3.530 \(\int \frac{x}{2+e^{-x}+e^x} \, dx\)

Optimal. Leaf size=20 \[ -\frac{x}{e^x+1}+x-\log \left (e^x+1\right ) \]

[Out]

x - x/(1 + E^x) - Log[1 + E^x]

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Rubi [A]  time = 0.13448, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2267, 6688, 2191, 2282, 36, 29, 31} \[ -\frac{x}{e^x+1}+x-\log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/(2 + E^(-x) + E^x),x]

[Out]

x - x/(1 + E^x) - Log[1 + E^x]

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

Mathematica [A]  time = 0.0251414, size = 20, normalized size = 1. \[ -\frac{x}{e^x+1}+x-\log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(2 + E^(-x) + E^x),x]

[Out]

x - x/(1 + E^x) - Log[1 + E^x]

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Maple [A]  time = 0.009, size = 19, normalized size = 1. \begin{align*} -\ln \left ( 1+{{\rm e}^{x}} \right ) +{\frac{{{\rm e}^{x}}x}{1+{{\rm e}^{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(2+exp(-x)+exp(x)),x)

[Out]

-ln(1+exp(x))+x*exp(x)/(1+exp(x))

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Maxima [A]  time = 0.987925, size = 24, normalized size = 1.2 \begin{align*} \frac{x e^{x}}{e^{x} + 1} - \log \left (e^{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(2+exp(-x)+exp(x)),x, algorithm="maxima")

[Out]

x*e^x/(e^x + 1) - log(e^x + 1)

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Fricas [A]  time = 1.383, size = 59, normalized size = 2.95 \begin{align*} \frac{x e^{x} -{\left (e^{x} + 1\right )} \log \left (e^{x} + 1\right )}{e^{x} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(2+exp(-x)+exp(x)),x, algorithm="fricas")

[Out]

(x*e^x - (e^x + 1)*log(e^x + 1))/(e^x + 1)

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Sympy [A]  time = 0.088783, size = 14, normalized size = 0.7 \begin{align*} x - \frac{x}{e^{x} + 1} - \log{\left (e^{x} + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(2+exp(-x)+exp(x)),x)

[Out]

x - x/(exp(x) + 1) - log(exp(x) + 1)

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Giac [A]  time = 1.42235, size = 38, normalized size = 1.9 \begin{align*} \frac{x e^{x} - e^{x} \log \left (e^{x} + 1\right ) - \log \left (e^{x} + 1\right )}{e^{x} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(2+exp(-x)+exp(x)),x, algorithm="giac")

[Out]

(x*e^x - e^x*log(e^x + 1) - log(e^x + 1))/(e^x + 1)