Integrand size = 10, antiderivative size = 81 \[ \int e^{\arcsin (a x)} x^3 \, dx=-\frac {e^{\arcsin (a x)} \cos (2 \arcsin (a x))}{10 a^4}+\frac {e^{\arcsin (a x)} \cos (4 \arcsin (a x))}{34 a^4}+\frac {e^{\arcsin (a x)} \sin (2 \arcsin (a x))}{20 a^4}-\frac {e^{\arcsin (a x)} \sin (4 \arcsin (a x))}{136 a^4} \] Output:
-1/10*exp(arcsin(a*x))*cos(2*arcsin(a*x))/a^4+1/34*exp(arcsin(a*x))*cos(4* arcsin(a*x))/a^4+1/20*exp(arcsin(a*x))*sin(2*arcsin(a*x))/a^4-1/136*exp(ar csin(a*x))*sin(4*arcsin(a*x))/a^4
Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int e^{\arcsin (a x)} x^3 \, dx=\frac {e^{\arcsin (a x)} (-68 \cos (2 \arcsin (a x))+20 \cos (4 \arcsin (a x))+34 \sin (2 \arcsin (a x))-5 \sin (4 \arcsin (a x)))}{680 a^4} \] Input:
Integrate[E^ArcSin[a*x]*x^3,x]
Output:
(E^ArcSin[a*x]*(-68*Cos[2*ArcSin[a*x]] + 20*Cos[4*ArcSin[a*x]] + 34*Sin[2* ArcSin[a*x]] - 5*Sin[4*ArcSin[a*x]]))/(680*a^4)
Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90, 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.
\(\displaystyle \int x^3 e^{\arcsin (a x)} \, dx\) |
\(\Big \downarrow \) 5335 |
\(\displaystyle \frac {\int e^{\arcsin (a x)} x^3 \sqrt {1-a^2 x^2}d\arcsin (a x)}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^3 e^{\arcsin (a x)} x^3 \sqrt {1-a^2 x^2}d\arcsin (a x)}{a^4}\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \frac {\int \left (\frac {1}{4} e^{\arcsin (a x)} \sin (2 \arcsin (a x))-\frac {1}{8} e^{\arcsin (a x)} \sin (4 \arcsin (a x))\right )d\arcsin (a x)}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{20} e^{\arcsin (a x)} \sin (2 \arcsin (a x))-\frac {1}{136} e^{\arcsin (a x)} \sin (4 \arcsin (a x))-\frac {1}{10} e^{\arcsin (a x)} \cos (2 \arcsin (a x))+\frac {1}{34} e^{\arcsin (a x)} \cos (4 \arcsin (a x))}{a^4}\) |
Input:
Int[E^ArcSin[a*x]*x^3,x]
Output:
(-1/10*(E^ArcSin[a*x]*Cos[2*ArcSin[a*x]]) + (E^ArcSin[a*x]*Cos[4*ArcSin[a* x]])/34 + (E^ArcSin[a*x]*Sin[2*ArcSin[a*x]])/20 - (E^ArcSin[a*x]*Sin[4*Arc Sin[a*x]])/136)/a^4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
\[\int {\mathrm e}^{\arcsin \left (a x \right )} x^{3}d x\]
Input:
int(exp(arcsin(a*x))*x^3,x)
Output:
int(exp(arcsin(a*x))*x^3,x)
Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.67 \[ \int e^{\arcsin (a x)} x^3 \, dx=\frac {{\left (20 \, a^{4} x^{4} - 3 \, a^{2} x^{2} + {\left (5 \, a^{3} x^{3} + 6 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} - 6\right )} e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a^{4}} \] Input:
integrate(exp(arcsin(a*x))*x^3,x, algorithm="fricas")
Output:
1/85*(20*a^4*x^4 - 3*a^2*x^2 + (5*a^3*x^3 + 6*a*x)*sqrt(-a^2*x^2 + 1) - 6) *e^(arcsin(a*x))/a^4
Time = 0.44 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.23 \[ \int e^{\arcsin (a x)} x^3 \, dx=\begin {cases} \frac {4 x^{4} e^{\operatorname {asin}{\left (a x \right )}}}{17} + \frac {x^{3} \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{17 a} - \frac {3 x^{2} e^{\operatorname {asin}{\left (a x \right )}}}{85 a^{2}} + \frac {6 x \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{85 a^{3}} - \frac {6 e^{\operatorname {asin}{\left (a x \right )}}}{85 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(exp(asin(a*x))*x**3,x)
Output:
Piecewise((4*x**4*exp(asin(a*x))/17 + x**3*sqrt(-a**2*x**2 + 1)*exp(asin(a *x))/(17*a) - 3*x**2*exp(asin(a*x))/(85*a**2) + 6*x*sqrt(-a**2*x**2 + 1)*e xp(asin(a*x))/(85*a**3) - 6*exp(asin(a*x))/(85*a**4), Ne(a, 0)), (x**4/4, True))
\[ \int e^{\arcsin (a x)} x^3 \, dx=\int { x^{3} e^{\left (\arcsin \left (a x\right )\right )} \,d x } \] Input:
integrate(exp(arcsin(a*x))*x^3,x, algorithm="maxima")
Output:
integrate(x^3*e^(arcsin(a*x)), x)
Time = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.20 \[ \int e^{\arcsin (a x)} x^3 \, dx=-\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x e^{\left (\arcsin \left (a x\right )\right )}}{17 \, a^{3}} + \frac {11 \, \sqrt {-a^{2} x^{2} + 1} x e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a^{3}} + \frac {4 \, {\left (a^{2} x^{2} - 1\right )}^{2} e^{\left (\arcsin \left (a x\right )\right )}}{17 \, a^{4}} + \frac {37 \, {\left (a^{2} x^{2} - 1\right )} e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a^{4}} + \frac {11 \, e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a^{4}} \] Input:
integrate(exp(arcsin(a*x))*x^3,x, algorithm="giac")
Output:
-1/17*(-a^2*x^2 + 1)^(3/2)*x*e^(arcsin(a*x))/a^3 + 11/85*sqrt(-a^2*x^2 + 1 )*x*e^(arcsin(a*x))/a^3 + 4/17*(a^2*x^2 - 1)^2*e^(arcsin(a*x))/a^4 + 37/85 *(a^2*x^2 - 1)*e^(arcsin(a*x))/a^4 + 11/85*e^(arcsin(a*x))/a^4
Timed out. \[ \int e^{\arcsin (a x)} x^3 \, dx=\int x^3\,{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )} \,d x \] Input:
int(x^3*exp(asin(a*x)),x)
Output:
int(x^3*exp(asin(a*x)), x)
\[ \int e^{\arcsin (a x)} x^3 \, dx=\int e^{\mathit {asin} \left (a x \right )} x^{3}d x \] Input:
int(exp(asin(a*x))*x^3,x)
Output:
int(e**asin(a*x)*x**3,x)