### 3.281 $$\int \frac{e^{2 x}}{1+e^x} \, dx$$

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

[Out]

E^x - Log[1 + E^x]

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Rubi [A]  time = 0.0196965, antiderivative size = 12, normalized size of antiderivative = 1., 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 = {2248, 43} $e^x-\log \left (e^x+1\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*x)/(1 + E^x),x]

[Out]

E^x - Log[1 + E^x]

Rule 2248

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]

Rule 43

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])

Rubi steps

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

Mathematica [A]  time = 0.0074975, size = 12, normalized size = 1. $e^x-\log \left (e^x+1\right )$

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., size = 11, normalized size = 0.9 \begin{align*}{{\rm e}^{x}}-\ln \left ( 1+{{\rm e}^{x}} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.954929, size = 14, normalized size = 1.17 \begin{align*} e^{x} - \log \left (e^{x} + 1\right ) \end{align*}

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 = 2.00244, size = 27, normalized size = 2.25 \begin{align*} e^{x} - \log \left (e^{x} + 1\right ) \end{align*}

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 = 0.082269, size = 8, normalized size = 0.67 \begin{align*} e^{x} - \log{\left (e^{x} + 1 \right )} \end{align*}

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 = 1.0726, size = 14, normalized size = 1.17 \begin{align*} e^{x} - \log \left (e^{x} + 1\right ) \end{align*}

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)