\(\int \frac {e^x}{a+b e^x} \, dx\) [2]

Optimal result

Integrand size = 13, antiderivative size = 12 \[ \int \frac {e^x}{a+b e^x} \, dx=\frac {\log \left (a+b e^x\right )}{b} \]

[Out]

ln(a+b*exp(x))/b

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2278, 31} \[ \int \frac {e^x}{a+b e^x} \, dx=\frac {\log \left (a+b e^x\right )}{b} \]

[In]

Int[E^x/(a + b*E^x),x]

[Out]

Log[a + b*E^x]/b

Rule 31

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

Rule 2278

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{a+b e^x} \, dx=\frac {\log \left (a+b e^x\right )}{b} \]

[In]

Integrate[E^x/(a + b*E^x),x]

[Out]

Log[a + b*E^x]/b

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \,{\mathrm e}^{x}\right )}{b}\)	\(12\)
default	\(\frac {\ln \left (a +b \,{\mathrm e}^{x}\right )}{b}\)	\(12\)
norman	\(\frac {\ln \left (a +b \,{\mathrm e}^{x}\right )}{b}\)	\(12\)
parallelrisch	\(\frac {\ln \left (a +b \,{\mathrm e}^{x}\right )}{b}\)	\(12\)
risch	\(\frac {\ln \left ({\mathrm e}^{x}+\frac {a}{b}\right )}{b}\)	\(14\)

[In]

int(exp(x)/(a+b*exp(x)),x,method=_RETURNVERBOSE)

[Out]

ln(a+b*exp(x))/b

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {e^x}{a+b e^x} \, dx=\frac {\log \left (b e^{x} + a\right )}{b} \]

[In]

integrate(exp(x)/(a+b*exp(x)),x, algorithm="fricas")

[Out]

log(b*e^x + a)/b

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {e^x}{a+b e^x} \, dx=\frac {\log {\left (\frac {a}{b} + e^{x} \right )}}{b} \]

[In]

integrate(exp(x)/(a+b*exp(x)),x)

[Out]

log(a/b + exp(x))/b

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {e^x}{a+b e^x} \, dx=\frac {\log \left (b e^{x} + a\right )}{b} \]

[In]

integrate(exp(x)/(a+b*exp(x)),x, algorithm="maxima")

[Out]

log(b*e^x + a)/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{a+b e^x} \, dx=\frac {\log \left ({\left | b e^{x} + a \right |}\right )}{b} \]

[In]

integrate(exp(x)/(a+b*exp(x)),x, algorithm="giac")

[Out]

log(abs(b*e^x + a))/b

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {e^x}{a+b e^x} \, dx=\frac {\ln \left (a+b\,{\mathrm {e}}^x\right )}{b} \]

[In]

int(exp(x)/(a + b*exp(x)),x)

[Out]

log(a + b*exp(x))/b