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

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

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Rubi [A]  time = 0.11, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.03, size = 52, normalized size = 0.80 \[ -\frac {\cos ^{-1}(x)^2}{4 x^4}-\frac {1}{12 x^2}+\frac {\sqrt {1-x^2} \left (2 x^2+1\right ) \cos ^{-1}(x)}{6 x^3}+\frac {\log (x)}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*1/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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{-1}(x)^2}{x^5} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.07, size = 44, normalized size = 0.68 \[ \frac {4 \, x^{4} \log \relax (x) + 2 \, {\left (2 \, x^{3} + x\right )} \sqrt {-x^{2} + 1} \arccos \relax (x) - x^{2} - 3 \, \arccos \relax (x)^{2}}{12 \, x^{4}} \]

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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giac [B]  time = 0.99, size = 104, normalized size = 1.60 \[ -\frac {1}{48} \, {\left (\frac {x^{3} {\left (\frac {9 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + 1\right )}}{{\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}} - \frac {9 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{x^{3}}\right )} \arccos \relax (x) - \frac {2 \, x^{2} + 1}{12 \, x^{2}} - \frac {\arccos \relax (x)^{2}}{4 \, x^{4}} + \frac {1}{6} \, \log \left (x^{2}\right ) \]

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)

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maple [A]  time = 0.08, size = 52, normalized size = 0.80

method	result	size
default	\(-\frac {1}{12 x^{2}}-\frac {\arccos \relax (x )^{2}}{4 x^{4}}+\frac {\ln \relax (x )}{3}+\frac {\arccos \relax (x ) \sqrt {-x^{2}+1}}{6 x^{3}}+\frac {\arccos \relax (x ) \sqrt {-x^{2}+1}}{3 x}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(x)^2/x^5,x,method=_RETURNVERBOSE)

[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 = 0.96, size = 51, normalized size = 0.78 \[ \frac {1}{6} \, {\left (\frac {2 \, \sqrt {-x^{2} + 1}}{x} + \frac {\sqrt {-x^{2} + 1}}{x^{3}}\right )} \arccos \relax (x) - \frac {1}{12 \, x^{2}} - \frac {\arccos \relax (x)^{2}}{4 \, x^{4}} + \frac {1}{3} \, \log \relax (x) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {acos}\relax (x)}^2}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acos(x)^2/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acos}^{2}{\relax (x )}}{x^{5}}\, dx \]

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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