3.645 \(\int \frac{\cos ^{-1}(x)^2}{x^5} \, dx\)

Optimal. Leaf size=65 \[ -\frac{1}{12 x^2}-\frac{\cos ^{-1}(x)^2}{4 x^4}+\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{3 x}+\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{6 x^3}+\frac{\log (x)}{3} \]

[Out]

-1/(12*x^2) + (Sqrt[1 - x^2]*ArcCos[x])/(6*x^3) + (Sqrt[1 - x^2]*ArcCos[x])/(3*x) - ArcCos[x]^2/(4*x^4) + Log[
x]/3

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Rubi [A]  time = 0.109141, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4628, 4702, 4682, 29, 30} \[ -\frac{1}{12 x^2}-\frac{\cos ^{-1}(x)^2}{4 x^4}+\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{3 x}+\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{6 x^3}+\frac{\log (x)}{3} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[x]^2/x^5,x]

[Out]

-1/(12*x^2) + (Sqrt[1 - x^2]*ArcCos[x])/(6*x^3) + (Sqrt[1 - x^2]*ArcCos[x])/(3*x) - ArcCos[x]^2/(4*x^4) + Log[
x]/3

Rule 4628

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

Rule 4702

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rule 4682

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n)/(d*f*(m + 1)), x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x
^2)^FracPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCo
s[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}(x)^2}{x^5} \, dx &=-\frac{\cos ^{-1}(x)^2}{4 x^4}-\frac{1}{2} \int \frac{\cos ^{-1}(x)}{x^4 \sqrt{1-x^2}} \, dx\\ &=\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{6 x^3}-\frac{\cos ^{-1}(x)^2}{4 x^4}+\frac{1}{6} \int \frac{1}{x^3} \, dx-\frac{1}{3} \int \frac{\cos ^{-1}(x)}{x^2 \sqrt{1-x^2}} \, dx\\ &=-\frac{1}{12 x^2}+\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{6 x^3}+\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{3 x}-\frac{\cos ^{-1}(x)^2}{4 x^4}+\frac{1}{3} \int \frac{1}{x} \, dx\\ &=-\frac{1}{12 x^2}+\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{6 x^3}+\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{3 x}-\frac{\cos ^{-1}(x)^2}{4 x^4}+\frac{\log (x)}{3}\\ \end{align*}

Mathematica [A]  time = 0.0321626, size = 52, normalized size = 0.8 \[ -\frac{1}{12 x^2}-\frac{\cos ^{-1}(x)^2}{4 x^4}+\frac{\sqrt{1-x^2} \left (2 x^2+1\right ) \cos ^{-1}(x)}{6 x^3}+\frac{\log (x)}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[x]^2/x^5,x]

[Out]

-1/(12*x^2) + (Sqrt[1 - x^2]*(1 + 2*x^2)*ArcCos[x])/(6*x^3) - ArcCos[x]^2/(4*x^4) + Log[x]/3

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Maple [A]  time = 0.033, size = 52, normalized size = 0.8 \begin{align*} -{\frac{1}{12\,{x}^{2}}}-{\frac{ \left ( \arccos \left ( x \right ) \right ) ^{2}}{4\,{x}^{4}}}+{\frac{\ln \left ( x \right ) }{3}}+{\frac{\arccos \left ( x \right ) }{6\,{x}^{3}}\sqrt{-{x}^{2}+1}}+{\frac{\arccos \left ( x \right ) }{3\,x}\sqrt{-{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(x)^2/x^5,x)

[Out]

-1/12/x^2-1/4*arccos(x)^2/x^4+1/3*ln(x)+1/6*arccos(x)*(-x^2+1)^(1/2)/x^3+1/3*arccos(x)*(-x^2+1)^(1/2)/x

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Maxima [A]  time = 1.42872, size = 69, normalized size = 1.06 \begin{align*} \frac{1}{6} \,{\left (\frac{2 \, \sqrt{-x^{2} + 1}}{x} + \frac{\sqrt{-x^{2} + 1}}{x^{3}}\right )} \arccos \left (x\right ) - \frac{1}{12 \, x^{2}} - \frac{\arccos \left (x\right )^{2}}{4 \, x^{4}} + \frac{1}{3} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x)^2/x^5,x, algorithm="maxima")

[Out]

1/6*(2*sqrt(-x^2 + 1)/x + sqrt(-x^2 + 1)/x^3)*arccos(x) - 1/12/x^2 - 1/4*arccos(x)^2/x^4 + 1/3*log(x)

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Fricas [A]  time = 2.81129, size = 119, normalized size = 1.83 \begin{align*} \frac{4 \, x^{4} \log \left (x\right ) + 2 \,{\left (2 \, x^{3} + x\right )} \sqrt{-x^{2} + 1} \arccos \left (x\right ) - x^{2} - 3 \, \arccos \left (x\right )^{2}}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x)^2/x^5,x, algorithm="fricas")

[Out]

1/12*(4*x^4*log(x) + 2*(2*x^3 + x)*sqrt(-x^2 + 1)*arccos(x) - x^2 - 3*arccos(x)^2)/x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acos}^{2}{\left (x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(x)**2/x**5,x)

[Out]

Integral(acos(x)**2/x**5, x)

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Giac [B]  time = 1.11555, size = 140, normalized size = 2.15 \begin{align*} -\frac{1}{48} \,{\left (\frac{x^{3}{\left (\frac{9 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + 1\right )}}{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{3}} - \frac{9 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}{x} - \frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{3}}{x^{3}}\right )} \arccos \left (x\right ) - \frac{2 \, x^{2} + 1}{12 \, x^{2}} - \frac{\arccos \left (x\right )^{2}}{4 \, x^{4}} + \frac{1}{6} \, \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x)^2/x^5,x, algorithm="giac")

[Out]

-1/48*(x^3*(9*(sqrt(-x^2 + 1) - 1)^2/x^2 + 1)/(sqrt(-x^2 + 1) - 1)^3 - 9*(sqrt(-x^2 + 1) - 1)/x - (sqrt(-x^2 +
1) - 1)^3/x^3)*arccos(x) - 1/12*(2*x^2 + 1)/x^2 - 1/4*arccos(x)^2/x^4 + 1/6*log(x^2)