\(\int e^{\arcsin (a x)} \, dx\) [442]

Optimal result

Integrand size = 6, antiderivative size = 39 \[ \int e^{\arcsin (a x)} \, dx=\frac {1}{2} e^{\arcsin (a x)} x+\frac {e^{\arcsin (a x)} \sqrt {1-a^2 x^2}}{2 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4920, 4518} \[ \int e^{\arcsin (a x)} \, dx=\frac {\sqrt {1-a^2 x^2} e^{\arcsin (a x)}}{2 a}+\frac {1}{2} x e^{\arcsin (a x)} \]

[In]

Int[E^ArcSin[a*x],x]

[Out]

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

Rule 4518

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(C
os[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] + Simp[e*F^(c*(a + b*x))*(Sin[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rule 4920

Int[(u_.)*(f_)^(ArcSin[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[(u /. x -> -a/b + Si
n[x]/b)*f^(c*x^n)*Cos[x], x], x, ArcSin[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int e^{\arcsin (a x)} \, dx=\frac {e^{\arcsin (a x)} \left (a x+\sqrt {1-a^2 x^2}\right )}{2 a} \]

[In]

Integrate[E^ArcSin[a*x],x]

[Out]

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

Maple [F]

\[\int {\mathrm e}^{\arcsin \left (a x \right )}d x\]

[In]

int(exp(arcsin(a*x)),x)

[Out]

int(exp(arcsin(a*x)),x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67 \[ \int e^{\arcsin (a x)} \, dx=\frac {{\left (a x + \sqrt {-a^{2} x^{2} + 1}\right )} e^{\left (\arcsin \left (a x\right )\right )}}{2 \, a} \]

[In]

integrate(exp(arcsin(a*x)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int e^{\arcsin (a x)} \, dx=\begin {cases} \frac {x e^{\operatorname {asin}{\left (a x \right )}}}{2} + \frac {\sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{2 a} & \text {for}\: a \neq 0 \\x & \text {otherwise} \end {cases} \]

[In]

integrate(exp(asin(a*x)),x)

[Out]

Piecewise((x*exp(asin(a*x))/2 + sqrt(-a**2*x**2 + 1)*exp(asin(a*x))/(2*a), Ne(a, 0)), (x, True))

Maxima [F]

\[ \int e^{\arcsin (a x)} \, dx=\int { e^{\left (\arcsin \left (a x\right )\right )} \,d x } \]

[In]

integrate(exp(arcsin(a*x)),x, algorithm="maxima")

[Out]

integrate(e^(arcsin(a*x)), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int e^{\arcsin (a x)} \, dx=\frac {1}{2} \, x e^{\left (\arcsin \left (a x\right )\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} e^{\left (\arcsin \left (a x\right )\right )}}{2 \, a} \]

[In]

integrate(exp(arcsin(a*x)),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int e^{\arcsin (a x)} \, dx=\int {\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )} \,d x \]

[In]

int(exp(asin(a*x)),x)

[Out]

int(exp(asin(a*x)), x)