\(\int \frac {x}{(b f^{-x}+a f^x)^2} \, dx\) [59]

Optimal result

Integrand size = 17, antiderivative size = 63 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {x}{2 a b \log (f)}-\frac {x}{2 a \left (b+a f^{2 x}\right ) \log (f)}-\frac {\log \left (b+a f^{2 x}\right )}{4 a b \log ^2(f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2321, 2222, 2320, 36, 29, 31} \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^2} \, dx=-\frac {\log \left (a f^{2 x}+b\right )}{4 a b \log ^2(f)}-\frac {x}{2 a \log (f) \left (a f^{2 x}+b\right )}+\frac {x}{2 a b \log (f)} \]

[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 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]

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 2222

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 2320

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 2321

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]

Rubi steps \begin{align*} \text {integral}& = \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 {\text {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 {\text {Subst}\left (\int \frac {1}{b+a x} \, dx,x,f^{2 x}\right )}{4 b \log ^2(f)}+\frac {\text {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] (verified)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.76 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {\frac {2 f^{2 x} x \log (f)}{b+a f^{2 x}}-\frac {\log \left (b+a f^{2 x}\right )}{a}}{4 b \log ^2(f)} \]

[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)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89

method	result	size
norman	\(\frac {x \,{\mathrm e}^{2 x \ln \left (f \right )}}{2 b \ln \left (f \right ) \left (a \,{\mathrm e}^{2 x \ln \left (f \right )}+b \right )}-\frac {\ln \left (a \,{\mathrm e}^{2 x \ln \left (f \right )}+b \right )}{4 \ln \left (f \right )^{2} a b}\)	\(56\)
risch	\(\frac {x}{2 a b \ln \left (f \right )}-\frac {x}{2 a \left (b +a \,f^{2 x}\right ) \ln \left (f \right )}-\frac {\ln \left (f^{2 x}+\frac {b}{a}\right )}{4 \ln \left (f \right )^{2} a b}\)	\(60\)
parallelrisch	\(-\frac {-2 f^{2 x} \ln \left (f \right ) a x +\ln \left (b +a \,f^{2 x}\right ) f^{2 x} a +\ln \left (b +a \,f^{2 x}\right ) b}{4 \ln \left (f \right )^{2} a b \left (b +a \,f^{2 x}\right )}\)	\(65\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {2 \, a f^{2 \, x} x \log \left (f\right ) - {\left (a f^{2 \, x} + b\right )} \log \left (a f^{2 \, x} + b\right )}{4 \, {\left (a^{2} b f^{2 \, x} \log \left (f\right )^{2} + a b^{2} \log \left (f\right )^{2}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {x}{2 a b \log {\left (f \right )} + 2 b^{2} f^{- 2 x} \log {\left (f \right )}} - \frac {x}{2 a b \log {\left (f \right )}} - \frac {\log {\left (\frac {a}{b} + f^{- 2 x} \right )}}{4 a b \log {\left (f \right )}^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {f^{2 \, x} x}{2 \, {\left (a b f^{2 \, x} \log \left (f\right ) + b^{2} \log \left (f\right )\right )}} - \frac {\log \left (\frac {a f^{2 \, x} + b}{a}\right )}{4 \, a b \log \left (f\right )^{2}} \]

[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)

Giac [F]

\[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^2} \, dx=\int { \frac {x}{{\left (a f^{x} + \frac {b}{f^{x}}\right )}^{2}} \,d x } \]

[In]

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

[Out]

integrate(x/(a*f^x + b/f^x)^2, x)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {f^{2\,x}\,x}{2\,\left (b^2\,\ln \left (f\right )+a\,b\,f^{2\,x}\,\ln \left (f\right )\right )}-\frac {\ln \left (b+a\,f^{2\,x}\right )}{4\,a\,b\,{\ln \left (f\right )}^2} \]

[In]

int(x/(b/f^x + a*f^x)^2,x)

[Out]

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