3.109 \(\int e^{\cos ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=82 \[ \frac {3 e^{\cos ^{-1}(a x)} \cos \left (3 \cos ^{-1}(a x)\right )}{40 a^3}-\frac {e^{\cos ^{-1}(a x)} \sin \left (3 \cos ^{-1}(a x)\right )}{40 a^3}+\frac {x e^{\cos ^{-1}(a x)}}{8 a^2}-\frac {\sqrt {1-a^2 x^2} e^{\cos ^{-1}(a x)}}{8 a^3} \]

[Out]

1/8*exp(arccos(a*x))*x/a^2+3/40*exp(arccos(a*x))*cos(3*arccos(a*x))/a^3-1/40*exp(arccos(a*x))*sin(3*arccos(a*x
))/a^3-1/8*exp(arccos(a*x))*(-a^2*x^2+1)^(1/2)/a^3

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Rubi [A]  time = 0.06, antiderivative size = 82, 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 = {4837, 12, 4469, 4432} \[ -\frac {\sqrt {1-a^2 x^2} e^{\cos ^{-1}(a x)}}{8 a^3}+\frac {x e^{\cos ^{-1}(a x)}}{8 a^2}+\frac {3 e^{\cos ^{-1}(a x)} \cos \left (3 \cos ^{-1}(a x)\right )}{40 a^3}-\frac {e^{\cos ^{-1}(a x)} \sin \left (3 \cos ^{-1}(a x)\right )}{40 a^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCos[a*x]*x^2,x]

[Out]

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

Rule 12

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

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]

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 4837

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

Rubi steps

\begin {align*} \int e^{\cos ^{-1}(a x)} x^2 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {e^x \cos ^2(x) \sin (x)}{a^2} \, dx,x,\cos ^{-1}(a x)\right )}{a}\\ &=-\frac {\operatorname {Subst}\left (\int e^x \cos ^2(x) \sin (x) \, dx,x,\cos ^{-1}(a x)\right )}{a^3}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{4} e^x \sin (x)+\frac {1}{4} e^x \sin (3 x)\right ) \, dx,x,\cos ^{-1}(a x)\right )}{a^3}\\ &=-\frac {\operatorname {Subst}\left (\int e^x \sin (x) \, dx,x,\cos ^{-1}(a x)\right )}{4 a^3}-\frac {\operatorname {Subst}\left (\int e^x \sin (3 x) \, dx,x,\cos ^{-1}(a x)\right )}{4 a^3}\\ &=\frac {e^{\cos ^{-1}(a x)} x}{8 a^2}-\frac {e^{\cos ^{-1}(a x)} \sqrt {1-a^2 x^2}}{8 a^3}+\frac {3 e^{\cos ^{-1}(a x)} \cos \left (3 \cos ^{-1}(a x)\right )}{40 a^3}-\frac {e^{\cos ^{-1}(a x)} \sin \left (3 \cos ^{-1}(a x)\right )}{40 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 50, normalized size = 0.61 \[ -\frac {e^{\cos ^{-1}(a x)} \left (5 \sqrt {1-a^2 x^2}-5 a x-3 \cos \left (3 \cos ^{-1}(a x)\right )+\sin \left (3 \cos ^{-1}(a x)\right )\right )}{40 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCos[a*x]*x^2,x]

[Out]

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

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fricas [A]  time = 0.46, size = 46, normalized size = 0.56 \[ \frac {{\left (3 \, a^{3} x^{3} - a x - {\left (a^{2} x^{2} + 1\right )} \sqrt {-a^{2} x^{2} + 1}\right )} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccos(a*x))*x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 69, normalized size = 0.84 \[ \frac {3}{10} \, x^{3} e^{\left (\arccos \left (a x\right )\right )} - \frac {\sqrt {-a^{2} x^{2} + 1} x^{2} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a} - \frac {x e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccos(a*x))*x^2,x, algorithm="giac")

[Out]

3/10*x^3*e^(arccos(a*x)) - 1/10*sqrt(-a^2*x^2 + 1)*x^2*e^(arccos(a*x))/a - 1/10*x*e^(arccos(a*x))/a^2 - 1/10*s
qrt(-a^2*x^2 + 1)*e^(arccos(a*x))/a^3

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\arccos \left (a x \right )} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arccos(a*x))*x^2,x)

[Out]

int(exp(arccos(a*x))*x^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} e^{\left (\arccos \left (a x\right )\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccos(a*x))*x^2,x, algorithm="maxima")

[Out]

integrate(x^2*e^(arccos(a*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {e}}^{\mathrm {acos}\left (a\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*exp(acos(a*x)),x)

[Out]

int(x^2*exp(acos(a*x)), x)

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sympy [A]  time = 1.19, size = 85, normalized size = 1.04 \[ \begin {cases} \frac {3 x^{3} e^{\operatorname {acos}{\left (a x \right )}}}{10} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {acos}{\left (a x \right )}}}{10 a} - \frac {x e^{\operatorname {acos}{\left (a x \right )}}}{10 a^{2}} - \frac {\sqrt {- a^{2} x^{2} + 1} e^{\operatorname {acos}{\left (a x \right )}}}{10 a^{3}} & \text {for}\: a \neq 0 \\\frac {x^{3} e^{\frac {\pi }{2}}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(acos(a*x))*x**2,x)

[Out]

Piecewise((3*x**3*exp(acos(a*x))/10 - x**2*sqrt(-a**2*x**2 + 1)*exp(acos(a*x))/(10*a) - x*exp(acos(a*x))/(10*a
**2) - sqrt(-a**2*x**2 + 1)*exp(acos(a*x))/(10*a**3), Ne(a, 0)), (x**3*exp(pi/2)/3, True))

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