3.7.66 \(\int \frac {\cos ^{-1}(x)}{x^4 \sqrt {1-x^2}} \, dx\) [666]

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

[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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Rubi [A]
time = 0.06, antiderivative size = 54, normalized size of antiderivative = 1.00, 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 = {4790, 4772, 29, 30} \begin {gather*} -\frac {2 \sqrt {1-x^2} \text {ArcCos}(x)}{3 x}-\frac {\sqrt {1-x^2} \text {ArcCos}(x)}{3 x^3}+\frac {1}{6 x^2}-\frac {2 \log (x)}{3} \end {gather*}

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 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 4772

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/(f*(m + 1)))*Simp[(d
+ e*x^2)^p/(1 - c^2*x^2)^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, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rule 4790

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/(f*(m + 1)))*Simp[(d + e*x
^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[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*}

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Mathematica [A]
time = 0.03, size = 38, normalized size = 0.70 \begin {gather*} \frac {x-2 \sqrt {1-x^2} \left (1+2 x^2\right ) \cos ^{-1}(x)-4 x^3 \log (x)}{6 x^3} \end {gather*}

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 = 0.11, size = 43, normalized size = 0.80

method	result	size
default	\(\frac {1}{6 x^{2}}-\frac {2 \ln \left (x \right )}{3}-\frac {\arccos \left (x \right ) \sqrt {-x^{2}+1}}{3 x^{3}}-\frac {2 \arccos \left (x \right ) \sqrt {-x^{2}+1}}{3 x}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(x)/x^4/(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[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 = 1.34, size = 42, normalized size = 0.78 \begin {gather*} -\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 {gather*}

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 = 0.52, size = 36, normalized size = 0.67 \begin {gather*} -\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 {gather*}

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 = 6.61, size = 49, normalized size = 0.91 \begin {gather*} \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 x^{2}} & \text {for}\: x < 1 \\\text {NaN} & \text {otherwise} \end {cases} \end {gather*}

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*x**2), x < 1), (nan, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (42) = 84\).
time = 0.78, size = 95, normalized size = 1.76 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {acos}\left (x\right )}{x^4\,\sqrt {1-x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(x)/(x^4*(1 - x^2)^(1/2)),x)

[Out]

int(acos(x)/(x^4*(1 - x^2)^(1/2)), x)

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