3.533 \(\int \frac{x}{2+f^{-c-d x}+f^{c+d x}} \, dx\)

Optimal. Leaf size=50 \[ -\frac{\log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac{x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac{x}{d \log (f)} \]

[Out]

x/(d*Log[f]) - x/(d*(1 + f^(c + d*x))*Log[f]) - Log[1 + f^(c + d*x)]/(d^2*Log[f]^2)

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Rubi [A]  time = 0.288148, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2267, 6688, 2191, 2282, 36, 29, 31} \[ -\frac{\log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac{x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac{x}{d \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[x/(2 + f^(-c - d*x) + f^(c + d*x)),x]

[Out]

x/(d*Log[f]) - x/(d*(1 + f^(c + d*x))*Log[f]) - Log[1 + f^(c + d*x)]/(d^2*Log[f]^2)

Rule 2267

Int[(u_)/((a_) + (b_.)*(F_)^(v_) + (c_.)*(F_)^(w_)), x_Symbol] :> Int[(u*F^v)/(c + a*F^v + b*F^(2*v)), x] /; F
reeQ[{F, a, b, c}, x] && EqQ[w, -v] && LinearQ[v, x] && If[RationalQ[Coefficient[v, x, 1]], GtQ[Coefficient[v,
x, 1], 0], LtQ[LeafCount[v], LeafCount[w]]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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}{2+f^{-c-d x}+f^{c+d x}} \, dx &=\int \frac{f^{c+d x} x}{1+2 f^{c+d x}+f^{2 (c+d x)}} \, dx\\ &=\int \frac{f^{c+d x} x}{\left (1+f^{c+d x}\right )^2} \, dx\\ &=-\frac{x}{d \left (1+f^{c+d x}\right ) \log (f)}+\frac{\int \frac{1}{1+f^{c+d x}} \, dx}{d \log (f)}\\ &=-\frac{x}{d \left (1+f^{c+d x}\right ) \log (f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{x (1+x)} \, dx,x,f^{c+d x}\right )}{d^2 \log ^2(f)}\\ &=-\frac{x}{d \left (1+f^{c+d x}\right ) \log (f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,f^{c+d x}\right )}{d^2 \log ^2(f)}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,f^{c+d x}\right )}{d^2 \log ^2(f)}\\ &=\frac{x}{d \log (f)}-\frac{x}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac{\log \left (1+f^{c+d x}\right )}{d^2 \log ^2(f)}\\ \end{align*}

Mathematica [A]  time = 0.0652724, size = 44, normalized size = 0.88 \[ \frac{\frac{d x \log (f) f^{c+d x}}{f^{c+d x}+1}-\log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(2 + f^(-c - d*x) + f^(c + d*x)),x]

[Out]

((d*f^(c + d*x)*x*Log[f])/(1 + f^(c + d*x)) - Log[1 + f^(c + d*x)])/(d^2*Log[f]^2)

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Maple [A]  time = 0.014, size = 64, normalized size = 1.3 \begin{align*} -{\frac{x{{\rm e}^{ \left ( -dx-c \right ) \ln \left ( f \right ) }}}{d\ln \left ( f \right ) \left ({{\rm e}^{ \left ( -dx-c \right ) \ln \left ( f \right ) }}+1 \right ) }}-{\frac{\ln \left ({{\rm e}^{ \left ( -dx-c \right ) \ln \left ( f \right ) }}+1 \right ) }{ \left ( \ln \left ( f \right ) \right ) ^{2}{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(2+f^(-d*x-c)+f^(d*x+c)),x)

[Out]

-1/d/ln(f)*x*exp((-d*x-c)*ln(f))/(exp((-d*x-c)*ln(f))+1)-1/d^2/ln(f)^2*ln(exp((-d*x-c)*ln(f))+1)

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Maxima [A]  time = 0.98975, size = 77, normalized size = 1.54 \begin{align*} \frac{f^{d x} f^{c} x}{d f^{d x} f^{c} \log \left (f\right ) + d \log \left (f\right )} - \frac{\log \left (\frac{f^{d x} f^{c} + 1}{f^{c}}\right )}{d^{2} \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="maxima")

[Out]

f^(d*x)*f^c*x/(d*f^(d*x)*f^c*log(f) + d*log(f)) - log((f^(d*x)*f^c + 1)/f^c)/(d^2*log(f)^2)

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Fricas [A]  time = 1.29051, size = 147, normalized size = 2.94 \begin{align*} \frac{d f^{d x + c} x \log \left (f\right ) -{\left (f^{d x + c} + 1\right )} \log \left (f^{d x + c} + 1\right )}{d^{2} f^{d x + c} \log \left (f\right )^{2} + d^{2} \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="fricas")

[Out]

(d*f^(d*x + c)*x*log(f) - (f^(d*x + c) + 1)*log(f^(d*x + c) + 1))/(d^2*f^(d*x + c)*log(f)^2 + d^2*log(f)^2)

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Sympy [A]  time = 0.137308, size = 42, normalized size = 0.84 \begin{align*} - \frac{x}{d f^{c + d x} \log{\left (f \right )} + d \log{\left (f \right )}} + \frac{x}{d \log{\left (f \right )}} - \frac{\log{\left (f^{c + d x} + 1 \right )}}{d^{2} \log{\left (f \right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(2+f**(-d*x-c)+f**(d*x+c)),x)

[Out]

-x/(d*f**(c + d*x)*log(f) + d*log(f)) + x/(d*log(f)) - log(f**(c + d*x) + 1)/(d**2*log(f)**2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{f^{d x + c} + f^{-d x - c} + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="giac")

[Out]

integrate(x/(f^(d*x + c) + f^(-d*x - c) + 2), x)