\(\int \frac {e^{-n x}}{(a+b e^{n x})^2} \, dx\) [31]

Optimal result

Integrand size = 18, antiderivative size = 61 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=-\frac {e^{-n x}}{a^2 n}-\frac {b}{a^2 \left (a+b e^{n x}\right ) n}-\frac {2 b x}{a^3}+\frac {2 b \log \left (a+b e^{n x}\right )}{a^3 n} \]

[Out]

-1/a^2/exp(n*x)/n-b/a^2/(a+b*exp(n*x))/n-2*b*x/a^3+2*b*ln(a+b*exp(n*x))/a^3/n

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, 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} \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {2 b \log \left (a+b e^{n x}\right )}{a^3 n}-\frac {2 b x}{a^3}-\frac {b}{a^2 n \left (a+b e^{n x}\right )}-\frac {e^{-n x}}{a^2 n} \]

[In]

Int[1/(E^(n*x)*(a + b*E^(n*x))^2),x]

[Out]

-(1/(a^2*E^(n*x)*n)) - b/(a^2*(a + b*E^(n*x))*n) - (2*b*x)/a^3 + (2*b*Log[a + b*E^(n*x)])/(a^3*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)^2} \, dx,x,e^{n x}\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{a^2 x^2}-\frac {2 b}{a^3 x}+\frac {b^2}{a^2 (a+b x)^2}+\frac {2 b^2}{a^3 (a+b x)}\right ) \, dx,x,e^{n x}\right )}{n} \\ & = -\frac {e^{-n x}}{a^2 n}-\frac {b}{a^2 \left (a+b e^{n x}\right ) n}-\frac {2 b x}{a^3}+\frac {2 b \log \left (a+b e^{n x}\right )}{a^3 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {-\frac {a b+a^2 e^{-n x}-b^2 e^{n x}}{a+b e^{n x}}+2 b \log \left (b+a e^{-n x}\right )}{a^3 n} \]

[In]

Integrate[1/(E^(n*x)*(a + b*E^(n*x))^2),x]

[Out]

(-((a*b + a^2/E^(n*x) - b^2*E^(n*x))/(a + b*E^(n*x))) + 2*b*Log[b + a/E^(n*x)])/(a^3*n)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {b}{a^{2} \left (a +b \,{\mathrm e}^{n x}\right )}+\frac {2 b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{3}}-\frac {{\mathrm e}^{-n x}}{a^{2}}-\frac {2 b \ln \left ({\mathrm e}^{n x}\right )}{a^{3}}}{n}\)	\(59\)
default	\(\frac {-\frac {b}{a^{2} \left (a +b \,{\mathrm e}^{n x}\right )}+\frac {2 b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{3}}-\frac {{\mathrm e}^{-n x}}{a^{2}}-\frac {2 b \ln \left ({\mathrm e}^{n x}\right )}{a^{3}}}{n}\)	\(59\)
risch	\(-\frac {{\mathrm e}^{-n x}}{a^{2} n}-\frac {2 b x}{a^{3}}-\frac {b}{a^{2} \left (a +b \,{\mathrm e}^{n x}\right ) n}+\frac {2 b \ln \left ({\mathrm e}^{n x}+\frac {a}{b}\right )}{a^{3} n}\)	\(62\)
norman	\(\frac {\left (-\frac {2 b \,{\mathrm e}^{n x}}{a^{2} n}-\frac {1}{a n}-\frac {2 b x \,{\mathrm e}^{n x}}{a^{2}}-\frac {2 b^{2} x \,{\mathrm e}^{2 n x}}{a^{3}}\right ) {\mathrm e}^{-n x}}{a +b \,{\mathrm e}^{n x}}+\frac {2 b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{3} n}\)	\(85\)
parallelrisch	\(\frac {\left (-2 x \,{\mathrm e}^{2 n x} b^{2} n +2 \ln \left (a +b \,{\mathrm e}^{n x}\right ) {\mathrm e}^{2 n x} b^{2}-2 x \,{\mathrm e}^{n x} a b n +2 \ln \left (a +b \,{\mathrm e}^{n x}\right ) {\mathrm e}^{n x} a b +2 \,{\mathrm e}^{2 n x} b^{2}-a^{2}\right ) {\mathrm e}^{-n x}}{a^{3} n \left (a +b \,{\mathrm e}^{n x}\right )}\)	\(101\)

[In]

int(1/exp(n*x)/(a+b*exp(n*x))^2,x,method=_RETURNVERBOSE)

[Out]

1/n*(-1/a^2*b/(a+b*exp(n*x))+2/a^3*b*ln(a+b*exp(n*x))-1/a^2/exp(n*x)-2/a^3*b*ln(exp(n*x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=-\frac {2 \, b^{2} n x e^{\left (2 \, n x\right )} + a^{2} + 2 \, {\left (a b n x + a b\right )} e^{\left (n x\right )} - 2 \, {\left (b^{2} e^{\left (2 \, n x\right )} + a b e^{\left (n x\right )}\right )} \log \left (b e^{\left (n x\right )} + a\right )}{a^{3} b n e^{\left (2 \, n x\right )} + a^{4} n e^{\left (n x\right )}} \]

[In]

integrate(1/exp(n*x)/(a+b*exp(n*x))^2,x, algorithm="fricas")

[Out]

-(2*b^2*n*x*e^(2*n*x) + a^2 + 2*(a*b*n*x + a*b)*e^(n*x) - 2*(b^2*e^(2*n*x) + a*b*e^(n*x))*log(b*e^(n*x) + a))/
(a^3*b*n*e^(2*n*x) + a^4*n*e^(n*x))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {b^{2}}{a^{4} n e^{- n x} + a^{3} b n} + \begin {cases} - \frac {e^{- n x}}{a^{2} n} & \text {for}\: a^{2} n \neq 0 \\\frac {x}{a^{2}} & \text {otherwise} \end {cases} + \frac {2 b \log {\left (e^{- n x} + \frac {b}{a} \right )}}{a^{3} n} \]

[In]

integrate(1/exp(n*x)/(a+b*exp(n*x))**2,x)

[Out]

b**2/(a**4*n*exp(-n*x) + a**3*b*n) + Piecewise((-exp(-n*x)/(a**2*n), Ne(a**2*n, 0)), (x/a**2, True)) + 2*b*log
(exp(-n*x) + b/a)/(a**3*n)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {b^{2}}{{\left (a^{4} e^{\left (-n x\right )} + a^{3} b\right )} n} - \frac {e^{\left (-n x\right )}}{a^{2} n} + \frac {2 \, b \log \left (a e^{\left (-n x\right )} + b\right )}{a^{3} n} \]

[In]

integrate(1/exp(n*x)/(a+b*exp(n*x))^2,x, algorithm="maxima")

[Out]

b^2/((a^4*e^(-n*x) + a^3*b)*n) - e^(-n*x)/(a^2*n) + 2*b*log(a*e^(-n*x) + b)/(a^3*n)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=-\frac {\frac {2 \, b n x}{a^{3}} - \frac {2 \, b \log \left ({\left | b e^{\left (n x\right )} + a \right |}\right )}{a^{3}} + \frac {2 \, b e^{\left (n x\right )} + a}{{\left (b e^{\left (2 \, n x\right )} + a e^{\left (n x\right )}\right )} a^{2}}}{n} \]

[In]

integrate(1/exp(n*x)/(a+b*exp(n*x))^2,x, algorithm="giac")

[Out]

-(2*b*n*x/a^3 - 2*b*log(abs(b*e^(n*x) + a))/a^3 + (2*b*e^(n*x) + a)/((b*e^(2*n*x) + a*e^(n*x))*a^2))/n

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {2\,b\,\ln \left (a+b\,{\mathrm {e}}^{n\,x}\right )}{a^3\,n}-\frac {\frac {1}{a\,n}+\frac {2\,b^2\,x\,{\mathrm {e}}^{2\,n\,x}}{a^3}-\frac {2\,b^2\,{\mathrm {e}}^{2\,n\,x}}{a^3\,n}+\frac {2\,b\,x\,{\mathrm {e}}^{n\,x}}{a^2}}{a\,{\mathrm {e}}^{n\,x}+b\,{\mathrm {e}}^{2\,n\,x}} \]

[In]

int(exp(-n*x)/(a + b*exp(n*x))^2,x)

[Out]

(2*b*log(a + b*exp(n*x)))/(a^3*n) - (1/(a*n) + (2*b^2*x*exp(2*n*x))/a^3 - (2*b^2*exp(2*n*x))/(a^3*n) + (2*b*x*
exp(n*x))/a^2)/(a*exp(n*x) + b*exp(2*n*x))