3.116 \(\int \frac {x}{\sqrt {1-x^2} \cos ^{-1}(x)} \, dx\)

Optimal. Leaf size=5 \[ -\text {Ci}\left (\cos ^{-1}(x)\right ) \]

[Out]

-Ci(arccos(x))

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Rubi [A]  time = 0.06, antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4724, 3302} \[ -\text {CosIntegral}\left (\cos ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/(Sqrt[1 - x^2]*ArcCos[x]),x]

[Out]

-CosIntegral[ArcCos[x]]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 4724

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Cos[x]^m*Sin[x]^(2*p + 1), x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 5, normalized size = 1.00 \[ -\text {Ci}\left (\cos ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(Sqrt[1 - x^2]*ArcCos[x]),x]

[Out]

-CosIntegral[ArcCos[x]]

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{2} + 1} x}{{\left (x^{2} - 1\right )} \arccos \relax (x)}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccos(x)/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-x^2 + 1)*x/((x^2 - 1)*arccos(x)), x)

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giac [A]  time = 0.20, size = 5, normalized size = 1.00 \[ -\operatorname {Ci}\left (\arccos \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccos(x)/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

-cos_integral(arccos(x))

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maple [A]  time = 0.19, size = 6, normalized size = 1.20 \[ -\Ci \left (\arccos \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/arccos(x)/(-x^2+1)^(1/2),x)

[Out]

-Ci(arccos(x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {-x^{2} + 1} \arccos \relax (x)}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccos(x)/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/(sqrt(-x^2 + 1)*arccos(x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.20 \[ \int \frac {x}{\mathrm {acos}\relax (x)\,\sqrt {1-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(acos(x)*(1 - x^2)^(1/2)),x)

[Out]

int(x/(acos(x)*(1 - x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \operatorname {acos}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/acos(x)/(-x**2+1)**(1/2),x)

[Out]

Integral(x/(sqrt(-(x - 1)*(x + 1))*acos(x)), x)

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