3.466 \(\int \frac{e^{\sin ^{-1}(a x)}}{\sqrt{1-a^2 x^2}} \, dx\)

Optimal. Leaf size=10 \[ \frac{e^{\sin ^{-1}(a x)}}{a} \]

[Out]

E^ArcSin[a*x]/a

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Rubi [A]  time = 0.214341, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4836, 6688, 6720, 2194} \[ \frac{e^{\sin ^{-1}(a x)}}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a*x]/Sqrt[1 - a^2*x^2],x]

[Out]

E^ArcSin[a*x]/a

Rule 4836

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

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

Mathematica [A]  time = 0.0153145, size = 10, normalized size = 1. \[ \frac{e^{\sin ^{-1}(a x)}}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSin[a*x]/Sqrt[1 - a^2*x^2],x]

[Out]

E^ArcSin[a*x]/a

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Maple [A]  time = 0.005, size = 10, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{\arcsin \left ( ax \right ) }}}{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(arcsin(a*x))/a

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Maxima [A]  time = 1.47392, size = 12, normalized size = 1.2 \begin{align*} \frac{e^{\left (\arcsin \left (a x\right )\right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(arcsin(a*x))/a

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Fricas [A]  time = 2.18016, size = 26, normalized size = 2.6 \begin{align*} \frac{e^{\left (\arcsin \left (a x\right )\right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(arcsin(a*x))/a

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Sympy [A]  time = 0.461056, size = 8, normalized size = 0.8 \begin{align*} \begin{cases} \frac{e^{\operatorname{asin}{\left (a x \right )}}}{a} & \text{for}\: a \neq 0 \\x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((exp(asin(a*x))/a, Ne(a, 0)), (x, True))

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Giac [A]  time = 1.16343, size = 12, normalized size = 1.2 \begin{align*} \frac{e^{\left (\arcsin \left (a x\right )\right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(arcsin(a*x))/a