\(\int \frac {\log (1-t)}{1-t} \, dt\) [65]

Optimal result

Integrand size = 14, antiderivative size = 12 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \log ^2(1-t) \]

[Out]

-1/2*ln(1-t)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2437, 2338} \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \log ^2(1-t) \]

[In]

Int[Log[1 - t]/(1 - t),t]

[Out]

-1/2*Log[1 - t]^2

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\log (t)}{t} \, dt,t,1-t\right ) \\ & = -\frac {1}{2} \log ^2(1-t) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \log ^2(1-t) \]

[In]

Integrate[Log[1 - t]/(1 - t),t]

[Out]

-1/2*Log[1 - t]^2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(-\frac {\ln \left (1-t \right )^{2}}{2}\)	\(11\)
default	\(-\frac {\ln \left (1-t \right )^{2}}{2}\)	\(11\)
norman	\(-\frac {\ln \left (1-t \right )^{2}}{2}\)	\(11\)
risch	\(-\frac {\ln \left (1-t \right )^{2}}{2}\)	\(11\)
parts	\(-\ln \left (-1+t \right ) \ln \left (1-t \right )+\frac {\ln \left (-1+t \right )^{2}}{2}\)	\(22\)

[In]

int(ln(1-t)/(1-t),t,method=_RETURNVERBOSE)

[Out]

-1/2*ln(1-t)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \, \log \left (-t + 1\right )^{2} \]

[In]

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

[Out]

-1/2*log(-t + 1)^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {\log (1-t)}{1-t} \, dt=- \frac {\log {\left (1 - t \right )}^{2}}{2} \]

[In]

integrate(ln(1-t)/(1-t),t)

[Out]

-log(1 - t)**2/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \, \log \left (-t + 1\right )^{2} \]

[In]

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

[Out]

-1/2*log(-t + 1)^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \, \log \left (-t + 1\right )^{2} \]

[In]

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

[Out]

-1/2*log(-t + 1)^2

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {{\ln \left (1-t\right )}^2}{2} \]

[In]

int(-log(1 - t)/(t - 1),t)

[Out]

-log(1 - t)^2/2