\(\int \frac {1}{\sqrt {x-x^2}} \, dx\) [56]

Optimal result

Integrand size = 11, antiderivative size = 8 \[ \int \frac {1}{\sqrt {x-x^2}} \, dx=-\arcsin (1-2 x) \]

[Out]

arcsin(-1+2*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {633, 222} \[ \int \frac {1}{\sqrt {x-x^2}} \, dx=-\arcsin (1-2 x) \]

[In]

Int[1/Sqrt[x - x^2],x]

[Out]

-ArcSin[1 - 2*x]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,1-2 x\right ) \\ & = -\arcsin (1-2 x) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(40\) vs. \(2(8)=16\).

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 5.00 \[ \int \frac {1}{\sqrt {x-x^2}} \, dx=-\frac {2 \sqrt {-1+x} \sqrt {x} \log \left (\sqrt {-1+x}-\sqrt {x}\right )}{\sqrt {-((-1+x) x)}} \]

[In]

Integrate[1/Sqrt[x - x^2],x]

[Out]

(-2*Sqrt[-1 + x]*Sqrt[x]*Log[Sqrt[-1 + x] - Sqrt[x]])/Sqrt[-((-1 + x)*x)]

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
default	\(\arcsin \left (2 x -1\right )\)	\(7\)
meijerg	\(2 \arcsin \left (\sqrt {x}\right )\)	\(7\)
pseudoelliptic	\(-2 \arctan \left (\frac {\sqrt {-x \left (-1+x \right )}}{x}\right )\)	\(16\)
trager	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {-x^{2}+x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )\)	\(36\)

[In]

int(1/(-x^2+x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

arcsin(2*x-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16 vs. \(2 (6) = 12\).

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\sqrt {x-x^2}} \, dx=-2 \, \arctan \left (\frac {\sqrt {-x^{2} + x}}{x}\right ) \]

[In]

integrate(1/(-x^2+x)^(1/2),x, algorithm="fricas")

[Out]

-2*arctan(sqrt(-x^2 + x)/x)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\sqrt {x-x^2}} \, dx=\operatorname {asin}{\left (2 x - 1 \right )} \]

[In]

integrate(1/(-x**2+x)**(1/2),x)

[Out]

asin(2*x - 1)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {x-x^2}} \, dx=\arcsin \left (2 \, x - 1\right ) \]

[In]

integrate(1/(-x^2+x)^(1/2),x, algorithm="maxima")

[Out]

arcsin(2*x - 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (6) = 12\).

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 3.12 \[ \int \frac {1}{\sqrt {x-x^2}} \, dx=\frac {1}{4} \, \sqrt {-x^{2} + x} {\left (2 \, x - 1\right )} + \frac {1}{8} \, \arcsin \left (2 \, x - 1\right ) \]

[In]

integrate(1/(-x^2+x)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {x-x^2}} \, dx=\mathrm {asin}\left (2\,x-1\right ) \]

[In]

int(1/(x - x^2)^(1/2),x)

[Out]

asin(2*x - 1)