\(\int \log (a+b e^x) \, dx\) [116]

Optimal result

Integrand size = 8, antiderivative size = 38 \[ \int \log \left (a+b e^x\right ) \, dx=x \log \left (a+b e^x\right )-x \log \left (1+\frac {b e^x}{a}\right )-\operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right ) \]

[Out]

x*ln(a+b*exp(x))-x*ln(1+b*exp(x)/a)-polylog(2,-b*exp(x)/a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2318, 2221, 2317, 2438} \[ \int \log \left (a+b e^x\right ) \, dx=-\operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+x \log \left (a+b e^x\right )-x \log \left (\frac {b e^x}{a}+1\right ) \]

[In]

Int[Log[a + b*E^x],x]

[Out]

x*Log[a + b*E^x] - x*Log[1 + (b*E^x)/a] - PolyLog[2, -((b*E^x)/a)]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2318

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[x*Log[a + b*(F^(e*(c + d*x)
))^n], x] - Dist[b*d*e*n*Log[F], Int[x*((F^(e*(c + d*x)))^n/(a + b*(F^(e*(c + d*x)))^n)), x], x] /; FreeQ[{F,
a, b, c, d, e, n}, x] &&  !GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps \begin{align*} \text {integral}& = x \log \left (a+b e^x\right )-b \int \frac {e^x x}{a+b e^x} \, dx \\ & = x \log \left (a+b e^x\right )-x \log \left (1+\frac {b e^x}{a}\right )+\int \log \left (1+\frac {b e^x}{a}\right ) \, dx \\ & = x \log \left (a+b e^x\right )-x \log \left (1+\frac {b e^x}{a}\right )+\text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a}\right )}{x} \, dx,x,e^x\right ) \\ & = x \log \left (a+b e^x\right )-x \log \left (1+\frac {b e^x}{a}\right )-\text {Li}_2\left (-\frac {b e^x}{a}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \log \left (a+b e^x\right ) \, dx=x \log \left (a+b e^x\right )-x \log \left (1+\frac {b e^x}{a}\right )-\operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right ) \]

[In]

Integrate[Log[a + b*E^x],x]

[Out]

x*Log[a + b*E^x] - x*Log[1 + (b*E^x)/a] - PolyLog[2, -((b*E^x)/a)]

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\operatorname {dilog}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+\ln \left (a +b \,{\mathrm e}^{x}\right ) \ln \left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\)	\(28\)
default	\(\operatorname {dilog}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+\ln \left (a +b \,{\mathrm e}^{x}\right ) \ln \left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\)	\(28\)
risch	\(-\ln \left (\frac {a +b \,{\mathrm e}^{x}}{a}\right ) x +x \ln \left (a +b \,{\mathrm e}^{x}\right )-\operatorname {dilog}\left (\frac {a +b \,{\mathrm e}^{x}}{a}\right )\)	\(38\)
parts	\(x \ln \left (a +b \,{\mathrm e}^{x}\right )-b \left (\frac {\operatorname {dilog}\left (\frac {a +b \,{\mathrm e}^{x}}{a}\right )}{b}+\frac {x \ln \left (\frac {a +b \,{\mathrm e}^{x}}{a}\right )}{b}\right )\)	\(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \log \left (a+b e^x\right ) \, dx=x \log \left (b e^{x} + a\right ) - x \log \left (\frac {b e^{x} + a}{a}\right ) - {\rm Li}_2\left (-\frac {b e^{x} + a}{a} + 1\right ) \]

[In]

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

[Out]

x*log(b*e^x + a) - x*log((b*e^x + a)/a) - dilog(-(b*e^x + a)/a + 1)

Sympy [F]

\[ \int \log \left (a+b e^x\right ) \, dx=- b \int \frac {x e^{x}}{a + b e^{x}}\, dx + x \log {\left (a + b e^{x} \right )} \]

[In]

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

[Out]

-b*Integral(x*exp(x)/(a + b*exp(x)), x) + x*log(a + b*exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \log \left (a+b e^x\right ) \, dx=\log \left (b e^{x} + a\right ) \log \left (-\frac {b e^{x} + a}{a} + 1\right ) + {\rm Li}_2\left (\frac {b e^{x} + a}{a}\right ) \]

[In]

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

[Out]

log(b*e^x + a)*log(-(b*e^x + a)/a + 1) + dilog((b*e^x + a)/a)

Giac [F]

\[ \int \log \left (a+b e^x\right ) \, dx=\int { \log \left (b e^{x} + a\right ) \,d x } \]

[In]

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

[Out]

integrate(log(b*e^x + a), x)

Mupad [B] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \log \left (a+b e^x\right ) \, dx=x\,\ln \left (a+b\,{\mathrm {e}}^x\right )-x\,\ln \left (\frac {b\,{\mathrm {e}}^x}{a}+1\right )-\mathrm {polylog}\left (2,-\frac {b\,{\mathrm {e}}^x}{a}\right ) \]

[In]

int(log(a + b*exp(x)),x)

[Out]

x*log(a + b*exp(x)) - x*log((b*exp(x))/a + 1) - polylog(2, -(b*exp(x))/a)