Integrand size = 18, antiderivative size = 61 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=-\frac {e^{-n x}}{a^2 n}-\frac {b}{a^2 \left (a+b e^{n x}\right ) n}-\frac {2 b x}{a^3}+\frac {2 b \log \left (a+b e^{n x}\right )}{a^3 n} \] Output:
-1/a^2/exp(n*x)/n-b/a^2/(a+b*exp(n*x))/n-2*b*x/a^3+2*b*ln(a+b*exp(n*x))/a^ 3/n
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {-\frac {a b+a^2 e^{-n x}-b^2 e^{n x}}{a+b e^{n x}}+2 b \log \left (b+a e^{-n x}\right )}{a^3 n} \] Input:
Integrate[1/(E^(n*x)*(a + b*E^(n*x))^2),x]
Output:
(-((a*b + a^2/E^(n*x) - b^2*E^(n*x))/(a + b*E^(n*x))) + 2*b*Log[b + a/E^(n *x)])/(a^3*n)
Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx\) |
\(\Big \downarrow \) 2678 |
\(\displaystyle \frac {\int \frac {e^{-2 n x}}{\left (a+b e^{n x}\right )^2}de^{n x}}{n}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {2 b^2}{a^3 \left (a+b e^{n x}\right )}+\frac {b^2}{a^2 \left (a+b e^{n x}\right )^2}-\frac {2 e^{-n x} b}{a^3}+\frac {e^{-2 n x}}{a^2}\right )de^{n x}}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 b \log \left (e^{n x}\right )}{a^3}+\frac {2 b \log \left (a+b e^{n x}\right )}{a^3}-\frac {b}{a^2 \left (a+b e^{n x}\right )}-\frac {e^{-n x}}{a^2}}{n}\) |
Input:
Int[1/(E^(n*x)*(a + b*E^(n*x))^2),x]
Output:
(-(1/(a^2*E^(n*x))) - b/(a^2*(a + b*E^(n*x))) - (2*b*Log[E^(n*x)])/a^3 + ( 2*b*Log[a + b*E^(n*x)])/a^3)/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 = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {-\frac {{\mathrm e}^{-x n}}{a^{2}}-\frac {2 b \ln \left ({\mathrm e}^{x n}\right )}{a^{3}}-\frac {b}{a^{2} \left (a +b \,{\mathrm e}^{x n}\right )}+\frac {2 b \ln \left (a +b \,{\mathrm e}^{x n}\right )}{a^{3}}}{n}\) | \(59\) |
default | \(\frac {-\frac {{\mathrm e}^{-x n}}{a^{2}}-\frac {2 b \ln \left ({\mathrm e}^{x n}\right )}{a^{3}}-\frac {b}{a^{2} \left (a +b \,{\mathrm e}^{x n}\right )}+\frac {2 b \ln \left (a +b \,{\mathrm e}^{x n}\right )}{a^{3}}}{n}\) | \(59\) |
risch | \(-\frac {{\mathrm e}^{-x n}}{a^{2} n}-\frac {2 b x}{a^{3}}-\frac {b}{a^{2} \left (a +b \,{\mathrm e}^{x n}\right ) n}+\frac {2 b \ln \left ({\mathrm e}^{x n}+\frac {a}{b}\right )}{a^{3} n}\) | \(62\) |
norman | \(\frac {\left (-\frac {2 b \,{\mathrm e}^{x n}}{a^{2} n}-\frac {1}{a n}-\frac {2 b x \,{\mathrm e}^{x n}}{a^{2}}-\frac {2 b^{2} x \,{\mathrm e}^{2 x n}}{a^{3}}\right ) {\mathrm e}^{-x n}}{a +b \,{\mathrm e}^{x n}}+\frac {2 b \ln \left (a +b \,{\mathrm e}^{x n}\right )}{a^{3} n}\) | \(85\) |
parallelrisch | \(\frac {\left (-2 x \,{\mathrm e}^{2 x n} b^{2} n +2 \ln \left (a +b \,{\mathrm e}^{x n}\right ) {\mathrm e}^{2 x n} b^{2}-2 x \,{\mathrm e}^{x n} a b n +2 \ln \left (a +b \,{\mathrm e}^{x n}\right ) {\mathrm e}^{x n} a b +2 \,{\mathrm e}^{2 x n} b^{2}-a^{2}\right ) {\mathrm e}^{-x n}}{a^{3} n \left (a +b \,{\mathrm e}^{x n}\right )}\) | \(101\) |
Input:
int(1/exp(x*n)/(a+b*exp(x*n))^2,x,method=_RETURNVERBOSE)
Output:
1/n*(-1/a^2/exp(x*n)-2/a^3*b*ln(exp(x*n))-b/a^2/(a+b*exp(x*n))+2/a^3*b*ln( a+b*exp(x*n)))
Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=-\frac {2 \, b^{2} n x e^{\left (2 \, n x\right )} + a^{2} + 2 \, {\left (a b n x + a b\right )} e^{\left (n x\right )} - 2 \, {\left (b^{2} e^{\left (2 \, n x\right )} + a b e^{\left (n x\right )}\right )} \log \left (b e^{\left (n x\right )} + a\right )}{a^{3} b n e^{\left (2 \, n x\right )} + a^{4} n e^{\left (n x\right )}} \] Input:
integrate(1/exp(n*x)/(a+b*exp(n*x))^2,x, algorithm="fricas")
Output:
-(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))
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {b^{2}}{a^{4} n e^{- n x} + a^{3} b n} + \begin {cases} - \frac {e^{- n x}}{a^{2} n} & \text {for}\: a^{2} n \neq 0 \\\frac {x}{a^{2}} & \text {otherwise} \end {cases} + \frac {2 b \log {\left (e^{- n x} + \frac {b}{a} \right )}}{a^{3} n} \] Input:
integrate(1/exp(n*x)/(a+b*exp(n*x))**2,x)
Output:
b**2/(a**4*n*exp(-n*x) + a**3*b*n) + Piecewise((-exp(-n*x)/(a**2*n), Ne(a* *2*n, 0)), (x/a**2, True)) + 2*b*log(exp(-n*x) + b/a)/(a**3*n)
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {b^{2}}{{\left (a^{4} e^{\left (-n x\right )} + a^{3} b\right )} n} - \frac {e^{\left (-n x\right )}}{a^{2} n} + \frac {2 \, b \log \left (a e^{\left (-n x\right )} + b\right )}{a^{3} n} \] Input:
integrate(1/exp(n*x)/(a+b*exp(n*x))^2,x, algorithm="maxima")
Output:
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)
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=-\frac {2 \, b x}{a^{3}} + \frac {2 \, b \log \left ({\left | b e^{\left (n x\right )} + a \right |}\right )}{a^{3} n} - \frac {2 \, b e^{\left (n x\right )} + a}{{\left (b e^{\left (2 \, n x\right )} + a e^{\left (n x\right )}\right )} a^{2} n} \] Input:
integrate(1/exp(n*x)/(a+b*exp(n*x))^2,x, algorithm="giac")
Output:
-2*b*x/a^3 + 2*b*log(abs(b*e^(n*x) + a))/(a^3*n) - (2*b*e^(n*x) + a)/((b*e ^(2*n*x) + a*e^(n*x))*a^2*n)
Time = 24.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {2\,b\,\ln \left (a+b\,{\mathrm {e}}^{n\,x}\right )}{a^3\,n}-\frac {\frac {1}{a\,n}+\frac {2\,b^2\,x\,{\mathrm {e}}^{2\,n\,x}}{a^3}-\frac {2\,b^2\,{\mathrm {e}}^{2\,n\,x}}{a^3\,n}+\frac {2\,b\,x\,{\mathrm {e}}^{n\,x}}{a^2}}{a\,{\mathrm {e}}^{n\,x}+b\,{\mathrm {e}}^{2\,n\,x}} \] Input:
int(exp(-n*x)/(a + b*exp(n*x))^2,x)
Output:
(2*b*log(a + b*exp(n*x)))/(a^3*n) - (1/(a*n) + (2*b^2*x*exp(2*n*x))/a^3 - (2*b^2*exp(2*n*x))/(a^3*n) + (2*b*x*exp(n*x))/a^2)/(a*exp(n*x) + b*exp(2*n *x))
Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.74 \[ \int \frac {e^{-n x}}{\left (a+b e^{n x}\right )^2} \, dx=\frac {2 e^{2 n x} \mathrm {log}\left (e^{n x} b +a \right ) b^{2}-2 e^{2 n x} b^{2} n x +2 e^{2 n x} b^{2}+2 e^{n x} \mathrm {log}\left (e^{n x} b +a \right ) a b -2 e^{n x} a b n x -a^{2}}{e^{n x} a^{3} n \left (e^{n x} b +a \right )} \] Input:
int(1/exp(n*x)/(a+b*exp(n*x))^2,x)
Output:
(2*e**(2*n*x)*log(e**(n*x)*b + a)*b**2 - 2*e**(2*n*x)*b**2*n*x + 2*e**(2*n *x)*b**2 + 2*e**(n*x)*log(e**(n*x)*b + a)*a*b - 2*e**(n*x)*a*b*n*x - a**2) /(e**(n*x)*a**3*n*(e**(n*x)*b + a))