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^(d*x+c))/ln(f)

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Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, 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 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]

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]]]

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*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.00 \[ -\frac {1}{d \log (f) \left (f^{c+d x}+1\right )} \]

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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fricas [A]  time = 0.41, size = 20, normalized size = 1.00 \[ -\frac {1}{d f^{d x + c} \log \relax (f) + d \log \relax (f)} \]

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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giac [A]  time = 0.38, size = 22, normalized size = 1.10 \[ -\frac {1}{{\left (f^{d x} f^{c} + 1\right )} d \log \relax (f)} \]

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]

-1/((f^(d*x)*f^c + 1)*d*log(f))

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maple [A]  time = 0.03, size = 25, normalized size = 1.25 \[ \frac {1}{\left ({\mathrm e}^{\left (-d x -c \right ) \ln \relax (f )}+1\right ) d \ln \relax (f )} \]

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 = 0.93, size = 22, normalized size = 1.10 \[ \frac {1}{d {\left (f^{-d x - c} + 1\right )} \log \relax (f)} \]

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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mupad [B]  time = 3.56, size = 20, normalized size = 1.00 \[ -\frac {1}{d\,\ln \relax (f)\,\left (f^{c+d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(d*log(f)*(f^(c + d*x) + 1))

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sympy [A]  time = 0.11, size = 19, normalized size = 0.95 \[ - \frac {1}{d f^{c + d x} \log {\relax (f )} + d \log {\relax (f )}} \]

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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