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

Optimal. Leaf size=59 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} \log (f)}+\frac{f^x}{2 a \log (f) \left (a+b f^{2 x}\right )} \]

[Out]

f^x/(2*a*(a + b*f^(2*x))*Log[f]) + ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]*Log[f])

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Rubi [A]  time = 0.0394861, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2249, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} \log (f)}+\frac{f^x}{2 a \log (f) \left (a+b f^{2 x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[f^x/(a + b*f^(2*x))^2,x]

[Out]

f^x/(2*a*(a + b*f^(2*x))*Log[f]) + ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]/(2*a^(3/2)*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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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}{\left (a+b f^{2 x}\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^2} \, dx,x,f^x\right )}{\log (f)}\\ &=\frac{f^x}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,f^x\right )}{2 a \log (f)}\\ &=\frac{f^x}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0501102, size = 53, normalized size = 0.9 \[ \frac{\frac{f^x}{a^2+a b f^{2 x}}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{a^{3/2} \sqrt{b}}}{2 \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^x/(a + b*f^(2*x))^2,x]

[Out]

(f^x/(a^2 + a*b*f^(2*x)) + ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]/(a^(3/2)*Sqrt[b]))/(2*Log[f])

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Maple [A]  time = 0.031, size = 82, normalized size = 1.4 \begin{align*}{\frac{{f}^{x}}{2\,\ln \left ( f \right ) a \left ( a+b \left ({f}^{x} \right ) ^{2} \right ) }}-{\frac{1}{4\,\ln \left ( f \right ) a}\ln \left ({f}^{x}-{a{\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{1}{4\,\ln \left ( f \right ) a}\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))^2,x)

[Out]

1/2/ln(f)/a*f^x/(a+b*(f^x)^2)-1/4/(-a*b)^(1/2)/a/ln(f)*ln(f^x-1/(-a*b)^(1/2)*a)+1/4/(-a*b)^(1/2)/a/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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53019, size = 373, normalized size = 6.32 \begin{align*} \left [\frac{2 \, a b f^{x} -{\left (\sqrt{-a b} b f^{2 \, x} + \sqrt{-a b} a\right )} \log \left (\frac{b f^{2 \, x} - 2 \, \sqrt{-a b} f^{x} - a}{b f^{2 \, x} + a}\right )}{4 \,{\left (a^{2} b^{2} f^{2 \, x} \log \left (f\right ) + a^{3} b \log \left (f\right )\right )}}, \frac{a b f^{x} -{\left (\sqrt{a b} b f^{2 \, x} + \sqrt{a b} a\right )} \arctan \left (\frac{\sqrt{a b}}{b f^{x}}\right )}{2 \,{\left (a^{2} b^{2} f^{2 \, x} \log \left (f\right ) + a^{3} b \log \left (f\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*a*b*f^x - (sqrt(-a*b)*b*f^(2*x) + sqrt(-a*b)*a)*log((b*f^(2*x) - 2*sqrt(-a*b)*f^x - a)/(b*f^(2*x) + a)
))/(a^2*b^2*f^(2*x)*log(f) + a^3*b*log(f)), 1/2*(a*b*f^x - (sqrt(a*b)*b*f^(2*x) + sqrt(a*b)*a)*arctan(sqrt(a*b
)/(b*f^x)))/(a^2*b^2*f^(2*x)*log(f) + a^3*b*log(f))]

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Sympy [A]  time = 0.378443, size = 53, normalized size = 0.9 \begin{align*} \frac{f^{x}}{2 a^{2} \log{\left (f \right )} + 2 a b f^{2 x} \log{\left (f \right )}} + \frac{\operatorname{RootSum}{\left (16 z^{2} a^{3} b + 1, \left ( i \mapsto i \log{\left (4 i a^{2} + 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))**2,x)

[Out]

f**x/(2*a**2*log(f) + 2*a*b*f**(2*x)*log(f)) + RootSum(16*_z**2*a**3*b + 1, Lambda(_i, _i*log(4*_i*a**2 + f**x
)))/log(f)

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Giac [A]  time = 1.17298, size = 66, normalized size = 1.12 \begin{align*} \frac{\arctan \left (\frac{b f^{x}}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a \log \left (f\right )} + \frac{f^{x}}{2 \,{\left (b f^{2 \, x} + a\right )} a \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))^2,x, algorithm="giac")

[Out]

1/2*arctan(b*f^x/sqrt(a*b))/(sqrt(a*b)*a*log(f)) + 1/2*f^x/((b*f^(2*x) + a)*a*log(f))