3.3 \(\int \frac {e^{d x}}{a+b e^{c+d x}} \, dx\)

Optimal. Leaf size=24 \[ \frac {e^{-c} \log \left (a+b e^{c+d x}\right )}{b d} \]

[Out]

ln(a+b*exp(d*x+c))/b/d/exp(c)

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Rubi [A]  time = 0.07, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2247, 2246, 31} \[ \frac {e^{-c} \log \left (a+b e^{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Int[E^(d*x)/(a + b*E^(c + d*x)),x]

[Out]

Log[a + b*E^(c + d*x)]/(b*d*E^c)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2246

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, x]

Rule 2247

Int[((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.)*((G_)^((h_.)*((f_.) + (g_.)*(x_))))^(m_.),
x_Symbol] :> Dist[(G^(h*(f + g*x)))^m/(F^(e*(c + d*x)))^n, Int[(F^(e*(c + d*x)))^n*(a + b*(F^(e*(c + d*x)))^n)
^p, x], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, m, n, p}, x] && EqQ[d*e*n*Log[F], g*h*m*Log[G]]

Rubi steps

\begin {align*} \int \frac {e^{d x}}{a+b e^{c+d x}} \, dx &=e^{-c} \int \frac {e^{c+d x}}{a+b e^{c+d x}} \, dx\\ &=\frac {e^{-c} \operatorname {Subst}\left (\int \frac {1}{a+b x} \, dx,x,e^{c+d x}\right )}{d}\\ &=\frac {e^{-c} \log \left (a+b e^{c+d x}\right )}{b d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \[ \frac {e^{-c} \log \left (a+b e^{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(d*x)/(a + b*E^(c + d*x)),x]

[Out]

Log[a + b*E^(c + d*x)]/(b*d*E^c)

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fricas [A]  time = 0.41, size = 22, normalized size = 0.92 \[ \frac {e^{\left (-c\right )} \log \left (b e^{\left (d x + c\right )} + a\right )}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x)/(a+b*exp(d*x+c)),x, algorithm="fricas")

[Out]

e^(-c)*log(b*e^(d*x + c) + a)/(b*d)

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giac [A]  time = 0.26, size = 23, normalized size = 0.96 \[ \frac {e^{\left (-c\right )} \log \left ({\left | b e^{\left (d x + c\right )} + a \right |}\right )}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x)/(a+b*exp(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 23, normalized size = 0.96 \[ \frac {{\mathrm e}^{-c} \ln \left (b \,{\mathrm e}^{c} {\mathrm e}^{d x}+a \right )}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(d*x)/(b*exp(d*x+c)+a),x)

[Out]

1/d*ln(b*exp(d*x)*exp(c)+a)/b/exp(c)

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maxima [A]  time = 0.44, size = 22, normalized size = 0.92 \[ \frac {e^{\left (-c\right )} \log \left (b e^{\left (d x + c\right )} + a\right )}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x)/(a+b*exp(d*x+c)),x, algorithm="maxima")

[Out]

e^(-c)*log(b*e^(d*x + c) + a)/(b*d)

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mupad [B]  time = 0.12, size = 22, normalized size = 0.92 \[ \frac {\ln \left (a+b\,{\mathrm {e}}^{c+d\,x}\right )\,{\mathrm {e}}^{-c}}{b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(d*x)/(a + b*exp(c + d*x)),x)

[Out]

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

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sympy [A]  time = 0.16, size = 19, normalized size = 0.79 \[ \frac {e^{- c} \log {\left (\frac {a e^{- c}}{b} + e^{d x} \right )}}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x)/(a+b*exp(d*x+c)),x)

[Out]

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

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