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

Optimal result

Integrand size = 10, antiderivative size = 82 \[ \int e^{\arcsin (a x)} x^2 \, dx=\frac {e^{\arcsin (a x)} x}{8 a^2}+\frac {e^{\arcsin (a x)} \sqrt {1-a^2 x^2}}{8 a^3}-\frac {e^{\arcsin (a x)} \cos (3 \arcsin (a x))}{40 a^3}-\frac {3 e^{\arcsin (a x)} \sin (3 \arcsin (a x))}{40 a^3} \] Output:

1/8*exp(arcsin(a*x))*x/a^2+1/8*exp(arcsin(a*x))*(-a^2*x^2+1)^(1/2)/a^3-1/4 
0*exp(arcsin(a*x))*cos(3*arcsin(a*x))/a^3-3/40*exp(arcsin(a*x))*sin(3*arcs 
in(a*x))/a^3
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.61 \[ \int e^{\arcsin (a x)} x^2 \, dx=-\frac {e^{\arcsin (a x)} \left (-5 a x-5 \sqrt {1-a^2 x^2}+\cos (3 \arcsin (a x))+3 \sin (3 \arcsin (a x))\right )}{40 a^3} \] Input:

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

Output:

-1/40*(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]]))/a^3
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5335, 27, 4972, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 e^{\arcsin (a x)} \, dx\)

\(\Big \downarrow \) 5335

\(\displaystyle \frac {\int e^{\arcsin (a x)} x^2 \sqrt {1-a^2 x^2}d\arcsin (a x)}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int a^2 e^{\arcsin (a x)} x^2 \sqrt {1-a^2 x^2}d\arcsin (a x)}{a^3}\)

\(\Big \downarrow \) 4972

\(\displaystyle \frac {\int \left (\frac {1}{4} e^{\arcsin (a x)} \sqrt {1-a^2 x^2}-\frac {1}{4} e^{\arcsin (a x)} \cos (3 \arcsin (a x))\right )d\arcsin (a x)}{a^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{8} \sqrt {1-a^2 x^2} e^{\arcsin (a x)}+\frac {1}{8} a x e^{\arcsin (a x)}-\frac {3}{40} e^{\arcsin (a x)} \sin (3 \arcsin (a x))-\frac {1}{40} e^{\arcsin (a x)} \cos (3 \arcsin (a x))}{a^3}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4972

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 5335

Int[(u_.)*(f_)^(ArcSin[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Simp[ 
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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.55 \[ \int e^{\arcsin (a x)} x^2 \, dx=\frac {{\left (3 \, a^{3} x^{3} - a x + {\left (a^{2} x^{2} + 1\right )} \sqrt {-a^{2} x^{2} + 1}\right )} e^{\left (\arcsin \left (a x\right )\right )}}{10 \, a^{3}} \] Input:

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

Output:

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
 

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int e^{\arcsin (a x)} x^2 \, dx=\begin {cases} \frac {3 x^{3} e^{\operatorname {asin}{\left (a x \right )}}}{10} + \frac {x^{2} \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{10 a} - \frac {x e^{\operatorname {asin}{\left (a x \right )}}}{10 a^{2}} + \frac {\sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{10 a^{3}} & \text {for}\: a \neq 0 \\\frac {x^{3}}{3} & \text {otherwise} \end {cases} \] Input:

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

Output:

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))
 

Maxima [F]

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

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

Output:

integrate(x^2*e^(arcsin(a*x)), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93 \[ \int e^{\arcsin (a x)} x^2 \, dx=\frac {3 \, {\left (a^{2} x^{2} - 1\right )} x e^{\left (\arcsin \left (a x\right )\right )}}{10 \, a^{2}} + \frac {x e^{\left (\arcsin \left (a x\right )\right )}}{5 \, a^{2}} - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} e^{\left (\arcsin \left (a x\right )\right )}}{10 \, a^{3}} + \frac {\sqrt {-a^{2} x^{2} + 1} e^{\left (\arcsin \left (a x\right )\right )}}{5 \, a^{3}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(e**asin(a*x)*x**2,x)