3.468 \(\int \frac{e^{\sin ^{-1}(a x)}}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{\left (\frac{2}{3}-\frac{4 i}{3}\right ) e^{(1+2 i) \sin ^{-1}(a x)} \text{Hypergeometric2F1}\left (1-\frac{i}{2},2,2-\frac{i}{2},-e^{2 i \sin ^{-1}(a x)}\right )}{a}+\frac{x e^{\sin ^{-1}(a x)}}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac{e^{\sin ^{-1}(a x)}}{6 a \left (1-a^2 x^2\right )} \]

[Out]

(E^ArcSin[a*x]*x)/(3*(1 - a^2*x^2)^(3/2)) - E^ArcSin[a*x]/(6*a*(1 - a^2*x^2)) + ((2/3 - (4*I)/3)*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 = 0.289181, antiderivative size = 96, 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, 4448, 4451} \[ \frac{x e^{\sin ^{-1}(a x)}}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac{e^{\sin ^{-1}(a x)}}{6 a \left (1-a^2 x^2\right )}+\frac{\left (\frac{2}{3}-\frac{4 i}{3}\right ) e^{(1+2 i) \sin ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},2;2-\frac{i}{2};-e^{2 i \sin ^{-1}(a x)}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a*x]/(1 - a^2*x^2)^(5/2),x]

[Out]

(E^ArcSin[a*x]*x)/(3*(1 - a^2*x^2)^(3/2)) - E^ArcSin[a*x]/(6*a*(1 - a^2*x^2)) + ((2/3 - (4*I)/3)*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 4448

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a + b
*x))*Sec[d + e*x]^(n - 2))/(e^2*(n - 1)*(n - 2)), x] + (Dist[(e^2*(n - 2)^2 + b^2*c^2*Log[F]^2)/(e^2*(n - 1)*(
n - 2)), Int[F^(c*(a + b*x))*Sec[d + e*x]^(n - 2), x], x] + Simp[(F^(c*(a + b*x))*Sec[d + e*x]^(n - 1)*Sin[d +
e*x])/(e*(n - 1)), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[b^2*c^2*Log[F]^2 + e^2*(n - 2)^2, 0] && GtQ[n,
1] && NeQ[n, 2]

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 )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^x \cos (x)}{\left (1-\sin ^2(x)\right )^{5/2}} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int \frac{e^x \cos (x)}{\cos ^2(x)^{5/2}} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int e^x \sec ^4(x) \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{e^{\sin ^{-1}(a x)} x}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac{e^{\sin ^{-1}(a x)}}{6 a \left (1-a^2 x^2\right )}+\frac{5 \operatorname{Subst}\left (\int e^x \sec ^2(x) \, dx,x,\sin ^{-1}(a x)\right )}{6 a}\\ &=\frac{e^{\sin ^{-1}(a x)} x}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac{e^{\sin ^{-1}(a x)}}{6 a \left (1-a^2 x^2\right )}+\frac{\left (\frac{2}{3}-\frac{4 i}{3}\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 = 0.170193, size = 84, normalized size = 0.88 \[ \frac{e^{\sin ^{-1}(a x)} \left ((1-2 i) \left (1+e^{2 i \sin ^{-1}(a x)}\right )^2 \text{Hypergeometric2F1}\left (1-\frac{i}{2},2,2-\frac{i}{2},-e^{2 i \sin ^{-1}(a x)}\right )+\frac{2 a x}{\sqrt{1-a^2 x^2}}-1\right )}{6 \left (a-a^3 x^2\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSin[a*x]/(1 - a^2*x^2)^(5/2),x]

[Out]

(E^ArcSin[a*x]*(-1 + (2*a*x)/Sqrt[1 - a^2*x^2] + (1 - 2*I)*(1 + E^((2*I)*ArcSin[a*x]))^2*Hypergeometric2F1[1 -
I/2, 2, 2 - I/2, -E^((2*I)*ArcSin[a*x])]))/(6*(a - a^3*x^2))

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{\arcsin \left ( ax \right ) }} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

int(exp(arcsin(a*x))/(-a^2*x^2+1)^(5/2),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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate(e^(arcsin(a*x))/(-a^2*x^2 + 1)^(5/2), 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^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*e^(arcsin(a*x))/(a^6*x^6 - 3*a^4*x^4 + 3*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{5}{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)**(5/2),x)

[Out]

Integral(exp(asin(a*x))/(-(a*x - 1)*(a*x + 1))**(5/2), 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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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