3.451 \(\int e^{\sin ^{-1}(a+b x)} x^3 \, dx\)

Optimal. Leaf size=309 \[ -\frac{a^3 (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^4}-\frac{a^3 \sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^4}+\frac{3 a^2 e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^4}-\frac{3 a^2 e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^4}-\frac{3 a (a+b x) e^{\sin ^{-1}(a+b x)}}{8 b^4}+\frac{9 a e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^4}-\frac{3 a \sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{8 b^4}+\frac{e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{20 b^4}-\frac{e^{\sin ^{-1}(a+b x)} \sin \left (4 \sin ^{-1}(a+b x)\right )}{136 b^4}+\frac{3 a e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^4}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{10 b^4}+\frac{e^{\sin ^{-1}(a+b x)} \cos \left (4 \sin ^{-1}(a+b x)\right )}{34 b^4} \]

[Out]

(-3*a*E^ArcSin[a + b*x]*(a + b*x))/(8*b^4) - (a^3*E^ArcSin[a + b*x]*(a + b*x))/(2*b^4) - (3*a*E^ArcSin[a + b*x
]*Sqrt[1 - (a + b*x)^2])/(8*b^4) - (a^3*E^ArcSin[a + b*x]*Sqrt[1 - (a + b*x)^2])/(2*b^4) - (E^ArcSin[a + b*x]*
Cos[2*ArcSin[a + b*x]])/(10*b^4) - (3*a^2*E^ArcSin[a + b*x]*Cos[2*ArcSin[a + b*x]])/(5*b^4) + (3*a*E^ArcSin[a
+ b*x]*Cos[3*ArcSin[a + b*x]])/(40*b^4) + (E^ArcSin[a + b*x]*Cos[4*ArcSin[a + b*x]])/(34*b^4) + (E^ArcSin[a +
b*x]*Sin[2*ArcSin[a + b*x]])/(20*b^4) + (3*a^2*E^ArcSin[a + b*x]*Sin[2*ArcSin[a + b*x]])/(10*b^4) + (9*a*E^Arc
Sin[a + b*x]*Sin[3*ArcSin[a + b*x]])/(40*b^4) - (E^ArcSin[a + b*x]*Sin[4*ArcSin[a + b*x]])/(136*b^4)

________________________________________________________________________________________

Rubi [A]  time = 0.527435, antiderivative size = 309, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {4836, 6741, 12, 6742, 4433, 4469, 4432} \[ -\frac{a^3 (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^4}-\frac{a^3 \sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^4}+\frac{3 a^2 e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^4}-\frac{3 a^2 e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^4}-\frac{3 a (a+b x) e^{\sin ^{-1}(a+b x)}}{8 b^4}+\frac{9 a e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^4}-\frac{3 a \sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{8 b^4}+\frac{e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{20 b^4}-\frac{e^{\sin ^{-1}(a+b x)} \sin \left (4 \sin ^{-1}(a+b x)\right )}{136 b^4}+\frac{3 a e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^4}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{10 b^4}+\frac{e^{\sin ^{-1}(a+b x)} \cos \left (4 \sin ^{-1}(a+b x)\right )}{34 b^4} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a + b*x]*x^3,x]

[Out]

(-3*a*E^ArcSin[a + b*x]*(a + b*x))/(8*b^4) - (a^3*E^ArcSin[a + b*x]*(a + b*x))/(2*b^4) - (3*a*E^ArcSin[a + b*x
]*Sqrt[1 - (a + b*x)^2])/(8*b^4) - (a^3*E^ArcSin[a + b*x]*Sqrt[1 - (a + b*x)^2])/(2*b^4) - (E^ArcSin[a + b*x]*
Cos[2*ArcSin[a + b*x]])/(10*b^4) - (3*a^2*E^ArcSin[a + b*x]*Cos[2*ArcSin[a + b*x]])/(5*b^4) + (3*a*E^ArcSin[a
+ b*x]*Cos[3*ArcSin[a + b*x]])/(40*b^4) + (E^ArcSin[a + b*x]*Cos[4*ArcSin[a + b*x]])/(34*b^4) + (E^ArcSin[a +
b*x]*Sin[2*ArcSin[a + b*x]])/(20*b^4) + (3*a^2*E^ArcSin[a + b*x]*Sin[2*ArcSin[a + b*x]])/(10*b^4) + (9*a*E^Arc
Sin[a + b*x]*Sin[3*ArcSin[a + b*x]])/(40*b^4) - (E^ArcSin[a + b*x]*Sin[4*ArcSin[a + b*x]])/(136*b^4)

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 6741

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

Rule 12

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

Rule 6742

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

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]

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 4432

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*S
in[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] - Simp[(e*F^(c*(a + b*x))*Cos[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+b x)} x^3 \, dx &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \left (-\frac{a}{b}+\frac{\sin (x)}{b}\right )^3 \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{e^x \cos (x) (-a+\sin (x))^3}{b^3} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) (-a+\sin (x))^3 \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a^3 e^x \cos (x)+3 a^2 e^x \cos (x) \sin (x)-3 a e^x \cos (x) \sin ^2(x)+e^x \cos (x) \sin ^3(x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \sin ^3(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}-\frac{(3 a) \operatorname{Subst}\left (\int e^x \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int e^x \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}-\frac{a^3 \operatorname{Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}\\ &=-\frac{a^3 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^4}-\frac{a^3 e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^4}+\frac{\operatorname{Subst}\left (\int \left (\frac{1}{4} e^x \sin (2 x)-\frac{1}{8} e^x \sin (4 x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}-\frac{(3 a) \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+b x)\right )}{b^4}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2} e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}\\ &=-\frac{a^3 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^4}-\frac{a^3 e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^4}-\frac{\operatorname{Subst}\left (\int e^x \sin (4 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}+\frac{\operatorname{Subst}\left (\int e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^4}-\frac{(3 a) \operatorname{Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^4}+\frac{(3 a) \operatorname{Subst}\left (\int e^x \cos (3 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^4}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^4}\\ &=-\frac{3 a e^{\sin ^{-1}(a+b x)} (a+b x)}{8 b^4}-\frac{a^3 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^4}-\frac{3 a e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{8 b^4}-\frac{a^3 e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^4}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{10 b^4}-\frac{3 a^2 e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^4}+\frac{3 a e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^4}+\frac{e^{\sin ^{-1}(a+b x)} \cos \left (4 \sin ^{-1}(a+b x)\right )}{34 b^4}+\frac{e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{20 b^4}+\frac{3 a^2 e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^4}+\frac{9 a e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^4}-\frac{e^{\sin ^{-1}(a+b x)} \sin \left (4 \sin ^{-1}(a+b x)\right )}{136 b^4}\\ \end{align*}

Mathematica [A]  time = 0.412054, size = 148, normalized size = 0.48 \[ \frac{e^{\sin ^{-1}(a+b x)} \left (-340 a^3 (a+b x)-85 \left (4 a^2+3\right ) a \sqrt{1-(a+b x)^2}+204 a^2 \sin \left (2 \sin ^{-1}(a+b x)\right )-68 \left (6 a^2+1\right ) \cos \left (2 \sin ^{-1}(a+b x)\right )-255 a (a+b x)+153 a \sin \left (3 \sin ^{-1}(a+b x)\right )+34 \sin \left (2 \sin ^{-1}(a+b x)\right )-5 \sin \left (4 \sin ^{-1}(a+b x)\right )+51 a \cos \left (3 \sin ^{-1}(a+b x)\right )+20 \cos \left (4 \sin ^{-1}(a+b x)\right )\right )}{680 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSin[a + b*x]*x^3,x]

[Out]

(E^ArcSin[a + b*x]*(-255*a*(a + b*x) - 340*a^3*(a + b*x) - 85*a*(3 + 4*a^2)*Sqrt[1 - (a + b*x)^2] - 68*(1 + 6*
a^2)*Cos[2*ArcSin[a + b*x]] + 51*a*Cos[3*ArcSin[a + b*x]] + 20*Cos[4*ArcSin[a + b*x]] + 34*Sin[2*ArcSin[a + b*
x]] + 204*a^2*Sin[2*ArcSin[a + b*x]] + 153*a*Sin[3*ArcSin[a + b*x]] - 5*Sin[4*ArcSin[a + b*x]]))/(680*b^4)

________________________________________________________________________________________

Maple [F]  time = 0.012, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{\arcsin \left ( bx+a \right ) }}{x}^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arcsin(b*x+a))*x^3,x)

[Out]

int(exp(arcsin(b*x+a))*x^3,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} e^{\left (\arcsin \left (b x + a\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsin(b*x+a))*x^3,x, algorithm="maxima")

[Out]

integrate(x^3*e^(arcsin(b*x + a)), x)

________________________________________________________________________________________

Fricas [A]  time = 2.23107, size = 308, normalized size = 1. \begin{align*} \frac{{\left (40 \, b^{4} x^{4} + 7 \, a b^{3} x^{3} - 3 \,{\left (5 \, a^{2} + 2\right )} b^{2} x^{2} + 6 \, a^{4} + 3 \,{\left (8 \, a^{3} + 13 \, a\right )} b x - 57 \, a^{2} +{\left (10 \, b^{3} x^{3} - 21 \, a b^{2} x^{2} - 24 \, a^{3} + 6 \,{\left (5 \, a^{2} + 2\right )} b x - 39 \, a\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} - 12\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{170 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsin(b*x+a))*x^3,x, algorithm="fricas")

[Out]

1/170*(40*b^4*x^4 + 7*a*b^3*x^3 - 3*(5*a^2 + 2)*b^2*x^2 + 6*a^4 + 3*(8*a^3 + 13*a)*b*x - 57*a^2 + (10*b^3*x^3
- 21*a*b^2*x^2 - 24*a^3 + 6*(5*a^2 + 2)*b*x - 39*a)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1) - 12)*e^(arcsin(b*x + a
))/b^4

________________________________________________________________________________________

Sympy [A]  time = 6.1733, size = 416, normalized size = 1.35 \begin{align*} \begin{cases} \frac{3 a^{4} e^{\operatorname{asin}{\left (a + b x \right )}}}{85 b^{4}} + \frac{12 a^{3} x e^{\operatorname{asin}{\left (a + b x \right )}}}{85 b^{3}} - \frac{12 a^{3} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a + b x \right )}}}{85 b^{4}} - \frac{3 a^{2} x^{2} e^{\operatorname{asin}{\left (a + b x \right )}}}{34 b^{2}} + \frac{3 a^{2} x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a + b x \right )}}}{17 b^{3}} - \frac{57 a^{2} e^{\operatorname{asin}{\left (a + b x \right )}}}{170 b^{4}} + \frac{7 a x^{3} e^{\operatorname{asin}{\left (a + b x \right )}}}{170 b} - \frac{21 a x^{2} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a + b x \right )}}}{170 b^{2}} + \frac{39 a x e^{\operatorname{asin}{\left (a + b x \right )}}}{170 b^{3}} - \frac{39 a \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a + b x \right )}}}{170 b^{4}} + \frac{4 x^{4} e^{\operatorname{asin}{\left (a + b x \right )}}}{17} + \frac{x^{3} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a + b x \right )}}}{17 b} - \frac{3 x^{2} e^{\operatorname{asin}{\left (a + b x \right )}}}{85 b^{2}} + \frac{6 x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a + b x \right )}}}{85 b^{3}} - \frac{6 e^{\operatorname{asin}{\left (a + b x \right )}}}{85 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} e^{\operatorname{asin}{\left (a \right )}}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(asin(b*x+a))*x**3,x)

[Out]

Piecewise((3*a**4*exp(asin(a + b*x))/(85*b**4) + 12*a**3*x*exp(asin(a + b*x))/(85*b**3) - 12*a**3*sqrt(-a**2 -
2*a*b*x - b**2*x**2 + 1)*exp(asin(a + b*x))/(85*b**4) - 3*a**2*x**2*exp(asin(a + b*x))/(34*b**2) + 3*a**2*x*s
qrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*exp(asin(a + b*x))/(17*b**3) - 57*a**2*exp(asin(a + b*x))/(170*b**4) + 7*
a*x**3*exp(asin(a + b*x))/(170*b) - 21*a*x**2*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*exp(asin(a + b*x))/(170*b*
*2) + 39*a*x*exp(asin(a + b*x))/(170*b**3) - 39*a*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*exp(asin(a + b*x))/(17
0*b**4) + 4*x**4*exp(asin(a + b*x))/17 + x**3*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*exp(asin(a + b*x))/(17*b)
- 3*x**2*exp(asin(a + b*x))/(85*b**2) + 6*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*exp(asin(a + b*x))/(85*b**3)
- 6*exp(asin(a + b*x))/(85*b**4), Ne(b, 0)), (x**4*exp(asin(a))/4, True))

________________________________________________________________________________________

Giac [A]  time = 1.26267, size = 451, normalized size = 1.46 \begin{align*} -\frac{{\left (b x + a\right )} a^{3} e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{4}} + \frac{3 \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} a^{2} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{4}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1} a^{3} e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{4}} - \frac{9 \,{\left ({\left (b x + a\right )}^{2} - 1\right )}{\left (b x + a\right )} a e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{4}} + \frac{6 \,{\left ({\left (b x + a\right )}^{2} - 1\right )} a^{2} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{4}} - \frac{{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}}{\left (b x + a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{17 \, b^{4}} + \frac{3 \,{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}} a e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{4}} + \frac{4 \,{\left ({\left (b x + a\right )}^{2} - 1\right )}^{2} e^{\left (\arcsin \left (b x + a\right )\right )}}{17 \, b^{4}} - \frac{3 \,{\left (b x + a\right )} a e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{4}} + \frac{3 \, a^{2} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{4}} + \frac{11 \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{85 \, b^{4}} - \frac{3 \, \sqrt{-{\left (b x + a\right )}^{2} + 1} a e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{4}} + \frac{37 \,{\left ({\left (b x + a\right )}^{2} - 1\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{85 \, b^{4}} + \frac{11 \, e^{\left (\arcsin \left (b x + a\right )\right )}}{85 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsin(b*x+a))*x^3,x, algorithm="giac")

[Out]

-1/2*(b*x + a)*a^3*e^(arcsin(b*x + a))/b^4 + 3/5*sqrt(-(b*x + a)^2 + 1)*(b*x + a)*a^2*e^(arcsin(b*x + a))/b^4
- 1/2*sqrt(-(b*x + a)^2 + 1)*a^3*e^(arcsin(b*x + a))/b^4 - 9/10*((b*x + a)^2 - 1)*(b*x + a)*a*e^(arcsin(b*x +
a))/b^4 + 6/5*((b*x + a)^2 - 1)*a^2*e^(arcsin(b*x + a))/b^4 - 1/17*(-(b*x + a)^2 + 1)^(3/2)*(b*x + a)*e^(arcsi
n(b*x + a))/b^4 + 3/10*(-(b*x + a)^2 + 1)^(3/2)*a*e^(arcsin(b*x + a))/b^4 + 4/17*((b*x + a)^2 - 1)^2*e^(arcsin
(b*x + a))/b^4 - 3/5*(b*x + a)*a*e^(arcsin(b*x + a))/b^4 + 3/5*a^2*e^(arcsin(b*x + a))/b^4 + 11/85*sqrt(-(b*x
+ a)^2 + 1)*(b*x + a)*e^(arcsin(b*x + a))/b^4 - 3/5*sqrt(-(b*x + a)^2 + 1)*a*e^(arcsin(b*x + a))/b^4 + 37/85*(
(b*x + a)^2 - 1)*e^(arcsin(b*x + a))/b^4 + 11/85*e^(arcsin(b*x + a))/b^4