\(\int e^x \sqrt {1-e^{2 x}} \, dx\) [642]

Optimal result

Integrand size = 17, antiderivative size = 29 \[ \int e^x \sqrt {1-e^{2 x}} \, dx=\frac {1}{2} e^x \sqrt {1-e^{2 x}}+\frac {\arcsin \left (e^x\right )}{2} \]

[Out]

1/2*arcsin(exp(x))+1/2*exp(x)*(1-exp(2*x))^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2281, 201, 222} \[ \int e^x \sqrt {1-e^{2 x}} \, dx=\frac {\arcsin \left (e^x\right )}{2}+\frac {1}{2} e^x \sqrt {1-e^{2 x}} \]

[In]

Int[E^x*Sqrt[1 - E^(2*x)],x]

[Out]

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

Rule 201

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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[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 \sqrt {1-x^2} \, dx,x,e^x\right ) \\ & = \frac {1}{2} e^x \sqrt {1-e^{2 x}}+\frac {1}{2} \text {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} \sin ^{-1}\left (e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int e^x \sqrt {1-e^{2 x}} \, dx=\frac {1}{2} e^x \sqrt {1-e^{2 x}}-\arctan \left (\frac {\sqrt {1-e^{2 x}}}{1+e^x}\right ) \]

[In]

Integrate[E^x*Sqrt[1 - E^(2*x)],x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {\arcsin \left ({\mathrm e}^{x}\right )}{2}+\frac {{\mathrm e}^{x} \sqrt {1-{\mathrm e}^{2 x}}}{2}\)	\(21\)
risch	\(-\frac {{\mathrm e}^{x} \left (-1+{\mathrm e}^{2 x}\right )}{2 \sqrt {1-{\mathrm e}^{2 x}}}+\frac {\arcsin \left ({\mathrm e}^{x}\right )}{2}\)	\(27\)

[In]

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

[Out]

1/2*exp(x)*(1-exp(x)^2)^(1/2)+1/2*arcsin(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int e^x \sqrt {1-e^{2 x}} \, dx=\frac {1}{2} \, \sqrt {-e^{\left (2 \, x\right )} + 1} e^{x} - \arctan \left ({\left (\sqrt {-e^{\left (2 \, x\right )} + 1} - 1\right )} e^{\left (-x\right )}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int e^x \sqrt {1-e^{2 x}} \, dx=\frac {\sqrt {1 - e^{2 x}} e^{x}}{2} + \frac {\operatorname {asin}{\left (e^{x} \right )}}{2} \]

[In]

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

[Out]

sqrt(1 - exp(2*x))*exp(x)/2 + asin(exp(x))/2

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int e^x \sqrt {1-e^{2 x}} \, dx=\frac {1}{2} \, \sqrt {-e^{\left (2 \, x\right )} + 1} e^{x} + \frac {1}{2} \, \arcsin \left (e^{x}\right ) \]

[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*arcsin(e^x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int e^x \sqrt {1-e^{2 x}} \, dx=\frac {1}{2} \, \sqrt {-e^{\left (2 \, x\right )} + 1} e^{x} + \frac {1}{2} \, \arcsin \left (e^{x}\right ) \]

[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*arcsin(e^x)

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int e^x \sqrt {1-e^{2 x}} \, dx=\frac {\mathrm {asin}\left ({\mathrm {e}}^x\right )}{2}+\frac {{\mathrm {e}}^x\,\sqrt {1-{\mathrm {e}}^{2\,x}}}{2} \]

[In]

int(exp(x)*(1 - exp(2*x))^(1/2),x)

[Out]

asin(exp(x))/2 + (exp(x)*(1 - exp(2*x))^(1/2))/2