\(\int \frac {1}{x^5 (1-2 x^4+x^8)} \, dx\) [298]

Optimal result

Integrand size = 16, antiderivative size = 37 \[ \int \frac {1}{x^5 \left (1-2 x^4+x^8\right )} \, dx=-\frac {1}{4 x^4}+\frac {1}{4 \left (1-x^4\right )}+2 \log (x)-\frac {1}{2} \log \left (1-x^4\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {28, 272, 46} \[ \int \frac {1}{x^5 \left (1-2 x^4+x^8\right )} \, dx=\frac {1}{4 \left (1-x^4\right )}-\frac {1}{4 x^4}-\frac {1}{2} \log \left (1-x^4\right )+2 \log (x) \]

[In]

Int[1/(x^5*(1 - 2*x^4 + x^8)),x]

[Out]

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

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 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^5 \left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {1}{(-1+x)^2 x^2} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{(-1+x)^2}-\frac {2}{-1+x}+\frac {1}{x^2}+\frac {2}{x}\right ) \, dx,x,x^4\right ) \\ & = -\frac {1}{4 x^4}+\frac {1}{4 \left (1-x^4\right )}+2 \log (x)-\frac {1}{2} \log \left (1-x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^5 \left (1-2 x^4+x^8\right )} \, dx=-\frac {1}{4 x^4}-\frac {1}{4 \left (-1+x^4\right )}+2 \log (x)-\frac {1}{2} \log \left (1-x^4\right ) \]

[In]

Integrate[1/(x^5*(1 - 2*x^4 + x^8)),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {\frac {1}{4}-\frac {x^{4}}{2}}{x^{4} \left (x^{4}-1\right )}+2 \ln \left (x \right )-\frac {\ln \left (x^{4}-1\right )}{2}\)	\(32\)
norman	\(\frac {\frac {1}{4}-\frac {x^{4}}{2}}{x^{4} \left (x^{4}-1\right )}+2 \ln \left (x \right )-\frac {\ln \left (x -1\right )}{2}-\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(44\)
default	\(-\frac {1}{4 x^{4}}+2 \ln \left (x \right )+\frac {1}{16 x +16}-\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}+\frac {1}{8 x^{2}+8}-\frac {1}{16 \left (x -1\right )}-\frac {\ln \left (x -1\right )}{2}\)	\(54\)
parallelrisch	\(\frac {8 \ln \left (x \right ) x^{8}-2 \ln \left (x -1\right ) x^{8}-2 \ln \left (x +1\right ) x^{8}-2 \ln \left (x^{2}+1\right ) x^{8}+1-8 \ln \left (x \right ) x^{4}+2 \ln \left (x -1\right ) x^{4}+2 \ln \left (x +1\right ) x^{4}+2 \ln \left (x^{2}+1\right ) x^{4}-2 x^{4}}{4 x^{4} \left (x^{4}-1\right )}\)	\(92\)

[In]

int(1/x^5/(x^8-2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x^5 \left (1-2 x^4+x^8\right )} \, dx=-\frac {2 \, x^{4} + 2 \, {\left (x^{8} - x^{4}\right )} \log \left (x^{4} - 1\right ) - 8 \, {\left (x^{8} - x^{4}\right )} \log \left (x\right ) - 1}{4 \, {\left (x^{8} - x^{4}\right )}} \]

[In]

integrate(1/x^5/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^5 \left (1-2 x^4+x^8\right )} \, dx=\frac {1 - 2 x^{4}}{4 x^{8} - 4 x^{4}} + 2 \log {\left (x \right )} - \frac {\log {\left (x^{4} - 1 \right )}}{2} \]

[In]

integrate(1/x**5/(x**8-2*x**4+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^5 \left (1-2 x^4+x^8\right )} \, dx=-\frac {2 \, x^{4} - 1}{4 \, {\left (x^{8} - x^{4}\right )}} - \frac {1}{2} \, \log \left (x^{4} - 1\right ) + \frac {1}{2} \, \log \left (x^{4}\right ) \]

[In]

integrate(1/x^5/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^5 \left (1-2 x^4+x^8\right )} \, dx=-\frac {2 \, x^{4} - 1}{4 \, {\left (x^{8} - x^{4}\right )}} + \frac {1}{2} \, \log \left (x^{4}\right ) - \frac {1}{2} \, \log \left ({\left | x^{4} - 1 \right |}\right ) \]

[In]

integrate(1/x^5/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^5 \left (1-2 x^4+x^8\right )} \, dx=2\,\ln \left (x\right )-\frac {\ln \left (x^4-1\right )}{2}+\frac {\frac {x^4}{2}-\frac {1}{4}}{x^4-x^8} \]

[In]

int(1/(x^5*(x^8 - 2*x^4 + 1)),x)

[Out]

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