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

Optimal. Leaf size=61 \[ -\frac{b}{a^2 n \left (a+b e^{n x}\right )}+\frac{2 b \log \left (a+b e^{n x}\right )}{a^3 n}-\frac{2 b x}{a^3}-\frac{e^{-n x}}{a^2 n} \]

[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)

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Rubi [A]  time = 0.0572723, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2248, 44} \[ -\frac{b}{a^2 n \left (a+b e^{n x}\right )}+\frac{2 b \log \left (a+b e^{n x}\right )}{a^3 n}-\frac{2 b x}{a^3}-\frac{e^{-n x}}{a^2 n} \]

Antiderivative was successfully verified.

[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 2248

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]

Rule 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^2} \, dx,x,e^{n x}\right )}{n}\\ &=\frac{\operatorname{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]  time = 0.107794, size = 49, normalized size = 0.8 \[ -\frac{a \left (\frac{b}{a+b e^{n x}}+e^{-n x}\right )-2 b \log \left (a+b e^{n x}\right )+2 b n x}{a^3 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 67, normalized size = 1.1 \begin{align*} -{\frac{1}{{a}^{2}{{\rm e}^{nx}}n}}-2\,{\frac{b\ln \left ({{\rm e}^{nx}} \right ) }{{a}^{3}n}}-{\frac{b}{{a}^{2} \left ( a+b{{\rm e}^{nx}} \right ) n}}+2\,{\frac{b\ln \left ( a+b{{\rm e}^{nx}} \right ) }{{a}^{3}n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.10071, size = 77, normalized size = 1.26 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.52061, size = 198, normalized size = 3.25 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [A]  time = 0.201925, size = 78, normalized size = 1.28 \begin{align*} - \frac{b}{a^{3} n + a^{2} b n e^{n x}} + \begin{cases} - \frac{e^{- n x}}{a^{2} n} & \text{for}\: a^{2} n \neq 0 \\x \left (\frac{2 b}{a^{3}} + \frac{a - 2 b}{a^{3}}\right ) & \text{otherwise} \end{cases} - \frac{2 b x}{a^{3}} + \frac{2 b \log{\left (\frac{a}{b} + e^{n x} \right )}}{a^{3} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28641, size = 80, normalized size = 1.31 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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