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]

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

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Rubi [A]  time = 0.12, antiderivative size = 101, normalized size of antiderivative = 1.00, 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 12

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

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 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 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]

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*}

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Mathematica [A]  time = 0.15, 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)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )}\,{d x} \]

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)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\arcsin \left (a x \right )^{2}} x^{3}\, dx \]

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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\mathrm {e}}^{{\mathrm {asin}\left (a\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*exp(asin(a*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} e^{\operatorname {asin}^{2}{\left (a x \right )}}\, dx \]

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)

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