\(\int \frac {f^x}{a+b f^{2 x}} \, dx\) [43]

Optimal result

Integrand size = 15, antiderivative size = 30 \[ \int \frac {f^x}{a+b f^{2 x}} \, dx=\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2281, 211} \[ \int \frac {f^x}{a+b f^{2 x}} \, dx=\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \]

[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 211

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

Rule 2281

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {f^x}{a+b f^{2 x}} \, dx=\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \]

[In]

Integrate[f^x/(a + b*f^(2*x)),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(52\) vs. \(2(22)=44\).

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77

method	result	size
risch	\(-\frac {\ln \left (f^{x}-\frac {a}{\sqrt {-a b}}\right )}{2 \sqrt {-a b}\, \ln \left (f \right )}+\frac {\ln \left (f^{x}+\frac {a}{\sqrt {-a b}}\right )}{2 \sqrt {-a b}\, \ln \left (f \right )}\)	\(53\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.87 \[ \int \frac {f^x}{a+b f^{2 x}} \, dx=\left [-\frac {\sqrt {-a b} \log \left (\frac {b f^{2 \, x} - 2 \, \sqrt {-a b} f^{x} - a}{b f^{2 \, x} + a}\right )}{2 \, a b \log \left (f\right )}, -\frac {\sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b f^{x}}\right )}{a b \log \left (f\right )}\right ] \]

[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))]

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {f^x}{a+b f^{2 x}} \, dx=\frac {\operatorname {RootSum} {\left (4 z^{2} a b + 1, \left ( i \mapsto i \log {\left (2 i a + f^{x} \right )} \right )\right )}}{\log {\left (f \right )}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {f^x}{a+b f^{2 x}} \, dx=\frac {\arctan \left (\frac {b f^{x}}{\sqrt {a b}}\right )}{\sqrt {a b} \log \left (f\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {f^x}{a+b f^{2 x}} \, dx=\frac {\arctan \left (\frac {b f^{x}}{\sqrt {a b}}\right )}{\sqrt {a b} \log \left (f\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {f^x}{a+b f^{2 x}} \, dx=\frac {\mathrm {atan}\left (\frac {b\,f^x}{\sqrt {a\,b}}\right )}{\ln \left (f\right )\,\sqrt {a\,b}} \]

[In]

int(f^x/(a + b*f^(2*x)),x)

[Out]

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