\(\int e^{\arcsin (a x)^2} x^3 \, dx\) [445]

   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 = 12, antiderivative size = 101 \[ \int e^{\arcsin (a x)^2} x^3 \, dx=\frac {e \sqrt {\pi } \text {erf}(1-i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2-i \arcsin (a x))}{32 a^4}+\frac {e \sqrt {\pi } \text {erf}(1+i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2+i \arcsin (a x))}{32 a^4} \]

[Out]

1/16*I*exp(1)*erfi(-I+arcsin(a*x))*Pi^(1/2)/a^4-1/16*I*exp(1)*erfi(I+arcsin(a*x))*Pi^(1/2)/a^4-1/32*I*exp(4)*e
rfi(-2*I+arcsin(a*x))*Pi^(1/2)/a^4+1/32*I*exp(4)*erfi(2*I+arcsin(a*x))*Pi^(1/2)/a^4

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4920, 12, 4562, 2266, 2235} \[ \int e^{\arcsin (a x)^2} x^3 \, dx=\frac {e \sqrt {\pi } \text {erf}(1-i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2-i \arcsin (a x))}{32 a^4}+\frac {e \sqrt {\pi } \text {erf}(1+i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2+i \arcsin (a x))}{32 a^4} \]

[In]

Int[E^ArcSin[a*x]^2*x^3,x]

[Out]

(E*Sqrt[Pi]*Erf[1 - I*ArcSin[a*x]])/(16*a^4) - (E^4*Sqrt[Pi]*Erf[2 - I*ArcSin[a*x]])/(32*a^4) + (E*Sqrt[Pi]*Er
f[1 + I*ArcSin[a*x]])/(16*a^4) - (E^4*Sqrt[Pi]*Erf[2 + I*ArcSin[a*x]])/(32*a^4)

Rule 12

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

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 4562

Int[Cos[v_]^(n_.)*(F_)^(u_)*Sin[v_]^(m_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sin[v]^m*Cos[v]^n, x], x] /;
FreeQ[F, x] && (LinearQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[m, 0] && IGtQ[n,
 0]

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 \frac {e^{x^2} \cos (x) \sin ^3(x)}{a^3} \, dx,x,\arcsin (a x)\right )}{a} \\ & = \frac {\text {Subst}\left (\int e^{x^2} \cos (x) \sin ^3(x) \, dx,x,\arcsin (a x)\right )}{a^4} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{8} i e^{-2 i x+x^2}-\frac {1}{8} i e^{2 i x+x^2}-\frac {1}{16} i e^{-4 i x+x^2}+\frac {1}{16} i e^{4 i x+x^2}\right ) \, dx,x,\arcsin (a x)\right )}{a^4} \\ & = -\frac {i \text {Subst}\left (\int e^{-4 i x+x^2} \, dx,x,\arcsin (a x)\right )}{16 a^4}+\frac {i \text {Subst}\left (\int e^{4 i x+x^2} \, dx,x,\arcsin (a x)\right )}{16 a^4}+\frac {i \text {Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\arcsin (a x)\right )}{8 a^4}-\frac {i \text {Subst}\left (\int e^{2 i x+x^2} \, dx,x,\arcsin (a x)\right )}{8 a^4} \\ & = \frac {(i e) \text {Subst}\left (\int e^{\frac {1}{4} (-2 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{8 a^4}-\frac {(i e) \text {Subst}\left (\int e^{\frac {1}{4} (2 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{8 a^4}-\frac {\left (i e^4\right ) \text {Subst}\left (\int e^{\frac {1}{4} (-4 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{16 a^4}+\frac {\left (i e^4\right ) \text {Subst}\left (\int e^{\frac {1}{4} (4 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{16 a^4} \\ & = \frac {e \sqrt {\pi } \text {erf}(1-i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2-i \arcsin (a x))}{32 a^4}+\frac {e \sqrt {\pi } \text {erf}(1+i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2+i \arcsin (a x))}{32 a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66 \[ \int e^{\arcsin (a x)^2} x^3 \, dx=\frac {e \sqrt {\pi } \left (2 (\text {erf}(1-i \arcsin (a x))+\text {erf}(1+i \arcsin (a x)))-e^3 (\text {erf}(2-i \arcsin (a x))+\text {erf}(2+i \arcsin (a x)))\right )}{32 a^4} \]

[In]

Integrate[E^ArcSin[a*x]^2*x^3,x]

[Out]

(E*Sqrt[Pi]*(2*(Erf[1 - I*ArcSin[a*x]] + Erf[1 + I*ArcSin[a*x]]) - E^3*(Erf[2 - I*ArcSin[a*x]] + Erf[2 + I*Arc
Sin[a*x]])))/(32*a^4)

Maple [F]

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

[In]

int(exp(arcsin(a*x)^2)*x^3,x)

[Out]

int(exp(arcsin(a*x)^2)*x^3,x)

Fricas [F]

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

[In]

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

[Out]

integral(x^3*e^(arcsin(a*x)^2), x)

Sympy [F]

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

[In]

integrate(exp(asin(a*x)**2)*x**3,x)

[Out]

Integral(x**3*exp(asin(a*x)**2), x)

Maxima [F]

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

[In]

integrate(exp(arcsin(a*x)^2)*x^3,x, algorithm="maxima")

[Out]

integrate(x^3*e^(arcsin(a*x)^2), x)

Giac [F]

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

[In]

integrate(exp(arcsin(a*x)^2)*x^3,x, algorithm="giac")

[Out]

integrate(x^3*e^(arcsin(a*x)^2), x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(x^3*exp(asin(a*x)^2), x)