3.154 \(\int \frac{1}{a+b e^{m x}} \, dx\)

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

[Out]

x/a - Log[a + b*E^(m*x)]/(a*m)

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Rubi [A]  time = 0.0154089, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {2282, 36, 29, 31} \[ \frac{x}{a}-\frac{\log \left (a+b e^{m x}\right )}{a m} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*E^(m*x))^(-1),x]

[Out]

x/a - Log[a + b*E^(m*x)]/(a*m)

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rubi steps

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

Mathematica [A]  time = 0.0054045, size = 24, normalized size = 1. \[ \frac{x}{a}-\frac{\log \left (a+b e^{m x}\right )}{a m} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*E^(m*x))^(-1),x]

[Out]

x/a - Log[a + b*E^(m*x)]/(a*m)

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Maple [A]  time = 0.004, size = 31, normalized size = 1.3 \begin{align*}{\frac{\ln \left ({{\rm e}^{mx}} \right ) }{ma}}-{\frac{\ln \left ( a+b{{\rm e}^{mx}} \right ) }{ma}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*exp(m*x)),x)

[Out]

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

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Maxima [A]  time = 0.928376, size = 31, normalized size = 1.29 \begin{align*} \frac{x}{a} - \frac{\log \left (b e^{\left (m x\right )} + a\right )}{a m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(m*x)),x, algorithm="maxima")

[Out]

x/a - log(b*e^(m*x) + a)/(a*m)

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Fricas [A]  time = 1.52804, size = 46, normalized size = 1.92 \begin{align*} \frac{m x - \log \left (b e^{\left (m x\right )} + a\right )}{a m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(m*x)),x, algorithm="fricas")

[Out]

(m*x - log(b*e^(m*x) + a))/(a*m)

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Sympy [A]  time = 0.168962, size = 15, normalized size = 0.62 \begin{align*} \frac{x}{a} - \frac{\log{\left (\frac{a}{b} + e^{m x} \right )}}{a m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(m*x)),x)

[Out]

x/a - log(a/b + exp(m*x))/(a*m)

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Giac [A]  time = 1.07444, size = 32, normalized size = 1.33 \begin{align*} \frac{x}{a} - \frac{\log \left ({\left | b e^{\left (m x\right )} + a \right |}\right )}{a m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(m*x)),x, algorithm="giac")

[Out]

x/a - log(abs(b*e^(m*x) + a))/(a*m)