3.63 \(\int \frac{x}{\cos ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=63 \[ \frac{\text{Si}\left (2 \cos ^{-1}(a x)\right )}{a^2}+\frac{x \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{1}{2 a^2 \cos ^{-1}(a x)}+\frac{x^2}{\cos ^{-1}(a x)} \]

[Out]

(x*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]^2) - 1/(2*a^2*ArcCos[a*x]) + x^2/ArcCos[a*x] + SinIntegral[2*ArcCos[a*x
]]/a^2

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Rubi [A]  time = 0.164639, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {4634, 4720, 4636, 4406, 12, 3299, 4642} \[ \frac{\text{Si}\left (2 \cos ^{-1}(a x)\right )}{a^2}+\frac{x \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{1}{2 a^2 \cos ^{-1}(a x)}+\frac{x^2}{\cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]^2) - 1/(2*a^2*ArcCos[a*x]) + x^2/ArcCos[a*x] + SinIntegral[2*ArcCos[a*x
]]/a^2

Rule 4634

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcCo
s[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcCos[c*x])^(n + 1
))/Sqrt[1 - c^2*x^2], x], x] + Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*
x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 4720

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp
[((f*x)^m*(a + b*ArcCos[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] + Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)
^(m - 1)*(a + b*ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4636

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

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 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 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 4642

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp[(a + b*ArcCos[c*x])
^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rubi steps

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

Mathematica [A]  time = 0.0386597, size = 63, normalized size = 1. \[ \frac{\text{Si}\left (2 \cos ^{-1}(a x)\right )}{a^2}+\frac{x \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}+\frac{2 a^2 x^2-1}{2 a^2 \cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]^2) + (-1 + 2*a^2*x^2)/(2*a^2*ArcCos[a*x]) + SinIntegral[2*ArcCos[a*x]]/
a^2

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Maple [A]  time = 0.044, size = 43, normalized size = 0.7 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{\sin \left ( 2\,\arccos \left ( ax \right ) \right ) }{4\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}}+{\frac{\cos \left ( 2\,\arccos \left ( ax \right ) \right ) }{2\,\arccos \left ( ax \right ) }}+{\it Si} \left ( 2\,\arccos \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(1/4/arccos(a*x)^2*sin(2*arccos(a*x))+1/2/arccos(a*x)*cos(2*arccos(a*x))+Si(2*arccos(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{4 \, a^{2} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2} \int \frac{x}{\arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}\,{d x} - \sqrt{a x + 1} \sqrt{-a x + 1} a x -{\left (2 \, a^{2} x^{2} - 1\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}{2 \, a^{2} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2*integrate(x/arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x
), x) - sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x - (2*a^2*x^2 - 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x))/(a^2*ar
ctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{\arccos \left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccos(a*x)^3,x, algorithm="fricas")

[Out]

integral(x/arccos(a*x)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{acos}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/acos(a*x)**3, x)

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Giac [A]  time = 1.17452, size = 77, normalized size = 1.22 \begin{align*} \frac{x^{2}}{\arccos \left (a x\right )} + \frac{\operatorname{Si}\left (2 \, \arccos \left (a x\right )\right )}{a^{2}} + \frac{\sqrt{-a^{2} x^{2} + 1} x}{2 \, a \arccos \left (a x\right )^{2}} - \frac{1}{2 \, a^{2} \arccos \left (a x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2/arccos(a*x) + sin_integral(2*arccos(a*x))/a^2 + 1/2*sqrt(-a^2*x^2 + 1)*x/(a*arccos(a*x)^2) - 1/2/(a^2*arcc
os(a*x))