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

Optimal. Leaf size=68 \[ \frac {4 \sqrt {1-x}}{45 x^{3/2}}+\frac {\sqrt {1-x}}{15 x^{5/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}} \]

[Out]

-1/3*arccos(x^(1/2))/x^3+1/15*(1-x)^(1/2)/x^(5/2)+4/45*(1-x)^(1/2)/x^(3/2)+8/45*(1-x)^(1/2)/x^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4843, 12, 45, 37} \[ \frac {4 \sqrt {1-x}}{45 x^{3/2}}+\frac {\sqrt {1-x}}{15 x^{5/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[Sqrt[x]]/x^4,x]

[Out]

Sqrt[1 - x]/(15*x^(5/2)) + (4*Sqrt[1 - x])/(45*x^(3/2)) + (8*Sqrt[1 - x])/(45*Sqrt[x]) - ArcCos[Sqrt[x]]/(3*x^
3)

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^4} \, dx &=-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {1}{3} \int \frac {1}{2 \sqrt {1-x} x^{7/2}} \, dx\\ &=-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {1}{6} \int \frac {1}{\sqrt {1-x} x^{7/2}} \, dx\\ &=\frac {\sqrt {1-x}}{15 x^{5/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {2}{15} \int \frac {1}{\sqrt {1-x} x^{5/2}} \, dx\\ &=\frac {\sqrt {1-x}}{15 x^{5/2}}+\frac {4 \sqrt {1-x}}{45 x^{3/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {4}{45} \int \frac {1}{\sqrt {1-x} x^{3/2}} \, dx\\ &=\frac {\sqrt {1-x}}{15 x^{5/2}}+\frac {4 \sqrt {1-x}}{45 x^{3/2}}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 37, normalized size = 0.54 \[ \frac {\sqrt {-((x-1) x)} \left (8 x^2+4 x+3\right )-15 \cos ^{-1}\left (\sqrt {x}\right )}{45 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[Sqrt[x]]/x^4,x]

[Out]

(Sqrt[-((-1 + x)*x)]*(3 + 4*x + 8*x^2) - 15*ArcCos[Sqrt[x]])/(45*x^3)

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fricas [A]  time = 0.43, size = 33, normalized size = 0.49 \[ \frac {{\left (8 \, x^{2} + 4 \, x + 3\right )} \sqrt {x} \sqrt {-x + 1} - 15 \, \arccos \left (\sqrt {x}\right )}{45 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*((8*x^2 + 4*x + 3)*sqrt(x)*sqrt(-x + 1) - 15*arccos(sqrt(x)))/x^3

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giac [B]  time = 1.04, size = 106, normalized size = 1.56 \[ \frac {{\left (\sqrt {-x + 1} - 1\right )}^{5}}{480 \, x^{\frac {5}{2}}} + \frac {5 \, {\left (\sqrt {-x + 1} - 1\right )}^{3}}{288 \, x^{\frac {3}{2}}} + \frac {5 \, {\left (\sqrt {-x + 1} - 1\right )}}{48 \, \sqrt {x}} - \frac {{\left (\frac {150 \, {\left (\sqrt {-x + 1} - 1\right )}^{4}}{x^{2}} + \frac {25 \, {\left (\sqrt {-x + 1} - 1\right )}^{2}}{x} + 3\right )} x^{\frac {5}{2}}}{1440 \, {\left (\sqrt {-x + 1} - 1\right )}^{5}} - \frac {\arccos \left (\sqrt {x}\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(sqrt(-x + 1) - 1)^5/x^(5/2) + 5/288*(sqrt(-x + 1) - 1)^3/x^(3/2) + 5/48*(sqrt(-x + 1) - 1)/sqrt(x) - 1/
1440*(150*(sqrt(-x + 1) - 1)^4/x^2 + 25*(sqrt(-x + 1) - 1)^2/x + 3)*x^(5/2)/(sqrt(-x + 1) - 1)^5 - 1/3*arccos(
sqrt(x))/x^3

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maple [A]  time = 0.01, size = 47, normalized size = 0.69 \[ -\frac {\arccos \left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {1-x}}{15 x^{\frac {5}{2}}}+\frac {4 \sqrt {1-x}}{45 x^{\frac {3}{2}}}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(x^(1/2))/x^4,x)

[Out]

-1/3*arccos(x^(1/2))/x^3+1/15*(1-x)^(1/2)/x^(5/2)+4/45*(1-x)^(1/2)/x^(3/2)+8/45*(1-x)^(1/2)/x^(1/2)

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maxima [A]  time = 0.40, size = 46, normalized size = 0.68 \[ \frac {8 \, \sqrt {-x + 1}}{45 \, \sqrt {x}} + \frac {4 \, \sqrt {-x + 1}}{45 \, x^{\frac {3}{2}}} + \frac {\sqrt {-x + 1}}{15 \, x^{\frac {5}{2}}} - \frac {\arccos \left (\sqrt {x}\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/45*sqrt(-x + 1)/sqrt(x) + 4/45*sqrt(-x + 1)/x^(3/2) + 1/15*sqrt(-x + 1)/x^(5/2) - 1/3*arccos(sqrt(x))/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (\sqrt {x}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(x^(1/2))/x^4,x)

[Out]

int(acos(x^(1/2))/x^4, x)

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sympy [A]  time = 15.86, size = 60, normalized size = 0.88 \[ - \frac {\begin {cases} - \frac {\sqrt {1 - x}}{\sqrt {x}} - \frac {2 \left (1 - x\right )^{\frac {3}{2}}}{3 x^{\frac {3}{2}}} - \frac {\left (1 - x\right )^{\frac {5}{2}}}{5 x^{\frac {5}{2}}} & \text {for}\: x \geq 0 \wedge x < 1 \end {cases}}{3} - \frac {\operatorname {acos}{\left (\sqrt {x} \right )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Piecewise((-sqrt(1 - x)/sqrt(x) - 2*(1 - x)**(3/2)/(3*x**(3/2)) - (1 - x)**(5/2)/(5*x**(5/2)), (x >= 0) & (x
< 1)))/3 - acos(sqrt(x))/(3*x**3)

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