\(\int e^{\arcsin (a+b x)} x^2 \, dx\) [327]

Optimal result

Integrand size = 12, antiderivative size = 205 \[ \int e^{\arcsin (a+b x)} x^2 \, dx=\frac {e^{\arcsin (a+b x)} (a+b x)}{8 b^3}+\frac {a^2 e^{\arcsin (a+b x)} (a+b x)}{2 b^3}+\frac {e^{\arcsin (a+b x)} \sqrt {1-(a+b x)^2}}{8 b^3}+\frac {a^2 e^{\arcsin (a+b x)} \sqrt {1-(a+b x)^2}}{2 b^3}+\frac {2 a e^{\arcsin (a+b x)} \cos (2 \arcsin (a+b x))}{5 b^3}-\frac {e^{\arcsin (a+b x)} \cos (3 \arcsin (a+b x))}{40 b^3}-\frac {a e^{\arcsin (a+b x)} \sin (2 \arcsin (a+b x))}{5 b^3}-\frac {3 e^{\arcsin (a+b x)} \sin (3 \arcsin (a+b x))}{40 b^3} \] Output:

1/8*exp(arcsin(b*x+a))*(b*x+a)/b^3+1/2*a^2*exp(arcsin(b*x+a))*(b*x+a)/b^3+ 
1/8*exp(arcsin(b*x+a))*(1-(b*x+a)^2)^(1/2)/b^3+1/2*a^2*exp(arcsin(b*x+a))* 
(1-(b*x+a)^2)^(1/2)/b^3+2/5*a*exp(arcsin(b*x+a))*cos(2*arcsin(b*x+a))/b^3- 
1/40*exp(arcsin(b*x+a))*cos(3*arcsin(b*x+a))/b^3-1/5*a*exp(arcsin(b*x+a))* 
sin(2*arcsin(b*x+a))/b^3-3/40*exp(arcsin(b*x+a))*sin(3*arcsin(b*x+a))/b^3
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.50 \[ \int e^{\arcsin (a+b x)} x^2 \, dx=\frac {e^{\arcsin (a+b x)} \left (5 (a+b x)+20 a^2 (a+b x)+5 \left (1+4 a^2\right ) \sqrt {1-(a+b x)^2}+16 a \cos (2 \arcsin (a+b x))-\cos (3 \arcsin (a+b x))-8 a \sin (2 \arcsin (a+b x))-3 \sin (3 \arcsin (a+b x))\right )}{40 b^3} \] Input:

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

Output:

(E^ArcSin[a + b*x]*(5*(a + b*x) + 20*a^2*(a + b*x) + 5*(1 + 4*a^2)*Sqrt[1 
- (a + b*x)^2] + 16*a*Cos[2*ArcSin[a + b*x]] - Cos[3*ArcSin[a + b*x]] - 8* 
a*Sin[2*ArcSin[a + b*x]] - 3*Sin[3*ArcSin[a + b*x]]))/(40*b^3)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5335, 7292, 27, 7293, 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.

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

\(\Big \downarrow \) 5335

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{2} a^2 (a+b x) e^{\arcsin (a+b x)}+\frac {1}{2} a^2 \sqrt {1-(a+b x)^2} e^{\arcsin (a+b x)}+\frac {1}{8} (a+b x) e^{\arcsin (a+b x)}+\frac {1}{8} \sqrt {1-(a+b x)^2} e^{\arcsin (a+b x)}-\frac {1}{5} a e^{\arcsin (a+b x)} \sin (2 \arcsin (a+b x))-\frac {3}{40} e^{\arcsin (a+b x)} \sin (3 \arcsin (a+b x))+\frac {2}{5} a e^{\arcsin (a+b x)} \cos (2 \arcsin (a+b x))-\frac {1}{40} e^{\arcsin (a+b x)} \cos (3 \arcsin (a+b x))}{b^3}\)

Input:

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

Output:

((E^ArcSin[a + b*x]*(a + b*x))/8 + (a^2*E^ArcSin[a + b*x]*(a + b*x))/2 + ( 
E^ArcSin[a + b*x]*Sqrt[1 - (a + b*x)^2])/8 + (a^2*E^ArcSin[a + b*x]*Sqrt[1 
 - (a + b*x)^2])/2 + (2*a*E^ArcSin[a + b*x]*Cos[2*ArcSin[a + b*x]])/5 - (E 
^ArcSin[a + b*x]*Cos[3*ArcSin[a + b*x]])/40 - (a*E^ArcSin[a + b*x]*Sin[2*A 
rcSin[a + b*x]])/5 - (3*E^ArcSin[a + b*x]*Sin[3*ArcSin[a + b*x]])/40)/b^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 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]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.41 \[ \int e^{\arcsin (a+b x)} x^2 \, dx=\frac {{\left (3 \, b^{3} x^{3} + a b^{2} x^{2} - {\left (2 \, a^{2} + 1\right )} b x + {\left (b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} + 1\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} + 3 \, a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{3}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19 \[ \int e^{\arcsin (a+b x)} x^2 \, dx=\begin {cases} - \frac {a^{2} x e^{\operatorname {asin}{\left (a + b x \right )}}}{5 b^{2}} + \frac {a^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{5 b^{3}} + \frac {a x^{2} e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b} - \frac {a x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{5 b^{2}} + \frac {3 a e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b^{3}} + \frac {3 x^{3} e^{\operatorname {asin}{\left (a + b x \right )}}}{10} + \frac {x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b} - \frac {x e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b^{2}} + \frac {\sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a + b x \right )}}}{10 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} e^{\operatorname {asin}{\left (a \right )}}}{3} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a**2*x*exp(asin(a + b*x))/(5*b**2) + a**2*sqrt(-a**2 - 2*a*b*x 
 - b**2*x**2 + 1)*exp(asin(a + b*x))/(5*b**3) + a*x**2*exp(asin(a + b*x))/ 
(10*b) - a*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*exp(asin(a + b*x))/(5*b 
**2) + 3*a*exp(asin(a + b*x))/(10*b**3) + 3*x**3*exp(asin(a + b*x))/10 + x 
**2*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*exp(asin(a + b*x))/(10*b) - x*ex 
p(asin(a + b*x))/(10*b**2) + sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*exp(asi 
n(a + b*x))/(10*b**3), Ne(b, 0)), (x**3*exp(asin(a))/3, True))
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.01 \[ \int e^{\arcsin (a+b x)} x^2 \, dx=\frac {{\left (b x + a\right )} a^{2} e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{3}} - \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a^{2} e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{3}} + \frac {3 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} {\left (b x + a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{3}} - \frac {4 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} a e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} - \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}} e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{3}} + \frac {{\left (b x + a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} - \frac {2 \, a e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{3}} \] Input:

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

Output:

1/2*(b*x + a)*a^2*e^(arcsin(b*x + a))/b^3 - 2/5*sqrt(-(b*x + a)^2 + 1)*(b* 
x + a)*a*e^(arcsin(b*x + a))/b^3 + 1/2*sqrt(-(b*x + a)^2 + 1)*a^2*e^(arcsi 
n(b*x + a))/b^3 + 3/10*((b*x + a)^2 - 1)*(b*x + a)*e^(arcsin(b*x + a))/b^3 
 - 4/5*((b*x + a)^2 - 1)*a*e^(arcsin(b*x + a))/b^3 - 1/10*(-(b*x + a)^2 + 
1)^(3/2)*e^(arcsin(b*x + a))/b^3 + 1/5*(b*x + a)*e^(arcsin(b*x + a))/b^3 - 
 2/5*a*e^(arcsin(b*x + a))/b^3 + 1/5*sqrt(-(b*x + a)^2 + 1)*e^(arcsin(b*x 
+ a))/b^3
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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