3.666 \(\int \frac{\cos ^{-1}(x)}{x^4 \sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=54 \[ \frac{1}{6 x^2}-\frac{2 \sqrt{1-x^2} \cos ^{-1}(x)}{3 x}-\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{3 x^3}-\frac{2 \log (x)}{3} \]

[Out]

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

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Rubi [A]  time = 0.0909131, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {4702, 4682, 29, 30} \[ \frac{1}{6 x^2}-\frac{2 \sqrt{1-x^2} \cos ^{-1}(x)}{3 x}-\frac{\sqrt{1-x^2} \cos ^{-1}(x)}{3 x^3}-\frac{2 \log (x)}{3} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[x]/(x^4*Sqrt[1 - x^2]),x]

[Out]

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

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

Mathematica [A]  time = 0.0430401, size = 38, normalized size = 0.7 \[ \frac{-4 x^3 \log (x)-2 \sqrt{1-x^2} \left (2 x^2+1\right ) \cos ^{-1}(x)+x}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[x]/(x^4*Sqrt[1 - x^2]),x]

[Out]

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

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Maple [A]  time = 0.043, size = 43, normalized size = 0.8 \begin{align*}{\frac{1}{6\,{x}^{2}}}-{\frac{2\,\ln \left ( x \right ) }{3}}-{\frac{\arccos \left ( x \right ) }{3\,{x}^{3}}\sqrt{-{x}^{2}+1}}-{\frac{2\,\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)/x^4/(-x^2+1)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.4098, size = 57, normalized size = 1.06 \begin{align*} -\frac{1}{3} \,{\left (\frac{2 \, \sqrt{-x^{2} + 1}}{x} + \frac{\sqrt{-x^{2} + 1}}{x^{3}}\right )} \arccos \left (x\right ) + \frac{1}{6 \, x^{2}} - \frac{2}{3} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x)/x^4/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.69828, size = 95, normalized size = 1.76 \begin{align*} -\frac{4 \, x^{3} \log \left (x\right ) + 2 \,{\left (2 \, x^{2} + 1\right )} \sqrt{-x^{2} + 1} \arccos \left (x\right ) - x}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x)/x^4/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 114.909, size = 60, normalized size = 1.11 \begin{align*} \left (\begin{cases} - \frac{\sqrt{1 - x^{2}}}{x} - \frac{\left (1 - x^{2}\right )^{\frac{3}{2}}}{3 x^{3}} & \text{for}\: x > -1 \wedge x < 1 \end{cases}\right ) \operatorname{acos}{\left (x \right )} + \begin{cases} \text{NaN} & \text{for}\: x < -1 \\- \frac{2 \log{\left (x \right )}}{3} - \frac{1}{6} + \frac{2 i \pi }{3} + \frac{1}{6 x^{2}} & \text{for}\: x < 1 \\\text{NaN} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(x)/x**4/(-x**2+1)**(1/2),x)

[Out]

Piecewise((-sqrt(1 - x**2)/x - (1 - x**2)**(3/2)/(3*x**3), (x > -1) & (x < 1)))*acos(x) + Piecewise((nan, x <
-1), (-2*log(x)/3 - 1/6 + 2*I*pi/3 + 1/(6*x**2), x < 1), (nan, True))

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Giac [B]  time = 1.10374, size = 128, normalized size = 2.37 \begin{align*} \frac{1}{24} \,{\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}{6 \, x^{2}} - \frac{1}{3} \, \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x)/x^4/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/24*(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) + 1/6*(2*x^2 + 1)/x^2 - 1/3*log(x^2)