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

Optimal. Leaf size=87 \[ -\frac{a^2}{12 x^2}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{6 x^3}+\frac{1}{3} a^4 \log (x)-\frac{\cos ^{-1}(a x)^2}{4 x^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.151741, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4628, 4702, 4682, 29, 30} \[ -\frac{a^2}{12 x^2}+\frac{a^3 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{3 x}+\frac{a \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{6 x^3}+\frac{1}{3} a^4 \log (x)-\frac{\cos ^{-1}(a x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0408128, size = 69, normalized size = 0.79 \[ -\frac{a^2}{12 x^2}+\frac{a \sqrt{1-a^2 x^2} \left (2 a^2 x^2+1\right ) \cos ^{-1}(a x)}{6 x^3}+\frac{1}{3} a^4 \log (x)-\frac{\cos ^{-1}(a x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.053, size = 76, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \arccos \left ( ax \right ) \right ) ^{2}}{4\,{x}^{4}}}+{\frac{a\arccos \left ( ax \right ) }{6\,{x}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{a}^{2}}{12\,{x}^{2}}}+{\frac{{a}^{3}\arccos \left ( ax \right ) }{3\,x}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{{a}^{4}\ln \left ( ax \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.49606, size = 100, normalized size = 1.15 \begin{align*} \frac{1}{12} \,{\left (4 \, a^{2} \log \left (x\right ) - \frac{1}{x^{2}}\right )} a^{2} + \frac{1}{6} \,{\left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1} a^{2}}{x} + \frac{\sqrt{-a^{2} x^{2} + 1}}{x^{3}}\right )} a \arccos \left (a x\right ) - \frac{\arccos \left (a x\right )^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.38229, size = 149, normalized size = 1.71 \begin{align*} \frac{4 \, a^{4} x^{4} \log \left (x\right ) - a^{2} x^{2} + 2 \,{\left (2 \, a^{3} x^{3} + a x\right )} \sqrt{-a^{2} x^{2} + 1} \arccos \left (a x\right ) - 3 \, \arccos \left (a x\right )^{2}}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a*x)**2/x**5, x)

________________________________________________________________________________________

Giac [B]  time = 1.31776, size = 223, normalized size = 2.56 \begin{align*} -\frac{1}{48} \,{\left ({\left (\frac{{\left (a^{4} + \frac{9 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{x^{2}}\right )} a^{6} x^{3}}{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}{\left | a \right |}} - \frac{\frac{9 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{4}}{x} + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{x^{3}}}{a^{2}{\left | a \right |}}\right )} \arccos \left (a x\right ) - \frac{4 \,{\left (2 \, a^{4} \log \left (a^{2} x^{2}\right ) - \frac{a^{2}}{x^{2}}\right )}}{a}\right )} a - \frac{\arccos \left (a x\right )^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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