\(\int e^x \log (a+b e^x) \, dx\) [284]

Optimal result

Integrand size = 12, antiderivative size = 25 \[ \int e^x \log \left (a+b e^x\right ) \, dx=-e^x+\frac {\left (a+b e^x\right ) \log \left (a+b e^x\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2225, 2634, 12, 2280, 45} \[ \int e^x \log \left (a+b e^x\right ) \, dx=e^x \log \left (a+b e^x\right )+\frac {a \log \left (a+b e^x\right )}{b}-e^x \]

[In]

Int[E^x*Log[a + b*E^x],x]

[Out]

-E^x + (a*Log[a + b*E^x])/b + E^x*Log[a + b*E^x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2280

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[g*h*(Log[G]/(d*e*Log[F]))]}, Dist[Denominator[m]*(G^(f*h - c*g*(h/d))/(d*e*Log[F])), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^(e*((c + d*x)/Denominator[m]))], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = e^x \log \left (a+b e^x\right )-\int \frac {b e^{2 x}}{a+b e^x} \, dx \\ & = e^x \log \left (a+b e^x\right )-b \int \frac {e^{2 x}}{a+b e^x} \, dx \\ & = e^x \log \left (a+b e^x\right )-b \text {Subst}\left (\int \frac {x}{a+b x} \, dx,x,e^x\right ) \\ & = e^x \log \left (a+b e^x\right )-b \text {Subst}\left (\int \left (\frac {1}{b}-\frac {a}{b (a+b x)}\right ) \, dx,x,e^x\right ) \\ & = -e^x+\frac {a \log \left (a+b e^x\right )}{b}+e^x \log \left (a+b e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int e^x \log \left (a+b e^x\right ) \, dx=-e^x+\frac {\left (a+b e^x\right ) \log \left (a+b e^x\right )}{b} \]

[In]

Integrate[E^x*Log[a + b*E^x],x]

[Out]

-E^x + ((a + b*E^x)*Log[a + b*E^x])/b

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {\left (a +b \,{\mathrm e}^{x}\right ) \ln \left (a +b \,{\mathrm e}^{x}\right )-b \,{\mathrm e}^{x}-a}{b}\)	\(28\)
default	\(\frac {\left (a +b \,{\mathrm e}^{x}\right ) \ln \left (a +b \,{\mathrm e}^{x}\right )-b \,{\mathrm e}^{x}-a}{b}\)	\(28\)
norman	\({\mathrm e}^{x} \ln \left (a +b \,{\mathrm e}^{x}\right )+\frac {a \ln \left (a +b \,{\mathrm e}^{x}\right )}{b}-{\mathrm e}^{x}\)	\(28\)
risch	\({\mathrm e}^{x} \ln \left (a +b \,{\mathrm e}^{x}\right )-{\mathrm e}^{x}+\frac {a \ln \left ({\mathrm e}^{x}+\frac {a}{b}\right )}{b}\)	\(30\)
parallelrisch	\(\frac {\ln \left (a +b \,{\mathrm e}^{x}\right ) {\mathrm e}^{x} b -b \,{\mathrm e}^{x}+\ln \left (a +b \,{\mathrm e}^{x}\right ) a +a}{b}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int e^x \log \left (a+b e^x\right ) \, dx=-\frac {b e^{x} - {\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right )}{b} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int e^x \log \left (a+b e^x\right ) \, dx=-\frac {b e^{x} - {\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right ) + a}{b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int e^x \log \left (a+b e^x\right ) \, dx=-\frac {b e^{x} - {\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right ) + a}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.82 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int e^x \log \left (a+b e^x\right ) \, dx={\mathrm {e}}^x\,\ln \left (a+b\,{\mathrm {e}}^x\right )-{\mathrm {e}}^x+\frac {a\,\ln \left (a+b\,{\mathrm {e}}^x\right )}{b} \]

[In]

int(exp(x)*log(a + b*exp(x)),x)

[Out]

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