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

Optimal. Leaf size=41 \[ \frac{e^{\cos ^{-1}(a x)} \cos \left (2 \cos ^{-1}(a x)\right )}{5 a^2}-\frac{e^{\cos ^{-1}(a x)} \sin \left (2 \cos ^{-1}(a x)\right )}{10 a^2} \]

[Out]

(E^ArcCos[a*x]*Cos[2*ArcCos[a*x]])/(5*a^2) - (E^ArcCos[a*x]*Sin[2*ArcCos[a*x]])/(10*a^2)

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Rubi [A]  time = 0.0340067, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4837, 12, 4469, 4432} \[ \frac{e^{\cos ^{-1}(a x)} \cos \left (2 \cos ^{-1}(a x)\right )}{5 a^2}-\frac{e^{\cos ^{-1}(a x)} \sin \left (2 \cos ^{-1}(a x)\right )}{10 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcCos[a*x]*Cos[2*ArcCos[a*x]])/(5*a^2) - (E^ArcCos[a*x]*Sin[2*ArcCos[a*x]])/(10*a^2)

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]

Rule 12

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

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 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]

Rubi steps

\begin{align*} \int e^{\cos ^{-1}(a x)} x \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{e^x \cos (x) \sin (x)}{a} \, dx,x,\cos ^{-1}(a x)\right )}{a}\\ &=-\frac{\operatorname{Subst}\left (\int e^x \cos (x) \sin (x) \, dx,x,\cos ^{-1}(a x)\right )}{a^2}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{2} e^x \sin (2 x) \, dx,x,\cos ^{-1}(a x)\right )}{a^2}\\ &=-\frac{\operatorname{Subst}\left (\int e^x \sin (2 x) \, dx,x,\cos ^{-1}(a x)\right )}{2 a^2}\\ &=\frac{e^{\cos ^{-1}(a x)} \cos \left (2 \cos ^{-1}(a x)\right )}{5 a^2}-\frac{e^{\cos ^{-1}(a x)} \sin \left (2 \cos ^{-1}(a x)\right )}{10 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0366191, size = 30, normalized size = 0.73 \[ -\frac{e^{\cos ^{-1}(a x)} \left (\sin \left (2 \cos ^{-1}(a x)\right )-2 \cos \left (2 \cos ^{-1}(a x)\right )\right )}{10 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^ArcCos[a*x]*(-2*Cos[2*ArcCos[a*x]] + Sin[2*ArcCos[a*x]]))/(10*a^2)

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Maple [F]  time = 0.007, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{\arccos \left ( ax \right ) }}x\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x e^{\left (\arccos \left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*e^(arccos(a*x)), x)

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Fricas [A]  time = 2.63661, size = 89, normalized size = 2.17 \begin{align*} \frac{{\left (2 \, a^{2} x^{2} - \sqrt{-a^{2} x^{2} + 1} a x - 1\right )} e^{\left (\arccos \left (a x\right )\right )}}{5 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(2*a^2*x^2 - sqrt(-a^2*x^2 + 1)*a*x - 1)*e^(arccos(a*x))/a^2

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Sympy [A]  time = 0.539324, size = 58, normalized size = 1.41 \begin{align*} \begin{cases} \frac{2 x^{2} e^{\operatorname{acos}{\left (a x \right )}}}{5} - \frac{x \sqrt{- a^{2} x^{2} + 1} e^{\operatorname{acos}{\left (a x \right )}}}{5 a} - \frac{e^{\operatorname{acos}{\left (a x \right )}}}{5 a^{2}} & \text{for}\: a \neq 0 \\\frac{x^{2} e^{\frac{\pi }{2}}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*x**2*exp(acos(a*x))/5 - x*sqrt(-a**2*x**2 + 1)*exp(acos(a*x))/(5*a) - exp(acos(a*x))/(5*a**2), Ne
(a, 0)), (x**2*exp(pi/2)/2, True))

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Giac [A]  time = 1.23812, size = 59, normalized size = 1.44 \begin{align*} \frac{2}{5} \, x^{2} e^{\left (\arccos \left (a x\right )\right )} - \frac{\sqrt{-a^{2} x^{2} + 1} x e^{\left (\arccos \left (a x\right )\right )}}{5 \, a} - \frac{e^{\left (\arccos \left (a x\right )\right )}}{5 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^2*e^(arccos(a*x)) - 1/5*sqrt(-a^2*x^2 + 1)*x*e^(arccos(a*x))/a - 1/5*e^(arccos(a*x))/a^2