3.4.67 \(\int \frac {\text {ArcSin}(\sqrt {x})}{x^4} \, dx\) [367]

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

[Out]

-1/3*arcsin(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 = {4926, 12, 47, 37} \begin {gather*} -\frac {\text {ArcSin}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*Sqrt[1 - x]/x^(5/2) - (4*Sqrt[1 - x])/(45*x^(3/2)) - (8*Sqrt[1 - x])/(45*Sqrt[x]) - ArcSin[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 4926

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((a + b*ArcSin[
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 {\sin ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx &=-\frac {\sin ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {1}{3} \int \frac {1}{2 \sqrt {1-x} x^{7/2}} \, dx\\ &=-\frac {\sin ^{-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 {\sin ^{-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 {\sin ^{-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 {\sin ^{-1}\left (\sqrt {x}\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 44, normalized size = 0.65 \begin {gather*} 2 \left (-\frac {\sqrt {1-x} \left (3+4 x+8 x^2\right )}{90 x^{5/2}}-\frac {\text {ArcSin}\left (\sqrt {x}\right )}{6 x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*(-1/90*(Sqrt[1 - x]*(3 + 4*x + 8*x^2))/x^(5/2) - ArcSin[Sqrt[x]]/(6*x^3))

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Maple [A]
time = 0.01, size = 47, normalized size = 0.69

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*arcsin(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 = 0.48, size = 46, normalized size = 0.68 \begin {gather*} -\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 {\arcsin \left (\sqrt {x}\right )}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(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*arcsin(sqrt(x))/x^3

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Fricas [A]
time = 2.01, size = 33, normalized size = 0.49 \begin {gather*} -\frac {{\left (8 \, x^{2} + 4 \, x + 3\right )} \sqrt {x} \sqrt {-x + 1} + 15 \, \arcsin \left (\sqrt {x}\right )}{45 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 11.90, size = 66, normalized size = 0.97 \begin {gather*} \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 {asin}{\left (\sqrt {x} \right )}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(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 - asin(sqrt(x))/(3*x**3)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (46) = 92\).
time = 0.41, size = 106, normalized size = 1.56 \begin {gather*} -\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 {\arcsin \left (\sqrt {x}\right )}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(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*arcsin
(sqrt(x))/x^3

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {asin}\left (\sqrt {x}\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asin(x^(1/2))/x^4, x)

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