\(\int \frac {x}{(a+b e^{c+d x})^2} \, dx\) [11]

Optimal result

Integrand size = 15, antiderivative size = 107 \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=-\frac {x}{a^2 d}+\frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}+\frac {\log \left (a+b e^{c+d x}\right )}{a^2 d^2}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}-\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {\frac {d x \left (a d x+b e^{c+d x} (-2+d x)\right )}{a+b e^{c+d x}}-2 (-1+d x) \log \left (1+\frac {b e^{c+d x}}{a}\right )-2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{2 a^2 d^2} \] Input:

Integrate[x/(a + b*E^(c + d*x))^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.25, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2616, 2615, 2620, 2621, 2715, 2720, 47, 14, 16, 2838}

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.

\(\displaystyle \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx\)

\(\Big \downarrow \) 2616

\(\displaystyle \frac {\int \frac {x}{a+b e^{c+d x}}dx}{a}-\frac {b \int \frac {e^{c+d x} x}{\left (a+b e^{c+d x}\right )^2}dx}{a}\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \int \frac {e^{c+d x} x}{a+b e^{c+d x}}dx}{a}}{a}-\frac {b \int \frac {e^{c+d x} x}{\left (a+b e^{c+d x}\right )^2}dx}{a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {\int \log \left (\frac {e^{c+d x} b}{a}+1\right )dx}{b d}\right )}{a}}{a}-\frac {b \int \frac {e^{c+d x} x}{\left (a+b e^{c+d x}\right )^2}dx}{a}\)

\(\Big \downarrow \) 2621

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {\int \log \left (\frac {e^{c+d x} b}{a}+1\right )dx}{b d}\right )}{a}}{a}-\frac {b \left (\frac {\int \frac {1}{a+b e^{c+d x}}dx}{b d}-\frac {x}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {\int e^{-c-d x} \log \left (\frac {e^{c+d x} b}{a}+1\right )de^{c+d x}}{b d^2}\right )}{a}}{a}-\frac {b \left (\frac {\int \frac {1}{a+b e^{c+d x}}dx}{b d}-\frac {x}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {\int e^{-c-d x} \log \left (\frac {e^{c+d x} b}{a}+1\right )de^{c+d x}}{b d^2}\right )}{a}}{a}-\frac {b \left (\frac {\int \frac {e^{-c-d x}}{a+b e^{c+d x}}de^{c+d x}}{b d^2}-\frac {x}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 47

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {\int e^{-c-d x} \log \left (\frac {e^{c+d x} b}{a}+1\right )de^{c+d x}}{b d^2}\right )}{a}}{a}-\frac {b \left (\frac {\frac {\int e^{-c-d x}de^{c+d x}}{a}-\frac {b \int \frac {1}{a+b e^{c+d x}}de^{c+d x}}{a}}{b d^2}-\frac {x}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {\int e^{-c-d x} \log \left (\frac {e^{c+d x} b}{a}+1\right )de^{c+d x}}{b d^2}\right )}{a}}{a}-\frac {b \left (\frac {\frac {\log \left (e^{c+d x}\right )}{a}-\frac {b \int \frac {1}{a+b e^{c+d x}}de^{c+d x}}{a}}{b d^2}-\frac {x}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {\int e^{-c-d x} \log \left (\frac {e^{c+d x} b}{a}+1\right )de^{c+d x}}{b d^2}\right )}{a}}{a}-\frac {b \left (\frac {\frac {\log \left (e^{c+d x}\right )}{a}-\frac {\log \left (a+b e^{c+d x}\right )}{a}}{b d^2}-\frac {x}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {x^2}{2 a}-\frac {b \left (\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{b d^2}+\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}\right )}{a}}{a}-\frac {b \left (\frac {\frac {\log \left (e^{c+d x}\right )}{a}-\frac {\log \left (a+b e^{c+d x}\right )}{a}}{b d^2}-\frac {x}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

Input:

Int[x/(a + b*E^(c + d*x))^2,x]
 

Output:

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

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 47

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c 
 - a*d)   Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d)   Int[1/(c + d*x), x 
], x] /; FreeQ[{a, b, c, d}, x]
 

rule 2615

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x 
_))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ 
b/a   Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] 
, x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2616

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + 
(d_.)*(x_))^(m_.), x_Symbol] :> Simp[1/a   Int[(c + d*x)^m*(a + b*(F^(g*(e 
+ f*x)))^n)^(p + 1), x], x] - Simp[b/a   Int[(c + d*x)^m*(F^(g*(e + f*x)))^ 
n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n 
}, x] && ILtQ[p, 0] && IGtQ[m, 0]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2621

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*( 
(e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> 
 Simp[(c + d*x)^m*((a + b*(F^(g*(e + f*x)))^n)^(p + 1)/(b*f*g*n*(p + 1)*Log 
[F])), x] - Simp[d*(m/(b*f*g*n*(p + 1)*Log[F]))   Int[(c + d*x)^(m - 1)*(a 
+ b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, 
m, n, p}, x] && NeQ[p, -1]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.44

method	result	size
derivativedivides	\(\frac {\frac {\left (d x +c \right )^{2}}{2 a^{2}}+\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}-\frac {b \left (d x +c \right ) {\mathrm e}^{d x +c}}{a^{2} \left (a +b \,{\mathrm e}^{d x +c}\right )}-\frac {\operatorname {dilog}\left (\frac {a +b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2}}-\frac {\left (d x +c \right ) \ln \left (\frac {a +b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2}}-c \left (\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{d x +c}\right )}-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}\right )}{d^{2}}\)	\(154\)
default	\(\frac {\frac {\left (d x +c \right )^{2}}{2 a^{2}}+\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}-\frac {b \left (d x +c \right ) {\mathrm e}^{d x +c}}{a^{2} \left (a +b \,{\mathrm e}^{d x +c}\right )}-\frac {\operatorname {dilog}\left (\frac {a +b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2}}-\frac {\left (d x +c \right ) \ln \left (\frac {a +b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2}}-c \left (\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{d x +c}\right )}-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}\right )}{d^{2}}\)	\(154\)
risch	\(\frac {x}{a d \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {x^{2}}{2 a^{2}}+\frac {c x}{a^{2} d}+\frac {c^{2}}{2 a^{2} d^{2}}-\frac {x \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d}-\frac {\ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right ) c}{a^{2} d^{2}}-\frac {\operatorname {polylog}\left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d^{2}}-\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a^{2} d^{2}}+\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2} d^{2}}-\frac {c \ln \left ({\mathrm e}^{d x +c}\right )}{a^{2} d^{2}}+\frac {c \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2} d^{2}}\)	\(186\)

Input:

int(x/(a+b*exp(d*x+c))^2,x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.64 \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {a d^{2} x^{2} - a c^{2} - 2 \, a c - 2 \, {\left (b e^{\left (d x + c\right )} + a\right )} {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )} + a}{a} + 1\right ) + {\left (b d^{2} x^{2} - b c^{2} - 2 \, b d x - 2 \, b c\right )} e^{\left (d x + c\right )} + 2 \, {\left (a c + {\left (b c + b\right )} e^{\left (d x + c\right )} + a\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 2 \, {\left (a d x + a c + {\left (b d x + b c\right )} e^{\left (d x + c\right )}\right )} \log \left (\frac {b e^{\left (d x + c\right )} + a}{a}\right )}{2 \, {\left (a^{2} b d^{2} e^{\left (d x + c\right )} + a^{3} d^{2}\right )}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {x}{a^{2} d + a b d e^{c + d x}} + \frac {\int \frac {d x}{a + b e^{c} e^{d x}}\, dx + \int \left (- \frac {1}{a + b e^{c} e^{d x}}\right )\, dx}{a d} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {x}{a b d e^{\left (d x + c\right )} + a^{2} d} + \frac {x^{2}}{2 \, a^{2}} - \frac {x}{a^{2} d} - \frac {d x \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right )}{a^{2} d^{2}} + \frac {\log \left (b e^{\left (d x + c\right )} + a\right )}{a^{2} d^{2}} \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\int { \frac {x}{{\left (b e^{\left (d x + c\right )} + a\right )}^{2}} \,d x } \] Input:

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

Output:

integrate(x/(b*e^(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\int \frac {x}{{\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\int \frac {x}{e^{2 d x +2 c} b^{2}+2 e^{d x +c} a b +a^{2}}d x \] Input:

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

Output:

int(x/(e**(2*c + 2*d*x)*b**2 + 2*e**(c + d*x)*a*b + a**2),x)