\(\int \frac {e^{-n x}}{(a+b e^{n x})^3} \, dx\) [32]

Optimal result

Integrand size = 18, antiderivative size = 83 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=-\frac {e^{-n x}}{a^3 n}-\frac {b}{2 a^2 \left (a+b e^{n x}\right )^2 n}-\frac {2 b}{a^3 \left (a+b e^{n x}\right ) n}-\frac {3 b x}{a^4}+\frac {3 b \log \left (a+b e^{n x}\right )}{a^4 n} \]

[Out]

-1/a^3/exp(n*x)/n-1/2*b/a^2/(a+b*exp(n*x))^2/n-2*b/a^3/(a+b*exp(n*x))/n-3*b*x/a^4+3*b*ln(a+b*exp(n*x))/a^4/n

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 83, 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} \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {3 b \log \left (a+b e^{n x}\right )}{a^4 n}-\frac {3 b x}{a^4}-\frac {2 b}{a^3 n \left (a+b e^{n x}\right )}-\frac {e^{-n x}}{a^3 n}-\frac {b}{2 a^2 n \left (a+b e^{n x}\right )^2} \]

[In]

Int[1/(E^(n*x)*(a + b*E^(n*x))^3),x]

[Out]

-(1/(a^3*E^(n*x)*n)) - b/(2*a^2*(a + b*E^(n*x))^2*n) - (2*b)/(a^3*(a + b*E^(n*x))*n) - (3*b*x)/a^4 + (3*b*Log[
a + b*E^(n*x)])/(a^4*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)^3} \, dx,x,e^{n x}\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{a^3 x^2}-\frac {3 b}{a^4 x}+\frac {b^2}{a^2 (a+b x)^3}+\frac {2 b^2}{a^3 (a+b x)^2}+\frac {3 b^2}{a^4 (a+b x)}\right ) \, dx,x,e^{n x}\right )}{n} \\ & = -\frac {e^{-n x}}{a^3 n}-\frac {b}{2 a^2 \left (a+b e^{n x}\right )^2 n}-\frac {2 b}{a^3 \left (a+b e^{n x}\right ) n}-\frac {3 b x}{a^4}+\frac {3 b \log \left (a+b e^{n x}\right )}{a^4 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {\frac {5 b^3-2 a^3 e^{-3 n x}-4 a^2 b e^{-2 n x}+4 a b^2 e^{-n x}}{\left (b+a e^{-n x}\right )^2}+6 b \log \left (b+a e^{-n x}\right )}{2 a^4 n} \]

[In]

Integrate[1/(E^(n*x)*(a + b*E^(n*x))^3),x]

[Out]

((5*b^3 - (2*a^3)/E^(3*n*x) - (4*a^2*b)/E^(2*n*x) + (4*a*b^2)/E^(n*x))/(b + a/E^(n*x))^2 + 6*b*Log[b + a/E^(n*
x)])/(2*a^4*n)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {{\mathrm e}^{-n x}}{a^{3} n}-\frac {3 b x}{a^{4}}-\frac {b \left (4 b \,{\mathrm e}^{n x}+5 a \right )}{2 a^{3} n \left (a +b \,{\mathrm e}^{n x}\right )^{2}}+\frac {3 b \ln \left ({\mathrm e}^{n x}+\frac {a}{b}\right )}{a^{4} n}\)	\(73\)
derivativedivides	\(\frac {-\frac {b}{2 a^{2} \left (a +b \,{\mathrm e}^{n x}\right )^{2}}+\frac {3 b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{4}}-\frac {2 b}{a^{3} \left (a +b \,{\mathrm e}^{n x}\right )}-\frac {{\mathrm e}^{-n x}}{a^{3}}-\frac {3 b \ln \left ({\mathrm e}^{n x}\right )}{a^{4}}}{n}\)	\(75\)
default	\(\frac {-\frac {b}{2 a^{2} \left (a +b \,{\mathrm e}^{n x}\right )^{2}}+\frac {3 b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{4}}-\frac {2 b}{a^{3} \left (a +b \,{\mathrm e}^{n x}\right )}-\frac {{\mathrm e}^{-n x}}{a^{3}}-\frac {3 b \ln \left ({\mathrm e}^{n x}\right )}{a^{4}}}{n}\)	\(75\)
norman	\(\frac {\left (-\frac {1}{a n}-\frac {3 b x \,{\mathrm e}^{n x}}{a^{2}}-\frac {6 b^{2} x \,{\mathrm e}^{2 n x}}{a^{3}}-\frac {3 b^{3} x \,{\mathrm e}^{3 n x}}{a^{4}}+\frac {6 b^{2} {\mathrm e}^{2 n x}}{a^{3} n}+\frac {9 b^{3} {\mathrm e}^{3 n x}}{2 a^{4} n}\right ) {\mathrm e}^{-n x}}{\left (a +b \,{\mathrm e}^{n x}\right )^{2}}+\frac {3 b \ln \left (a +b \,{\mathrm e}^{n x}\right )}{a^{4} n}\)	\(121\)
parallelrisch	\(\frac {\left (-6 x \,{\mathrm e}^{3 n x} b^{3} n +6 \ln \left (a +b \,{\mathrm e}^{n x}\right ) {\mathrm e}^{3 n x} b^{3}-12 x \,{\mathrm e}^{2 n x} a \,b^{2} n +12 \ln \left (a +b \,{\mathrm e}^{n x}\right ) {\mathrm e}^{2 n x} a \,b^{2}-6 x \,{\mathrm e}^{n x} a^{2} b n +9 \,{\mathrm e}^{3 n x} b^{3}+6 \ln \left (a +b \,{\mathrm e}^{n x}\right ) {\mathrm e}^{n x} a^{2} b +12 \,{\mathrm e}^{2 n x} a \,b^{2}-2 a^{3}\right ) {\mathrm e}^{-n x}}{2 a^{4} n \left (a +b \,{\mathrm e}^{n x}\right )^{2}}\)	\(153\)

[In]

int(1/exp(n*x)/(a+b*exp(n*x))^3,x,method=_RETURNVERBOSE)

[Out]

-1/a^3/exp(n*x)/n-3*b*x/a^4-1/2*b*(4*b*exp(n*x)+5*a)/a^3/n/(a+b*exp(n*x))^2+3*b/a^4/n*ln(exp(n*x)+a/b)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.69 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=-\frac {6 \, b^{3} n x e^{\left (3 \, n x\right )} + 2 \, a^{3} + 6 \, {\left (2 \, a b^{2} n x + a b^{2}\right )} e^{\left (2 \, n x\right )} + 3 \, {\left (2 \, a^{2} b n x + 3 \, a^{2} b\right )} e^{\left (n x\right )} - 6 \, {\left (b^{3} e^{\left (3 \, n x\right )} + 2 \, a b^{2} e^{\left (2 \, n x\right )} + a^{2} b e^{\left (n x\right )}\right )} \log \left (b e^{\left (n x\right )} + a\right )}{2 \, {\left (a^{4} b^{2} n e^{\left (3 \, n x\right )} + 2 \, a^{5} b n e^{\left (2 \, n x\right )} + a^{6} n e^{\left (n x\right )}\right )}} \]

[In]

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

[Out]

-1/2*(6*b^3*n*x*e^(3*n*x) + 2*a^3 + 6*(2*a*b^2*n*x + a*b^2)*e^(2*n*x) + 3*(2*a^2*b*n*x + 3*a^2*b)*e^(n*x) - 6*
(b^3*e^(3*n*x) + 2*a*b^2*e^(2*n*x) + a^2*b*e^(n*x))*log(b*e^(n*x) + a))/(a^4*b^2*n*e^(3*n*x) + 2*a^5*b*n*e^(2*
n*x) + a^6*n*e^(n*x))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {6 a b^{2} e^{- n x} + 5 b^{3}}{2 a^{6} n e^{- 2 n x} + 4 a^{5} b n e^{- n x} + 2 a^{4} b^{2} n} + \begin {cases} - \frac {e^{- n x}}{a^{3} n} & \text {for}\: a^{3} n \neq 0 \\\frac {x}{a^{3}} & \text {otherwise} \end {cases} + \frac {3 b \log {\left (e^{- n x} + \frac {b}{a} \right )}}{a^{4} n} \]

[In]

integrate(1/exp(n*x)/(a+b*exp(n*x))**3,x)

[Out]

(6*a*b**2*exp(-n*x) + 5*b**3)/(2*a**6*n*exp(-2*n*x) + 4*a**5*b*n*exp(-n*x) + 2*a**4*b**2*n) + Piecewise((-exp(
-n*x)/(a**3*n), Ne(a**3*n, 0)), (x/a**3, True)) + 3*b*log(exp(-n*x) + b/a)/(a**4*n)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {6 \, a b^{2} e^{\left (-n x\right )} + 5 \, b^{3}}{2 \, {\left (2 \, a^{5} b e^{\left (-n x\right )} + a^{6} e^{\left (-2 \, n x\right )} + a^{4} b^{2}\right )} n} - \frac {e^{\left (-n x\right )}}{a^{3} n} + \frac {3 \, b \log \left (a e^{\left (-n x\right )} + b\right )}{a^{4} n} \]

[In]

integrate(1/exp(n*x)/(a+b*exp(n*x))^3,x, algorithm="maxima")

[Out]

1/2*(6*a*b^2*e^(-n*x) + 5*b^3)/((2*a^5*b*e^(-n*x) + a^6*e^(-2*n*x) + a^4*b^2)*n) - e^(-n*x)/(a^3*n) + 3*b*log(
a*e^(-n*x) + b)/(a^4*n)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=-\frac {\frac {6 \, b n x}{a^{4}} - \frac {6 \, b \log \left ({\left | b e^{\left (n x\right )} + a \right |}\right )}{a^{4}} + \frac {{\left (6 \, a b^{2} e^{\left (2 \, n x\right )} + 9 \, a^{2} b e^{\left (n x\right )} + 2 \, a^{3}\right )} e^{\left (-n x\right )}}{{\left (b e^{\left (n x\right )} + a\right )}^{2} a^{4}}}{2 \, n} \]

[In]

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

[Out]

-1/2*(6*b*n*x/a^4 - 6*b*log(abs(b*e^(n*x) + a))/a^4 + (6*a*b^2*e^(2*n*x) + 9*a^2*b*e^(n*x) + 2*a^3)*e^(-n*x)/(
(b*e^(n*x) + a)^2*a^4))/n

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {\frac {6\,b^2\,{\mathrm {e}}^{2\,n\,x}}{a^3\,n}-\frac {1}{a\,n}+\frac {9\,b^3\,{\mathrm {e}}^{3\,n\,x}}{2\,a^4\,n}}{{\mathrm {e}}^{n\,x}\,a^2+2\,{\mathrm {e}}^{2\,n\,x}\,a\,b+{\mathrm {e}}^{3\,n\,x}\,b^2}-\frac {3\,b\,\ln \left ({\mathrm {e}}^{n\,x}\right )}{a^4\,n}+\frac {3\,b\,\ln \left (a+b\,{\mathrm {e}}^{n\,x}\right )}{a^4\,n} \]

[In]

int(exp(-n*x)/(a + b*exp(n*x))^3,x)

[Out]

((6*b^2*exp(2*n*x))/(a^3*n) - 1/(a*n) + (9*b^3*exp(3*n*x))/(2*a^4*n))/(a^2*exp(n*x) + b^2*exp(3*n*x) + 2*a*b*e
xp(2*n*x)) - (3*b*log(exp(n*x)))/(a^4*n) + (3*b*log(a + b*exp(n*x)))/(a^4*n)