\(\int \frac {e^x}{(1+e^x) \log (1+e^x)} \, dx\) [281]

Optimal result

Integrand size = 19, antiderivative size = 7 \[ \int \frac {e^x}{\left (1+e^x\right ) \log \left (1+e^x\right )} \, dx=\log \left (\log \left (1+e^x\right )\right ) \]

[Out]

ln(ln(exp(x)+1))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2320, 2437, 2339, 29} \[ \int \frac {e^x}{\left (1+e^x\right ) \log \left (1+e^x\right )} \, dx=\log \left (\log \left (e^x+1\right )\right ) \]

[In]

Int[E^x/((1 + E^x)*Log[1 + E^x]),x]

[Out]

Log[Log[1 + E^x]]

Rule 29

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

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 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{(1+x) \log (1+x)} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,1+e^x\right ) \\ & = \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (1+e^x\right )\right ) \\ & = \log \left (\log \left (1+e^x\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{\left (1+e^x\right ) \log \left (1+e^x\right )} \, dx=\log \left (\log \left (1+e^x\right )\right ) \]

[In]

Integrate[E^x/((1 + E^x)*Log[1 + E^x]),x]

[Out]

Log[Log[1 + E^x]]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\ln \left (\ln \left ({\mathrm e}^{x}+1\right )\right )\)	\(7\)
default	\(\ln \left (\ln \left ({\mathrm e}^{x}+1\right )\right )\)	\(7\)
norman	\(\ln \left (\ln \left ({\mathrm e}^{x}+1\right )\right )\)	\(7\)
risch	\(\ln \left (\ln \left ({\mathrm e}^{x}+1\right )\right )\)	\(7\)
parallelrisch	\(\ln \left (\ln \left ({\mathrm e}^{x}+1\right )\right )\)	\(7\)

[In]

int(exp(x)/(exp(x)+1)/ln(exp(x)+1),x,method=_RETURNVERBOSE)

[Out]

ln(ln(exp(x)+1))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.86 \[ \int \frac {e^x}{\left (1+e^x\right ) \log \left (1+e^x\right )} \, dx=\log \left (\log \left (e^{x} + 1\right )\right ) \]

[In]

integrate(exp(x)/(exp(x)+1)/log(exp(x)+1),x, algorithm="fricas")

[Out]

log(log(e^x + 1))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{\left (1+e^x\right ) \log \left (1+e^x\right )} \, dx=\log {\left (\log {\left (e^{x} + 1 \right )} \right )} \]

[In]

integrate(exp(x)/(exp(x)+1)/ln(exp(x)+1),x)

[Out]

log(log(exp(x) + 1))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.86 \[ \int \frac {e^x}{\left (1+e^x\right ) \log \left (1+e^x\right )} \, dx=\log \left (\log \left (e^{x} + 1\right )\right ) \]

[In]

integrate(exp(x)/(exp(x)+1)/log(exp(x)+1),x, algorithm="maxima")

[Out]

log(log(e^x + 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.86 \[ \int \frac {e^x}{\left (1+e^x\right ) \log \left (1+e^x\right )} \, dx=\log \left (\log \left (e^{x} + 1\right )\right ) \]

[In]

integrate(exp(x)/(exp(x)+1)/log(exp(x)+1),x, algorithm="giac")

[Out]

log(log(e^x + 1))

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.86 \[ \int \frac {e^x}{\left (1+e^x\right ) \log \left (1+e^x\right )} \, dx=\ln \left (\ln \left ({\mathrm {e}}^x+1\right )\right ) \]

[In]

int(exp(x)/(log(exp(x) + 1)*(exp(x) + 1)),x)

[Out]

log(log(exp(x) + 1))