3.43 \(\int \frac{f^x}{a+b f^{2 x}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.029278, 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 = {2249, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^x/(a + b*f^(2*x)),x]

[Out]

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

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, 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{f^x}{a+b f^{2 x}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,f^x\right )}{\log (f)}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0083495, size = 30, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^x/(a + b*f^(2*x)),x]

[Out]

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

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Maple [B]  time = 0.028, size = 53, normalized size = 1.8 \begin{align*} -{\frac{1}{2\,\ln \left ( f \right ) }\ln \left ({f}^{x}-{a{\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{1}{2\,\ln \left ( f \right ) }\ln \left ({f}^{x}+{a{\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^x/(a+b*f^(2*x)),x)

[Out]

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

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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(f^x/(a+b*f^(2*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56626, size = 190, normalized size = 6.33 \begin{align*} \left [-\frac{\sqrt{-a b} \log \left (\frac{b f^{2 \, x} - 2 \, \sqrt{-a b} f^{x} - a}{b f^{2 \, x} + a}\right )}{2 \, a b \log \left (f\right )}, -\frac{\sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b f^{x}}\right )}{a b \log \left (f\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^x/(a+b*f^(2*x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.465495, size = 24, normalized size = 0.8 \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(f**x/(a+b*f**(2*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 = 1.24852, size = 28, normalized size = 0.93 \begin{align*} \frac{\arctan \left (\frac{b 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(f^x/(a+b*f^(2*x)),x, algorithm="giac")

[Out]

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