3.6.22 \(\int \frac {1}{b+a e^{n x}} \, dx\) [522]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, 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 = {2320, 36, 29, 31} \begin {gather*} \frac {x}{b}-\frac {\log \left (a e^{n x}+b\right )}{b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*E^(n*x))^(-1),x]

[Out]

x/b - Log[b + a*E^(n*x)]/(b*n)

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]

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 2320

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]]]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 38, normalized size = 1.58 \begin {gather*} \frac {\log \left (e^{n x}\right )}{b n}-\frac {\log \left (b^2 n+a b e^{n x} n\right )}{b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*E^(n*x))^(-1),x]

[Out]

Log[E^(n*x)]/(b*n) - Log[b^2*n + a*b*E^(n*x)*n]/(b*n)

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Maple [A]
time = 0.02, size = 29, normalized size = 1.21

method	result	size
norman	\(\frac {x}{b}-\frac {\ln \left (b +a \,{\mathrm e}^{n x}\right )}{b n}\)	\(24\)
risch	\(\frac {x}{b}-\frac {\ln \left ({\mathrm e}^{n x}+\frac {b}{a}\right )}{b n}\)	\(26\)
derivativedivides	\(\frac {-\frac {\ln \left (b +a \,{\mathrm e}^{n x}\right )}{b}+\frac {\ln \left ({\mathrm e}^{n x}\right )}{b}}{n}\)	\(29\)
default	\(\frac {-\frac {\ln \left (b +a \,{\mathrm e}^{n x}\right )}{b}+\frac {\ln \left ({\mathrm e}^{n x}\right )}{b}}{n}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b+a*exp(n*x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 3.52, size = 23, normalized size = 0.96 \begin {gather*} \frac {x}{b} - \frac {\log \left (a e^{\left (n x\right )} + b\right )}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b - log(a*e^(n*x) + b)/(b*n)

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Fricas [A]
time = 0.47, size = 22, normalized size = 0.92 \begin {gather*} \frac {n x - \log \left (a e^{\left (n x\right )} + b\right )}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(n*x - log(a*e^(n*x) + b))/(b*n)

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Sympy [A]
time = 0.04, size = 15, normalized size = 0.62 \begin {gather*} \frac {x}{b} - \frac {\log {\left (e^{n x} + \frac {b}{a} \right )}}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b - log(exp(n*x) + b/a)/(b*n)

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Giac [A]
time = 1.39, size = 26, normalized size = 1.08 \begin {gather*} \frac {\frac {n x}{b} - \frac {\log \left ({\left | a e^{\left (n x\right )} + b \right |}\right )}{b}}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(n*x/b - log(abs(a*e^(n*x) + b))/b)/n

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Mupad [B]
time = 0.33, size = 22, normalized size = 0.92 \begin {gather*} -\frac {\ln \left (b+a\,{\mathrm {e}}^{n\,x}\right )-n\,x}{b\,n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b + a*exp(n*x)),x)

[Out]

-(log(b + a*exp(n*x)) - n*x)/(b*n)

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