Integrand size = 18, antiderivative size = 83 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=-\frac {e^{-n x}}{a^3 n}-\frac {b}{2 a^2 \left (a+b e^{n x}\right )^2 n}-\frac {2 b}{a^3 \left (a+b e^{n x}\right ) n}-\frac {3 b x}{a^4}+\frac {3 b \log \left (a+b e^{n x}\right )}{a^4 n} \] Output:
-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
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {\frac {5 b^3-2 a^3 e^{-3 n x}-4 a^2 b e^{-2 n x}+4 a b^2 e^{-n x}}{\left (b+a e^{-n x}\right )^2}+6 b \log \left (b+a e^{-n x}\right )}{2 a^4 n} \] Input:
Integrate[1/(E^(n*x)*(a + b*E^(n*x))^3),x]
Output:
((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)
Time = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2678, 54, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx\) |
\(\Big \downarrow \) 2678 |
\(\displaystyle \frac {\int \frac {e^{-2 n x}}{\left (a+b e^{n x}\right )^3}de^{n x}}{n}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {3 b^2}{a^4 \left (a+b e^{n x}\right )}+\frac {2 b^2}{a^3 \left (a+b e^{n x}\right )^2}+\frac {b^2}{a^2 \left (a+b e^{n x}\right )^3}-\frac {3 e^{-n x} b}{a^4}+\frac {e^{-2 n x}}{a^3}\right )de^{n x}}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {3 b \log \left (e^{n x}\right )}{a^4}+\frac {3 b \log \left (a+b e^{n x}\right )}{a^4}-\frac {2 b}{a^3 \left (a+b e^{n x}\right )}-\frac {e^{-n x}}{a^3}-\frac {b}{2 a^2 \left (a+b e^{n x}\right )^2}}{n}\) |
Input:
Int[1/(E^(n*x)*(a + b*E^(n*x))^3),x]
Output:
(-(1/(a^3*E^(n*x))) - b/(2*a^2*(a + b*E^(n*x))^2) - (2*b)/(a^3*(a + b*E^(n *x))) - (3*b*Log[E^(n*x)])/a^4 + (3*b*Log[a + b*E^(n*x)])/a^4)/n
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_ .) + (g_.)*(x_))), x_Symbol] :> With[{m = FullSimplify[g*h*(Log[G]/(d*e*Log [F]))]}, Simp[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)/De nominator[m]))], x] /; LeQ[m, -1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-\frac {{\mathrm e}^{-x n}}{a^{3} n}-\frac {3 b x}{a^{4}}-\frac {b \left (4 b \,{\mathrm e}^{x n}+5 a \right )}{2 a^{3} n \left (a +b \,{\mathrm e}^{x n}\right )^{2}}+\frac {3 b \ln \left ({\mathrm e}^{x n}+\frac {a}{b}\right )}{a^{4} n}\) | \(73\) |
derivativedivides | \(\frac {-\frac {{\mathrm e}^{-x n}}{a^{3}}-\frac {3 b \ln \left ({\mathrm e}^{x n}\right )}{a^{4}}-\frac {b}{2 a^{2} \left (a +b \,{\mathrm e}^{x n}\right )^{2}}+\frac {3 b \ln \left (a +b \,{\mathrm e}^{x n}\right )}{a^{4}}-\frac {2 b}{a^{3} \left (a +b \,{\mathrm e}^{x n}\right )}}{n}\) | \(75\) |
default | \(\frac {-\frac {{\mathrm e}^{-x n}}{a^{3}}-\frac {3 b \ln \left ({\mathrm e}^{x n}\right )}{a^{4}}-\frac {b}{2 a^{2} \left (a +b \,{\mathrm e}^{x n}\right )^{2}}+\frac {3 b \ln \left (a +b \,{\mathrm e}^{x n}\right )}{a^{4}}-\frac {2 b}{a^{3} \left (a +b \,{\mathrm e}^{x n}\right )}}{n}\) | \(75\) |
norman | \(\frac {\left (-\frac {1}{a n}-\frac {3 b x \,{\mathrm e}^{x n}}{a^{2}}-\frac {6 b^{2} x \,{\mathrm e}^{2 x n}}{a^{3}}-\frac {3 b^{3} x \,{\mathrm e}^{3 x n}}{a^{4}}+\frac {6 b^{2} {\mathrm e}^{2 x n}}{a^{3} n}+\frac {9 b^{3} {\mathrm e}^{3 x n}}{2 a^{4} n}\right ) {\mathrm e}^{-x n}}{\left (a +b \,{\mathrm e}^{x n}\right )^{2}}+\frac {3 b \ln \left (a +b \,{\mathrm e}^{x n}\right )}{a^{4} n}\) | \(121\) |
parallelrisch | \(\frac {\left (-6 x \,{\mathrm e}^{3 x n} b^{3} n +6 \ln \left (a +b \,{\mathrm e}^{x n}\right ) {\mathrm e}^{3 x n} b^{3}-12 x \,{\mathrm e}^{2 x n} a \,b^{2} n +12 \ln \left (a +b \,{\mathrm e}^{x n}\right ) {\mathrm e}^{2 x n} a \,b^{2}-6 x \,{\mathrm e}^{x n} a^{2} b n +9 \,{\mathrm e}^{3 x n} b^{3}+6 \ln \left (a +b \,{\mathrm e}^{x n}\right ) {\mathrm e}^{x n} a^{2} b +12 \,{\mathrm e}^{2 x n} a \,b^{2}-2 a^{3}\right ) {\mathrm e}^{-x n}}{2 a^{4} n \left (a +b \,{\mathrm e}^{x n}\right )^{2}}\) | \(153\) |
Input:
int(1/exp(x*n)/(a+b*exp(x*n))^3,x,method=_RETURNVERBOSE)
Output:
-1/a^3/exp(x*n)/n-3*b*x/a^4-1/2*b*(4*b*exp(x*n)+5*a)/a^3/n/(a+b*exp(x*n))^ 2+3*b/a^4/n*ln(exp(x*n)+1/b*a)
Time = 0.07 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.69 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=-\frac {6 \, b^{3} n x e^{\left (3 \, n x\right )} + 2 \, a^{3} + 6 \, {\left (2 \, a b^{2} n x + a b^{2}\right )} e^{\left (2 \, n x\right )} + 3 \, {\left (2 \, a^{2} b n x + 3 \, a^{2} b\right )} e^{\left (n x\right )} - 6 \, {\left (b^{3} e^{\left (3 \, n x\right )} + 2 \, a b^{2} e^{\left (2 \, n x\right )} + a^{2} b e^{\left (n x\right )}\right )} \log \left (b e^{\left (n x\right )} + a\right )}{2 \, {\left (a^{4} b^{2} n e^{\left (3 \, n x\right )} + 2 \, a^{5} b n e^{\left (2 \, n x\right )} + a^{6} n e^{\left (n x\right )}\right )}} \] Input:
integrate(1/exp(n*x)/(a+b*exp(n*x))^3,x, algorithm="fricas")
Output:
-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))
Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {6 a b^{2} e^{- n x} + 5 b^{3}}{2 a^{6} n e^{- 2 n x} + 4 a^{5} b n e^{- n x} + 2 a^{4} b^{2} n} + \begin {cases} - \frac {e^{- n x}}{a^{3} n} & \text {for}\: a^{3} n \neq 0 \\\frac {x}{a^{3}} & \text {otherwise} \end {cases} + \frac {3 b \log {\left (e^{- n x} + \frac {b}{a} \right )}}{a^{4} n} \] Input:
integrate(1/exp(n*x)/(a+b*exp(n*x))**3,x)
Output:
(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)
Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {6 \, a b^{2} e^{\left (-n x\right )} + 5 \, b^{3}}{2 \, {\left (2 \, a^{5} b e^{\left (-n x\right )} + a^{6} e^{\left (-2 \, n x\right )} + a^{4} b^{2}\right )} n} - \frac {e^{\left (-n x\right )}}{a^{3} n} + \frac {3 \, b \log \left (a e^{\left (-n x\right )} + b\right )}{a^{4} n} \] Input:
integrate(1/exp(n*x)/(a+b*exp(n*x))^3,x, algorithm="maxima")
Output:
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)
Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=-\frac {3 \, b x}{a^{4}} + \frac {3 \, b \log \left ({\left | b e^{\left (n x\right )} + a \right |}\right )}{a^{4} n} - \frac {{\left (6 \, a b^{2} e^{\left (2 \, n x\right )} + 9 \, a^{2} b e^{\left (n x\right )} + 2 \, a^{3}\right )} e^{\left (-n x\right )}}{2 \, {\left (b e^{\left (n x\right )} + a\right )}^{2} a^{4} n} \] Input:
integrate(1/exp(n*x)/(a+b*exp(n*x))^3,x, algorithm="giac")
Output:
-3*b*x/a^4 + 3*b*log(abs(b*e^(n*x) + a))/(a^4*n) - 1/2*(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)
Time = 23.75 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {\frac {6\,b^2\,{\mathrm {e}}^{2\,n\,x}}{a^3\,n}-\frac {1}{a\,n}+\frac {9\,b^3\,{\mathrm {e}}^{3\,n\,x}}{2\,a^4\,n}}{{\mathrm {e}}^{n\,x}\,a^2+2\,{\mathrm {e}}^{2\,n\,x}\,a\,b+{\mathrm {e}}^{3\,n\,x}\,b^2}-\frac {3\,b\,\ln \left ({\mathrm {e}}^{n\,x}\right )}{a^4\,n}+\frac {3\,b\,\ln \left (a+b\,{\mathrm {e}}^{n\,x}\right )}{a^4\,n} \] Input:
int(exp(-n*x)/(a + b*exp(n*x))^3,x)
Output:
((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*exp(2*n*x)) - (3*b*log(exp(n*x)))/(a^4* n) + (3*b*log(a + b*exp(n*x)))/(a^4*n)
Time = 0.15 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.07 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^3} \, dx=\frac {6 e^{3 n x} \mathrm {log}\left (e^{n x} b +a \right ) b^{3}-6 e^{3 n x} b^{3} n x +3 e^{3 n x} b^{3}+12 e^{2 n x} \mathrm {log}\left (e^{n x} b +a \right ) a \,b^{2}-12 e^{2 n x} a \,b^{2} n x +6 e^{n x} \mathrm {log}\left (e^{n x} b +a \right ) a^{2} b -6 e^{n x} a^{2} b n x -6 e^{n x} a^{2} b -2 a^{3}}{2 e^{n x} a^{4} n \left (e^{2 n x} b^{2}+2 e^{n x} a b +a^{2}\right )} \] Input:
int(1/exp(n*x)/(a+b*exp(n*x))^3,x)
Output:
(6*e**(3*n*x)*log(e**(n*x)*b + a)*b**3 - 6*e**(3*n*x)*b**3*n*x + 3*e**(3*n *x)*b**3 + 12*e**(2*n*x)*log(e**(n*x)*b + a)*a*b**2 - 12*e**(2*n*x)*a*b**2 *n*x + 6*e**(n*x)*log(e**(n*x)*b + a)*a**2*b - 6*e**(n*x)*a**2*b*n*x - 6*e **(n*x)*a**2*b - 2*a**3)/(2*e**(n*x)*a**4*n*(e**(2*n*x)*b**2 + 2*e**(n*x)* a*b + a**2))