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

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 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]

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}{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*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \]

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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fricas [A]  time = 0.41, size = 86, normalized size = 2.87 \[ \left [-\frac {\sqrt {-a b} \log \left (\frac {a f^{2 \, x} - 2 \, \sqrt {-a b} f^{x} - b}{a f^{2 \, x} + b}\right )}{2 \, a b \log \relax (f)}, -\frac {\sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{a f^{x}}\right )}{a b \log \relax (f)}\right ] \]

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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giac [A]  time = 0.29, size = 21, normalized size = 0.70 \[ \frac {\arctan \left (\frac {a f^{x}}{\sqrt {a b}}\right )}{\sqrt {a b} \log \relax (f)} \]

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))

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maple [A]  time = 0.01, size = 22, normalized size = 0.73 \[ \frac {\arctan \left (\frac {a \,f^{x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, \ln \relax (f )} \]

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 [A]  time = 0.97, size = 24, normalized size = 0.80 \[ -\frac {\arctan \left (\frac {b}{\sqrt {a b} f^{x}}\right )}{\sqrt {a b} \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.51, size = 21, normalized size = 0.70 \[ \frac {\mathrm {atan}\left (\frac {a\,f^x}{\sqrt {a\,b}}\right )}{\ln \relax (f)\,\sqrt {a\,b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/f^x + a*f^x),x)

[Out]

atan((a*f^x)/(a*b)^(1/2))/(log(f)*(a*b)^(1/2))

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sympy [A]  time = 0.19, size = 24, normalized size = 0.80 \[ \frac {\operatorname {RootSum} {\left (4 z^{2} a b + 1, \left (i \mapsto i \log {\left (2 i b + f^{x} \right )} \right )\right )}}{\log {\relax (f )}} \]

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*b + f**x)))/log(f)

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