\(\int \frac {1}{x (1+2 x^4+x^8)} \, dx\) [277]

Optimal result

Integrand size = 16, antiderivative size = 24 \[ \int \frac {1}{x \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{4 \left (1+x^4\right )}+\log (x)-\frac {1}{4} \log \left (1+x^4\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, 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} \[ \int \frac {1}{x \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{4 \left (x^4+1\right )}-\frac {1}{4} \log \left (x^4+1\right )+\log (x) \]

[In]

Int[1/(x*(1 + 2*x^4 + x^8)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{4 \left (1+x^4\right )}+\log (x)-\frac {1}{4} \log \left (1+x^4\right ) \]

[In]

Integrate[1/(x*(1 + 2*x^4 + x^8)),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {1}{4 x^{4}+4}+\ln \left (x \right )-\frac {\ln \left (x^{4}+1\right )}{4}\)	\(21\)
norman	\(\frac {1}{4 x^{4}+4}+\ln \left (x \right )-\frac {\ln \left (x^{4}+1\right )}{4}\)	\(21\)
risch	\(\frac {1}{4 x^{4}+4}+\ln \left (x \right )-\frac {\ln \left (x^{4}+1\right )}{4}\)	\(21\)
parallelrisch	\(\frac {4 \ln \left (x \right ) x^{4}-\ln \left (x^{4}+1\right ) x^{4}+1+4 \ln \left (x \right )-\ln \left (x^{4}+1\right )}{4 x^{4}+4}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x \left (1+2 x^4+x^8\right )} \, dx=-\frac {{\left (x^{4} + 1\right )} \log \left (x^{4} + 1\right ) - 4 \, {\left (x^{4} + 1\right )} \log \left (x\right ) - 1}{4 \, {\left (x^{4} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x \left (1+2 x^4+x^8\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{4} + 1 \right )}}{4} + \frac {1}{4 x^{4} + 4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{4 \, {\left (x^{4} + 1\right )}} - \frac {1}{4} \, \log \left (x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x \left (1+2 x^4+x^8\right )} \, dx=\frac {x^{4} + 2}{4 \, {\left (x^{4} + 1\right )}} - \frac {1}{4} \, \log \left (x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x \left (1+2 x^4+x^8\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^4+1\right )}{4}+\frac {1}{4\,\left (x^4+1\right )} \]

[In]

int(1/(x*(2*x^4 + x^8 + 1)),x)

[Out]

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