3.22 \(\int \frac{1}{(b e^{-p x}+a e^{p x})^2} \, dx\)

Optimal. Leaf size=22 \[ -\frac{1}{2 a p \left (a e^{2 p x}+b\right )} \]

[Out]

-1/(2*a*(b + a*E^(2*p*x))*p)

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Rubi [A]  time = 0.0235268, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2282, 261} \[ -\frac{1}{2 a p \left (a e^{2 p x}+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[(b/E^(p*x) + a*E^(p*x))^(-2),x]

[Out]

-1/(2*a*(b + a*E^(2*p*x))*p)

Rule 2282

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0202344, size = 22, normalized size = 1. \[ -\frac{1}{2 a p \left (a e^{2 p x}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*a*(b + a*E^(2*p*x))*p)

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Maple [A]  time = 0.002, size = 21, normalized size = 1. \begin{align*} -{\frac{1}{2\,pa \left ( a \left ({{\rm e}^{px}} \right ) ^{2}+b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.935151, size = 27, normalized size = 1.23 \begin{align*} \frac{1}{2 \,{\left (b^{2} e^{\left (-2 \, p x\right )} + a b\right )} p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/((b^2*e^(-2*p*x) + a*b)*p)

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Fricas [A]  time = 1.95579, size = 43, normalized size = 1.95 \begin{align*} -\frac{1}{2 \,{\left (a^{2} p e^{\left (2 \, p x\right )} + a b p\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(a^2*p*e^(2*p*x) + a*b*p)

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Sympy [A]  time = 0.102672, size = 20, normalized size = 0.91 \begin{align*} \frac{1}{2 a b p + 2 b^{2} p e^{- 2 p x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(2*a*b*p + 2*b**2*p*exp(-2*p*x))

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Giac [A]  time = 1.07501, size = 26, normalized size = 1.18 \begin{align*} -\frac{1}{2 \,{\left (a e^{\left (2 \, p x\right )} + b\right )} a p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/((a*e^(2*p*x) + b)*a*p)