3.1.21 \(\int \frac {1}{(a+b e^{p x})^2} \, dx\) [21]

Optimal. Leaf size=42 \[ \frac {1}{a \left (a+b e^{p x}\right ) p}+\frac {x}{a^2}-\frac {\log \left (a+b e^{p x}\right )}{a^2 p} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2320, 46} \begin {gather*} -\frac {\log \left (a+b e^{p x}\right )}{a^2 p}+\frac {x}{a^2}+\frac {1}{a p \left (a+b e^{p x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*E^(p*x))^(-2),x]

[Out]

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

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

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Mathematica [A]
time = 0.04, size = 39, normalized size = 0.93 \begin {gather*} \frac {\frac {a}{a+b e^{p x}}+\log \left (e^{p x}\right )-\log \left (a+b e^{p x}\right )}{a^2 p} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*E^(p*x))^(-2),x]

[Out]

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

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Mathics [A]
time = 1.94, size = 57, normalized size = 1.36 \begin {gather*} \frac {a+p x \left (a+b E^{p x}\right )-\text {Log}\left [\frac {a+b E^{p x}}{b}\right ] \left (a+b E^{p x}\right )}{a^2 p \left (a+b E^{p x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(a + b*E^(p*x))^2,x]')

[Out]

(a + p x (a + b E ^ (p x)) - Log[(a + b E ^ (p x)) / b] (a + b E ^ (p x))) / (a ^ 2 p (a + b E ^ (p x)))

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Maple [A]
time = 0.02, size = 43, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (a +b \,{\mathrm e}^{p x}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{p x}\right )}+\frac {\ln \left ({\mathrm e}^{p x}\right )}{a^{2}}}{p}\)	\(43\)
default	\(\frac {-\frac {\ln \left (a +b \,{\mathrm e}^{p x}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{p x}\right )}+\frac {\ln \left ({\mathrm e}^{p x}\right )}{a^{2}}}{p}\)	\(43\)
risch	\(\frac {x}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{p x}\right ) p}-\frac {\ln \left ({\mathrm e}^{p x}+\frac {a}{b}\right )}{a^{2} p}\)	\(43\)
norman	\(\frac {\frac {x}{a}+\frac {b x \,{\mathrm e}^{p x}}{a^{2}}-\frac {b \,{\mathrm e}^{p x}}{a^{2} p}}{a +b \,{\mathrm e}^{p x}}-\frac {\ln \left (a +b \,{\mathrm e}^{p x}\right )}{a^{2} p}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.24, size = 40, normalized size = 0.95 \begin {gather*} \frac {x}{a^{2}} + \frac {1}{{\left (a b e^{\left (p x\right )} + a^{2}\right )} p} - \frac {\log \left (b e^{\left (p x\right )} + a\right )}{a^{2} p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.30, size = 52, normalized size = 1.24 \begin {gather*} \frac {b p x e^{\left (p x\right )} + a p x - {\left (b e^{\left (p x\right )} + a\right )} \log \left (b e^{\left (p x\right )} + a\right ) + a}{a^{2} b p e^{\left (p x\right )} + a^{3} p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.08, size = 36, normalized size = 0.86 \begin {gather*} \frac {1}{a^{2} p + a b p e^{p x}} + \frac {x}{a^{2}} - \frac {\log {\left (\frac {a}{b} + e^{p x} \right )}}{a^{2} p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 45, normalized size = 1.07 \begin {gather*} \frac {\frac {p x}{a^{2}}-\frac {b \ln \left |\mathrm {e}^{p x} b+a\right |}{b a^{2}}+\frac {a}{a^{2} \left (\mathrm {e}^{p x} b+a\right )}}{p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*exp(p*x))^2,x)

[Out]

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

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Mupad [B]
time = 0.41, size = 58, normalized size = 1.38 \begin {gather*} \frac {\frac {x}{a}+\frac {b\,x\,{\mathrm {e}}^{p\,x}}{a^2}-\frac {b\,{\mathrm {e}}^{p\,x}}{a^2\,p}}{a+b\,{\mathrm {e}}^{p\,x}}-\frac {\ln \left (a+b\,{\mathrm {e}}^{p\,x}\right )}{a^2\,p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*exp(p*x))^2,x)

[Out]

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

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