3.9 \(\int \frac{\cos ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=56 \[ \frac{a \sqrt{1-a^2 x^2}}{6 x^2}+\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{\cos ^{-1}(a x)}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0354313, antiderivative size = 56, 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, 266, 51, 63, 208} \[ \frac{a \sqrt{1-a^2 x^2}}{6 x^2}+\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{\cos ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]/x^4,x]

[Out]

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

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 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 \frac{\cos ^{-1}(a x)}{x^4} \, dx &=-\frac{\cos ^{-1}(a x)}{3 x^3}-\frac{1}{3} a \int \frac{1}{x^3 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{\cos ^{-1}(a x)}{3 x^3}-\frac{1}{6} a \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=\frac{a \sqrt{1-a^2 x^2}}{6 x^2}-\frac{\cos ^{-1}(a x)}{3 x^3}-\frac{1}{12} a^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=\frac{a \sqrt{1-a^2 x^2}}{6 x^2}-\frac{\cos ^{-1}(a x)}{3 x^3}+\frac{1}{6} a \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )\\ &=\frac{a \sqrt{1-a^2 x^2}}{6 x^2}-\frac{\cos ^{-1}(a x)}{3 x^3}+\frac{1}{6} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0237518, size = 67, normalized size = 1.2 \[ \frac{a \sqrt{1-a^2 x^2}}{6 x^2}+\frac{1}{6} a^3 \log \left (\sqrt{1-a^2 x^2}+1\right )-\frac{1}{6} a^3 \log (x)-\frac{\cos ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a*x]/x^4,x]

[Out]

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

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Maple [A]  time = 0.004, size = 53, normalized size = 1. \begin{align*}{a}^{3} \left ( -{\frac{\arccos \left ( ax \right ) }{3\,{a}^{3}{x}^{3}}}+{\frac{1}{6\,{a}^{2}{x}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{1}{6}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a*x)/x^4,x)

[Out]

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

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Maxima [A]  time = 1.45277, size = 81, normalized size = 1.45 \begin{align*} \frac{1}{6} \,{\left (a^{2} \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\sqrt{-a^{2} x^{2} + 1}}{x^{2}}\right )} a - \frac{\arccos \left (a x\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.49061, size = 259, normalized size = 4.62 \begin{align*} \frac{a^{3} x^{3} \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right ) - a^{3} x^{3} \log \left (\sqrt{-a^{2} x^{2} + 1} - 1\right ) - 4 \, x^{3} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} a x}{a^{2} x^{2} - 1}\right ) + 2 \, \sqrt{-a^{2} x^{2} + 1} a x + 4 \,{\left (x^{3} - 1\right )} \arccos \left (a x\right )}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 3.00703, size = 110, normalized size = 1.96 \begin{align*} - \frac{a \left (\begin{cases} - \frac{a^{2} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{2} - \frac{a \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac{i a^{2} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{2} - \frac{i a}{2 x \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i}{2 a x^{3} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} & \text{otherwise} \end{cases}\right )}{3} - \frac{\operatorname{acos}{\left (a x \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x)/x**4,x)

[Out]

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

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Giac [A]  time = 1.13791, size = 95, normalized size = 1.7 \begin{align*} \frac{1}{12} \, a^{3}{\left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{a^{2} x^{2}} + \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right ) - \log \left (-\sqrt{-a^{2} x^{2} + 1} + 1\right )\right )} - \frac{\arccos \left (a x\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/x^4,x, algorithm="giac")

[Out]

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