\(\int \frac {e^{-n x}}{a+b e^{n x}} \, dx\) [30]

Optimal result

Integrand size = 18, antiderivative size = 40 \[ \int \frac {e^{-n x}}{a+b e^{n x}} \, dx=-\frac {e^{-n x}}{a n}-\frac {b x}{a^2}+\frac {b \log \left (a+b e^{n x}\right )}{a^2 n} \]

[Out]

-1/a/exp(n*x)/n-b*x/a^2+b*ln(a+b*exp(n*x))/a^2/n

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2280, 46} \[ \int \frac {e^{-n x}}{a+b e^{n x}} \, dx=\frac {b \log \left (a+b e^{n x}\right )}{a^2 n}-\frac {b x}{a^2}-\frac {e^{-n x}}{a n} \]

[In]

Int[1/(E^(n*x)*(a + b*E^(n*x))),x]

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2280

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^2 (a+b x)} \, dx,x,e^{n x}\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{a x^2}-\frac {b}{a^2 x}+\frac {b^2}{a^2 (a+b x)}\right ) \, dx,x,e^{n x}\right )}{n} \\ & = -\frac {e^{-n x}}{a n}-\frac {b x}{a^2}+\frac {b \log \left (a+b e^{n x}\right )}{a^2 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-n x}}{a+b e^{n x}} \, dx=-\frac {e^{-n x}}{a n}+\frac {b \log \left (a b n+a^2 e^{-n x} n\right )}{a^2 n} \]

[In]

Integrate[1/(E^(n*x)*(a + b*E^(n*x))),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\frac {b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{2}}-\frac {{\mathrm e}^{-n x}}{a}-\frac {b \ln \left ({\mathrm e}^{n x}\right )}{a^{2}}}{n}\)	\(42\)
default	\(\frac {\frac {b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{2}}-\frac {{\mathrm e}^{-n x}}{a}-\frac {b \ln \left ({\mathrm e}^{n x}\right )}{a^{2}}}{n}\)	\(42\)
risch	\(-\frac {{\mathrm e}^{-n x}}{a n}-\frac {b x}{a^{2}}+\frac {b \ln \left ({\mathrm e}^{n x}+\frac {a}{b}\right )}{a^{2} n}\)	\(42\)
parallelrisch	\(\frac {\left (-x \,{\mathrm e}^{n x} b n +b \ln \left (a +b \,{\mathrm e}^{n x}\right ) {\mathrm e}^{n x}-a \right ) {\mathrm e}^{-n x}}{a^{2} n}\)	\(42\)
norman	\(\left (-\frac {1}{a n}-\frac {b x \,{\mathrm e}^{n x}}{a^{2}}\right ) {\mathrm e}^{-n x}+\frac {b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{2} n}\)	\(46\)

[In]

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

[Out]

1/n*(1/a^2*b*ln(a+b*exp(n*x))-1/a/exp(n*x)-1/a^2*b*ln(exp(n*x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.98 \[ \int \frac {e^{-n x}}{a+b e^{n x}} \, dx=-\frac {{\left (b n x e^{\left (n x\right )} - b e^{\left (n x\right )} \log \left (b e^{\left (n x\right )} + a\right ) + a\right )} e^{\left (-n x\right )}}{a^{2} n} \]

[In]

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

[Out]

-(b*n*x*e^(n*x) - b*e^(n*x)*log(b*e^(n*x) + a) + a)*e^(-n*x)/(a^2*n)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-n x}}{a+b e^{n x}} \, dx=\begin {cases} - \frac {e^{- n x}}{a n} & \text {for}\: a n \neq 0 \\\frac {x}{a} & \text {otherwise} \end {cases} + \frac {b \log {\left (e^{- n x} + \frac {b}{a} \right )}}{a^{2} n} \]

[In]

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

[Out]

Piecewise((-exp(-n*x)/(a*n), Ne(a*n, 0)), (x/a, True)) + b*log(exp(-n*x) + b/a)/(a**2*n)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-n x}}{a+b e^{n x}} \, dx=-\frac {e^{\left (-n x\right )}}{a n} + \frac {b \log \left (a e^{\left (-n x\right )} + b\right )}{a^{2} n} \]

[In]

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

[Out]

-e^(-n*x)/(a*n) + b*log(a*e^(-n*x) + b)/(a^2*n)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-n x}}{a+b e^{n x}} \, dx=-\frac {\frac {b n x}{a^{2}} + \frac {e^{\left (-n x\right )}}{a} - \frac {b \log \left ({\left | b e^{\left (n x\right )} + a \right |}\right )}{a^{2}}}{n} \]

[In]

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

[Out]

-(b*n*x/a^2 + e^(-n*x)/a - b*log(abs(b*e^(n*x) + a))/a^2)/n

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-n x}}{a+b e^{n x}} \, dx=\frac {b\,\ln \left (a+b\,{\mathrm {e}}^{n\,x}\right )}{a^2\,n}-\frac {b\,x}{a^2}-\frac {{\mathrm {e}}^{-n\,x}}{a\,n} \]

[In]

int(exp(-n*x)/(a + b*exp(n*x)),x)

[Out]

(b*log(a + b*exp(n*x)))/(a^2*n) - (b*x)/a^2 - exp(-n*x)/(a*n)