\(\int e^x \log (1+e^x) \, dx\) [365]

Optimal result

Integrand size = 10, antiderivative size = 18 \[ \int e^x \log \left (1+e^x\right ) \, dx=-e^x+\left (1+e^x\right ) \log \left (1+e^x\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2225, 2634, 2280, 45} \[ \int e^x \log \left (1+e^x\right ) \, dx=-e^x+e^x \log \left (e^x+1\right )+\log \left (e^x+1\right ) \]

[In]

Int[E^x*Log[1 + E^x],x]

[Out]

-E^x + Log[1 + E^x] + E^x*Log[1 + E^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 (1+e^x\right )-\int \frac {e^{2 x}}{1+e^x} \, dx \\ & = e^x \log \left (1+e^x\right )-\text {Subst}\left (\int \frac {x}{1+x} \, dx,x,e^x\right ) \\ & = e^x \log \left (1+e^x\right )-\text {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,e^x\right ) \\ & = -e^x+\log \left (1+e^x\right )+e^x \log \left (1+e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^x \log \left (1+e^x\right ) \, dx=-e^x+\left (1+e^x\right ) \log \left (1+e^x\right ) \]

[In]

Integrate[E^x*Log[1 + E^x],x]

[Out]

-E^x + (1 + E^x)*Log[1 + E^x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\left (1+{\mathrm e}^{x}\right ) \ln \left (1+{\mathrm e}^{x}\right )-1-{\mathrm e}^{x}\)	\(17\)
default	\(\left (1+{\mathrm e}^{x}\right ) \ln \left (1+{\mathrm e}^{x}\right )-1-{\mathrm e}^{x}\)	\(17\)
norman	\({\mathrm e}^{x} \ln \left (1+{\mathrm e}^{x}\right )-{\mathrm e}^{x}+\ln \left (1+{\mathrm e}^{x}\right )\)	\(19\)
risch	\({\mathrm e}^{x} \ln \left (1+{\mathrm e}^{x}\right )-{\mathrm e}^{x}+\ln \left (1+{\mathrm e}^{x}\right )\)	\(19\)
parallelrisch	\({\mathrm e}^{x} \ln \left (1+{\mathrm e}^{x}\right )-{\mathrm e}^{x}+\ln \left (1+{\mathrm e}^{x}\right )+1\)	\(20\)

[In]

int(exp(x)*ln(1+exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int e^x \log \left (1+e^x\right ) \, dx={\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} \]

[In]

integrate(exp(x)*log(1+exp(x)),x, algorithm="fricas")

[Out]

(e^x + 1)*log(e^x + 1) - e^x

Sympy [F(-1)]

Timed out. \[ \int e^x \log \left (1+e^x\right ) \, dx=\text {Timed out} \]

[In]

integrate(exp(x)*ln(1+exp(x)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^x \log \left (1+e^x\right ) \, dx={\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} - 1 \]

[In]

integrate(exp(x)*log(1+exp(x)),x, algorithm="maxima")

[Out]

(e^x + 1)*log(e^x + 1) - e^x - 1

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^x \log \left (1+e^x\right ) \, dx={\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} - 1 \]

[In]

integrate(exp(x)*log(1+exp(x)),x, algorithm="giac")

[Out]

(e^x + 1)*log(e^x + 1) - e^x - 1

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^x \log \left (1+e^x\right ) \, dx=\ln \left ({\mathrm {e}}^x+1\right )-{\mathrm {e}}^x+{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+1\right ) \]

[In]

int(exp(x)*log(exp(x) + 1),x)

[Out]

log(exp(x) + 1) - exp(x) + exp(x)*log(exp(x) + 1)