3.537 \(\int \frac{1}{a+b e^{-x}+c e^x} \, dx\)

Optimal. Leaf size=36 \[ -\frac{2 \tanh ^{-1}\left (\frac{a+2 c e^x}{\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}} \]

[Out]

(-2*ArcTanh[(a + 2*c*E^x)/Sqrt[a^2 - 4*b*c]])/Sqrt[a^2 - 4*b*c]

________________________________________________________________________________________

Rubi [A]  time = 0.0568149, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2282, 1386, 618, 206} \[ -\frac{2 \tanh ^{-1}\left (\frac{a+2 c e^x}{\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/E^x + c*E^x)^(-1),x]

[Out]

(-2*ArcTanh[(a + 2*c*E^x)/Sqrt[a^2 - 4*b*c]])/Sqrt[a^2 - 4*b*c]

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 1386

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n_.) + (b_.)*(x_)^(mn_))^(p_.), x_Symbol] :> Int[x^(m - n*p)*(b + a*x^n + c
*x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[mn, -n] && IntegerQ[p] && PosQ[n]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{a+b e^{-x}+c e^x} \, dx &=\operatorname{Subst}\left (\int \frac{1}{x \left (a+\frac{b}{x}+c x\right )} \, dx,x,e^x\right )\\ &=\operatorname{Subst}\left (\int \frac{1}{b+a x+c x^2} \, dx,x,e^x\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{a^2-4 b c-x^2} \, dx,x,a+2 c e^x\right )\right )\\ &=-\frac{2 \tanh ^{-1}\left (\frac{a+2 c e^x}{\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}\\ \end{align*}

Mathematica [A]  time = 0.0235092, size = 36, normalized size = 1. \[ -\frac{2 \tanh ^{-1}\left (\frac{a+2 c e^x}{\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/E^x + c*E^x)^(-1),x]

[Out]

(-2*ArcTanh[(a + 2*c*E^x)/Sqrt[a^2 - 4*b*c]])/Sqrt[a^2 - 4*b*c]

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 36, normalized size = 1. \begin{align*} 2\,{\frac{1}{\sqrt{-{a}^{2}+4\,bc}}\arctan \left ({\frac{a+2\,c{{\rm e}^{x}}}{\sqrt{-{a}^{2}+4\,bc}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/exp(x)+c*exp(x)),x)

[Out]

2/(-a^2+4*b*c)^(1/2)*arctan((a+2*c*exp(x))/(-a^2+4*b*c)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.26063, size = 298, normalized size = 8.28 \begin{align*} \left [\frac{\log \left (\frac{2 \, c^{2} e^{\left (2 \, x\right )} + 2 \, a c e^{x} + a^{2} - 2 \, b c - \sqrt{a^{2} - 4 \, b c}{\left (2 \, c e^{x} + a\right )}}{c e^{\left (2 \, x\right )} + a e^{x} + b}\right )}{\sqrt{a^{2} - 4 \, b c}}, -\frac{2 \, \sqrt{-a^{2} + 4 \, b c} \arctan \left (-\frac{\sqrt{-a^{2} + 4 \, b c}{\left (2 \, c e^{x} + a\right )}}{a^{2} - 4 \, b c}\right )}{a^{2} - 4 \, b c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[log((2*c^2*e^(2*x) + 2*a*c*e^x + a^2 - 2*b*c - sqrt(a^2 - 4*b*c)*(2*c*e^x + a))/(c*e^(2*x) + a*e^x + b))/sqrt
(a^2 - 4*b*c), -2*sqrt(-a^2 + 4*b*c)*arctan(-sqrt(-a^2 + 4*b*c)*(2*c*e^x + a)/(a^2 - 4*b*c))/(a^2 - 4*b*c)]

________________________________________________________________________________________

Sympy [A]  time = 0.229591, size = 36, normalized size = 1. \begin{align*} \operatorname{RootSum}{\left (z^{2} \left (a^{2} - 4 b c\right ) - 1, \left ( i \mapsto i \log{\left (e^{x} + \frac{- i a^{2} + 4 i b c + a}{2 c} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(_z**2*(a**2 - 4*b*c) - 1, Lambda(_i, _i*log(exp(x) + (-_i*a**2 + 4*_i*b*c + a)/(2*c))))

________________________________________________________________________________________

Giac [A]  time = 1.14394, size = 47, normalized size = 1.31 \begin{align*} \frac{2 \, \arctan \left (\frac{2 \, c e^{x} + a}{\sqrt{-a^{2} + 4 \, b c}}\right )}{\sqrt{-a^{2} + 4 \, b c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan((2*c*e^x + a)/sqrt(-a^2 + 4*b*c))/sqrt(-a^2 + 4*b*c)