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

Optimal. Leaf size=107 \[ -\frac {x}{a^2 d}+\frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}+\frac {\log \left (a+b e^{c+d x}\right )}{a^2 d^2}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}-\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a^2 d^2} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2216, 2215, 2221, 2317, 2438, 2222, 2320, 36, 29, 31} \begin {gather*} -\frac {\text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac {\log \left (a+b e^{c+d x}\right )}{a^2 d^2}-\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{a^2 d}-\frac {x}{a^2 d}+\frac {x^2}{2 a^2}+\frac {x}{a d \left (a+b e^{c+d x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*E^(c + d*x))^2,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2215

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] - Dist[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 2216

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[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 2221

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]
- Dist[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 2222

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] - Dist[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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2320

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 2438

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]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx &=\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}\\ &=\frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}-\frac {b \int \frac {e^{c+d x} x}{a+b e^{c+d x}} \, dx}{a^2}-\frac {\int \frac {1}{a+b e^{c+d x}} \, dx}{a d}\\ &=\frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{x (a+b x)} \, dx,x,e^{c+d x}\right )}{a d^2}+\frac {\int \log \left (1+\frac {b e^{c+d x}}{a}\right ) \, dx}{a^2 d}\\ &=\frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,e^{c+d x}\right )}{a^2 d^2}+\frac {\text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a}\right )}{x} \, dx,x,e^{c+d x}\right )}{a^2 d^2}+\frac {b \text {Subst}\left (\int \frac {1}{a+b x} \, dx,x,e^{c+d x}\right )}{a^2 d^2}\\ &=-\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 {\text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a^2 d^2}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 85, normalized size = 0.79 \begin {gather*} \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 \text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{2 a^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*E^(c + d*x))^2,x]

[Out]

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

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Maple [A]
time = 0.02, size = 154, normalized size = 1.44

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 95, normalized size = 0.89 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Fricas [A]
time = 0.44, size = 176, normalized size = 1.64 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*e^(d*x + c) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*exp(c + d*x))^2,x)

[Out]

int(x/(a + b*exp(c + d*x))^2, x)

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