\(\int e^{\arccos (a x)} x^2 \, dx\) [109]

Optimal result

Integrand size = 10, antiderivative size = 82 \[ \int e^{\arccos (a x)} x^2 \, dx=\frac {e^{\arccos (a x)} x}{8 a^2}-\frac {e^{\arccos (a x)} \sqrt {1-a^2 x^2}}{8 a^3}+\frac {3 e^{\arccos (a x)} \cos (3 \arccos (a x))}{40 a^3}-\frac {e^{\arccos (a x)} \sin (3 \arccos (a x))}{40 a^3} \]

[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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4921, 12, 4557, 4517} \[ \int e^{\arccos (a x)} x^2 \, dx=-\frac {e^{\arccos (a x)} \sin (3 \arccos (a x))}{40 a^3}+\frac {3 e^{\arccos (a x)} \cos (3 \arccos (a x))}{40 a^3}+\frac {x e^{\arccos (a x)}}{8 a^2}-\frac {\sqrt {1-a^2 x^2} e^{\arccos (a x)}}{8 a^3} \]

[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 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 4921

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.61 \[ \int e^{\arccos (a x)} x^2 \, dx=-\frac {e^{\arccos (a x)} \left (-5 a x+5 \sqrt {1-a^2 x^2}-3 \cos (3 \arccos (a x))+\sin (3 \arccos (a x))\right )}{40 a^3} \]

[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

Maple [F]

\[\int {\mathrm e}^{\arccos \left (a x \right )} x^{2}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.56 \[ \int e^{\arccos (a x)} x^2 \, dx=\frac {{\left (3 \, a^{3} x^{3} - a x - {\left (a^{2} x^{2} + 1\right )} \sqrt {-a^{2} x^{2} + 1}\right )} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{3}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int e^{\arccos (a x)} x^2 \, dx=\begin {cases} \frac {3 x^{3} e^{\operatorname {acos}{\left (a x \right )}}}{10} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {acos}{\left (a x \right )}}}{10 a} - \frac {x e^{\operatorname {acos}{\left (a x \right )}}}{10 a^{2}} - \frac {\sqrt {- a^{2} x^{2} + 1} e^{\operatorname {acos}{\left (a x \right )}}}{10 a^{3}} & \text {for}\: a \neq 0 \\\frac {x^{3} e^{\frac {\pi }{2}}}{3} & \text {otherwise} \end {cases} \]

[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))

Maxima [F]

\[ \int e^{\arccos (a x)} x^2 \, dx=\int { x^{2} e^{\left (\arccos \left (a x\right )\right )} \,d x } \]

[In]

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

[Out]

integrate(x^2*e^(arccos(a*x)), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int e^{\arccos (a x)} x^2 \, dx=\frac {3}{10} \, x^{3} e^{\left (\arccos \left (a x\right )\right )} - \frac {\sqrt {-a^{2} x^{2} + 1} x^{2} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a} - \frac {x e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{3}} \]

[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

Mupad [F(-1)]

Timed out. \[ \int e^{\arccos (a x)} x^2 \, dx=\int x^2\,{\mathrm {e}}^{\mathrm {acos}\left (a\,x\right )} \,d x \]

[In]

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

[Out]

int(x^2*exp(acos(a*x)), x)