3.532 \(\int \frac{1}{2+f^{-c-d x}+f^{c+d x}} \, dx\)

Optimal. Leaf size=20 \[ -\frac{1}{d \log (f) \left (f^{c+d x}+1\right )} \]

[Out]

-(1/(d*(1 + f^(c + d*x))*Log[f]))

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Rubi [A]  time = 0.0203003, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2282, 32} \[ -\frac{1}{d \log (f) \left (f^{c+d x}+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[(2 + f^(-c - d*x) + f^(c + d*x))^(-1),x]

[Out]

-(1/(d*(1 + f^(c + d*x))*Log[f]))

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0172876, size = 20, normalized size = 1. \[ -\frac{1}{d \log (f) \left (f^{c+d x}+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + f^(-c - d*x) + f^(c + d*x))^(-1),x]

[Out]

-(1/(d*(1 + f^(c + d*x))*Log[f]))

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Maple [A]  time = 0.011, size = 25, normalized size = 1.3 \begin{align*}{\frac{1}{d\ln \left ( f \right ) \left ({{\rm e}^{ \left ( -dx-c \right ) \ln \left ( f \right ) }}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2+f^(-d*x-c)+f^(d*x+c)),x)

[Out]

1/d/ln(f)/(exp((-d*x-c)*ln(f))+1)

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Maxima [A]  time = 0.953765, size = 30, normalized size = 1.5 \begin{align*} \frac{1}{d{\left (f^{-d x - c} + 1\right )} \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="maxima")

[Out]

1/(d*(f^(-d*x - c) + 1)*log(f))

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Fricas [A]  time = 1.25229, size = 51, normalized size = 2.55 \begin{align*} -\frac{1}{d f^{d x + c} \log \left (f\right ) + d \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="fricas")

[Out]

-1/(d*f^(d*x + c)*log(f) + d*log(f))

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Sympy [A]  time = 0.102194, size = 19, normalized size = 0.95 \begin{align*} - \frac{1}{d f^{c + d x} \log{\left (f \right )} + d \log{\left (f \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2+f**(-d*x-c)+f**(d*x+c)),x)

[Out]

-1/(d*f**(c + d*x)*log(f) + d*log(f))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{f^{d x + c} + f^{-d x - c} + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/(f^(d*x + c) + f^(-d*x - c) + 2), x)