\(\int \frac {1}{x (1+x^5)} \, dx\) [1298]

Optimal result

Integrand size = 11, antiderivative size = 13 \[ \int \frac {1}{x \left (1+x^5\right )} \, dx=\log (x)-\frac {1}{5} \log \left (1+x^5\right ) \]

[Out]

ln(x)-1/5*ln(x^5+1)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {272, 36, 29, 31} \[ \int \frac {1}{x \left (1+x^5\right )} \, dx=\log (x)-\frac {1}{5} \log \left (x^5+1\right ) \]

[In]

Int[1/(x*(1 + x^5)),x]

[Out]

Log[x] - Log[1 + x^5]/5

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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}& = \frac {1}{5} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^5\right ) \\ & = \frac {1}{5} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^5\right )-\frac {1}{5} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^5\right ) \\ & = \log (x)-\frac {1}{5} \log \left (1+x^5\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+x^5\right )} \, dx=\log (x)-\frac {1}{5} \log \left (1+x^5\right ) \]

[In]

Integrate[1/(x*(1 + x^5)),x]

[Out]

Log[x] - Log[1 + x^5]/5

Maple [A] (verified)

Time = 4.39 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
meijerg	\(\ln \left (x \right )-\frac {\ln \left (x^{5}+1\right )}{5}\)	\(12\)
risch	\(\ln \left (x \right )-\frac {\ln \left (x^{5}+1\right )}{5}\)	\(12\)
default	\(\ln \left (x \right )-\frac {\ln \left (1+x \right )}{5}-\frac {\ln \left (x^{4}-x^{3}+x^{2}-x +1\right )}{5}\)	\(29\)
norman	\(\ln \left (x \right )-\frac {\ln \left (1+x \right )}{5}-\frac {\ln \left (x^{4}-x^{3}+x^{2}-x +1\right )}{5}\)	\(29\)
parallelrisch	\(\ln \left (x \right )-\frac {\ln \left (1+x \right )}{5}-\frac {\ln \left (x^{4}-x^{3}+x^{2}-x +1\right )}{5}\)	\(29\)

[In]

int(1/x/(x^5+1),x,method=_RETURNVERBOSE)

[Out]

ln(x)-1/5*ln(x^5+1)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x \left (1+x^5\right )} \, dx=-\frac {1}{5} \, \log \left (x^{5} + 1\right ) + \log \left (x\right ) \]

[In]

integrate(1/x/(x^5+1),x, algorithm="fricas")

[Out]

-1/5*log(x^5 + 1) + log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x \left (1+x^5\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{5} + 1 \right )}}{5} \]

[In]

integrate(1/x/(x**5+1),x)

[Out]

log(x) - log(x**5 + 1)/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x \left (1+x^5\right )} \, dx=-\frac {1}{5} \, \log \left (x^{5} + 1\right ) + \frac {1}{5} \, \log \left (x^{5}\right ) \]

[In]

integrate(1/x/(x^5+1),x, algorithm="maxima")

[Out]

-1/5*log(x^5 + 1) + 1/5*log(x^5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+x^5\right )} \, dx=-\frac {1}{5} \, \log \left ({\left | x^{5} + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/x/(x^5+1),x, algorithm="giac")

[Out]

-1/5*log(abs(x^5 + 1)) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 5.41 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x \left (1+x^5\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^5+1\right )}{5} \]

[In]

int(1/(x*(x^5 + 1)),x)

[Out]

log(x) - log(x^5 + 1)/5