\(\int \frac {1}{(b e^{-p x}+a e^{p x})^2} \, dx\) [22]

Optimal result

Integrand size = 18, antiderivative size = 22 \[ \int \frac {1}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx=-\frac {1}{2 a \left (b+a e^{2 p x}\right ) p} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2320, 267} \[ \int \frac {1}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx=-\frac {1}{2 a p \left (a e^{2 p x}+b\right )} \]

[In]

Int[(b/E^(p*x) + a*E^(p*x))^(-2),x]

[Out]

-1/2*1/(a*(b + a*E^(2*p*x))*p)

Rule 267

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]

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*} \text {integral}& = \frac {\text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx=-\frac {1}{2 a \left (b+a e^{2 p x}\right ) p} \]

[In]

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

[Out]

-1/2*1/(a*(b + a*E^(2*p*x))*p)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {1}{2 a \left (b +a \,{\mathrm e}^{2 p x}\right ) p}\)	\(20\)
derivativedivides	\(-\frac {1}{2 a \left (b +a \,{\mathrm e}^{2 p x}\right ) p}\)	\(21\)
default	\(-\frac {1}{2 a \left (b +a \,{\mathrm e}^{2 p x}\right ) p}\)	\(21\)
norman	\(-\frac {1}{2 a \left (b +a \,{\mathrm e}^{2 p x}\right ) p}\)	\(21\)
parallelrisch	\(-\frac {1}{2 a \left (b +a \,{\mathrm e}^{2 p x}\right ) p}\)	\(21\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx=-\frac {1}{2 \, {\left (a^{2} p e^{\left (2 \, p x\right )} + a b p\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx=\frac {1}{2 a b p + 2 b^{2} p e^{- 2 p x}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx=\frac {1}{2 \, {\left (b^{2} e^{\left (-2 \, p x\right )} + a b\right )} p} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx=-\frac {1}{2 \, {\left (a e^{\left (2 \, p x\right )} + b\right )} a p} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx=\frac {{\mathrm {e}}^{2\,p\,x}}{2\,b\,p\,\left (b+a\,{\mathrm {e}}^{2\,p\,x}\right )} \]

[In]

int(1/(a*exp(p*x) + b*exp(-p*x))^2,x)

[Out]

exp(2*p*x)/(2*b*p*(b + a*exp(2*p*x)))