3.5.39 \(\int e^{\text {ArcSin}(a x)} x^3 \, dx\) [439]

Optimal. Leaf size=81 \[ -\frac {e^{\text {ArcSin}(a x)} \cos (2 \text {ArcSin}(a x))}{10 a^4}+\frac {e^{\text {ArcSin}(a x)} \cos (4 \text {ArcSin}(a x))}{34 a^4}+\frac {e^{\text {ArcSin}(a x)} \sin (2 \text {ArcSin}(a x))}{20 a^4}-\frac {e^{\text {ArcSin}(a x)} \sin (4 \text {ArcSin}(a x))}{136 a^4} \]

[Out]

-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(arcsin(a*x))*sin(4*arcsin(a*x))/a^4

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Rubi [A]
time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4920, 12, 4557, 4517} \begin {gather*} \frac {e^{\text {ArcSin}(a x)} \sin (2 \text {ArcSin}(a x))}{20 a^4}-\frac {e^{\text {ArcSin}(a x)} \sin (4 \text {ArcSin}(a x))}{136 a^4}-\frac {e^{\text {ArcSin}(a x)} \cos (2 \text {ArcSin}(a x))}{10 a^4}+\frac {e^{\text {ArcSin}(a x)} \cos (4 \text {ArcSin}(a x))}{34 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a*x]*x^3,x]

[Out]

-1/10*(E^ArcSin[a*x]*Cos[2*ArcSin[a*x]])/a^4 + (E^ArcSin[a*x]*Cos[4*ArcSin[a*x]])/(34*a^4) + (E^ArcSin[a*x]*Si
n[2*ArcSin[a*x]])/(20*a^4) - (E^ArcSin[a*x]*Sin[4*ArcSin[a*x]])/(136*a^4)

Rule 12

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

Rule 4517

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]

Rule 4557

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 4920

Int[(u_.)*(f_)^(ArcSin[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[(u /. x -> -a/b + Si
n[x]/b)*f^(c*x^n)*Cos[x], x], x, ArcSin[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int e^{\sin ^{-1}(a x)} x^3 \, dx &=\frac {\text {Subst}\left (\int \frac {e^x \cos (x) \sin ^3(x)}{a^3} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac {\text {Subst}\left (\int e^x \cos (x) \sin ^3(x) \, dx,x,\sin ^{-1}(a x)\right )}{a^4}\\ &=\frac {\text {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 x)\right )}{a^4}\\ &=-\frac {\text {Subst}\left (\int e^x \sin (4 x) \, dx,x,\sin ^{-1}(a x)\right )}{8 a^4}+\frac {\text {Subst}\left (\int e^x \sin (2 x) \, dx,x,\sin ^{-1}(a x)\right )}{4 a^4}\\ &=-\frac {e^{\sin ^{-1}(a x)} \cos \left (2 \sin ^{-1}(a x)\right )}{10 a^4}+\frac {e^{\sin ^{-1}(a x)} \cos \left (4 \sin ^{-1}(a x)\right )}{34 a^4}+\frac {e^{\sin ^{-1}(a x)} \sin \left (2 \sin ^{-1}(a x)\right )}{20 a^4}-\frac {e^{\sin ^{-1}(a x)} \sin \left (4 \sin ^{-1}(a x)\right )}{136 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 50, normalized size = 0.62 \begin {gather*} \frac {e^{\text {ArcSin}(a x)} (-68 \cos (2 \text {ArcSin}(a x))+20 \cos (4 \text {ArcSin}(a x))+34 \sin (2 \text {ArcSin}(a x))-5 \sin (4 \text {ArcSin}(a x)))}{680 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSin[a*x]*x^3,x]

[Out]

(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)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{\arcsin \left (a x \right )} x^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*e^(arcsin(a*x)), x)

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Fricas [A]
time = 2.86, size = 54, normalized size = 0.67 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.54, size = 100, normalized size = 1.23 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*exp(asin(a*x))/(85*a**3) - 6*exp(asin(a*x))/(85*a**4), Ne(a, 0)), (x**4/4
, True))

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Giac [A]
time = 0.40, size = 97, normalized size = 1.20 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*exp(asin(a*x)),x)

[Out]

int(x^3*exp(asin(a*x)), x)

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