\frac {5 \arctan \left (x \right )}{8}\)	\(36\)
default	\(-\frac {1}{x}-\frac {1}{16 \left (x +1\right )}+\frac {5 \ln \left (x +1\right )}{16}-\frac {x}{8 \left (x^{2}+1\right )}-\frac {5 \arctan \left (x \right )}{8}-\frac {1}{16 \left (x -1\right )}-\frac {5 \ln \left (x -1\right )}{16}\)	\(47\)
parallelrisch	\(-\frac {-5 i \ln \left (x -i\right ) x^{5}+5 i \ln \left (x +i\right ) x^{5}+5 \ln \left (x -1\right ) x^{5}-5 \ln \left (x +1\right ) x^{5}-16+20 x^{4}+5 i \ln \left (x -i\right ) x -5 i \ln \left (x +i\right ) x -5 \ln \left (x -1\right ) x +5 \ln \left (x +1\right ) x}{16 x \left (x^{4}-1\right )}\)	\(92\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.53 \[ \int \frac {1}{x^2 \left (1-2 x^4+x^8\right )} \, dx=-\frac {20 \, x^{4} + 10 \, {\left (x^{5} - x\right )} \arctan \left (x\right ) - 5 \, {\left (x^{5} - x\right )} \log \left (x + 1\right ) + 5 \, {\left (x^{5} - x\right )} \log \left (x - 1\right ) - 16}{16 \, {\left (x^{5} - x\right )}} \]

[In]

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

[Out]

-1/16*(20*x^4 + 10*(x^5 - x)*arctan(x) - 5*(x^5 - x)*log(x + 1) + 5*(x^5 - x)*log(x - 1) - 16)/(x^5 - x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^2 \left (1-2 x^4+x^8\right )} \, dx=\frac {4 - 5 x^{4}}{4 x^{5} - 4 x} - \frac {5 \log {\left (x - 1 \right )}}{16} + \frac {5 \log {\left (x + 1 \right )}}{16} - \frac {5 \operatorname {atan}{\left (x \right )}}{8} \]

[In]

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

[Out]

(4 - 5*x**4)/(4*x**5 - 4*x) - 5*log(x - 1)/16 + 5*log(x + 1)/16 - 5*atan(x)/8

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^2 \left (1-2 x^4+x^8\right )} \, dx=-\frac {5 \, x^{4} - 4}{4 \, {\left (x^{5} - x\right )}} - \frac {5}{8} \, \arctan \left (x\right ) + \frac {5}{16} \, \log \left (x + 1\right ) - \frac {5}{16} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^2 \left (1-2 x^4+x^8\right )} \, dx=-\frac {5 \, x^{4} - 4}{4 \, {\left (x^{5} - x\right )}} - \frac {5}{8} \, \arctan \left (x\right ) + \frac {5}{16} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {5}{16} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^2 \left (1-2 x^4+x^8\right )} \, dx=\frac {5\,\mathrm {atanh}\left (x\right )}{8}-\frac {5\,\mathrm {atan}\left (x\right )}{8}+\frac {\frac {5\,x^4}{4}-1}{x-x^5} \]

[In]

int(1/(x^2*(x^8 - 2*x^4 + 1)),x)

[Out]

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