3.12 \(\int \frac{1}{(a+b e^{c+d x})^2} \, dx\)

Optimal. Leaf size=46 \[ -\frac{\log \left (a+b e^{c+d x}\right )}{a^2 d}+\frac{x}{a^2}+\frac{1}{a d \left (a+b e^{c+d x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.0328909, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2282, 44} \[ -\frac{\log \left (a+b e^{c+d x}\right )}{a^2 d}+\frac{x}{a^2}+\frac{1}{a d \left (a+b e^{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*E^(c + d*x))^(-2),x]

[Out]

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

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 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{1}{\left (a+b e^{c+d x}\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^2} \, dx,x,e^{c+d x}\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a^2 x}-\frac{b}{a (a+b x)^2}-\frac{b}{a^2 (a+b x)}\right ) \, dx,x,e^{c+d x}\right )}{d}\\ &=\frac{1}{a d \left (a+b e^{c+d x}\right )}+\frac{x}{a^2}-\frac{\log \left (a+b e^{c+d x}\right )}{a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.039304, size = 40, normalized size = 0.87 \[ \frac{\frac{a}{a+b e^{c+d x}}-\log \left (a+b e^{c+d x}\right )+d x}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*E^(c + d*x))^(-2),x]

[Out]

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

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Maple [A]  time = 0.001, size = 54, normalized size = 1.2 \begin{align*}{\frac{\ln \left ({{\rm e}^{dx+c}} \right ) }{{a}^{2}d}}-{\frac{\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{a}^{2}d}}+{\frac{1}{ad \left ( a+b{{\rm e}^{dx+c}} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05732, size = 69, normalized size = 1.5 \begin{align*} \frac{1}{{\left (a b e^{\left (d x + c\right )} + a^{2}\right )} d} + \frac{d x + c}{a^{2} d} - \frac{\log \left (b e^{\left (d x + c\right )} + a\right )}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/((a*b*e^(d*x + c) + a^2)*d) + (d*x + c)/(a^2*d) - log(b*e^(d*x + c) + a)/(a^2*d)

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Fricas [A]  time = 1.48526, size = 146, normalized size = 3.17 \begin{align*} \frac{b d x e^{\left (d x + c\right )} + a d x -{\left (b e^{\left (d x + c\right )} + a\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) + a}{a^{2} b d e^{\left (d x + c\right )} + a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*d*x*e^(d*x + c) + a*d*x - (b*e^(d*x + c) + a)*log(b*e^(d*x + c) + a) + a)/(a^2*b*d*e^(d*x + c) + a^3*d)

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Sympy [A]  time = 0.200763, size = 39, normalized size = 0.85 \begin{align*} \frac{1}{a^{2} d + a b d e^{c + d x}} + \frac{x}{a^{2}} - \frac{\log{\left (\frac{a}{b} + e^{c + d x} \right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a**2*d + a*b*d*exp(c + d*x)) + x/a**2 - log(a/b + exp(c + d*x))/(a**2*d)

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Giac [A]  time = 1.23734, size = 70, normalized size = 1.52 \begin{align*} \frac{d x + c}{a^{2} d} - \frac{\log \left ({\left | b e^{\left (d x + c\right )} + a \right |}\right )}{a^{2} d} + \frac{1}{{\left (b e^{\left (d x + c\right )} + a\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x + c)/(a^2*d) - log(abs(b*e^(d*x + c) + a))/(a^2*d) + 1/((b*e^(d*x + c) + a)*a*d)