\(\int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx\) [657]

Optimal result

Integrand size = 15, antiderivative size = 4 \[ \int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx=\text {arcsinh}\left (e^x\right ) \]

[Out]

arcsinh(exp(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2281, 221} \[ \int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx=\text {arcsinh}\left (e^x\right ) \]

[In]

Int[E^x/Sqrt[1 + E^(2*x)],x]

[Out]

ArcSinh[E^x]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 2281

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(20\) vs. \(2(4)=8\).

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 5.00 \[ \int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx=-\log \left (-e^x+\sqrt {1+e^{2 x}}\right ) \]

[In]

Integrate[E^x/Sqrt[1 + E^(2*x)],x]

[Out]

-Log[-E^x + Sqrt[1 + E^(2*x)]]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00

method	result	size
default	\(\operatorname {arcsinh}\left ({\mathrm e}^{x}\right )\)	\(4\)

[In]

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

[Out]

arcsinh(exp(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16 vs. \(2 (3) = 6\).

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 4.00 \[ \int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx=-\log \left (\sqrt {e^{\left (2 \, x\right )} + 1} - e^{x}\right ) \]

[In]

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

[Out]

-log(sqrt(e^(2*x) + 1) - e^x)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx=\operatorname {asinh}{\left (e^{x} \right )} \]

[In]

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

[Out]

asinh(exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx=\operatorname {arsinh}\left (e^{x}\right ) \]

[In]

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

[Out]

arcsinh(e^x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16 vs. \(2 (3) = 6\).

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 4.00 \[ \int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx=-\log \left (\sqrt {e^{\left (2 \, x\right )} + 1} - e^{x}\right ) \]

[In]

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

[Out]

-log(sqrt(e^(2*x) + 1) - e^x)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{\sqrt {1+e^{2 x}}} \, dx=\mathrm {asinh}\left ({\mathrm {e}}^x\right ) \]

[In]

int(exp(x)/(exp(2*x) + 1)^(1/2),x)

[Out]

asinh(exp(x))