3.2.55 \(\int \frac {e^{2 x}}{1+e^x} \, dx\) [155]

Optimal. Leaf size=12 \[ e^x-\log \left (1+e^x\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2280, 45} \begin {gather*} e^x-\log \left (e^x+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(1 + E^x),x]

[Out]

E^x - Log[1 + E^x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2280

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[g*h*(Log[G]/(d*e*Log[F]))]}, Dist[Denominator[m]*(G^(f*h - c*g*(h/d))/(d*e*Log[F])), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^(e*((c + d*x)/Denominator[m]))], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} e^x-\log \left (1+e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(1 + E^x),x]

[Out]

E^x - Log[1 + E^x]

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Maple [A]
time = 0.01, size = 11, normalized size = 0.92

method	result	size
default	\({\mathrm e}^{x}-\ln \left (1+{\mathrm e}^{x}\right )\)	\(11\)
norman	\({\mathrm e}^{x}-\ln \left (1+{\mathrm e}^{x}\right )\)	\(11\)
risch	\({\mathrm e}^{x}-\ln \left (1+{\mathrm e}^{x}\right )\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(1+exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 3.15, size = 10, normalized size = 0.83 \begin {gather*} e^{x} - \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(1+exp(x)),x, algorithm="maxima")

[Out]

e^x - log(e^x + 1)

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Fricas [A]
time = 0.86, size = 10, normalized size = 0.83 \begin {gather*} e^{x} - \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(1+exp(x)),x, algorithm="fricas")

[Out]

e^x - log(e^x + 1)

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Sympy [A]
time = 0.02, size = 8, normalized size = 0.67 \begin {gather*} e^{x} - \log {\left (e^{x} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(1+exp(x)),x)

[Out]

exp(x) - log(exp(x) + 1)

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Giac [A]
time = 0.79, size = 10, normalized size = 0.83 \begin {gather*} e^{x} - \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(1+exp(x)),x, algorithm="giac")

[Out]

e^x - log(e^x + 1)

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Mupad [B]
time = 0.04, size = 10, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^x-\ln \left ({\mathrm {e}}^x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(exp(x) + 1),x)

[Out]

exp(x) - log(exp(x) + 1)

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