3.65 \(\int \frac{\cos ^{-1}(\sqrt{x})}{x^3} \, dx\)

Optimal. Leaf size=50 \[ \frac{\sqrt{1-x}}{6 x^{3/2}}-\frac{\cos ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{\sqrt{1-x}}{3 \sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.0176729, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4843, 12, 45, 37} \[ \frac{\sqrt{1-x}}{6 x^{3/2}}-\frac{\cos ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{\sqrt{1-x}}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[Sqrt[x]]/x^3,x]

[Out]

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

Rule 4843

Int[((a_.) + ArcCos[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcCos[
u]))/(d*(m + 1)), x] + Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

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*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}\left (\sqrt{x}\right )}{x^3} \, dx &=-\frac{\cos ^{-1}\left (\sqrt{x}\right )}{2 x^2}-\frac{1}{2} \int \frac{1}{2 \sqrt{1-x} x^{5/2}} \, dx\\ &=-\frac{\cos ^{-1}\left (\sqrt{x}\right )}{2 x^2}-\frac{1}{4} \int \frac{1}{\sqrt{1-x} x^{5/2}} \, dx\\ &=\frac{\sqrt{1-x}}{6 x^{3/2}}-\frac{\cos ^{-1}\left (\sqrt{x}\right )}{2 x^2}-\frac{1}{6} \int \frac{1}{\sqrt{1-x} x^{3/2}} \, dx\\ &=\frac{\sqrt{1-x}}{6 x^{3/2}}+\frac{\sqrt{1-x}}{3 \sqrt{x}}-\frac{\cos ^{-1}\left (\sqrt{x}\right )}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0214374, size = 43, normalized size = 0.86 \[ \left (\frac{1}{6 x^{3/2}}+\frac{1}{3 \sqrt{x}}\right ) \sqrt{1-x}-\frac{\cos ^{-1}\left (\sqrt{x}\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[Sqrt[x]]/x^3,x]

[Out]

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

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Maple [A]  time = 0.004, size = 35, normalized size = 0.7 \begin{align*} -{\frac{1}{2\,{x}^{2}}\arccos \left ( \sqrt{x} \right ) }+{\frac{1}{6}\sqrt{1-x}{x}^{-{\frac{3}{2}}}}+{\frac{1}{3}\sqrt{1-x}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(x^(1/2))/x^3,x)

[Out]

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

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Maxima [A]  time = 1.47328, size = 46, normalized size = 0.92 \begin{align*} \frac{\sqrt{-x + 1}}{3 \, \sqrt{x}} + \frac{\sqrt{-x + 1}}{6 \, x^{\frac{3}{2}}} - \frac{\arccos \left (\sqrt{x}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x^(1/2))/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.6279, size = 84, normalized size = 1.68 \begin{align*} \frac{{\left (2 \, x + 1\right )} \sqrt{x} \sqrt{-x + 1} - 3 \, \arccos \left (\sqrt{x}\right )}{6 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x^(1/2))/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 24.9142, size = 44, normalized size = 0.88 \begin{align*} - \frac{\begin{cases} - \frac{\sqrt{1 - x}}{\sqrt{x}} - \frac{\left (1 - x\right )^{\frac{3}{2}}}{3 x^{\frac{3}{2}}} & \text{for}\: x \geq 0 \wedge x < 1 \end{cases}}{2} - \frac{\operatorname{acos}{\left (\sqrt{x} \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(x**(1/2))/x**3,x)

[Out]

-Piecewise((-sqrt(1 - x)/sqrt(x) - (1 - x)**(3/2)/(3*x**(3/2)), (x >= 0) & (x < 1)))/2 - acos(sqrt(x))/(2*x**2
)

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Giac [B]  time = 1.28093, size = 100, normalized size = 2. \begin{align*} \frac{{\left (\sqrt{-x + 1} - 1\right )}^{3}}{48 \, x^{\frac{3}{2}}} + \frac{3 \,{\left (\sqrt{-x + 1} - 1\right )}}{16 \, \sqrt{x}} - \frac{x^{\frac{3}{2}}{\left (\frac{9 \,{\left (\sqrt{-x + 1} - 1\right )}^{2}}{x} + 1\right )}}{48 \,{\left (\sqrt{-x + 1} - 1\right )}^{3}} - \frac{\arccos \left (\sqrt{x}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x^(1/2))/x^3,x, algorithm="giac")

[Out]

1/48*(sqrt(-x + 1) - 1)^3/x^(3/2) + 3/16*(sqrt(-x + 1) - 1)/sqrt(x) - 1/48*x^(3/2)*(9*(sqrt(-x + 1) - 1)^2/x +
1)/(sqrt(-x + 1) - 1)^3 - 1/2*arccos(sqrt(x))/x^2