\(\int \log (\frac {-1+x}{1+x}) \, dx\) [275]

Optimal result

Integrand size = 10, antiderivative size = 27 \[ \int \log \left (\frac {-1+x}{1+x}\right ) \, dx=-\left ((1-x) \log \left (-\frac {1-x}{1+x}\right )\right )-2 \log (1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2535, 31} \[ \int \log \left (\frac {-1+x}{1+x}\right ) \, dx=-\left ((1-x) \log \left (-\frac {1-x}{x+1}\right )\right )-2 \log (x+1) \]

[In]

Int[Log[(-1 + x)/(1 + x)],x]

[Out]

-((1 - x)*Log[-((1 - x)/(1 + x))]) - 2*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 2535

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \log \left (\frac {-1+x}{1+x}\right ) \, dx=(-1+x) \log \left (\frac {-1+x}{1+x}\right )-2 \log (1+x) \]

[In]

Integrate[Log[(-1 + x)/(1 + x)],x]

[Out]

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

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
risch	\(x \ln \left (\frac {-1+x}{x +1}\right )-\ln \left (x^{2}-1\right )\)	\(22\)
parts	\(x \ln \left (\frac {-1+x}{x +1}\right )-\ln \left (\left (-1+x \right ) \left (x +1\right )\right )\)	\(24\)
parallelrisch	\(x \ln \left (\frac {-1+x}{x +1}\right )-2 \ln \left (-1+x \right )+\ln \left (\frac {-1+x}{x +1}\right )\)	\(30\)
derivativedivides	\(2 \ln \left (-\frac {2}{x +1}\right )+\ln \left (1-\frac {2}{x +1}\right ) \left (1-\frac {2}{x +1}\right ) \left (x +1\right )\)	\(35\)
default	\(2 \ln \left (-\frac {2}{x +1}\right )+\ln \left (1-\frac {2}{x +1}\right ) \left (1-\frac {2}{x +1}\right ) \left (x +1\right )\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \log \left (\frac {-1+x}{1+x}\right ) \, dx=x \log \left (\frac {x - 1}{x + 1}\right ) - \log \left (x^{2} - 1\right ) \]

[In]

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

[Out]

x*log((x - 1)/(x + 1)) - log(x^2 - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \log \left (\frac {-1+x}{1+x}\right ) \, dx=x \log {\left (\frac {x - 1}{x + 1} \right )} - \log {\left (x^{2} - 1 \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \log \left (\frac {-1+x}{1+x}\right ) \, dx=x \log \left (\frac {x - 1}{x + 1}\right ) - \log \left (x + 1\right ) - \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (21) = 42\).

Time = 0.33 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.81 \[ \int \log \left (\frac {-1+x}{1+x}\right ) \, dx=-\frac {2 \, \log \left (\frac {\frac {\frac {x - 1}{x + 1} + 1}{\frac {x - 1}{x + 1} - 1} + 1}{\frac {\frac {x - 1}{x + 1} + 1}{\frac {x - 1}{x + 1} - 1} - 1}\right )}{\frac {x - 1}{x + 1} - 1} - 2 \, \log \left (\frac {{\left | x - 1 \right |}}{{\left | x + 1 \right |}}\right ) + 2 \, \log \left ({\left | \frac {x - 1}{x + 1} - 1 \right |}\right ) \]

[In]

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

[Out]

-2*log((((x - 1)/(x + 1) + 1)/((x - 1)/(x + 1) - 1) + 1)/(((x - 1)/(x + 1) + 1)/((x - 1)/(x + 1) - 1) - 1))/((
x - 1)/(x + 1) - 1) - 2*log(abs(x - 1)/abs(x + 1)) + 2*log(abs((x - 1)/(x + 1) - 1))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \log \left (\frac {-1+x}{1+x}\right ) \, dx=x\,\ln \left (\frac {x-1}{x+1}\right )-\ln \left (x^2-1\right ) \]

[In]

int(log((x - 1)/(x + 1)),x)

[Out]

x*log((x - 1)/(x + 1)) - log(x^2 - 1)