3.10 \(\int \frac{e^{\sin ^{-1}(x)} x^3}{\sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=44 \[ \frac{1}{10} \left (x^3-3 \sqrt{1-x^2} x^2-3 \sqrt{1-x^2}+3 x\right ) e^{\sin ^{-1}(x)} \]

[Out]

(E^ArcSin[x]*(3*x + x^3 - 3*Sqrt[1 - x^2] - 3*x^2*Sqrt[1 - x^2]))/10

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Rubi [A]  time = 0.67604, antiderivative size = 62, normalized size of antiderivative = 1.41, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {4836, 6741, 6720, 4434, 4432} \[ \frac{1}{10} x^3 e^{\sin ^{-1}(x)}-\frac{3}{10} \sqrt{1-x^2} x^2 e^{\sin ^{-1}(x)}-\frac{3}{10} \sqrt{1-x^2} e^{\sin ^{-1}(x)}+\frac{3}{10} x e^{\sin ^{-1}(x)} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcSin[x]*x^3)/Sqrt[1 - x^2],x]

[Out]

(3*E^ArcSin[x]*x)/10 + (E^ArcSin[x]*x^3)/10 - (3*E^ArcSin[x]*Sqrt[1 - x^2])/10 - (3*E^ArcSin[x]*x^2*Sqrt[1 - x
^2])/10

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 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rule 4434

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*
x))*Sin[d + e*x]^n)/(e^2*n^2 + b^2*c^2*Log[F]^2), x] + (Dist[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x] - Simp[(e*n*F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x]^(n - 1))/(
e^2*n^2 + b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
n, 1]

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 \frac{e^{\sin ^{-1}(x)} x^3}{\sqrt{1-x^2}} \, dx &=\operatorname{Subst}\left (\int \frac{e^x \cos (x) \sin ^3(x)}{\sqrt{1-\sin ^2(x)}} \, dx,x,\sin ^{-1}(x)\right )\\ &=\operatorname{Subst}\left (\int \frac{e^x \cos (x) \sin ^3(x)}{\sqrt{\cos ^2(x)}} \, dx,x,\sin ^{-1}(x)\right )\\ &=1 \operatorname{Subst}\left (\int e^x \sin ^3(x) \, dx,x,\sin ^{-1}(x)\right )\\ &=\frac{1}{10} e^{\sin ^{-1}(x)} x^3-\frac{3}{10} e^{\sin ^{-1}(x)} x^2 \sqrt{1-x^2}+\frac{3}{5} \operatorname{Subst}\left (\int e^x \sin (x) \, dx,x,\sin ^{-1}(x)\right )\\ &=\frac{3}{10} e^{\sin ^{-1}(x)} x+\frac{1}{10} e^{\sin ^{-1}(x)} x^3-\frac{3}{10} e^{\sin ^{-1}(x)} \sqrt{1-x^2}-\frac{3}{10} e^{\sin ^{-1}(x)} x^2 \sqrt{1-x^2}\\ \end{align*}

Mathematica [A]  time = 0.157857, size = 38, normalized size = 0.86 \[ -\frac{1}{40} e^{\sin ^{-1}(x)} \left (15 \left (\sqrt{1-x^2}-x\right )+\sin \left (3 \sin ^{-1}(x)\right )-3 \cos \left (3 \sin ^{-1}(x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcSin[x]*x^3)/Sqrt[1 - x^2],x]

[Out]

-(E^ArcSin[x]*(15*(-x + Sqrt[1 - x^2]) - 3*Cos[3*ArcSin[x]] + Sin[3*ArcSin[x]]))/40

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{\arcsin \left ( x \right ) }}{x}^{3}{\frac{1}{\sqrt{-{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arcsin(x))*x^3/(-x^2+1)^(1/2),x)

[Out]

int(exp(arcsin(x))*x^3/(-x^2+1)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} e^{\arcsin \left (x\right )}}{\sqrt{-x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsin(x))*x^3/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3*e^arcsin(x)/sqrt(-x^2 + 1), x)

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Fricas [A]  time = 2.72596, size = 80, normalized size = 1.82 \begin{align*} \frac{1}{10} \,{\left (x^{3} - 3 \,{\left (x^{2} + 1\right )} \sqrt{-x^{2} + 1} + 3 \, x\right )} e^{\arcsin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsin(x))*x^3/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.14133, size = 56, normalized size = 1.27 \begin{align*} \frac{x^{3} e^{\operatorname{asin}{\left (x \right )}}}{10} - \frac{3 x^{2} \sqrt{1 - x^{2}} e^{\operatorname{asin}{\left (x \right )}}}{10} + \frac{3 x e^{\operatorname{asin}{\left (x \right )}}}{10} - \frac{3 \sqrt{1 - x^{2}} e^{\operatorname{asin}{\left (x \right )}}}{10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(asin(x))*x**3/(-x**2+1)**(1/2),x)

[Out]

x**3*exp(asin(x))/10 - 3*x**2*sqrt(1 - x**2)*exp(asin(x))/10 + 3*x*exp(asin(x))/10 - 3*sqrt(1 - x**2)*exp(asin
(x))/10

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Giac [A]  time = 1.15492, size = 62, normalized size = 1.41 \begin{align*} \frac{1}{10} \,{\left (x^{2} - 1\right )} x e^{\arcsin \left (x\right )} + \frac{3}{10} \,{\left (-x^{2} + 1\right )}^{\frac{3}{2}} e^{\arcsin \left (x\right )} + \frac{2}{5} \, x e^{\arcsin \left (x\right )} - \frac{3}{5} \, \sqrt{-x^{2} + 1} e^{\arcsin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsin(x))*x^3/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/10*(x^2 - 1)*x*e^arcsin(x) + 3/10*(-x^2 + 1)^(3/2)*e^arcsin(x) + 2/5*x*e^arcsin(x) - 3/5*sqrt(-x^2 + 1)*e^ar
csin(x)