3.1.12 \(\int \frac {1}{(a+b e^{c+d x})^2} \, dx\) [12]

Optimal. Leaf size=46 \[ \frac {1}{a d \left (a+b e^{c+d x}\right )}+\frac {x}{a^2}-\frac {\log \left (a+b e^{c+d x}\right )}{a^2 d} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2320, 46} \begin {gather*} -\frac {\log \left (a+b e^{c+d x}\right )}{a^2 d}+\frac {x}{a^2}+\frac {1}{a d \left (a+b e^{c+d x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*E^(c + d*x))^(-2),x]

[Out]

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

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 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]]]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 45, normalized size = 0.98 \begin {gather*} \frac {\frac {a}{a+b e^{c+d x}}+\log \left (e^{c+d x}\right )-\log \left (a+b e^{c+d x}\right )}{a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*E^(c + d*x))^(-2),x]

[Out]

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

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Maple [A]
time = 0.01, size = 49, normalized size = 1.07

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a^{2}}}{d}\)	\(49\)
default	\(\frac {-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a^{2}}}{d}\)	\(49\)
risch	\(\frac {x}{a^{2}}+\frac {c}{a^{2} d}+\frac {1}{a d \left (a +b \,{\mathrm e}^{d x +c}\right )}-\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {a}{b}\right )}{a^{2} d}\)	\(55\)
norman	\(\frac {\frac {x}{a}+\frac {b x \,{\mathrm e}^{d x +c}}{a^{2}}-\frac {b \,{\mathrm e}^{d x +c}}{a^{2} d}}{a +b \,{\mathrm e}^{d x +c}}-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2} d}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*exp(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 51, normalized size = 1.11 \begin {gather*} \frac {1}{{\left (a b e^{\left (d x + c\right )} + a^{2}\right )} d} + \frac {d x + c}{a^{2} d} - \frac {\log \left (b e^{\left (d x + c\right )} + a\right )}{a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 60, normalized size = 1.30 \begin {gather*} \frac {b d x e^{\left (d x + c\right )} + a d x - {\left (b e^{\left (d x + c\right )} + a\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) + a}{a^{2} b d e^{\left (d x + c\right )} + a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.06, size = 39, normalized size = 0.85 \begin {gather*} \frac {1}{a^{2} d + a b d e^{c + d x}} + \frac {x}{a^{2}} - \frac {\log {\left (\frac {a}{b} + e^{c + d x} \right )}}{a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.95, size = 51, normalized size = 1.11 \begin {gather*} \frac {b {\left (\frac {\log \left ({\left | -\frac {a}{b e^{\left (d x + c\right )} + a} + 1 \right |}\right )}{a^{2} b} + \frac {1}{{\left (b e^{\left (d x + c\right )} + a\right )} a b}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.56, size = 66, normalized size = 1.43 \begin {gather*} \frac {\frac {x}{a}+\frac {b\,x\,{\mathrm {e}}^{c+d\,x}}{a^2}-\frac {b\,{\mathrm {e}}^{c+d\,x}}{a^2\,d}}{a+b\,{\mathrm {e}}^{c+d\,x}}-\frac {\ln \left (a+b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*exp(c + d*x))^2,x)

[Out]

(x/a + (b*x*exp(c + d*x))/a^2 - (b*exp(c + d*x))/(a^2*d))/(a + b*exp(c + d*x)) - log(a + b*exp(d*x)*exp(c))/(a
^2*d)

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