3.530 \(\int \frac{e^x}{\sqrt{a^2+e^{2 x}}} \, dx\)

Optimal. Leaf size=18 \[ \tanh ^{-1}\left (\frac{e^x}{\sqrt{a^2+e^{2 x}}}\right ) \]

[Out]

ArcTanh[E^x/Sqrt[a^2 + E^(2*x)]]

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Rubi [A]  time = 0.0282844, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2249, 217, 206} \[ \tanh ^{-1}\left (\frac{e^x}{\sqrt{a^2+e^{2 x}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x/Sqrt[a^2 + E^(2*x)],x]

[Out]

ArcTanh[E^x/Sqrt[a^2 + E^(2*x)]]

Rule 2249

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]

Rule 217

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0053018, size = 18, normalized size = 1. \[ \tanh ^{-1}\left (\frac{e^x}{\sqrt{a^2+e^{2 x}}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/Sqrt[a^2 + E^(2*x)],x]

[Out]

ArcTanh[E^x/Sqrt[a^2 + E^(2*x)]]

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Maple [A]  time = 0.01, size = 15, normalized size = 0.8 \begin{align*} \ln \left ({{\rm e}^{x}}+\sqrt{{a}^{2}+ \left ({{\rm e}^{x}} \right ) ^{2}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(exp(x)+(a^2+exp(x)^2)^(1/2))

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Maxima [A]  time = 0.930315, size = 12, normalized size = 0.67 \begin{align*} \operatorname{arsinh}\left (\frac{e^{x}}{\sqrt{a^{2}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsinh(e^x/sqrt(a^2))

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Fricas [A]  time = 1.88472, size = 45, normalized size = 2.5 \begin{align*} -\log \left (\sqrt{a^{2} + e^{\left (2 \, x\right )}} - e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sqrt(a^2 + e^(2*x)) - e^x)

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Sympy [A]  time = 0.629821, size = 31, normalized size = 1.72 \begin{align*} \begin{cases} \operatorname{asinh}{\left (\sqrt{\frac{1}{a^{2}}} e^{x} \right )} & \text{for}\: a^{2} > 0 \\\operatorname{acosh}{\left (\sqrt{- \frac{1}{a^{2}}} e^{x} \right )} & \text{for}\: a^{2} < 0 \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((asinh(sqrt(a**(-2))*exp(x)), a**2 > 0), (acosh(sqrt(-1/a**2)*exp(x)), a**2 < 0))

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Giac [A]  time = 1.13445, size = 24, normalized size = 1.33 \begin{align*} -\log \left (\sqrt{a^{2} + e^{\left (2 \, x\right )}} - e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sqrt(a^2 + e^(2*x)) - e^x)