3.442 \(\int e^{\sin ^{-1}(a x)} \, dx\)

Optimal. Leaf size=39 \[ \frac {\sqrt {1-a^2 x^2} e^{\sin ^{-1}(a x)}}{2 a}+\frac {1}{2} x e^{\sin ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4836, 4433} \[ \frac {\sqrt {1-a^2 x^2} e^{\sin ^{-1}(a x)}}{2 a}+\frac {1}{2} x e^{\sin ^{-1}(a x)} \]

Antiderivative was successfully verified.

[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 4433

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.79 \[ \frac {\left (\sqrt {1-a^2 x^2}+a x\right ) e^{\sin ^{-1}(a x)}}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 26, normalized size = 0.67 \[ \frac {{\left (a x + \sqrt {-a^{2} x^{2} + 1}\right )} e^{\left (\arcsin \left (a x\right )\right )}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.18, size = 31, normalized size = 0.79 \[ \frac {1}{2} \, x e^{\left (\arcsin \left (a x\right )\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} e^{\left (\arcsin \left (a x\right )\right )}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\arcsin \left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left (\arcsin \left (a x\right )\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(arcsin(a*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(asin(a*x)), x)

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sympy [A]  time = 0.16, size = 32, normalized size = 0.82 \[ \begin {cases} \frac {x e^{\operatorname {asin}{\left (a x \right )}}}{2} + \frac {\sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{2 a} & \text {for}\: a \neq 0 \\x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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