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]

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

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Rubi [A]  time = 0.05, 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 12

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

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 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 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]

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*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.00 \[ \frac {\left (a+b e^x\right ) \log \left (a+b e^x\right )}{b}-e^x \]

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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fricas [A]  time = 0.40, size = 25, normalized size = 1.00 \[ -\frac {b e^{x} - {\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right )}{b} \]

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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giac [A]  time = 0.16, size = 26, normalized size = 1.04 \[ -\frac {b e^{x} - {\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right ) + a}{b} \]

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

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maple [A]  time = 0.06, size = 34, normalized size = 1.36 \[ {\mathrm e}^{x} \ln \left (b \,{\mathrm e}^{x}+a \right )+\frac {a \ln \left (b \,{\mathrm e}^{x}+a \right )}{b}-{\mathrm e}^{x}-\frac {a}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.74, size = 26, normalized size = 1.04 \[ -\frac {b e^{x} - {\left (b e^{x} + a\right )} \log \left (b e^{x} + a\right ) + a}{b} \]

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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mupad [B]  time = 0.66, size = 27, normalized size = 1.08 \[ {\mathrm {e}}^x\,\ln \left (a+b\,{\mathrm {e}}^x\right )-{\mathrm {e}}^x+\frac {a\,\ln \left (a+b\,{\mathrm {e}}^x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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