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

Optimal. Leaf size=82 \[ \frac{\sqrt{1-a^2 x^2} e^{\sin ^{-1}(a x)}}{8 a^3}+\frac{x e^{\sin ^{-1}(a x)}}{8 a^2}-\frac{3 e^{\sin ^{-1}(a x)} \sin \left (3 \sin ^{-1}(a x)\right )}{40 a^3}-\frac{e^{\sin ^{-1}(a x)} \cos \left (3 \sin ^{-1}(a x)\right )}{40 a^3} \]

[Out]

(E^ArcSin[a*x]*x)/(8*a^2) + (E^ArcSin[a*x]*Sqrt[1 - a^2*x^2])/(8*a^3) - (E^ArcSin[a*x]*Cos[3*ArcSin[a*x]])/(40
*a^3) - (3*E^ArcSin[a*x]*Sin[3*ArcSin[a*x]])/(40*a^3)

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Rubi [A]  time = 0.062202, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4836, 12, 4469, 4433} \[ \frac{\sqrt{1-a^2 x^2} e^{\sin ^{-1}(a x)}}{8 a^3}+\frac{x e^{\sin ^{-1}(a x)}}{8 a^2}-\frac{3 e^{\sin ^{-1}(a x)} \sin \left (3 \sin ^{-1}(a x)\right )}{40 a^3}-\frac{e^{\sin ^{-1}(a x)} \cos \left (3 \sin ^{-1}(a x)\right )}{40 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcSin[a*x]*x)/(8*a^2) + (E^ArcSin[a*x]*Sqrt[1 - a^2*x^2])/(8*a^3) - (E^ArcSin[a*x]*Cos[3*ArcSin[a*x]])/(40
*a^3) - (3*E^ArcSin[a*x]*Sin[3*ArcSin[a*x]])/(40*a^3)

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 4469

Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :
> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g
}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 4433

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*C
os[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin{align*} \int e^{\sin ^{-1}(a x)} x^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^x \cos (x) \sin ^2(x)}{a^2} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(a x)\right )}{a^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{4} e^x \cos (x)-\frac{1}{4} e^x \cos (3 x)\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a^3}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a x)\right )}{4 a^3}-\frac{\operatorname{Subst}\left (\int e^x \cos (3 x) \, dx,x,\sin ^{-1}(a x)\right )}{4 a^3}\\ &=\frac{e^{\sin ^{-1}(a x)} x}{8 a^2}+\frac{e^{\sin ^{-1}(a x)} \sqrt{1-a^2 x^2}}{8 a^3}-\frac{e^{\sin ^{-1}(a x)} \cos \left (3 \sin ^{-1}(a x)\right )}{40 a^3}-\frac{3 e^{\sin ^{-1}(a x)} \sin \left (3 \sin ^{-1}(a x)\right )}{40 a^3}\\ \end{align*}

Mathematica [A]  time = 0.113662, size = 50, normalized size = 0.61 \[ -\frac{e^{\sin ^{-1}(a x)} \left (-5 \sqrt{1-a^2 x^2}-5 a x+3 \sin \left (3 \sin ^{-1}(a x)\right )+\cos \left (3 \sin ^{-1}(a x)\right )\right )}{40 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^ArcSin[a*x]*(-5*a*x - 5*Sqrt[1 - a^2*x^2] + Cos[3*ArcSin[a*x]] + 3*Sin[3*ArcSin[a*x]]))/(40*a^3)

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Maple [F]  time = 0.007, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{\arcsin \left ( ax \right ) }}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*e^(arcsin(a*x)), x)

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Fricas [A]  time = 2.0602, size = 107, normalized size = 1.3 \begin{align*} \frac{{\left (3 \, a^{3} x^{3} - a x +{\left (a^{2} x^{2} + 1\right )} \sqrt{-a^{2} x^{2} + 1}\right )} e^{\left (\arcsin \left (a x\right )\right )}}{10 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(3*a^3*x^3 - a*x + (a^2*x^2 + 1)*sqrt(-a^2*x^2 + 1))*e^(arcsin(a*x))/a^3

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Sympy [A]  time = 1.81857, size = 80, normalized size = 0.98 \begin{align*} \begin{cases} \frac{3 x^{3} e^{\operatorname{asin}{\left (a x \right )}}}{10} + \frac{x^{2} \sqrt{- a^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a x \right )}}}{10 a} - \frac{x e^{\operatorname{asin}{\left (a x \right )}}}{10 a^{2}} + \frac{\sqrt{- a^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a x \right )}}}{10 a^{3}} & \text{for}\: a \neq 0 \\\frac{x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*x**3*exp(asin(a*x))/10 + x**2*sqrt(-a**2*x**2 + 1)*exp(asin(a*x))/(10*a) - x*exp(asin(a*x))/(10*a
**2) + sqrt(-a**2*x**2 + 1)*exp(asin(a*x))/(10*a**3), Ne(a, 0)), (x**3/3, True))

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Giac [A]  time = 1.19897, size = 103, normalized size = 1.26 \begin{align*} \frac{3 \,{\left (a^{2} x^{2} - 1\right )} x e^{\left (\arcsin \left (a x\right )\right )}}{10 \, a^{2}} + \frac{x e^{\left (\arcsin \left (a x\right )\right )}}{5 \, a^{2}} - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} e^{\left (\arcsin \left (a x\right )\right )}}{10 \, a^{3}} + \frac{\sqrt{-a^{2} x^{2} + 1} e^{\left (\arcsin \left (a x\right )\right )}}{5 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(a^2*x^2 - 1)*x*e^(arcsin(a*x))/a^2 + 1/5*x*e^(arcsin(a*x))/a^2 - 1/10*(-a^2*x^2 + 1)^(3/2)*e^(arcsin(a*x
))/a^3 + 1/5*sqrt(-a^2*x^2 + 1)*e^(arcsin(a*x))/a^3