3.467 \(\int \frac{e^{\sin ^{-1}(a x)}}{(1-a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=45 \[ \frac{\left (\frac{4}{5}-\frac{8 i}{5}\right ) e^{(1+2 i) \sin ^{-1}(a x)} \text{Hypergeometric2F1}\left (1-\frac{i}{2},2,2-\frac{i}{2},-e^{2 i \sin ^{-1}(a x)}\right )}{a} \]

[Out]

((4/5 - (8*I)/5)*E^((1 + 2*I)*ArcSin[a*x])*Hypergeometric2F1[1 - I/2, 2, 2 - I/2, -E^((2*I)*ArcSin[a*x])])/a

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Rubi [A]  time = 0.258082, antiderivative size = 45, 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, 4451} \[ \frac{\left (\frac{4}{5}-\frac{8 i}{5}\right ) e^{(1+2 i) \sin ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},2;2-\frac{i}{2};-e^{2 i \sin ^{-1}(a x)}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a*x]/(1 - a^2*x^2)^(3/2),x]

[Out]

((4/5 - (8*I)/5)*E^((1 + 2*I)*ArcSin[a*x])*Hypergeometric2F1[1 - I/2, 2, 2 - I/2, -E^((2*I)*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 4451

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(I*n*(d + e*x))*
F^(c*(a + b*x))*Hypergeometric2F1[n, n/2 - (I*b*c*Log[F])/(2*e), 1 + n/2 - (I*b*c*Log[F])/(2*e), -E^(2*I*(d +
e*x))])/(I*e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{e^{\sin ^{-1}(a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^x \cos (x)}{\left (1-\sin ^2(x)\right )^{3/2}} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int \frac{e^x \cos (x)}{\cos ^2(x)^{3/2}} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int e^x \sec ^2(x) \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{\left (\frac{4}{5}-\frac{8 i}{5}\right ) e^{(1+2 i) \sin ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},2;2-\frac{i}{2};-e^{2 i \sin ^{-1}(a x)}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0680029, size = 45, normalized size = 1. \[ \frac{\left (\frac{4}{5}-\frac{8 i}{5}\right ) e^{(1+2 i) \sin ^{-1}(a x)} \text{Hypergeometric2F1}\left (1-\frac{i}{2},2,2-\frac{i}{2},-e^{2 i \sin ^{-1}(a x)}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSin[a*x]/(1 - a^2*x^2)^(3/2),x]

[Out]

((4/5 - (8*I)/5)*E^((1 + 2*I)*ArcSin[a*x])*Hypergeometric2F1[1 - I/2, 2, 2 - I/2, -E^((2*I)*ArcSin[a*x])])/a

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Maple [F]  time = 0.08, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{\arcsin \left ( ax \right ) }} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arcsin(a*x))/(-a^2*x^2+1)^(3/2),x)

[Out]

int(exp(arcsin(a*x))/(-a^2*x^2+1)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (\arcsin \left (a x\right )\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

integrate(e^(arcsin(a*x))/(-a^2*x^2 + 1)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} x^{2} + 1} e^{\left (\arcsin \left (a x\right )\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsin(a*x))/(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)*e^(arcsin(a*x))/(a^4*x^4 - 2*a^2*x^2 + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\operatorname{asin}{\left (a x \right )}}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(asin(a*x))/(-a**2*x**2+1)**(3/2),x)

[Out]

Integral(exp(asin(a*x))/(-(a*x - 1)*(a*x + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (\arcsin \left (a x\right )\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

integrate(e^(arcsin(a*x))/(-a^2*x^2 + 1)^(3/2), x)