\(\int \frac {1}{x^3 (1-2 x^4+x^8)} \, dx\) [297]

Optimal result

Integrand size = 16, antiderivative size = 32 \[ \int \frac {1}{x^3 \left (1-2 x^4+x^8\right )} \, dx=-\frac {3}{4 x^2}+\frac {1}{4 x^2 \left (1-x^4\right )}+\frac {3 \text {arctanh}\left (x^2\right )}{4} \]

[Out]

-3/4/x^2+1/4/x^2/(-x^4+1)+3/4*arctanh(x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {28, 281, 296, 331, 213} \[ \int \frac {1}{x^3 \left (1-2 x^4+x^8\right )} \, dx=\frac {3 \text {arctanh}\left (x^2\right )}{4}-\frac {3}{4 x^2}+\frac {1}{4 x^2 \left (1-x^4\right )} \]

[In]

Int[1/(x^3*(1 - 2*x^4 + x^8)),x]

[Out]

-3/(4*x^2) + 1/(4*x^2*(1 - x^4)) + (3*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^3 \left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (-1+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{4 x^2 \left (1-x^4\right )}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{x^2 \left (-1+x^2\right )} \, dx,x,x^2\right ) \\ & = -\frac {3}{4 x^2}+\frac {1}{4 x^2 \left (1-x^4\right )}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,x^2\right ) \\ & = -\frac {3}{4 x^2}+\frac {1}{4 x^2 \left (1-x^4\right )}+\frac {3}{4} \tanh ^{-1}\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x^3 \left (1-2 x^4+x^8\right )} \, dx=\frac {1}{8} \left (\frac {4-6 x^4}{x^2 \left (-1+x^4\right )}-3 \log \left (1-x^2\right )+3 \log \left (1+x^2\right )\right ) \]

[In]

Integrate[1/(x^3*(1 - 2*x^4 + x^8)),x]

[Out]

((4 - 6*x^4)/(x^2*(-1 + x^4)) - 3*Log[1 - x^2] + 3*Log[1 + x^2])/8

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {\frac {1}{2}-\frac {3 x^{4}}{4}}{x^{2} \left (x^{4}-1\right )}-\frac {3 \ln \left (x^{2}-1\right )}{8}+\frac {3 \ln \left (x^{2}+1\right )}{8}\)	\(36\)
norman	\(\frac {\frac {1}{2}-\frac {3 x^{4}}{4}}{x^{2} \left (x^{4}-1\right )}-\frac {3 \ln \left (x -1\right )}{8}-\frac {3 \ln \left (x +1\right )}{8}+\frac {3 \ln \left (x^{2}+1\right )}{8}\)	\(40\)
default	\(-\frac {1}{2 x^{2}}+\frac {1}{16 x +16}-\frac {3 \ln \left (x +1\right )}{8}+\frac {3 \ln \left (x^{2}+1\right )}{8}-\frac {1}{8 \left (x^{2}+1\right )}-\frac {1}{16 \left (x -1\right )}-\frac {3 \ln \left (x -1\right )}{8}\)	\(50\)
parallelrisch	\(-\frac {3 \ln \left (x -1\right ) x^{6}+3 \ln \left (x +1\right ) x^{6}-3 \ln \left (x^{2}+1\right ) x^{6}-4+6 x^{4}-3 \ln \left (x -1\right ) x^{2}-3 \ln \left (x +1\right ) x^{2}+3 \ln \left (x^{2}+1\right ) x^{2}}{8 x^{2} \left (x^{4}-1\right )}\)	\(78\)

[In]

int(1/x^3/(x^8-2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

(1/2-3/4*x^4)/x^2/(x^4-1)-3/8*ln(x^2-1)+3/8*ln(x^2+1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (24) = 48\).

Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int \frac {1}{x^3 \left (1-2 x^4+x^8\right )} \, dx=-\frac {6 \, x^{4} - 3 \, {\left (x^{6} - x^{2}\right )} \log \left (x^{2} + 1\right ) + 3 \, {\left (x^{6} - x^{2}\right )} \log \left (x^{2} - 1\right ) - 4}{8 \, {\left (x^{6} - x^{2}\right )}} \]

[In]

integrate(1/x^3/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

-1/8*(6*x^4 - 3*(x^6 - x^2)*log(x^2 + 1) + 3*(x^6 - x^2)*log(x^2 - 1) - 4)/(x^6 - x^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^3 \left (1-2 x^4+x^8\right )} \, dx=\frac {2 - 3 x^{4}}{4 x^{6} - 4 x^{2}} - \frac {3 \log {\left (x^{2} - 1 \right )}}{8} + \frac {3 \log {\left (x^{2} + 1 \right )}}{8} \]

[In]

integrate(1/x**3/(x**8-2*x**4+1),x)

[Out]

(2 - 3*x**4)/(4*x**6 - 4*x**2) - 3*log(x**2 - 1)/8 + 3*log(x**2 + 1)/8

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^3 \left (1-2 x^4+x^8\right )} \, dx=-\frac {3 \, x^{4} - 2}{4 \, {\left (x^{6} - x^{2}\right )}} + \frac {3}{8} \, \log \left (x^{2} + 1\right ) - \frac {3}{8} \, \log \left (x^{2} - 1\right ) \]

[In]

integrate(1/x^3/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

-1/4*(3*x^4 - 2)/(x^6 - x^2) + 3/8*log(x^2 + 1) - 3/8*log(x^2 - 1)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^3 \left (1-2 x^4+x^8\right )} \, dx=-\frac {3 \, x^{4} - 2}{4 \, {\left (x^{6} - x^{2}\right )}} + \frac {3}{8} \, \log \left (x^{2} + 1\right ) - \frac {3}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

[In]

integrate(1/x^3/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

-1/4*(3*x^4 - 2)/(x^6 - x^2) + 3/8*log(x^2 + 1) - 3/8*log(abs(x^2 - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^3 \left (1-2 x^4+x^8\right )} \, dx=\frac {3\,\mathrm {atanh}\left (x^2\right )}{4}+\frac {\frac {3\,x^4}{4}-\frac {1}{2}}{x^2-x^6} \]

[In]

int(1/(x^3*(x^8 - 2*x^4 + 1)),x)

[Out]

(3*atanh(x^2))/4 + ((3*x^4)/4 - 1/2)/(x^2 - x^6)