3.116 \(\int \log (a+b e^x) \, dx\)

Optimal. Leaf size=38 \[ -\text {Li}_2\left (-\frac {b e^x}{a}\right )+x \log \left (a+b e^x\right )-x \log \left (\frac {b e^x}{a}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2280, 2190, 2279, 2391} \[ -\text {PolyLog}\left (2,-\frac {b e^x}{a}\right )+x \log \left (a+b e^x\right )-x \log \left (\frac {b e^x}{a}+1\right ) \]

Antiderivative was successfully verified.

[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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2280

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 2391

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*} \int \log \left (a+b e^x\right ) \, dx &=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 )+\operatorname {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*}

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Mathematica [A]  time = 0.00, size = 38, normalized size = 1.00 \[ -\text {Li}_2\left (-\frac {b e^x}{a}\right )+x \log \left (a+b e^x\right )-x \log \left (\frac {b e^x}{a}+1\right ) \]

Antiderivative was successfully verified.

[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)]

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fricas [A]  time = 1.16, size = 40, normalized size = 1.05 \[ x \log \left (b e^{x} + a\right ) - x \log \left (\frac {b e^{x} + a}{a}\right ) - {\rm Li}_2\left (-\frac {b e^{x} + a}{a} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (b e^{x} + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(b*e^x + a), x)

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maple [A]  time = 0.08, size = 28, normalized size = 0.74 \[ \ln \left (-\frac {b \,{\mathrm e}^{x}}{a}\right ) \ln \left (b \,{\mathrm e}^{x}+a \right )+\dilog \left (-\frac {b \,{\mathrm e}^{x}}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

dilog(-1/a*b*exp(x))+ln(b*exp(x)+a)*ln(-1/a*b*exp(x))

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maxima [A]  time = 0.69, size = 34, normalized size = 0.89 \[ \log \left (b e^{x} + a\right ) \log \left (-\frac {b e^{x} + a}{a} + 1\right ) + {\rm Li}_2\left (\frac {b e^{x} + a}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.38, size = 35, normalized size = 0.92 \[ x\,\ln \left (a+b\,{\mathrm {e}}^x\right )-x\,\ln \left (\frac {b\,{\mathrm {e}}^x}{a}+1\right )-\mathrm {polylog}\left (2,-\frac {b\,{\mathrm {e}}^x}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - b \int \frac {x e^{x}}{a + b e^{x}}\, dx + x \log {\left (a + b e^{x} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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