\(\int \frac {1}{\sqrt {5-4 x-x^2}} \, dx\) [295]

Optimal result

Integrand size = 14, antiderivative size = 12 \[ \int \frac {1}{\sqrt {5-4 x-x^2}} \, dx=-\arcsin \left (\frac {1}{3} (-2-x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {633, 222} \[ \int \frac {1}{\sqrt {5-4 x-x^2}} \, dx=-\arcsin \left (\frac {1}{3} (-x-2)\right ) \]

[In]

Int[1/Sqrt[5 - 4*x - x^2],x]

[Out]

-ArcSin[(-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}{6} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{36}}} \, dx,x,-4-2 x\right )\right ) \\ & = -\arcsin \left (\frac {1}{3} (-2-x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.92 \[ \int \frac {1}{\sqrt {5-4 x-x^2}} \, dx=-2 \arctan \left (\frac {\sqrt {5-4 x-x^2}}{5+x}\right ) \]

[In]

Integrate[1/Sqrt[5 - 4*x - x^2],x]

[Out]

-2*ArcTan[Sqrt[5 - 4*x - x^2]/(5 + x)]

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.58

method	result	size
default	\(\arcsin \left (\frac {2}{3}+\frac {x}{3}\right )\)	\(7\)
trager	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +\sqrt {-x^{2}-4 x +5}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )\)	\(39\)

[In]

int(1/(-x^2-4*x+5)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (6) = 12\).

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 2.42 \[ \int \frac {1}{\sqrt {5-4 x-x^2}} \, dx=-\arctan \left (\frac {\sqrt {-x^{2} - 4 \, x + 5} {\left (x + 2\right )}}{x^{2} + 4 \, x - 5}\right ) \]

[In]

integrate(1/(-x^2-4*x+5)^(1/2),x, algorithm="fricas")

[Out]

-arctan(sqrt(-x^2 - 4*x + 5)*(x + 2)/(x^2 + 4*x - 5))

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\sqrt {5-4 x-x^2}} \, dx=\operatorname {asin}{\left (\frac {x}{3} + \frac {2}{3} \right )} \]

[In]

integrate(1/(-x**2-4*x+5)**(1/2),x)

[Out]

asin(x/3 + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {5-4 x-x^2}} \, dx=-\arcsin \left (-\frac {1}{3} \, x - \frac {2}{3}\right ) \]

[In]

integrate(1/(-x^2-4*x+5)^(1/2),x, algorithm="maxima")

[Out]

-arcsin(-1/3*x - 2/3)

Giac [B] (verification not implemented)

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

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.17 \[ \int \frac {1}{\sqrt {5-4 x-x^2}} \, dx=\frac {1}{2} \, \sqrt {-x^{2} - 4 \, x + 5} {\left (x + 2\right )} + \frac {9}{2} \, \arcsin \left (\frac {1}{3} \, x + \frac {2}{3}\right ) \]

[In]

integrate(1/(-x^2-4*x+5)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(-x^2 - 4*x + 5)*(x + 2) + 9/2*arcsin(1/3*x + 2/3)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt {5-4 x-x^2}} \, dx=\mathrm {asin}\left (\frac {x}{3}+\frac {2}{3}\right ) \]

[In]

int(1/(5 - x^2 - 4*x)^(1/2),x)

[Out]

asin(x/3 + 2/3)