3.661 \(\int e^x \sqrt {1+e^{2 x}} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{2} e^x \sqrt {e^{2 x}+1}+\frac {1}{2} \sinh ^{-1}\left (e^x\right ) \]

[Out]

1/2*arcsinh(exp(x))+1/2*exp(x)*(1+exp(2*x))^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2249, 195, 215} \[ \frac {1}{2} e^x \sqrt {e^{2 x}+1}+\frac {1}{2} \sinh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sqrt[1 + E^(2*x)],x]

[Out]

(E^x*Sqrt[1 + E^(2*x)])/2 + ArcSinh[E^x]/2

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.89 \[ \frac {1}{2} \left (e^x \sqrt {e^{2 x}+1}+\sinh ^{-1}\left (e^x\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sqrt[1 + E^(2*x)],x]

[Out]

(E^x*Sqrt[1 + E^(2*x)] + ArcSinh[E^x])/2

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fricas [A]  time = 0.41, size = 29, normalized size = 1.07 \[ \frac {1}{2} \, \sqrt {e^{\left (2 \, x\right )} + 1} e^{x} - \frac {1}{2} \, \log \left (\sqrt {e^{\left (2 \, x\right )} + 1} - e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(e^(2*x) + 1)*e^x - 1/2*log(sqrt(e^(2*x) + 1) - e^x)

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giac [A]  time = 0.23, size = 29, normalized size = 1.07 \[ \frac {1}{2} \, \sqrt {e^{\left (2 \, x\right )} + 1} e^{x} - \frac {1}{2} \, \log \left (\sqrt {e^{\left (2 \, x\right )} + 1} - e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(e^(2*x) + 1)*e^x - 1/2*log(sqrt(e^(2*x) + 1) - e^x)

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maple [A]  time = 0.03, size = 19, normalized size = 0.70 \[ \frac {\arcsinh \left ({\mathrm e}^{x}\right )}{2}+\frac {\sqrt {{\mathrm e}^{2 x}+1}\, {\mathrm e}^{x}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*exp(x)*(1+exp(x)^2)^(1/2)+1/2*arcsinh(exp(x))

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maxima [A]  time = 1.69, size = 18, normalized size = 0.67 \[ \frac {1}{2} \, \sqrt {e^{\left (2 \, x\right )} + 1} e^{x} + \frac {1}{2} \, \operatorname {arsinh}\left (e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(e^(2*x) + 1)*e^x + 1/2*arcsinh(e^x)

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mupad [B]  time = 3.55, size = 18, normalized size = 0.67 \[ \frac {\mathrm {asinh}\left ({\mathrm {e}}^x\right )}{2}+\frac {{\mathrm {e}}^x\,\sqrt {{\mathrm {e}}^{2\,x}+1}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*(exp(2*x) + 1)^(1/2),x)

[Out]

asinh(exp(x))/2 + (exp(x)*(exp(2*x) + 1)^(1/2))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {e^{2 x} + 1} e^{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(exp(2*x) + 1)*exp(x), x)

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