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]

-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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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 12

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

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]

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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.42, size = 28, normalized size = 0.56 \[ \frac {{\left (2 \, x + 1\right )} \sqrt {x} \sqrt {-x + 1} - 3 \, \arccos \left (\sqrt {x}\right )}{6 \, x^{2}} \]

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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giac [B]  time = 3.99, size = 74, normalized size = 1.48 \[ \frac {{\left (\sqrt {-x + 1} - 1\right )}^{3}}{48 \, x^{\frac {3}{2}}} + \frac {3 \, {\left (\sqrt {-x + 1} - 1\right )}}{16 \, \sqrt {x}} - \frac {x^{\frac {3}{2}} {\left (\frac {9 \, {\left (\sqrt {-x + 1} - 1\right )}^{2}}{x} + 1\right )}}{48 \, {\left (\sqrt {-x + 1} - 1\right )}^{3}} - \frac {\arccos \left (\sqrt {x}\right )}{2 \, x^{2}} \]

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

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maple [A]  time = 0.00, size = 35, normalized size = 0.70 \[ -\frac {\arccos \left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {1-x}}{6 x^{\frac {3}{2}}}+\frac {\sqrt {1-x}}{3 \sqrt {x}} \]

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 = 0.41, size = 34, normalized size = 0.68 \[ \frac {\sqrt {-x + 1}}{3 \, \sqrt {x}} + \frac {\sqrt {-x + 1}}{6 \, x^{\frac {3}{2}}} - \frac {\arccos \left (\sqrt {x}\right )}{2 \, x^{2}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acos}\left (\sqrt {x}\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acos(x^(1/2))/x^3, x)

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sympy [A]  time = 7.25, size = 44, normalized size = 0.88 \[ - \frac {\begin {cases} - \frac {\sqrt {1 - x}}{\sqrt {x}} - \frac {\left (1 - x\right )^{\frac {3}{2}}}{3 x^{\frac {3}{2}}} & \text {for}\: x \geq 0 \wedge x < 1 \end {cases}}{2} - \frac {\operatorname {acos}{\left (\sqrt {x} \right )}}{2 x^{2}} \]

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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