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

Optimal. Leaf size=7 \[ \log \left (\log \left (1+e^x\right )\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 7, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2320, 2437, 2339, 29} \begin {gather*} \log \left (\log \left (e^x+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 7, normalized size = 1.00 \begin {gather*} \log \left (\log \left (1+e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Log[1 + E^x]]

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Maple [A]
time = 0.03, size = 7, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.33, size = 6, normalized size = 0.86 \begin {gather*} \log \left (\log \left (e^{x} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(e^x + 1))

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Fricas [A]
time = 0.56, size = 6, normalized size = 0.86 \begin {gather*} \log \left (\log \left (e^{x} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(e^x + 1))

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Sympy [A]
time = 0.05, size = 7, normalized size = 1.00 \begin {gather*} \log {\left (\log {\left (e^{x} + 1 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(exp(x) + 1))

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Giac [A]
time = 0.44, size = 6, normalized size = 0.86 \begin {gather*} \log \left (\log \left (e^{x} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(e^x + 1))

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Mupad [B]
time = 0.14, size = 6, normalized size = 0.86 \begin {gather*} \ln \left (\ln \left ({\mathrm {e}}^x+1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(exp(x) + 1))

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Chatgpt [A]
time = 1.00, size = 6, normalized size = 0.86 \begin {gather*} \ln \left (\ln \left ({\mathrm e}^{x}+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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