3.108 \(\int e^{\cos ^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=81 \[ \frac {e^{\cos ^{-1}(a x)} \cos \left (2 \cos ^{-1}(a x)\right )}{10 a^4}+\frac {e^{\cos ^{-1}(a x)} \cos \left (4 \cos ^{-1}(a x)\right )}{34 a^4}-\frac {e^{\cos ^{-1}(a x)} \sin \left (2 \cos ^{-1}(a x)\right )}{20 a^4}-\frac {e^{\cos ^{-1}(a x)} \sin \left (4 \cos ^{-1}(a x)\right )}{136 a^4} \]

[Out]

1/10*exp(arccos(a*x))*cos(2*arccos(a*x))/a^4+1/34*exp(arccos(a*x))*cos(4*arccos(a*x))/a^4-1/20*exp(arccos(a*x)
)*sin(2*arccos(a*x))/a^4-1/136*exp(arccos(a*x))*sin(4*arccos(a*x))/a^4

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Rubi [A]  time = 0.06, 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 = {4837, 12, 4469, 4432} \[ \frac {e^{\cos ^{-1}(a x)} \cos \left (2 \cos ^{-1}(a x)\right )}{10 a^4}+\frac {e^{\cos ^{-1}(a x)} \cos \left (4 \cos ^{-1}(a x)\right )}{34 a^4}-\frac {e^{\cos ^{-1}(a x)} \sin \left (2 \cos ^{-1}(a x)\right )}{20 a^4}-\frac {e^{\cos ^{-1}(a x)} \sin \left (4 \cos ^{-1}(a x)\right )}{136 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcCos[a*x]*Cos[2*ArcCos[a*x]])/(10*a^4) + (E^ArcCos[a*x]*Cos[4*ArcCos[a*x]])/(34*a^4) - (E^ArcCos[a*x]*Sin
[2*ArcCos[a*x]])/(20*a^4) - (E^ArcCos[a*x]*Sin[4*ArcCos[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 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^3 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {e^x \cos ^3(x) \sin (x)}{a^3} \, dx,x,\cos ^{-1}(a x)\right )}{a}\\ &=-\frac {\operatorname {Subst}\left (\int e^x \cos ^3(x) \sin (x) \, dx,x,\cos ^{-1}(a x)\right )}{a^4}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{4} e^x \sin (2 x)+\frac {1}{8} e^x \sin (4 x)\right ) \, dx,x,\cos ^{-1}(a x)\right )}{a^4}\\ &=-\frac {\operatorname {Subst}\left (\int e^x \sin (4 x) \, dx,x,\cos ^{-1}(a x)\right )}{8 a^4}-\frac {\operatorname {Subst}\left (\int e^x \sin (2 x) \, dx,x,\cos ^{-1}(a x)\right )}{4 a^4}\\ &=\frac {e^{\cos ^{-1}(a x)} \cos \left (2 \cos ^{-1}(a x)\right )}{10 a^4}+\frac {e^{\cos ^{-1}(a x)} \cos \left (4 \cos ^{-1}(a x)\right )}{34 a^4}-\frac {e^{\cos ^{-1}(a x)} \sin \left (2 \cos ^{-1}(a x)\right )}{20 a^4}-\frac {e^{\cos ^{-1}(a x)} \sin \left (4 \cos ^{-1}(a x)\right )}{136 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 50, normalized size = 0.62 \[ -\frac {e^{\cos ^{-1}(a x)} \left (-68 \cos \left (2 \cos ^{-1}(a x)\right )-20 \cos \left (4 \cos ^{-1}(a x)\right )+34 \sin \left (2 \cos ^{-1}(a x)\right )+5 \sin \left (4 \cos ^{-1}(a x)\right )\right )}{680 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/680*(E^ArcCos[a*x]*(-68*Cos[2*ArcCos[a*x]] - 20*Cos[4*ArcCos[a*x]] + 34*Sin[2*ArcCos[a*x]] + 5*Sin[4*ArcCos
[a*x]]))/a^4

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fricas [A]  time = 0.47, size = 55, normalized size = 0.68 \[ \frac {{\left (20 \, a^{4} x^{4} - 3 \, a^{2} x^{2} - {\left (5 \, a^{3} x^{3} + 6 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} - 6\right )} e^{\left (\arccos \left (a x\right )\right )}}{85 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccos(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^(arccos(a*x))/a^4

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giac [A]  time = 0.21, size = 82, normalized size = 1.01 \[ \frac {4}{17} \, x^{4} e^{\left (\arccos \left (a x\right )\right )} - \frac {\sqrt {-a^{2} x^{2} + 1} x^{3} e^{\left (\arccos \left (a x\right )\right )}}{17 \, a} - \frac {3 \, x^{2} e^{\left (\arccos \left (a x\right )\right )}}{85 \, a^{2}} - \frac {6 \, \sqrt {-a^{2} x^{2} + 1} x e^{\left (\arccos \left (a x\right )\right )}}{85 \, a^{3}} - \frac {6 \, e^{\left (\arccos \left (a x\right )\right )}}{85 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/17*x^4*e^(arccos(a*x)) - 1/17*sqrt(-a^2*x^2 + 1)*x^3*e^(arccos(a*x))/a - 3/85*x^2*e^(arccos(a*x))/a^2 - 6/85
*sqrt(-a^2*x^2 + 1)*x*e^(arccos(a*x))/a^3 - 6/85*e^(arccos(a*x))/a^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} e^{\left (\arccos \left (a x\right )\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*e^(arccos(a*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\mathrm {e}}^{\mathrm {acos}\left (a\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*exp(acos(a*x)), x)

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sympy [A]  time = 2.93, size = 105, normalized size = 1.30 \[ \begin {cases} \frac {4 x^{4} e^{\operatorname {acos}{\left (a x \right )}}}{17} - \frac {x^{3} \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {acos}{\left (a x \right )}}}{17 a} - \frac {3 x^{2} e^{\operatorname {acos}{\left (a x \right )}}}{85 a^{2}} - \frac {6 x \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {acos}{\left (a x \right )}}}{85 a^{3}} - \frac {6 e^{\operatorname {acos}{\left (a x \right )}}}{85 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4} e^{\frac {\pi }{2}}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((4*x**4*exp(acos(a*x))/17 - x**3*sqrt(-a**2*x**2 + 1)*exp(acos(a*x))/(17*a) - 3*x**2*exp(acos(a*x))/
(85*a**2) - 6*x*sqrt(-a**2*x**2 + 1)*exp(acos(a*x))/(85*a**3) - 6*exp(acos(a*x))/(85*a**4), Ne(a, 0)), (x**4*e
xp(pi/2)/4, True))

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