3.445 \(\int e^{\sin ^{-1}(a x)^2} x^3 \, dx\)

Optimal. Leaf size=101 \[ \frac{e \sqrt{\pi } \text{Erf}\left (1-i \sin ^{-1}(a x)\right )}{16 a^4}-\frac{e^4 \sqrt{\pi } \text{Erf}\left (2-i \sin ^{-1}(a x)\right )}{32 a^4}+\frac{e \sqrt{\pi } \text{Erf}\left (1+i \sin ^{-1}(a x)\right )}{16 a^4}-\frac{e^4 \sqrt{\pi } \text{Erf}\left (2+i \sin ^{-1}(a x)\right )}{32 a^4} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.123416, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {4836, 12, 4474, 2234, 2204} \[ \frac{e \sqrt{\pi } \text{Erf}\left (1-i \sin ^{-1}(a x)\right )}{16 a^4}-\frac{e^4 \sqrt{\pi } \text{Erf}\left (2-i \sin ^{-1}(a x)\right )}{32 a^4}+\frac{e \sqrt{\pi } \text{Erf}\left (1+i \sin ^{-1}(a x)\right )}{16 a^4}-\frac{e^4 \sqrt{\pi } \text{Erf}\left (2+i \sin ^{-1}(a x)\right )}{32 a^4} \]

Antiderivative was successfully verified.

[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 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 12

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

Rule 4474

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 2234

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 2204

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]

Rubi steps

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

Mathematica [A]  time = 0.0779962, size = 67, normalized size = 0.66 \[ \frac{e \sqrt{\pi } \left (2 \left (\text{Erf}\left (1-i \sin ^{-1}(a x)\right )+\text{Erf}\left (1+i \sin ^{-1}(a x)\right )\right )-e^3 \left (\text{Erf}\left (2-i \sin ^{-1}(a x)\right )+\text{Erf}\left (2+i \sin ^{-1}(a x)\right )\right )\right )}{32 a^4} \]

Antiderivative was successfully verified.

[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]  time = 0.01, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{ \left ( \arcsin \left ( ax \right ) \right ) ^{2}}}{x}^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*e^(arcsin(a*x)^2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*e^(arcsin(a*x)^2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} e^{\operatorname{asin}^{2}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*exp(asin(a*x)**2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*e^(arcsin(a*x)^2), x)