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

Optimal. Leaf size=205 \[ \frac{a^2 (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^3}+\frac{a^2 \sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^3}-\frac{a e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}+\frac{(a+b x) e^{\sin ^{-1}(a+b x)}}{8 b^3}-\frac{3 e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3}+\frac{\sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{8 b^3}+\frac{2 a e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3} \]

[Out]

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

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Rubi [A]  time = 0.364148, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 13, 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} \[ \frac{a^2 (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^3}+\frac{a^2 \sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^3}-\frac{a e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}+\frac{(a+b x) e^{\sin ^{-1}(a+b x)}}{8 b^3}-\frac{3 e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3}+\frac{\sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{8 b^3}+\frac{2 a e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a + b*x]*x^2,x]

[Out]

(E^ArcSin[a + b*x]*(a + b*x))/(8*b^3) + (a^2*E^ArcSin[a + b*x]*(a + b*x))/(2*b^3) + (E^ArcSin[a + b*x]*Sqrt[1
- (a + b*x)^2])/(8*b^3) + (a^2*E^ArcSin[a + b*x]*Sqrt[1 - (a + b*x)^2])/(2*b^3) + (2*a*E^ArcSin[a + b*x]*Cos[2
*ArcSin[a + b*x]])/(5*b^3) - (E^ArcSin[a + b*x]*Cos[3*ArcSin[a + b*x]])/(40*b^3) - (a*E^ArcSin[a + b*x]*Sin[2*
ArcSin[a + b*x]])/(5*b^3) - (3*E^ArcSin[a + b*x]*Sin[3*ArcSin[a + b*x]])/(40*b^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 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^2 \, dx &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \left (-\frac{a}{b}+\frac{\sin (x)}{b}\right )^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{e^x \cos (x) (a-\sin (x))^2}{b^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) (a-\sin (x))^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2 e^x \cos (x)-2 a e^x \cos (x) \sin (x)+e^x \cos (x) \sin ^2(x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac{(2 a) \operatorname{Subst}\left (\int e^x \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}+\frac{a^2 \operatorname{Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{a^2 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^3}+\frac{a^2 e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^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+b x)\right )}{b^3}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{2} e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{a^2 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^3}+\frac{a^2 e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^3}+\frac{\operatorname{Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{\operatorname{Subst}\left (\int e^x \cos (3 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{a \operatorname{Subst}\left (\int e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{e^{\sin ^{-1}(a+b x)} (a+b x)}{8 b^3}+\frac{a^2 e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^3}+\frac{e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{8 b^3}+\frac{a^2 e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^3}+\frac{2 a e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3}-\frac{a e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{5 b^3}-\frac{3 e^{\sin ^{-1}(a+b x)} \sin \left (3 \sin ^{-1}(a+b x)\right )}{40 b^3}\\ \end{align*}

Mathematica [A]  time = 0.220149, size = 103, normalized size = 0.5 \[ \frac{e^{\sin ^{-1}(a+b x)} \left (20 a^2 (a+b x)+5 \left (4 a^2+1\right ) \sqrt{1-(a+b x)^2}+5 (a+b x)-8 a \sin \left (2 \sin ^{-1}(a+b x)\right )-3 \sin \left (3 \sin ^{-1}(a+b x)\right )+16 a \cos \left (2 \sin ^{-1}(a+b x)\right )-\cos \left (3 \sin ^{-1}(a+b x)\right )\right )}{40 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSin[a + b*x]*x^2,x]

[Out]

(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)

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Maple [F]  time = 0.01, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{\arcsin \left ( bx+a \right ) }}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^2*e^(arcsin(b*x + a)), x)

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Fricas [A]  time = 2.0188, size = 198, normalized size = 0.97 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.33011, size = 243, normalized size = 1.19 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19495, size = 281, normalized size = 1.37 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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^(arcsin(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