3.54 \(\int \frac{1}{b f^{-x}+a f^x} \, dx\)

Optimal. Leaf size=30 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

[Out]

ArcTan[(Sqrt[a]*f^x)/Sqrt[b]]/(Sqrt[a]*Sqrt[b]*Log[f])

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Rubi [A]  time = 0.0211419, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2282, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[(b/f^x + a*f^x)^(-1),x]

[Out]

ArcTan[(Sqrt[a]*f^x)/Sqrt[b]]/(Sqrt[a]*Sqrt[b]*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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0077143, size = 30, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b/f^x + a*f^x)^(-1),x]

[Out]

ArcTan[(Sqrt[a]*f^x)/Sqrt[b]]/(Sqrt[a]*Sqrt[b]*Log[f])

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Maple [A]  time = 0.004, size = 22, normalized size = 0.7 \begin{align*}{\frac{1}{\ln \left ( f \right ) }\arctan \left ({a{f}^{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/(f^x)+a*f^x),x)

[Out]

1/ln(f)/(a*b)^(1/2)*arctan(a*f^x/(a*b)^(1/2))

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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/(b/(f^x)+a*f^x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62191, size = 190, normalized size = 6.33 \begin{align*} \left [-\frac{\sqrt{-a b} \log \left (\frac{a f^{2 \, x} - 2 \, \sqrt{-a b} f^{x} - b}{a f^{2 \, x} + b}\right )}{2 \, a b \log \left (f\right )}, -\frac{\sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{a f^{x}}\right )}{a b \log \left (f\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-a*b)*log((a*f^(2*x) - 2*sqrt(-a*b)*f^x - b)/(a*f^(2*x) + b))/(a*b*log(f)), -sqrt(a*b)*arctan(sqrt(
a*b)/(a*f^x))/(a*b*log(f))]

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Sympy [A]  time = 0.263538, size = 26, normalized size = 0.87 \begin{align*} \frac{\operatorname{RootSum}{\left (4 z^{2} a b + 1, \left ( i \mapsto i \log{\left (- 2 i a + f^{- x} \right )} \right )\right )}}{\log{\left (f \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b/(f**x)+a*f**x),x)

[Out]

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

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Giac [A]  time = 1.21153, size = 28, normalized size = 0.93 \begin{align*} \frac{\arctan \left (\frac{a f^{x}}{\sqrt{a b}}\right )}{\sqrt{a b} \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(a*f^x/sqrt(a*b))/(sqrt(a*b)*log(f))