\(\int \log (-\frac {x}{1+x}) \, dx\) [274]

Optimal result

Integrand size = 9, antiderivative size = 18 \[ \int \log \left (-\frac {x}{1+x}\right ) \, dx=x \log \left (-\frac {x}{1+x}\right )-\log (1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2536, 31} \[ \int \log \left (-\frac {x}{1+x}\right ) \, dx=x \log \left (-\frac {x}{x+1}\right )-\log (x+1) \]

[In]

Int[Log[-(x/(1 + x))],x]

[Out]

x*Log[-(x/(1 + x))] - Log[1 + x]

Rule 31

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

Rule 2536

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.), x_Symbol] :> Simp[
(a + b*x)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^p/b), x] - Dist[B*n*p*((b*c - a*d)/b), Int[(A + B*Log[e*((
a + b*x)^n/(c + d*x)^n)])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && EqQ[n + mn, 0] &&
NeQ[b*c - a*d, 0] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \log \left (-\frac {x}{1+x}\right ) \, dx=x \log \left (-\frac {x}{1+x}\right )-\log (1+x) \]

[In]

Integrate[Log[-(x/(1 + x))],x]

[Out]

x*Log[-(x/(1 + x))] - Log[1 + x]

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
risch	\(x \ln \left (-\frac {x}{x +1}\right )-\ln \left (x +1\right )\)	\(19\)
parts	\(x \ln \left (-\frac {x}{x +1}\right )-\ln \left (x +1\right )\)	\(19\)
parallelrisch	\(x \ln \left (-\frac {x}{x +1}\right )-\ln \left (x \right )+\ln \left (-\frac {x}{x +1}\right )\)	\(26\)
derivativedivides	\(\ln \left (\frac {1}{x +1}\right )-\ln \left (-1+\frac {1}{x +1}\right ) \left (-1+\frac {1}{x +1}\right ) \left (x +1\right )\)	\(28\)
default	\(\ln \left (\frac {1}{x +1}\right )-\ln \left (-1+\frac {1}{x +1}\right ) \left (-1+\frac {1}{x +1}\right ) \left (x +1\right )\)	\(28\)

[In]

int(ln(-x/(x+1)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \log \left (-\frac {x}{1+x}\right ) \, dx=x \log \left (-\frac {x}{x + 1}\right ) - \log \left (x + 1\right ) \]

[In]

integrate(log(-x/(1+x)),x, algorithm="fricas")

[Out]

x*log(-x/(x + 1)) - log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \log \left (-\frac {x}{1+x}\right ) \, dx=x \log {\left (- \frac {x}{x + 1} \right )} - \log {\left (x + 1 \right )} \]

[In]

integrate(ln(-x/(1+x)),x)

[Out]

x*log(-x/(x + 1)) - log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \log \left (-\frac {x}{1+x}\right ) \, dx=x \log \left (-\frac {x}{x + 1}\right ) - \log \left (x + 1\right ) \]

[In]

integrate(log(-x/(1+x)),x, algorithm="maxima")

[Out]

x*log(-x/(x + 1)) - log(x + 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (18) = 36\).

Time = 0.34 (sec) , antiderivative size = 80, normalized size of antiderivative = 4.44 \[ \int \log \left (-\frac {x}{1+x}\right ) \, dx=-\frac {\log \left (-\frac {x}{{\left (x + 1\right )} {\left (\frac {x}{x + 1} - 1\right )} {\left (\frac {x}{{\left (x + 1\right )} {\left (\frac {x}{x + 1} - 1\right )}} - 1\right )}}\right )}{\frac {x}{x + 1} - 1} - \log \left (\frac {{\left | x \right |}}{{\left | x + 1 \right |}}\right ) + \log \left ({\left | -\frac {x}{x + 1} + 1 \right |}\right ) \]

[In]

integrate(log(-x/(1+x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \log \left (-\frac {x}{1+x}\right ) \, dx=x\,\ln \left (-\frac {x}{x+1}\right )-\ln \left (x+1\right ) \]

[In]

int(log(-x/(x + 1)),x)

[Out]

x*log(-x/(x + 1)) - log(x + 1)