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

Optimal. Leaf size=62 \[ \frac{2}{5} x \sqrt{1-a^2 x^2} e^{\sin ^{-1}(a x)}+\frac{\left (1-a^2 x^2\right ) e^{\sin ^{-1}(a x)}}{5 a}+\frac{2 e^{\sin ^{-1}(a x)}}{5 a} \]

[Out]

(2*E^ArcSin[a*x])/(5*a) + (2*E^ArcSin[a*x]*x*Sqrt[1 - a^2*x^2])/5 + (E^ArcSin[a*x]*(1 - a^2*x^2))/(5*a)

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Rubi [A]  time = 0.20668, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4836, 6688, 6720, 4435, 2194} \[ \frac{2}{5} x \sqrt{1-a^2 x^2} e^{\sin ^{-1}(a x)}+\frac{\left (1-a^2 x^2\right ) e^{\sin ^{-1}(a x)}}{5 a}+\frac{2 e^{\sin ^{-1}(a x)}}{5 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^ArcSin[a*x])/(5*a) + (2*E^ArcSin[a*x]*x*Sqrt[1 - a^2*x^2])/5 + (E^ArcSin[a*x]*(1 - a^2*x^2))/(5*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 4435

Int[Cos[(d_.) + (e_.)*(x_)]^(m_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*
x))*Cos[d + e*x]^m)/(e^2*m^2 + b^2*c^2*Log[F]^2), x] + (Dist[(m*(m - 1)*e^2)/(e^2*m^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Cos[d + e*x]^(m - 2), x], x] + Simp[(e*m*F^(c*(a + b*x))*Sin[d + e*x]*Cos[d + e*x]^(m - 1))/(
e^2*m^2 + b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*m^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
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 e^{\sin ^{-1}(a x)} \sqrt{1-a^2 x^2} \, dx &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \sqrt{1-\sin ^2(x)} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \sqrt{\cos ^2(x)} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos ^2(x) \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{2}{5} e^{\sin ^{-1}(a x)} x \sqrt{1-a^2 x^2}+\frac{e^{\sin ^{-1}(a x)} \left (1-a^2 x^2\right )}{5 a}+\frac{2 \operatorname{Subst}\left (\int e^x \, dx,x,\sin ^{-1}(a x)\right )}{5 a}\\ &=\frac{2 e^{\sin ^{-1}(a x)}}{5 a}+\frac{2}{5} e^{\sin ^{-1}(a x)} x \sqrt{1-a^2 x^2}+\frac{e^{\sin ^{-1}(a x)} \left (1-a^2 x^2\right )}{5 a}\\ \end{align*}

Mathematica [A]  time = 0.0732716, size = 31, normalized size = 0.5 \[ \frac{e^{\sin ^{-1}(a x)} \left (2 \sin \left (2 \sin ^{-1}(a x)\right )+\cos \left (2 \sin ^{-1}(a x)\right )+5\right )}{10 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcSin[a*x]*(5 + Cos[2*ArcSin[a*x]] + 2*Sin[2*ArcSin[a*x]]))/(10*a)

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Maple [F]  time = 0.08, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{\arcsin \left ( ax \right ) }}\sqrt{-{a}^{2}{x}^{2}+1}\, dx \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]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a^{2} x^{2} + 1} e^{\left (\arcsin \left (a x\right )\right )}\,{d x} \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]

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

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Fricas [A]  time = 2.03805, size = 88, normalized size = 1.42 \begin{align*} -\frac{{\left (a^{2} x^{2} - 2 \, \sqrt{-a^{2} x^{2} + 1} a x - 3\right )} e^{\left (\arcsin \left (a x\right )\right )}}{5 \, 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]

-1/5*(a^2*x^2 - 2*sqrt(-a^2*x^2 + 1)*a*x - 3)*e^(arcsin(a*x))/a

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Sympy [A]  time = 0.542048, size = 49, normalized size = 0.79 \begin{align*} \begin{cases} - \frac{a x^{2} e^{\operatorname{asin}{\left (a x \right )}}}{5} + \frac{2 x \sqrt{- a^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a x \right )}}}{5} + \frac{3 e^{\operatorname{asin}{\left (a x \right )}}}{5 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((-a*x**2*exp(asin(a*x))/5 + 2*x*sqrt(-a**2*x**2 + 1)*exp(asin(a*x))/5 + 3*exp(asin(a*x))/(5*a), Ne(a
, 0)), (x, True))

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Giac [A]  time = 1.22359, size = 68, normalized size = 1.1 \begin{align*} \frac{2}{5} \, \sqrt{-a^{2} x^{2} + 1} x e^{\left (\arcsin \left (a x\right )\right )} - \frac{{\left (a^{2} x^{2} - 1\right )} e^{\left (\arcsin \left (a x\right )\right )}}{5 \, a} + \frac{2 \, e^{\left (\arcsin \left (a x\right )\right )}}{5 \, 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]

2/5*sqrt(-a^2*x^2 + 1)*x*e^(arcsin(a*x)) - 1/5*(a^2*x^2 - 1)*e^(arcsin(a*x))/a + 2/5*e^(arcsin(a*x))/a