\(\int \frac {e^{\arcsin (a x)}}{(1-a^2 x^2)^{3/2}} \, dx\) [467]

Optimal result

Integrand size = 21, antiderivative size = 45 \[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {\left (\frac {4}{5}-\frac {8 i}{5}\right ) e^{(1+2 i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \arcsin (a x)}\right )}{a} \]

[Out]

(4/5-8/5*I)*exp((1+2*I)*arcsin(a*x))*hypergeom([2, 1-1/2*I],[2-1/2*I],-(I*a*x+(-a^2*x^2+1)^(1/2))^2)/a

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4920, 6820, 6852, 4536} \[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {\left (\frac {4}{5}-\frac {8 i}{5}\right ) e^{(1+2 i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \arcsin (a x)}\right )}{a} \]

[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 4536

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))/(I*e*n + b*c*Log[F]))*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))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rule 4920

Int[(u_.)*(f_)^(ArcSin[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[(u /. x -> -a/b + Si
n[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 6820

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

Rule 6852

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {e^x \cos (x)}{\left (1-\sin ^2(x)\right )^{3/2}} \, dx,x,\arcsin (a x)\right )}{a} \\ & = \frac {\text {Subst}\left (\int \frac {e^x \cos (x)}{\cos ^2(x)^{3/2}} \, dx,x,\arcsin (a x)\right )}{a} \\ & = \frac {\text {Subst}\left (\int e^x \sec ^2(x) \, dx,x,\arcsin (a x)\right )}{a} \\ & = \frac {\left (\frac {4}{5}-\frac {8 i}{5}\right ) e^{(1+2 i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \arcsin (a x)}\right )}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {\left (\frac {4}{5}-\frac {8 i}{5}\right ) e^{(1+2 i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \arcsin (a x)}\right )}{a} \]

[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

Maple [F]

\[\int \frac {{\mathrm e}^{\arcsin \left (a x \right )}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}d x\]

[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)

Fricas [F]

\[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int \frac {e^{\operatorname {asin}{\left (a x \right )}}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int \frac {{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )}}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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