3.20 \(\int \frac{1}{(a+b e^{c+d x})^3} \, dx\)

Optimal. Leaf size=69 \[ \frac{1}{a^2 d \left (a+b e^{c+d x}\right )}-\frac{\log \left (a+b e^{c+d x}\right )}{a^3 d}+\frac{x}{a^3}+\frac{1}{2 a d \left (a+b e^{c+d x}\right )^2} \]

[Out]

1/(2*a*d*(a + b*E^(c + d*x))^2) + 1/(a^2*d*(a + b*E^(c + d*x))) + x/a^3 - Log[a + b*E^(c + d*x)]/(a^3*d)

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Rubi [A]  time = 0.0418567, antiderivative size = 69, 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} \[ \frac{1}{a^2 d \left (a+b e^{c+d x}\right )}-\frac{\log \left (a+b e^{c+d x}\right )}{a^3 d}+\frac{x}{a^3}+\frac{1}{2 a d \left (a+b e^{c+d x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*E^(c + d*x))^(-3),x]

[Out]

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

Mathematica [A]  time = 0.059463, size = 62, normalized size = 0.9 \[ \frac{\frac{a^2}{\left (a+b e^{c+d x}\right )^2}+\frac{2 a}{a+b e^{c+d x}}-2 \log \left (a+b e^{c+d x}\right )+2 d x}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*E^(c + d*x))^(-3),x]

[Out]

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

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Maple [A]  time = 0.003, size = 74, normalized size = 1.1 \begin{align*}{\frac{\ln \left ({{\rm e}^{dx+c}} \right ) }{d{a}^{3}}}-{\frac{\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{d{a}^{3}}}+{\frac{1}{{a}^{2}d \left ( a+b{{\rm e}^{dx+c}} \right ) }}+{\frac{1}{2\,ad \left ( a+b{{\rm e}^{dx+c}} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*exp(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.05004, size = 113, normalized size = 1.64 \begin{align*} \frac{2 \, b e^{\left (d x + c\right )} + 3 \, a}{2 \,{\left (a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} b e^{\left (d x + c\right )} + a^{4}\right )} d} + \frac{d x + c}{a^{3} d} - \frac{\log \left (b e^{\left (d x + c\right )} + a\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))^3,x, algorithm="maxima")

[Out]

1/2*(2*b*e^(d*x + c) + 3*a)/((a^2*b^2*e^(2*d*x + 2*c) + 2*a^3*b*e^(d*x + c) + a^4)*d) + (d*x + c)/(a^3*d) - lo
g(b*e^(d*x + c) + a)/(a^3*d)

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Fricas [A]  time = 1.52389, size = 300, normalized size = 4.35 \begin{align*} \frac{2 \, b^{2} d x e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} d x + 3 \, a^{2} + 2 \,{\left (2 \, a b d x + a b\right )} e^{\left (d x + c\right )} - 2 \,{\left (b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} + a^{2}\right )} \log \left (b e^{\left (d x + c\right )} + a\right )}{2 \,{\left (a^{3} b^{2} d e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{4} b d e^{\left (d x + c\right )} + a^{5} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(2*b^2*d*x*e^(2*d*x + 2*c) + 2*a^2*d*x + 3*a^2 + 2*(2*a*b*d*x + a*b)*e^(d*x + c) - 2*(b^2*e^(2*d*x + 2*c)
+ 2*a*b*e^(d*x + c) + a^2)*log(b*e^(d*x + c) + a))/(a^3*b^2*d*e^(2*d*x + 2*c) + 2*a^4*b*d*e^(d*x + c) + a^5*d)

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Sympy [A]  time = 0.335967, size = 76, normalized size = 1.1 \begin{align*} \frac{3 a + 2 b e^{c + d x}}{2 a^{4} d + 4 a^{3} b d e^{c + d x} + 2 a^{2} b^{2} d e^{2 c + 2 d x}} + \frac{x}{a^{3}} - \frac{\log{\left (\frac{a}{b} + e^{c + d x} \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))**3,x)

[Out]

(3*a + 2*b*exp(c + d*x))/(2*a**4*d + 4*a**3*b*d*exp(c + d*x) + 2*a**2*b**2*d*exp(2*c + 2*d*x)) + x/a**3 - log(
a/b + exp(c + d*x))/(a**3*d)

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Giac [A]  time = 1.21957, size = 93, normalized size = 1.35 \begin{align*} \frac{d x + c}{a^{3} d} - \frac{\log \left ({\left | b e^{\left (d x + c\right )} + a \right |}\right )}{a^{3} d} + \frac{2 \, a b e^{\left (d x + c\right )} + 3 \, a^{2}}{2 \,{\left (b e^{\left (d x + c\right )} + a\right )}^{2} a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(d*x+c))^3,x, algorithm="giac")

[Out]

(d*x + c)/(a^3*d) - log(abs(b*e^(d*x + c) + a))/(a^3*d) + 1/2*(2*a*b*e^(d*x + c) + 3*a^2)/((b*e^(d*x + c) + a)
^2*a^3*d)