Integrand size = 21, antiderivative size = 96 \[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {e^{\arcsin (a x)} x}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {e^{\arcsin (a x)}}{6 a \left (1-a^2 x^2\right )}+\frac {\left (\frac {2}{3}-\frac {4 i}{3}\right ) e^{(1+2 i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \arcsin (a x)}\right )}{a} \]
1/3*exp(arcsin(a*x))*x/(-a^2*x^2+1)^(3/2)-1/6*exp(arcsin(a*x))/a/(-a^2*x^2 +1)+(2/3-4/3*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
Time = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {e^{\arcsin (a x)} \left (-1+\frac {2 a x}{\sqrt {1-a^2 x^2}}+(1-2 i) \left (1+e^{2 i \arcsin (a x)}\right )^2 \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \arcsin (a x)}\right )\right )}{6 \left (a-a^3 x^2\right )} \]
(E^ArcSin[a*x]*(-1 + (2*a*x)/Sqrt[1 - a^2*x^2] + (1 - 2*I)*(1 + E^((2*I)*A rcSin[a*x]))^2*Hypergeometric2F1[1 - I/2, 2, 2 - I/2, -E^((2*I)*ArcSin[a*x ])]))/(6*(a - a^3*x^2))
Time = 0.58 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5335, 7292, 7271, 4948, 4951}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5335 |
\(\displaystyle \frac {\int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^2}d\arcsin (a x)}{a}\) |
\(\Big \downarrow \) 4948 |
\(\displaystyle \frac {\frac {5}{6} \int \frac {e^{\arcsin (a x)}}{1-a^2 x^2}d\arcsin (a x)+\frac {a x e^{\arcsin (a x)}}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {e^{\arcsin (a x)}}{6 \left (1-a^2 x^2\right )}}{a}\) |
\(\Big \downarrow \) 4951 |
\(\displaystyle \frac {\frac {a x e^{\arcsin (a x)}}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {e^{\arcsin (a x)}}{6 \left (1-a^2 x^2\right )}+\left (\frac {2}{3}-\frac {4 i}{3}\right ) e^{(1+2 i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \arcsin (a x)}\right )}{a}\) |
((a*E^ArcSin[a*x]*x)/(3*(1 - a^2*x^2)^(3/2)) - E^ArcSin[a*x]/(6*(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
3.5.68.3.1 Defintions of rubi rules used
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_), x_Symbo l] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Sec[d + e*x]^(n - 2)/(e^2*(n - 1) *(n - 2))), x] + (Simp[F^(c*(a + b*x))*Sec[d + e*x]^(n - 1)*(Sin[d + e*x]/( e*(n - 1))), x] + Simp[(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]) /; 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] && N eQ[n, 2]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symb ol] :> Simp[2^n*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hy pergeometric2F1[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]
Int[(u_.)*(f_)^(ArcSin[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Simp[ 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]
Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[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]) && !(Eq Q[v, x] && EqQ[m, 1])
\[\int \frac {{\mathrm e}^{\arcsin \left (a x \right )}}{\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}d x\]
\[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int \frac {e^{\operatorname {asin}{\left (a x \right )}}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {e^{\arcsin (a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int \frac {{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )}}{{\left (1-a^2\,x^2\right )}^{5/2}} \,d x \]