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

Optimal. Leaf size=12 \[ -\frac {1}{2} \log ^2(1-t) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2437, 2338} \begin {gather*} -\frac {1}{2} \log ^2(1-t) \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.00, size = 12, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \log ^2(1-t) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[1 - t]/(1 - t),t]

[Out]

-1/2*Log[1 - t]^2

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Mathics [A]
time = 1.64, size = 10, normalized size = 0.83 \begin {gather*} -\frac {\text {Log}\left [1-t\right ]^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[Log[1 - t]/(1 - t),t]')

[Out]

-Log[1 - t] ^ 2 / 2

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Maple [A]
time = 0.07, size = 11, normalized size = 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	\(-\frac {\ln \left (1-t \right )^{2}}{2}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 10, normalized size = 0.83 \begin {gather*} -\frac {1}{2} \, \log \left (-t + 1\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(-t + 1)^2

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Fricas [A]
time = 0.34, size = 10, normalized size = 0.83 \begin {gather*} -\frac {1}{2} \, \log \left (-t + 1\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(-t + 1)^2

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Sympy [A]
time = 0.04, size = 8, normalized size = 0.67 \begin {gather*} - \frac {\log {\left (1 - t \right )}^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(1 - t)**2/2

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Giac [A]
time = 0.00, size = 11, normalized size = 0.92 \begin {gather*} -\frac {\ln ^{2}\left (-t+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(-t + 1)^2

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Mupad [B]
time = 0.35, size = 10, normalized size = 0.83 \begin {gather*} -\frac {{\ln \left (1-t\right )}^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-log(1 - t)/(t - 1),t)

[Out]

-log(1 - t)^2/2

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