3.62 \(\int \frac{1}{(b f^{-x}+a f^x)^3} \, dx\)

Optimal. Leaf size=87 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac{f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2} \]

[Out]

-f^x/(4*a*(b + a*f^(2*x))^2*Log[f]) + f^x/(8*a*b*(b + a*f^(2*x))*Log[f]) + ArcTan[(Sqrt[a]*f^x)/Sqrt[b]]/(8*a^
(3/2)*b^(3/2)*Log[f])

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Rubi [A]  time = 0.0463815, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2282, 288, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac{f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(b/f^x + a*f^x)^(-3),x]

[Out]

-f^x/(4*a*(b + a*f^(2*x))^2*Log[f]) + f^x/(8*a*b*(b + a*f^(2*x))*Log[f]) + ArcTan[(Sqrt[a]*f^x)/Sqrt[b]]/(8*a^
(3/2)*b^(3/2)*Log[f])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, 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{1}{\left (b f^{-x}+a f^x\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (b+a x^2\right )^3} \, dx,x,f^x\right )}{\log (f)}\\ &=-\frac{f^x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (b+a x^2\right )^2} \, dx,x,f^x\right )}{4 a \log (f)}\\ &=-\frac{f^x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{f^x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,f^x\right )}{8 a b \log (f)}\\ &=-\frac{f^x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{f^x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0529376, size = 70, normalized size = 0.8 \[ \frac{\frac{\sqrt{a} \sqrt{b} f^x \left (a f^{2 x}-b\right )}{\left (a f^{2 x}+b\right )^2}+\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b/f^x + a*f^x)^(-3),x]

[Out]

((Sqrt[a]*Sqrt[b]*f^x*(-b + a*f^(2*x)))/(b + a*f^(2*x))^2 + ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/(8*a^(3/2)*b^(3/2)*
Log[f])

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Maple [A]  time = 0.008, size = 78, normalized size = 0.9 \begin{align*}{\frac{ \left ({f}^{x} \right ) ^{3}}{8\,\ln \left ( f \right ) \left ( a \left ({f}^{x} \right ) ^{2}+b \right ) ^{2}b}}-{\frac{{f}^{x}}{8\,\ln \left ( f \right ) \left ( a \left ({f}^{x} \right ) ^{2}+b \right ) ^{2}a}}+{\frac{1}{8\,b\ln \left ( f \right ) a}\arctan \left ({a{f}^{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/(f^x)+a*f^x)^3,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55599, size = 586, normalized size = 6.74 \begin{align*} \left [\frac{2 \, a^{2} b f^{3 \, x} - 2 \, a b^{2} f^{x} -{\left (\sqrt{-a b} a^{2} f^{4 \, x} + 2 \, \sqrt{-a b} a b f^{2 \, x} + \sqrt{-a b} b^{2}\right )} \log \left (\frac{a f^{2 \, x} - 2 \, \sqrt{-a b} f^{x} - b}{a f^{2 \, x} + b}\right )}{16 \,{\left (a^{4} b^{2} f^{4 \, x} \log \left (f\right ) + 2 \, a^{3} b^{3} f^{2 \, x} \log \left (f\right ) + a^{2} b^{4} \log \left (f\right )\right )}}, \frac{a^{2} b f^{3 \, x} - a b^{2} f^{x} -{\left (\sqrt{a b} a^{2} f^{4 \, x} + 2 \, \sqrt{a b} a b f^{2 \, x} + \sqrt{a b} b^{2}\right )} \arctan \left (\frac{\sqrt{a b}}{a f^{x}}\right )}{8 \,{\left (a^{4} b^{2} f^{4 \, x} \log \left (f\right ) + 2 \, a^{3} b^{3} f^{2 \, x} \log \left (f\right ) + a^{2} b^{4} \log \left (f\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b/(f^x)+a*f^x)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.299246, size = 87, normalized size = 1. \begin{align*} \frac{a f^{- x} - b f^{- 3 x}}{8 a^{3} b \log{\left (f \right )} + 16 a^{2} b^{2} f^{- 2 x} \log{\left (f \right )} + 8 a b^{3} f^{- 4 x} \log{\left (f \right )}} + \frac{\operatorname{RootSum}{\left (256 z^{2} a^{3} b^{3} + 1, \left ( i \mapsto i \log{\left (- 16 i a^{2} b + f^{- x} \right )} \right )\right )}}{\log{\left (f \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b/(f**x)+a*f**x)**3,x)

[Out]

(a*f**(-x) - b*f**(-3*x))/(8*a**3*b*log(f) + 16*a**2*b**2*f**(-2*x)*log(f) + 8*a*b**3*f**(-4*x)*log(f)) + Root
Sum(256*_z**2*a**3*b**3 + 1, Lambda(_i, _i*log(-16*_i*a**2*b + f**(-x))))/log(f)

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Giac [A]  time = 1.24434, size = 89, normalized size = 1.02 \begin{align*} \frac{\arctan \left (\frac{a f^{x}}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a b \log \left (f\right )} + \frac{a f^{3 \, x} - b f^{x}}{8 \,{\left (a f^{2 \, x} + b\right )}^{2} a b \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b/(f^x)+a*f^x)^3,x, algorithm="giac")

[Out]

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