3.5.64 \(\int e^{\text {ArcSin}(a x)} (1-a^2 x^2)^{3/2} \, dx\) [464]

Optimal. Leaf size=112 \[ \frac {24 e^{\text {ArcSin}(a x)}}{85 a}+\frac {24}{85} e^{\text {ArcSin}(a x)} x \sqrt {1-a^2 x^2}+\frac {12 e^{\text {ArcSin}(a x)} \left (1-a^2 x^2\right )}{85 a}+\frac {4}{17} e^{\text {ArcSin}(a x)} x \left (1-a^2 x^2\right )^{3/2}+\frac {e^{\text {ArcSin}(a x)} \left (1-a^2 x^2\right )^2}{17 a} \]

[Out]

24/85*exp(arcsin(a*x))/a+12/85*exp(arcsin(a*x))*(-a^2*x^2+1)/a+4/17*exp(arcsin(a*x))*x*(-a^2*x^2+1)^(3/2)+1/17
*exp(arcsin(a*x))*(-a^2*x^2+1)^2/a+24/85*exp(arcsin(a*x))*x*(-a^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4920, 6820, 6852, 4520, 2225} \begin {gather*} \frac {\left (1-a^2 x^2\right )^2 e^{\text {ArcSin}(a x)}}{17 a}+\frac {4}{17} x \left (1-a^2 x^2\right )^{3/2} e^{\text {ArcSin}(a x)}+\frac {12 \left (1-a^2 x^2\right ) e^{\text {ArcSin}(a x)}}{85 a}+\frac {24}{85} x \sqrt {1-a^2 x^2} e^{\text {ArcSin}(a x)}+\frac {24 e^{\text {ArcSin}(a x)}}{85 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(24*E^ArcSin[a*x])/(85*a) + (24*E^ArcSin[a*x]*x*Sqrt[1 - a^2*x^2])/85 + (12*E^ArcSin[a*x]*(1 - a^2*x^2))/(85*a
) + (4*E^ArcSin[a*x]*x*(1 - a^2*x^2)^(3/2))/17 + (E^ArcSin[a*x]*(1 - a^2*x^2)^2)/(17*a)

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 4520

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

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]

Rule 6820

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

Rule 6852

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])

Rubi steps

\begin {align*} \int e^{\sin ^{-1}(a x)} \left (1-a^2 x^2\right )^{3/2} \, dx &=\frac {\text {Subst}\left (\int e^x \cos (x) \left (1-\sin ^2(x)\right )^{3/2} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac {\text {Subst}\left (\int e^x \cos (x) \cos ^2(x)^{3/2} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac {\text {Subst}\left (\int e^x \cos ^4(x) \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=\frac {4}{17} e^{\sin ^{-1}(a x)} x \left (1-a^2 x^2\right )^{3/2}+\frac {e^{\sin ^{-1}(a x)} \left (1-a^2 x^2\right )^2}{17 a}+\frac {12 \text {Subst}\left (\int e^x \cos ^2(x) \, dx,x,\sin ^{-1}(a x)\right )}{17 a}\\ &=\frac {24}{85} e^{\sin ^{-1}(a x)} x \sqrt {1-a^2 x^2}+\frac {12 e^{\sin ^{-1}(a x)} \left (1-a^2 x^2\right )}{85 a}+\frac {4}{17} e^{\sin ^{-1}(a x)} x \left (1-a^2 x^2\right )^{3/2}+\frac {e^{\sin ^{-1}(a x)} \left (1-a^2 x^2\right )^2}{17 a}+\frac {24 \text {Subst}\left (\int e^x \, dx,x,\sin ^{-1}(a x)\right )}{85 a}\\ &=\frac {24 e^{\sin ^{-1}(a x)}}{85 a}+\frac {24}{85} e^{\sin ^{-1}(a x)} x \sqrt {1-a^2 x^2}+\frac {12 e^{\sin ^{-1}(a x)} \left (1-a^2 x^2\right )}{85 a}+\frac {4}{17} e^{\sin ^{-1}(a x)} x \left (1-a^2 x^2\right )^{3/2}+\frac {e^{\sin ^{-1}(a x)} \left (1-a^2 x^2\right )^2}{17 a}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 51, normalized size = 0.46 \begin {gather*} \frac {e^{\text {ArcSin}(a x)} (255+68 \cos (2 \text {ArcSin}(a x))+5 \cos (4 \text {ArcSin}(a x))+136 \sin (2 \text {ArcSin}(a x))+20 \sin (4 \text {ArcSin}(a x)))}{680 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcSin[a*x]*(255 + 68*Cos[2*ArcSin[a*x]] + 5*Cos[4*ArcSin[a*x]] + 136*Sin[2*ArcSin[a*x]] + 20*Sin[4*ArcSin[
a*x]]))/(680*a)

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{\arcsin \left (a x \right )} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 2.25, size = 55, normalized size = 0.49 \begin {gather*} \frac {{\left (5 \, a^{4} x^{4} - 22 \, a^{2} x^{2} - 4 \, {\left (5 \, a^{3} x^{3} - 11 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} + 41\right )} e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/85*(5*a^4*x^4 - 22*a^2*x^2 - 4*(5*a^3*x^3 - 11*a*x)*sqrt(-a^2*x^2 + 1) + 41)*e^(arcsin(a*x))/a

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Sympy [A]
time = 1.74, size = 95, normalized size = 0.85 \begin {gather*} \begin {cases} \frac {a^{3} x^{4} e^{\operatorname {asin}{\left (a x \right )}}}{17} - \frac {4 a^{2} x^{3} \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{17} - \frac {22 a x^{2} e^{\operatorname {asin}{\left (a x \right )}}}{85} + \frac {44 x \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{85} + \frac {41 e^{\operatorname {asin}{\left (a x \right )}}}{85 a} & \text {for}\: a \neq 0 \\x & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x**4*exp(asin(a*x))/17 - 4*a**2*x**3*sqrt(-a**2*x**2 + 1)*exp(asin(a*x))/17 - 22*a*x**2*exp(as
in(a*x))/85 + 44*x*sqrt(-a**2*x**2 + 1)*exp(asin(a*x))/85 + 41*exp(asin(a*x))/(85*a), Ne(a, 0)), (x, True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )}\,{\left (1-a^2\,x^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(asin(a*x))*(1 - a^2*x^2)^(3/2),x)

[Out]

int(exp(asin(a*x))*(1 - a^2*x^2)^(3/2), x)

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