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

Optimal. Leaf size=58 \[ -\frac{1}{6} a x^2 \sqrt{1-\frac{a^2}{x^2}}-\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-\frac{a^2}{x^2}}\right )+\frac{1}{3} x^3 \sec ^{-1}\left (\frac{x}{a}\right ) \]

[Out]

-(a*Sqrt[1 - a^2/x^2]*x^2)/6 + (x^3*ArcSec[x/a])/3 - (a^3*ArcTanh[Sqrt[1 - a^2/x^2]])/6

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Rubi [A]  time = 0.036197, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4833, 5220, 266, 51, 63, 208} \[ -\frac{1}{6} a x^2 \sqrt{1-\frac{a^2}{x^2}}-\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-\frac{a^2}{x^2}}\right )+\frac{1}{3} x^3 \sec ^{-1}\left (\frac{x}{a}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Sqrt[1 - a^2/x^2]*x^2)/6 + (x^3*ArcSec[x/a])/3 - (a^3*ArcTanh[Sqrt[1 - a^2/x^2]])/6

Rule 4833

Int[ArcCos[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcSec[a/c + (b*x^n)/c]^m, x] /;
FreeQ[{a, b, c, n, m}, x]

Rule 5220

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int x^2 \cos ^{-1}\left (\frac{a}{x}\right ) \, dx &=\int x^2 \sec ^{-1}\left (\frac{x}{a}\right ) \, dx\\ &=\frac{1}{3} x^3 \sec ^{-1}\left (\frac{x}{a}\right )-\frac{1}{3} a \int \frac{x}{\sqrt{1-\frac{a^2}{x^2}}} \, dx\\ &=\frac{1}{3} x^3 \sec ^{-1}\left (\frac{x}{a}\right )+\frac{1}{6} a \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-a^2 x}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{1}{6} a \sqrt{1-\frac{a^2}{x^2}} x^2+\frac{1}{3} x^3 \sec ^{-1}\left (\frac{x}{a}\right )+\frac{1}{12} a^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{1}{6} a \sqrt{1-\frac{a^2}{x^2}} x^2+\frac{1}{3} x^3 \sec ^{-1}\left (\frac{x}{a}\right )-\frac{1}{6} a \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-\frac{a^2}{x^2}}\right )\\ &=-\frac{1}{6} a \sqrt{1-\frac{a^2}{x^2}} x^2+\frac{1}{3} x^3 \sec ^{-1}\left (\frac{x}{a}\right )-\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-\frac{a^2}{x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0505763, size = 61, normalized size = 1.05 \[ \frac{1}{3} x^3 \cos ^{-1}\left (\frac{a}{x}\right )-\frac{1}{6} a \left (x^2 \sqrt{1-\frac{a^2}{x^2}}+a^2 \log \left (x \left (\sqrt{1-\frac{a^2}{x^2}}+1\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 56, normalized size = 1. \begin{align*} -{a}^{3} \left ( -{\frac{{x}^{3}}{3\,{a}^{3}}\arccos \left ({\frac{a}{x}} \right ) }+{\frac{{x}^{2}}{6\,{a}^{2}}\sqrt{1-{\frac{{a}^{2}}{{x}^{2}}}}}+{\frac{1}{6}{\it Artanh} \left ({\frac{1}{\sqrt{1-{\frac{{a}^{2}}{{x}^{2}}}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*(-1/3/a^3*x^3*arccos(a/x)+1/6/a^2*x^2*(1-a^2/x^2)^(1/2)+1/6*arctanh(1/(1-a^2/x^2)^(1/2)))

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Maxima [A]  time = 1.46236, size = 97, normalized size = 1.67 \begin{align*} \frac{1}{3} \, x^{3} \arccos \left (\frac{a}{x}\right ) - \frac{1}{12} \,{\left (a^{2} \log \left (\sqrt{-\frac{a^{2}}{x^{2}} + 1} + 1\right ) - a^{2} \log \left (\sqrt{-\frac{a^{2}}{x^{2}} + 1} - 1\right ) + 2 \, x^{2} \sqrt{-\frac{a^{2}}{x^{2}} + 1}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.53943, size = 207, normalized size = 3.57 \begin{align*} \frac{1}{6} \, a^{3} \log \left (x \sqrt{-\frac{a^{2} - x^{2}}{x^{2}}} - x\right ) - \frac{1}{6} \, a x^{2} \sqrt{-\frac{a^{2} - x^{2}}{x^{2}}} + \frac{1}{3} \,{\left (x^{3} - 1\right )} \arccos \left (\frac{a}{x}\right ) + \frac{2}{3} \, \arctan \left (\frac{x \sqrt{-\frac{a^{2} - x^{2}}{x^{2}}} - x}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 5.29622, size = 99, normalized size = 1.71 \begin{align*} - \frac{a \left (\begin{cases} \frac{a^{2} \operatorname{acosh}{\left (\frac{x}{a} \right )}}{2} + \frac{a x \sqrt{-1 + \frac{x^{2}}{a^{2}}}}{2} & \text{for}\: \frac{\left |{x^{2}}\right |}{\left |{a^{2}}\right |} > 1 \\- \frac{i a^{2} \operatorname{asin}{\left (\frac{x}{a} \right )}}{2} + \frac{i a x}{2 \sqrt{1 - \frac{x^{2}}{a^{2}}}} - \frac{i x^{3}}{2 a \sqrt{1 - \frac{x^{2}}{a^{2}}}} & \text{otherwise} \end{cases}\right )}{3} + \frac{x^{3} \operatorname{acos}{\left (\frac{a}{x} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*Piecewise((a**2*acosh(x/a)/2 + a*x*sqrt(-1 + x**2/a**2)/2, Abs(x**2)/Abs(a**2) > 1), (-I*a**2*asin(x/a)/2 +
I*a*x/(2*sqrt(1 - x**2/a**2)) - I*x**3/(2*a*sqrt(1 - x**2/a**2)), True))/3 + x**3*acos(a/x)/3

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Giac [A]  time = 1.29039, size = 95, normalized size = 1.64 \begin{align*} \frac{1}{3} \, x^{3} \arccos \left (\frac{a}{x}\right ) - \frac{1}{12} \,{\left (a^{2} \log \left (a^{2}\right ) \mathrm{sgn}\left (x\right ) - \frac{2 \, a^{2} \log \left ({\left | -x + \sqrt{-a^{2} + x^{2}} \right |}\right )}{\mathrm{sgn}\left (x\right )} + \frac{2 \, \sqrt{-a^{2} + x^{2}} x}{\mathrm{sgn}\left (x\right )}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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