\(\int \frac {\arccos (\sqrt {x})}{x^4} \, dx\) [66]

Optimal result

Integrand size = 10, antiderivative size = 68 \[ \int \frac {\arccos \left (\sqrt {x}\right )}{x^4} \, dx=\frac {\sqrt {1-x}}{15 x^{5/2}}+\frac {4 \sqrt {1-x}}{45 x^{3/2}}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}}-\frac {\arccos \left (\sqrt {x}\right )}{3 x^3} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4927, 12, 47, 37} \[ \int \frac {\arccos \left (\sqrt {x}\right )}{x^4} \, dx=-\frac {\arccos \left (\sqrt {x}\right )}{3 x^3}+\frac {4 \sqrt {1-x}}{45 x^{3/2}}+\frac {\sqrt {1-x}}{15 x^{5/2}}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}} \]

[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 47

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 4927

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*} \text {integral}& = -\frac {\arccos \left (\sqrt {x}\right )}{3 x^3}-\frac {1}{3} \int \frac {1}{2 \sqrt {1-x} x^{7/2}} \, dx \\ & = -\frac {\arccos \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 {\arccos \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 {\arccos \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 {\arccos \left (\sqrt {x}\right )}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.54 \[ \int \frac {\arccos \left (\sqrt {x}\right )}{x^4} \, dx=\frac {\sqrt {-((-1+x) x)} \left (3+4 x+8 x^2\right )-15 \arccos \left (\sqrt {x}\right )}{45 x^3} \]

[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)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(-\frac {\arccos \left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {1-x}}{15 x^{\frac {5}{2}}}+\frac {4 \sqrt {1-x}}{45 x^{\frac {3}{2}}}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}}\)	\(47\)
default	\(-\frac {\arccos \left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {1-x}}{15 x^{\frac {5}{2}}}+\frac {4 \sqrt {1-x}}{45 x^{\frac {3}{2}}}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}}\)	\(47\)
parts	\(-\frac {\arccos \left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {1-x}}{15 x^{\frac {5}{2}}}+\frac {4 \sqrt {1-x}}{45 x^{\frac {3}{2}}}+\frac {8 \sqrt {1-x}}{45 \sqrt {x}}\)	\(47\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.49 \[ \int \frac {\arccos \left (\sqrt {x}\right )}{x^4} \, dx=\frac {{\left (8 \, x^{2} + 4 \, x + 3\right )} \sqrt {x} \sqrt {-x + 1} - 15 \, \arccos \left (\sqrt {x}\right )}{45 \, x^{3}} \]

[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

Sympy [A] (verification not implemented)

Time = 8.48 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {\arccos \left (\sqrt {x}\right )}{x^4} \, dx=- \frac {\begin {cases} - \frac {\sqrt {1 - x}}{\sqrt {x}} - \frac {2 \left (1 - x\right )^{\frac {3}{2}}}{3 x^{\frac {3}{2}}} - \frac {\left (1 - x\right )^{\frac {5}{2}}}{5 x^{\frac {5}{2}}} & \text {for}\: \sqrt {x} > -1 \wedge \sqrt {x} < 1 \end {cases}}{3} - \frac {\operatorname {acos}{\left (\sqrt {x} \right )}}{3 x^{3}} \]

[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)), (sqrt(x) > -1)
& (sqrt(x) < 1)))/3 - acos(sqrt(x))/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {\arccos \left (\sqrt {x}\right )}{x^4} \, dx=\frac {8 \, \sqrt {-x + 1}}{45 \, \sqrt {x}} + \frac {4 \, \sqrt {-x + 1}}{45 \, x^{\frac {3}{2}}} + \frac {\sqrt {-x + 1}}{15 \, x^{\frac {5}{2}}} - \frac {\arccos \left (\sqrt {x}\right )}{3 \, x^{3}} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (46) = 92\).

Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.56 \[ \int \frac {\arccos \left (\sqrt {x}\right )}{x^4} \, dx=\frac {{\left (\sqrt {-x + 1} - 1\right )}^{5}}{480 \, x^{\frac {5}{2}}} + \frac {5 \, {\left (\sqrt {-x + 1} - 1\right )}^{3}}{288 \, x^{\frac {3}{2}}} + \frac {5 \, {\left (\sqrt {-x + 1} - 1\right )}}{48 \, \sqrt {x}} - \frac {{\left (\frac {150 \, {\left (\sqrt {-x + 1} - 1\right )}^{4}}{x^{2}} + \frac {25 \, {\left (\sqrt {-x + 1} - 1\right )}^{2}}{x} + 3\right )} x^{\frac {5}{2}}}{1440 \, {\left (\sqrt {-x + 1} - 1\right )}^{5}} - \frac {\arccos \left (\sqrt {x}\right )}{3 \, x^{3}} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {\arccos \left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\mathrm {acos}\left (\sqrt {x}\right )}{x^4} \,d x \]

[In]

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

[Out]

int(acos(x^(1/2))/x^4, x)