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

Optimal. Leaf size=63 \[ -\frac{\log \left (a f^{2 x}+b\right )}{4 a b \log ^2(f)}-\frac{x}{2 a \log (f) \left (a f^{2 x}+b\right )}+\frac{x}{2 a b \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.0783578, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {2283, 2191, 2282, 36, 29, 31} \[ -\frac{\log \left (a f^{2 x}+b\right )}{4 a b \log ^2(f)}-\frac{x}{2 a \log (f) \left (a f^{2 x}+b\right )}+\frac{x}{2 a b \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[x/(b/f^x + a*f^x)^2,x]

[Out]

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

Rule 2283

Int[(u_.)*((a_.)*(F_)^(v_) + (b_.)*(F_)^(w_))^(n_), x_Symbol] :> Int[u*F^(n*v)*(a + b*F^ExpandToSum[w - v, x])
^n, x] /; FreeQ[{F, a, b, n}, x] && ILtQ[n, 0] && LinearQ[{v, w}, x]

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{x}{\left (b f^{-x}+a f^x\right )^2} \, dx &=\int \frac{f^{2 x} x}{\left (b+a f^{2 x}\right )^2} \, dx\\ &=-\frac{x}{2 a \left (b+a f^{2 x}\right ) \log (f)}+\frac{\int \frac{1}{b+a f^{2 x}} \, dx}{2 a \log (f)}\\ &=-\frac{x}{2 a \left (b+a f^{2 x}\right ) \log (f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{x (b+a x)} \, dx,x,f^{2 x}\right )}{4 a \log ^2(f)}\\ &=-\frac{x}{2 a \left (b+a f^{2 x}\right ) \log (f)}-\frac{\operatorname{Subst}\left (\int \frac{1}{b+a x} \, dx,x,f^{2 x}\right )}{4 b \log ^2(f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,f^{2 x}\right )}{4 a b \log ^2(f)}\\ &=\frac{x}{2 a b \log (f)}-\frac{x}{2 a \left (b+a f^{2 x}\right ) \log (f)}-\frac{\log \left (b+a f^{2 x}\right )}{4 a b \log ^2(f)}\\ \end{align*}

Mathematica [A]  time = 0.0572333, size = 48, normalized size = 0.76 \[ \frac{\frac{2 x f^{2 x} \log (f)}{a f^{2 x}+b}-\frac{\log \left (a f^{2 x}+b\right )}{a}}{4 b \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(b/f^x + a*f^x)^2,x]

[Out]

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

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Maple [A]  time = 0.016, size = 56, normalized size = 0.9 \begin{align*}{\frac{x \left ({{\rm e}^{x\ln \left ( f \right ) }} \right ) ^{2}}{2\,b\ln \left ( f \right ) \left ( \left ({{\rm e}^{x\ln \left ( f \right ) }} \right ) ^{2}a+b \right ) }}-{\frac{\ln \left ( \left ({{\rm e}^{x\ln \left ( f \right ) }} \right ) ^{2}a+b \right ) }{4\, \left ( \ln \left ( f \right ) \right ) ^{2}ab}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06419, size = 73, normalized size = 1.16 \begin{align*} \frac{f^{2 \, x} x}{2 \,{\left (a b f^{2 \, x} \log \left (f\right ) + b^{2} \log \left (f\right )\right )}} - \frac{\log \left (\frac{a f^{2 \, x} + b}{a}\right )}{4 \, a b \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.50443, size = 144, normalized size = 2.29 \begin{align*} \frac{2 \, a f^{2 \, x} x \log \left (f\right ) -{\left (a f^{2 \, x} + b\right )} \log \left (a f^{2 \, x} + b\right )}{4 \,{\left (a^{2} b f^{2 \, x} \log \left (f\right )^{2} + a b^{2} \log \left (f\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.296525, size = 54, normalized size = 0.86 \begin{align*} \frac{x}{2 a b \log{\left (f \right )} + 2 b^{2} f^{- 2 x} \log{\left (f \right )}} - \frac{x}{2 a b \log{\left (f \right )}} - \frac{\log{\left (\frac{a}{b} + f^{- 2 x} \right )}}{4 a b \log{\left (f \right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*a*b*log(f) + 2*b**2*f**(-2*x)*log(f)) - x/(2*a*b*log(f)) - log(a/b + f**(-2*x))/(4*a*b*log(f)**2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a f^{x} + \frac{b}{f^{x}}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(a*f^x + b/f^x)^2, x)