3.67 \(\int \frac {\cos ^{-1}(\sqrt {x})}{x^5} \, dx\)

Optimal. Leaf size=86 \[ \frac {2 \sqrt {1-x}}{35 x^{3/2}}+\frac {3 \sqrt {1-x}}{70 x^{5/2}}+\frac {\sqrt {1-x}}{28 x^{7/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{4 x^4}+\frac {4 \sqrt {1-x}}{35 \sqrt {x}} \]

[Out]

-1/4*arccos(x^(1/2))/x^4+1/28*(1-x)^(1/2)/x^(7/2)+3/70*(1-x)^(1/2)/x^(5/2)+2/35*(1-x)^(1/2)/x^(3/2)+4/35*(1-x)
^(1/2)/x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4843, 12, 45, 37} \[ \frac {2 \sqrt {1-x}}{35 x^{3/2}}+\frac {3 \sqrt {1-x}}{70 x^{5/2}}+\frac {\sqrt {1-x}}{28 x^{7/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{4 x^4}+\frac {4 \sqrt {1-x}}{35 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[Sqrt[x]]/x^5,x]

[Out]

Sqrt[1 - x]/(28*x^(7/2)) + (3*Sqrt[1 - x])/(70*x^(5/2)) + (2*Sqrt[1 - x])/(35*x^(3/2)) + (4*Sqrt[1 - x])/(35*S
qrt[x]) - ArcCos[Sqrt[x]]/(4*x^4)

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^5} \, dx &=-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{4 x^4}-\frac {1}{4} \int \frac {1}{2 \sqrt {1-x} x^{9/2}} \, dx\\ &=-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{4 x^4}-\frac {1}{8} \int \frac {1}{\sqrt {1-x} x^{9/2}} \, dx\\ &=\frac {\sqrt {1-x}}{28 x^{7/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{4 x^4}-\frac {3}{28} \int \frac {1}{\sqrt {1-x} x^{7/2}} \, dx\\ &=\frac {\sqrt {1-x}}{28 x^{7/2}}+\frac {3 \sqrt {1-x}}{70 x^{5/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{4 x^4}-\frac {3}{35} \int \frac {1}{\sqrt {1-x} x^{5/2}} \, dx\\ &=\frac {\sqrt {1-x}}{28 x^{7/2}}+\frac {3 \sqrt {1-x}}{70 x^{5/2}}+\frac {2 \sqrt {1-x}}{35 x^{3/2}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{4 x^4}-\frac {2}{35} \int \frac {1}{\sqrt {1-x} x^{3/2}} \, dx\\ &=\frac {\sqrt {1-x}}{28 x^{7/2}}+\frac {3 \sqrt {1-x}}{70 x^{5/2}}+\frac {2 \sqrt {1-x}}{35 x^{3/2}}+\frac {4 \sqrt {1-x}}{35 \sqrt {x}}-\frac {\cos ^{-1}\left (\sqrt {x}\right )}{4 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 42, normalized size = 0.49 \[ \frac {\sqrt {-((x-1) x)} \left (16 x^3+8 x^2+6 x+5\right )-35 \cos ^{-1}\left (\sqrt {x}\right )}{140 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[Sqrt[x]]/x^5,x]

[Out]

(Sqrt[-((-1 + x)*x)]*(5 + 6*x + 8*x^2 + 16*x^3) - 35*ArcCos[Sqrt[x]])/(140*x^4)

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fricas [A]  time = 0.43, size = 38, normalized size = 0.44 \[ \frac {{\left (16 \, x^{3} + 8 \, x^{2} + 6 \, x + 5\right )} \sqrt {x} \sqrt {-x + 1} - 35 \, \arccos \left (\sqrt {x}\right )}{140 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x^(1/2))/x^5,x, algorithm="fricas")

[Out]

1/140*((16*x^3 + 8*x^2 + 6*x + 5)*sqrt(x)*sqrt(-x + 1) - 35*arccos(sqrt(x)))/x^4

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giac [B]  time = 0.21, size = 138, normalized size = 1.60 \[ \frac {{\left (\sqrt {-x + 1} - 1\right )}^{7}}{3584 \, x^{\frac {7}{2}}} + \frac {7 \, {\left (\sqrt {-x + 1} - 1\right )}^{5}}{2560 \, x^{\frac {5}{2}}} + \frac {7 \, {\left (\sqrt {-x + 1} - 1\right )}^{3}}{512 \, x^{\frac {3}{2}}} + \frac {35 \, {\left (\sqrt {-x + 1} - 1\right )}}{512 \, \sqrt {x}} - \frac {{\left (\frac {1225 \, {\left (\sqrt {-x + 1} - 1\right )}^{6}}{x^{3}} + \frac {245 \, {\left (\sqrt {-x + 1} - 1\right )}^{4}}{x^{2}} + \frac {49 \, {\left (\sqrt {-x + 1} - 1\right )}^{2}}{x} + 5\right )} x^{\frac {7}{2}}}{17920 \, {\left (\sqrt {-x + 1} - 1\right )}^{7}} - \frac {\arccos \left (\sqrt {x}\right )}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x^(1/2))/x^5,x, algorithm="giac")

[Out]

1/3584*(sqrt(-x + 1) - 1)^7/x^(7/2) + 7/2560*(sqrt(-x + 1) - 1)^5/x^(5/2) + 7/512*(sqrt(-x + 1) - 1)^3/x^(3/2)
+ 35/512*(sqrt(-x + 1) - 1)/sqrt(x) - 1/17920*(1225*(sqrt(-x + 1) - 1)^6/x^3 + 245*(sqrt(-x + 1) - 1)^4/x^2 +
49*(sqrt(-x + 1) - 1)^2/x + 5)*x^(7/2)/(sqrt(-x + 1) - 1)^7 - 1/4*arccos(sqrt(x))/x^4

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maple [A]  time = 0.00, size = 59, normalized size = 0.69 \[ -\frac {\arccos \left (\sqrt {x}\right )}{4 x^{4}}+\frac {\sqrt {1-x}}{28 x^{\frac {7}{2}}}+\frac {3 \sqrt {1-x}}{70 x^{\frac {5}{2}}}+\frac {2 \sqrt {1-x}}{35 x^{\frac {3}{2}}}+\frac {4 \sqrt {1-x}}{35 \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(x^(1/2))/x^5,x)

[Out]

-1/4*arccos(x^(1/2))/x^4+1/28*(1-x)^(1/2)/x^(7/2)+3/70*(1-x)^(1/2)/x^(5/2)+2/35*(1-x)^(1/2)/x^(3/2)+4/35*(1-x)
^(1/2)/x^(1/2)

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maxima [A]  time = 0.40, size = 58, normalized size = 0.67 \[ \frac {4 \, \sqrt {-x + 1}}{35 \, \sqrt {x}} + \frac {2 \, \sqrt {-x + 1}}{35 \, x^{\frac {3}{2}}} + \frac {3 \, \sqrt {-x + 1}}{70 \, x^{\frac {5}{2}}} + \frac {\sqrt {-x + 1}}{28 \, x^{\frac {7}{2}}} - \frac {\arccos \left (\sqrt {x}\right )}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(x^(1/2))/x^5,x, algorithm="maxima")

[Out]

4/35*sqrt(-x + 1)/sqrt(x) + 2/35*sqrt(-x + 1)/x^(3/2) + 3/70*sqrt(-x + 1)/x^(5/2) + 1/28*sqrt(-x + 1)/x^(7/2)
- 1/4*arccos(sqrt(x))/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (\sqrt {x}\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(x^(1/2))/x^5,x)

[Out]

int(acos(x^(1/2))/x^5, x)

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sympy [A]  time = 36.69, size = 71, normalized size = 0.83 \[ - \frac {\begin {cases} - \frac {\sqrt {1 - x}}{\sqrt {x}} - \frac {\left (1 - x\right )^{\frac {3}{2}}}{x^{\frac {3}{2}}} - \frac {3 \left (1 - x\right )^{\frac {5}{2}}}{5 x^{\frac {5}{2}}} - \frac {\left (1 - x\right )^{\frac {7}{2}}}{7 x^{\frac {7}{2}}} & \text {for}\: x \geq 0 \wedge x < 1 \end {cases}}{4} - \frac {\operatorname {acos}{\left (\sqrt {x} \right )}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(x**(1/2))/x**5,x)

[Out]

-Piecewise((-sqrt(1 - x)/sqrt(x) - (1 - x)**(3/2)/x**(3/2) - 3*(1 - x)**(5/2)/(5*x**(5/2)) - (1 - x)**(7/2)/(7
*x**(7/2)), (x >= 0) & (x < 1)))/4 - acos(sqrt(x))/(4*x**4)

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