### 3.284 $$\int e^x \log (a+b e^x) \, dx$$

Optimal. Leaf size=25 $\frac{\left (a+b e^x\right ) \log \left (a+b e^x\right )}{b}-e^x$

[Out]

-E^x + ((a + b*E^x)*Log[a + b*E^x])/b

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Rubi [A]  time = 0.0525146, antiderivative size = 31, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 5, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.417, Rules used = {2194, 2554, 12, 2248, 43} $e^x \log \left (a+b e^x\right )+\frac{a \log \left (a+b e^x\right )}{b}-e^x$

Antiderivative was successfully veriﬁed.

[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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 (a+b e^x\right ) \, dx &=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 \operatorname{Subst}\left (\int \frac{x}{a+b x} \, dx,x,e^x\right )\\ &=e^x \log \left (a+b e^x\right )-b \operatorname{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]  time = 0.0160492, size = 25, normalized size = 1. $\frac{\left (a+b e^x\right ) \log \left (a+b e^x\right )}{b}-e^x$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^x*Log[a + b*E^x],x]

[Out]

-E^x + ((a + b*E^x)*Log[a + b*E^x])/b

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Maple [A]  time = 0.003, size = 34, normalized size = 1.4 \begin{align*}{{\rm e}^{x}}\ln \left ( a+b{{\rm e}^{x}} \right ) -{{\rm e}^{x}}+{\frac{\ln \left ( a+b{{\rm e}^{x}} \right ) a}{b}}-{\frac{a}{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05244, size = 35, normalized size = 1.4 \begin{align*} -\frac{b e^{x} -{\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right ) + a}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 2.05948, size = 55, normalized size = 2.2 \begin{align*} -\frac{b e^{x} -{\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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

[Out]

Timed out

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Giac [A]  time = 1.19577, size = 35, normalized size = 1.4 \begin{align*} -\frac{b e^{x} -{\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right ) + a}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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