\(\int \frac {f^x}{(a+b f^{2 x})^3} \, dx\) [51]

Optimal result

Integrand size = 15, antiderivative size = 84 \[ \int \frac {f^x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {f^x}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac {3 f^x}{8 a^2 \left (a+b f^{2 x}\right ) \log (f)}+\frac {3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)} \]

[Out]

1/4*f^x/a/(a+b*f^(2*x))^2/ln(f)+3/8*f^x/a^2/(a+b*f^(2*x))/ln(f)+3/8*arctan(f^x*b^(1/2)/a^(1/2))/a^(5/2)/ln(f)/
b^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2281, 205, 211} \[ \int \frac {f^x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)}+\frac {3 f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}+\frac {f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2} \]

[In]

Int[f^x/(a + b*f^(2*x))^3,x]

[Out]

f^x/(4*a*(a + b*f^(2*x))^2*Log[f]) + (3*f^x)/(8*a^2*(a + b*f^(2*x))*Log[f]) + (3*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]
)/(8*a^(5/2)*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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^3} \, dx,x,f^x\right )}{\log (f)} \\ & = \frac {f^x}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac {3 \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^2} \, dx,x,f^x\right )}{4 a \log (f)} \\ & = \frac {f^x}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac {3 f^x}{8 a^2 \left (a+b f^{2 x}\right ) \log (f)}+\frac {3 \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,f^x\right )}{8 a^2 \log (f)} \\ & = \frac {f^x}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac {3 f^x}{8 a^2 \left (a+b f^{2 x}\right ) \log (f)}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {f^x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {\frac {5 a f^x+3 b f^{3 x}}{8 a^2 \left (a+b f^{2 x}\right )^2}+\frac {3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b}}}{\log (f)} \]

[In]

Integrate[f^x/(a + b*f^(2*x))^3,x]

[Out]

((5*a*f^x + 3*b*f^(3*x))/(8*a^2*(a + b*f^(2*x))^2) + (3*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(8*a^(5/2)*Sqrt[b]))/Lo
g[f]

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {f^{x} \left (3 b \,f^{2 x}+5 a \right )}{8 \ln \left (f \right ) a^{2} \left (a +b \,f^{2 x}\right )^{2}}-\frac {3 \ln \left (f^{x}-\frac {a}{\sqrt {-a b}}\right )}{16 \sqrt {-a b}\, a^{2} \ln \left (f \right )}+\frac {3 \ln \left (f^{x}+\frac {a}{\sqrt {-a b}}\right )}{16 \sqrt {-a b}\, a^{2} \ln \left (f \right )}\)	\(94\)

[In]

int(f^x/(a+b*f^(2*x))^3,x,method=_RETURNVERBOSE)

[Out]

1/8*f^x*(3*b*(f^x)^2+5*a)/ln(f)/a^2/(a+b*(f^x)^2)^2-3/16/(-a*b)^(1/2)/a^2/ln(f)*ln(f^x-1/(-a*b)^(1/2)*a)+3/16/
(-a*b)^(1/2)/a^2/ln(f)*ln(f^x+1/(-a*b)^(1/2)*a)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.07 \[ \int \frac {f^x}{\left (a+b f^{2 x}\right )^3} \, dx=\left [\frac {6 \, a b^{2} f^{3 \, x} + 10 \, a^{2} b f^{x} - 3 \, {\left (\sqrt {-a b} b^{2} f^{4 \, x} + 2 \, \sqrt {-a b} a b f^{2 \, x} + \sqrt {-a b} a^{2}\right )} \log \left (\frac {b f^{2 \, x} - 2 \, \sqrt {-a b} f^{x} - a}{b f^{2 \, x} + a}\right )}{16 \, {\left (a^{3} b^{3} f^{4 \, x} \log \left (f\right ) + 2 \, a^{4} b^{2} f^{2 \, x} \log \left (f\right ) + a^{5} b \log \left (f\right )\right )}}, \frac {3 \, a b^{2} f^{3 \, x} + 5 \, a^{2} b f^{x} - 3 \, {\left (\sqrt {a b} b^{2} f^{4 \, x} + 2 \, \sqrt {a b} a b f^{2 \, x} + \sqrt {a b} a^{2}\right )} \arctan \left (\frac {\sqrt {a b}}{b f^{x}}\right )}{8 \, {\left (a^{3} b^{3} f^{4 \, x} \log \left (f\right ) + 2 \, a^{4} b^{2} f^{2 \, x} \log \left (f\right ) + a^{5} b \log \left (f\right )\right )}}\right ] \]

[In]

integrate(f^x/(a+b*f^(2*x))^3,x, algorithm="fricas")

[Out]

[1/16*(6*a*b^2*f^(3*x) + 10*a^2*b*f^x - 3*(sqrt(-a*b)*b^2*f^(4*x) + 2*sqrt(-a*b)*a*b*f^(2*x) + sqrt(-a*b)*a^2)
*log((b*f^(2*x) - 2*sqrt(-a*b)*f^x - a)/(b*f^(2*x) + a)))/(a^3*b^3*f^(4*x)*log(f) + 2*a^4*b^2*f^(2*x)*log(f) +
a^5*b*log(f)), 1/8*(3*a*b^2*f^(3*x) + 5*a^2*b*f^x - 3*(sqrt(a*b)*b^2*f^(4*x) + 2*sqrt(a*b)*a*b*f^(2*x) + sqrt
(a*b)*a^2)*arctan(sqrt(a*b)/(b*f^x)))/(a^3*b^3*f^(4*x)*log(f) + 2*a^4*b^2*f^(2*x)*log(f) + a^5*b*log(f))]

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01 \[ \int \frac {f^x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {5 a f^{x} + 3 b f^{3 x}}{8 a^{4} \log {\left (f \right )} + 16 a^{3} b f^{2 x} \log {\left (f \right )} + 8 a^{2} b^{2} f^{4 x} \log {\left (f \right )}} + \frac {\operatorname {RootSum} {\left (256 z^{2} a^{5} b + 9, \left ( i \mapsto i \log {\left (\frac {16 i a^{3}}{3} + f^{x} \right )} \right )\right )}}{\log {\left (f \right )}} \]

[In]

integrate(f**x/(a+b*f**(2*x))**3,x)

[Out]

(5*a*f**x + 3*b*f**(3*x))/(8*a**4*log(f) + 16*a**3*b*f**(2*x)*log(f) + 8*a**2*b**2*f**(4*x)*log(f)) + RootSum(
256*_z**2*a**5*b + 9, Lambda(_i, _i*log(16*_i*a**3/3 + f**x)))/log(f)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90 \[ \int \frac {f^x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {3 \, b f^{3 \, x} + 5 \, a f^{x}}{8 \, {\left (a^{2} b^{2} f^{4 \, x} + 2 \, a^{3} b f^{2 \, x} + a^{4}\right )} \log \left (f\right )} + \frac {3 \, \arctan \left (\frac {b f^{x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} \log \left (f\right )} \]

[In]

integrate(f^x/(a+b*f^(2*x))^3,x, algorithm="maxima")

[Out]

1/8*(3*b*f^(3*x) + 5*a*f^x)/((a^2*b^2*f^(4*x) + 2*a^3*b*f^(2*x) + a^4)*log(f)) + 3/8*arctan(b*f^x/sqrt(a*b))/(
sqrt(a*b)*a^2*log(f))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {f^x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {3 \, \arctan \left (\frac {b f^{x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} \log \left (f\right )} + \frac {3 \, b f^{3 \, x} + 5 \, a f^{x}}{8 \, {\left (b f^{2 \, x} + a\right )}^{2} a^{2} \log \left (f\right )} \]

[In]

integrate(f^x/(a+b*f^(2*x))^3,x, algorithm="giac")

[Out]

3/8*arctan(b*f^x/sqrt(a*b))/(sqrt(a*b)*a^2*log(f)) + 1/8*(3*b*f^(3*x) + 5*a*f^x)/((b*f^(2*x) + a)^2*a^2*log(f)
)

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \frac {f^x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {\frac {5\,f^x}{8\,a\,\ln \left (f\right )}+\frac {3\,b\,f^{3\,x}}{8\,a^2\,\ln \left (f\right )}}{b^2\,f^{4\,x}+a^2+2\,a\,b\,f^{2\,x}}+\frac {3\,\mathrm {atan}\left (\frac {b\,f^x}{\sqrt {a\,b}}\right )}{8\,a^2\,\ln \left (f\right )\,\sqrt {a\,b}} \]

[In]

int(f^x/(a + b*f^(2*x))^3,x)

[Out]

((5*f^x)/(8*a*log(f)) + (3*b*f^(3*x))/(8*a^2*log(f)))/(b^2*f^(4*x) + a^2 + 2*a*b*f^(2*x)) + (3*atan((b*f^x)/(a
*b)^(1/2)))/(8*a^2*log(f)*(a*b)^(1/2))