\(\int \frac {e^{\arcsin (a x)}}{x^2} \, dx\) [444]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 83 \[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=(1-i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right )-(2-2 i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},2,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right ) \]

[Out]

(1-I)*a*exp((1+I)*arcsin(a*x))*hypergeom([1, 1/2-1/2*I],[3/2-1/2*I],(I*a*x+(-a^2*x^2+1)^(1/2))^2)+(-2+2*I)*a*e
xp((1+I)*arcsin(a*x))*hypergeom([2, 1/2-1/2*I],[3/2-1/2*I],(I*a*x+(-a^2*x^2+1)^(1/2))^2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4920, 12, 4559, 2283} \[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=(1-i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right )-(2-2 i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},2,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right ) \]

[In]

Int[E^ArcSin[a*x]/x^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2283

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[a^p*(G^(h*(f + g*x))/(g*h*Log[G]))*Hypergeometric2F1[-p, g*h*(Log[G]/(d*e*Log[F])), g*h*(Log[G]/(d*e*Log[F]))
 + 1, Simplify[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] || G
tQ[a, 0])

Rule 4559

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol]
 :> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
IGtQ[m, 0] && IGtQ[n, 0] && TrigQ[G] && TrigQ[H]

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]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=-\frac {e^{\arcsin (a x)}+(1+i) a e^{(1+i) \arcsin (a x)} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right )}{x} \]

[In]

Integrate[E^ArcSin[a*x]/x^2,x]

[Out]

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

Maple [F]

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

[In]

int(exp(arcsin(a*x))/x^2,x)

[Out]

int(exp(arcsin(a*x))/x^2,x)

Fricas [F]

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

[In]

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

[Out]

integral(e^(arcsin(a*x))/x^2, x)

Sympy [F]

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

[In]

integrate(exp(asin(a*x))/x**2,x)

[Out]

Integral(exp(asin(a*x))/x**2, x)

Maxima [F]

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

[In]

integrate(exp(arcsin(a*x))/x^2,x, algorithm="maxima")

[Out]

integrate(e^(arcsin(a*x))/x^2, x)

Giac [F]

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

[In]

integrate(exp(arcsin(a*x))/x^2,x, algorithm="giac")

[Out]

integrate(e^(arcsin(a*x))/x^2, x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(exp(asin(a*x))/x^2, x)