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]

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

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Rubi [A]  time = 0.08, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.42, size = 61, normalized size = 0.97 \[ \frac {2 \, a f^{2 \, x} x \log \relax (f) - {\left (a f^{2 \, x} + b\right )} \log \left (a f^{2 \, x} + b\right )}{4 \, {\left (a^{2} b f^{2 \, x} \log \relax (f)^{2} + a b^{2} \log \relax (f)^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (a f^{x} + \frac {b}{f^{x}}\right )}^{2}}\,{d x} \]

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)

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maple [A]  time = 0.03, size = 56, normalized size = 0.89 \[ \frac {x \,{\mathrm e}^{2 x \ln \relax (f )}}{2 \left (a \,{\mathrm e}^{2 x \ln \relax (f )}+b \right ) b \ln \relax (f )}-\frac {\ln \left (a \,{\mathrm e}^{2 x \ln \relax (f )}+b \right )}{4 a b \ln \relax (f )^{2}} \]

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 = 0.50, size = 54, normalized size = 0.86 \[ \frac {f^{2 \, x} x}{2 \, {\left (a b f^{2 \, x} \log \relax (f) + b^{2} \log \relax (f)\right )}} - \frac {\log \left (\frac {a f^{2 \, x} + b}{a}\right )}{4 \, a b \log \relax (f)^{2}} \]

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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mupad [B]  time = 3.61, size = 51, normalized size = 0.81 \[ \frac {f^{2\,x}\,x}{2\,\left (b^2\,\ln \relax (f)+a\,b\,f^{2\,x}\,\ln \relax (f)\right )}-\frac {\ln \left (b+a\,f^{2\,x}\right )}{4\,a\,b\,{\ln \relax (f)}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.19, size = 53, normalized size = 0.84 \[ - \frac {x}{2 a^{2} f^{2 x} \log {\relax (f )} + 2 a b \log {\relax (f )}} + \frac {x}{2 a b \log {\relax (f )}} - \frac {\log {\left (f^{2 x} + \frac {b}{a} \right )}}{4 a b \log {\relax (f )}^{2}} \]

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

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