\(\int \frac {1}{x^5 (1+2 x^4+x^8)} \, dx\) [279]

Optimal result

Integrand size = 16, antiderivative size = 33 \[ \int \frac {1}{x^5 \left (1+2 x^4+x^8\right )} \, dx=-\frac {1}{4 x^4}-\frac {1}{4 \left (1+x^4\right )}-2 \log (x)+\frac {1}{2} \log \left (1+x^4\right ) \]

[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 = 33, 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} \[ \int \frac {1}{x^5 \left (1+2 x^4+x^8\right )} \, dx=-\frac {1}{4 \left (x^4+1\right )}-\frac {1}{4 x^4}+\frac {1}{2} \log \left (x^4+1\right )-2 \log (x) \]

[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}{x^2 (1+x)^2} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {2}{x}+\frac {1}{(1+x)^2}+\frac {2}{1+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 = 33, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (1+2 x^4+x^8\right )} \, dx=-\frac {1}{4 x^4}-\frac {1}{4 \left (1+x^4\right )}-2 \log (x)+\frac {1}{2} \log \left (1+x^4\right ) \]

[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 = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {1}{4 x^{4}}-\frac {1}{4 \left (x^{4}+1\right )}-2 \ln \left (x \right )+\frac {\ln \left (x^{4}+1\right )}{2}\)	\(28\)
norman	\(\frac {-\frac {1}{4}-\frac {x^{4}}{2}}{x^{4} \left (x^{4}+1\right )}-2 \ln \left (x \right )+\frac {\ln \left (x^{4}+1\right )}{2}\)	\(32\)
risch	\(\frac {-\frac {1}{4}-\frac {x^{4}}{2}}{x^{4} \left (x^{4}+1\right )}-2 \ln \left (x \right )+\frac {\ln \left (x^{4}+1\right )}{2}\)	\(32\)
parallelrisch	\(-\frac {8 \ln \left (x \right ) x^{8}-2 \ln \left (x^{4}+1\right ) x^{8}+1+8 \ln \left (x \right ) x^{4}-2 \ln \left (x^{4}+1\right ) x^{4}+2 x^{4}}{4 x^{4} \left (x^{4}+1\right )}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x^5 \left (1+2 x^4+x^8\right )} \, dx=-\frac {2 \, x^{4} - 2 \, {\left (x^{8} + x^{4}\right )} \log \left (x^{4} + 1\right ) + 8 \, {\left (x^{8} + x^{4}\right )} \log \left (x\right ) + 1}{4 \, {\left (x^{8} + x^{4}\right )}} \]

[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 = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^5 \left (1+2 x^4+x^8\right )} \, dx=\frac {- 2 x^{4} - 1}{4 x^{8} + 4 x^{4}} - 2 \log {\left (x \right )} + \frac {\log {\left (x^{4} + 1 \right )}}{2} \]

[In]

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

[Out]

(-2*x**4 - 1)/(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 = 33, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (1+2 x^4+x^8\right )} \, dx=-\frac {2 \, x^{4} + 1}{4 \, {\left (x^{8} + x^{4}\right )}} + \frac {1}{2} \, \log \left (x^{4} + 1\right ) - \frac {1}{2} \, \log \left (x^{4}\right ) \]

[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 = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (1+2 x^4+x^8\right )} \, dx=-\frac {2 \, x^{4} + 1}{4 \, {\left (x^{8} + x^{4}\right )}} + \frac {1}{2} \, \log \left (x^{4} + 1\right ) - \frac {1}{2} \, \log \left (x^{4}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^5 \left (1+2 x^4+x^8\right )} \, dx=\frac {\ln \left (x^4+1\right )}{2}-2\,\ln \left (x\right )-\frac {\frac {x^4}{2}+\frac {1}{4}}{x^8+x^4} \]

[In]

int(1/(x^5*(2*x^4 + x^8 + 1)),x)

[Out]

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