### 3.365 $$\int e^x \log (1+e^x) \, dx$$

Optimal. Leaf size=18 $\left (e^x+1\right ) \log \left (e^x+1\right )-e^x$

[Out]

-E^x + (1 + E^x)*Log[1 + E^x]

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Rubi [A]  time = 0.0295601, antiderivative size = 22, normalized size of antiderivative = 1.22, number of steps used = 4, number of rules used = 4, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.4, Rules used = {2194, 2554, 2248, 43} $-e^x+e^x \log \left (e^x+1\right )+\log \left (e^x+1\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[E^x*Log[1 + E^x],x]

[Out]

-E^x + Log[1 + E^x] + E^x*Log[1 + E^x]

Rule 2194

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 2554

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]

Rule 2248

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 43

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])

Rubi steps

\begin{align*} \int e^x \log \left (1+e^x\right ) \, dx &=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 )-\operatorname{Subst}\left (\int \frac{x}{1+x} \, dx,x,e^x\right )\\ &=e^x \log \left (1+e^x\right )-\operatorname{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]  time = 0.005832, size = 18, normalized size = 1. $\left (e^x+1\right ) \log \left (e^x+1\right )-e^x$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^x*Log[1 + E^x],x]

[Out]

-E^x + (1 + E^x)*Log[1 + E^x]

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Maple [A]  time = 0.002, size = 17, normalized size = 0.9 \begin{align*} \left ( 1+{{\rm e}^{x}} \right ) \ln \left ( 1+{{\rm e}^{x}} \right ) -1-{{\rm e}^{x}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.931435, size = 22, normalized size = 1.22 \begin{align*}{\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} - 1 \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^x + 1)*log(e^x + 1) - e^x - 1

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Fricas [A]  time = 1.92823, size = 41, normalized size = 2.28 \begin{align*}{\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^x + 1)*log(e^x + 1) - e^x

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.04817, size = 22, normalized size = 1.22 \begin{align*}{\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right ) - e^{x} - 1 \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^x + 1)*log(e^x + 1) - e^x - 1