3.1.47 \(\int \frac {f^x}{(a+b f^{2 x})^2} \, dx\) [47]

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

[Out]

1/2*f^x/a/(a+b*f^(2*x))/ln(f)+1/2*arctan(f^x*b^(1/2)/a^(1/2))/a^(3/2)/ln(f)/b^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2281, 205, 211} \begin {gather*} \frac {\text {ArcTan}\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 )} \end {gather*}

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 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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*} \int \frac {f^x}{\left (a+b f^{2 x}\right )^2} \, dx &=\frac {\text {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 {\text {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*}

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Mathematica [A]
time = 0.09, size = 53, normalized size = 0.90 \begin {gather*} \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)} \end {gather*}

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 = 0.03, size = 82, normalized size = 1.39

method	result	size
risch	\(\frac {f^{x}}{2 a \left (a +b \,f^{2 x}\right ) \ln \left (f \right )}-\frac {\ln \left (f^{x}-\frac {a}{\sqrt {-b a}}\right )}{4 \sqrt {-b a}\, a \ln \left (f \right )}+\frac {\ln \left (f^{x}+\frac {a}{\sqrt {-b a}}\right )}{4 \sqrt {-b a}\, a \ln \left (f \right )}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/ln(f)/a*f^x/(a+b*(f^x)^2)-1/4/(-b*a)^(1/2)/a/ln(f)*ln(f^x-1/(-b*a)^(1/2)*a)+1/4/(-b*a)^(1/2)/a/ln(f)*ln(f^
x+1/(-b*a)^(1/2)*a)

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Maxima [A]
time = 0.48, size = 49, normalized size = 0.83 \begin {gather*} \frac {f^{x}}{2 \, {\left (a b f^{2 \, x} + a^{2}\right )} \log \left (f\right )} + \frac {\arctan \left (\frac {b f^{x}}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a \log \left (f\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 164, normalized size = 2.78 \begin {gather*} \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 {gather*}

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 = 0.12, size = 53, normalized size = 0.90 \begin {gather*} \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 {gather*}

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 = 2.85, size = 49, normalized size = 0.83 \begin {gather*} \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 {gather*}

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))

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Mupad [B]
time = 3.50, size = 49, normalized size = 0.83 \begin {gather*} \frac {f^x}{2\,a\,\ln \left (f\right )\,\left (a+b\,f^{2\,x}\right )}+\frac {\mathrm {atan}\left (\frac {b\,f^x}{\sqrt {a\,b}}\right )}{2\,a\,\ln \left (f\right )\,\sqrt {a\,b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^x/(a + b*f^(2*x))^2,x)

[Out]

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

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