\(\int \frac {1}{1-e^{-x}} \, dx\) [84]

Optimal result

Integrand size = 11, antiderivative size = 6 \[ \int \frac {1}{1-e^{-x}} \, dx=\log \left (-1+e^x\right ) \]

[Out]

ln(-1+exp(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 2.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2320, 36, 31, 29} \[ \int \frac {1}{1-e^{-x}} \, dx=x+\log \left (1-e^{-x}\right ) \]

[In]

Int[(1 - E^(-x))^(-1),x]

[Out]

x + Log[1 - E^(-x)]

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1-e^{-x}} \, dx=\log \left (-1+e^x\right ) \]

[In]

Integrate[(1 - E^(-x))^(-1),x]

[Out]

Log[-1 + E^x]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.67

method	result	size
norman	\(x +\ln \left (-1+{\mathrm e}^{-x}\right )\)	\(10\)
risch	\(x +\ln \left (-1+{\mathrm e}^{-x}\right )\)	\(10\)
parallelrisch	\(x +\ln \left (-1+{\mathrm e}^{-x}\right )\)	\(10\)
derivativedivides	\(-\ln \left ({\mathrm e}^{-x}\right )+\ln \left (-1+{\mathrm e}^{-x}\right )\)	\(16\)
default	\(-\ln \left ({\mathrm e}^{-x}\right )+\ln \left (-1+{\mathrm e}^{-x}\right )\)	\(16\)

[In]

int(1/(1-exp(-x)),x,method=_RETURNVERBOSE)

[Out]

x+ln(-1+exp(-x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.50 \[ \int \frac {1}{1-e^{-x}} \, dx=x + \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

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

[Out]

x + log(e^(-x) - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.33 \[ \int \frac {1}{1-e^{-x}} \, dx=x + \log {\left (-1 + e^{- x} \right )} \]

[In]

integrate(1/(1-exp(-x)),x)

[Out]

x + log(-1 + exp(-x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.50 \[ \int \frac {1}{1-e^{-x}} \, dx=x + \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

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

[Out]

x + log(e^(-x) - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.67 \[ \int \frac {1}{1-e^{-x}} \, dx=x + \log \left ({\left | e^{\left (-x\right )} - 1 \right |}\right ) \]

[In]

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

[Out]

x + log(abs(e^(-x) - 1))

Mupad [B] (verification not implemented)

Time = 15.70 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.17 \[ \int \frac {1}{1-e^{-x}} \, dx=\ln \left (1-{\mathrm {e}}^x\right ) \]

[In]

int(-1/(exp(-x) - 1),x)

[Out]

log(1 - exp(x))