\(\int \frac {1}{\sqrt {2+x-x^2}} \, dx\) [54]

Optimal result

Integrand size = 12, antiderivative size = 12 \[ \int \frac {1}{\sqrt {2+x-x^2}} \, dx=-\arcsin \left (\frac {1}{3} (1-2 x)\right ) \]

[Out]

arcsin(-1/3+2/3*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {633, 222} \[ \int \frac {1}{\sqrt {2+x-x^2}} \, dx=-\arcsin \left (\frac {1}{3} (1-2 x)\right ) \]

[In]

Int[1/Sqrt[2 + x - x^2],x]

[Out]

-ArcSin[(1 - 2*x)/3]

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}& = -\left (\frac {1}{3} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{9}}} \, dx,x,1-2 x\right )\right ) \\ & = -\arcsin \left (\frac {1}{3} (1-2 x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.75 \[ \int \frac {1}{\sqrt {2+x-x^2}} \, dx=-2 \arctan \left (\frac {\sqrt {2+x-x^2}}{1+x}\right ) \]

[In]

Integrate[1/Sqrt[2 + x - x^2],x]

[Out]

-2*ArcTan[Sqrt[2 + x - x^2]/(1 + x)]

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.58

method	result	size
default	\(\arcsin \left (-\frac {1}{3}+\frac {2 x}{3}\right )\)	\(7\)
trager	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {-x^{2}+x +2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )\)	\(37\)

[In]

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

[Out]

arcsin(-1/3+2/3*x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (6) = 12\).

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.50 \[ \int \frac {1}{\sqrt {2+x-x^2}} \, dx=-\arctan \left (\frac {\sqrt {-x^{2} + x + 2} {\left (2 \, x - 1\right )}}{2 \, {\left (x^{2} - x - 2\right )}}\right ) \]

[In]

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

[Out]

-arctan(1/2*sqrt(-x^2 + x + 2)*(2*x - 1)/(x^2 - x - 2))

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {2+x-x^2}} \, dx=\operatorname {asin}{\left (\frac {2 x}{3} - \frac {1}{3} \right )} \]

[In]

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

[Out]

asin(2*x/3 - 1/3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {2+x-x^2}} \, dx=-\arcsin \left (-\frac {2}{3} \, x + \frac {1}{3}\right ) \]

[In]

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

[Out]

-arcsin(-2/3*x + 1/3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (6) = 12\).

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.17 \[ \int \frac {1}{\sqrt {2+x-x^2}} \, dx=\frac {1}{4} \, \sqrt {-x^{2} + x + 2} {\left (2 \, x - 1\right )} + \frac {9}{8} \, \arcsin \left (\frac {2}{3} \, x - \frac {1}{3}\right ) \]

[In]

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

[Out]

1/4*sqrt(-x^2 + x + 2)*(2*x - 1) + 9/8*arcsin(2/3*x - 1/3)

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt {2+x-x^2}} \, dx=\mathrm {asin}\left (\frac {2\,x}{3}-\frac {1}{3}\right ) \]

[In]

int(1/(x - x^2 + 2)^(1/2),x)

[Out]

asin((2*x)/3 - 1/3)