\(\int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx\) [57]

Optimal result

Integrand size = 18, antiderivative size = 27 \[ \int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx=-2 \sqrt {2+x-x^2}-2 \arcsin \left (\frac {1}{3} (1-2 x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {654, 633, 222} \[ \int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx=-2 \arcsin \left (\frac {1}{3} (1-2 x)\right )-2 \sqrt {-x^2+x+2} \]

[In]

Int[(1 + 2*x)/Sqrt[2 + x - x^2],x]

[Out]

-2*Sqrt[2 + x - x^2] - 2*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]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx=-2 \sqrt {2+x-x^2}-4 \arctan \left (\frac {\sqrt {2+x-x^2}}{1+x}\right ) \]

[In]

Integrate[(1 + 2*x)/Sqrt[2 + x - x^2],x]

[Out]

-2*Sqrt[2 + x - x^2] - 4*ArcTan[Sqrt[2 + x - x^2]/(1 + x)]

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
default	\(2 \arcsin \left (-\frac {1}{3}+\frac {2 x}{3}\right )-2 \sqrt {-x^{2}+x +2}\)	\(22\)
risch	\(\frac {2 x^{2}-2 x -4}{\sqrt {-x^{2}+x +2}}+2 \arcsin \left (-\frac {1}{3}+\frac {2 x}{3}\right )\)	\(30\)
trager	\(-2 \sqrt {-x^{2}+x +2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {-x^{2}+x +2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )\)	\(51\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (21) = 42\).

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx=-2 \, \sqrt {-x^{2} + x + 2} - 2 \, \arctan \left (\frac {\sqrt {-x^{2} + x + 2} {\left (2 \, x - 1\right )}}{2 \, {\left (x^{2} - x - 2\right )}}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx=- 2 \sqrt {- x^{2} + x + 2} + 2 \operatorname {asin}{\left (\frac {2 x}{3} - \frac {1}{3} \right )} \]

[In]

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

[Out]

-2*sqrt(-x**2 + x + 2) + 2*asin(2*x/3 - 1/3)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx=-2 \, \sqrt {-x^{2} + x + 2} - 2 \, \arcsin \left (-\frac {2}{3} \, x + \frac {1}{3}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx=-2 \, \sqrt {-x^{2} + x + 2} + 2 \, \arcsin \left (\frac {2}{3} \, x - \frac {1}{3}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {1+2 x}{\sqrt {2+x-x^2}} \, dx=\mathrm {asin}\left (\frac {2\,x}{3}-\frac {1}{3}\right )-2\,\sqrt {-x^2+x+2}-\ln \left (x\,1{}\mathrm {i}+\sqrt {-x^2+x+2}-\frac {1}{2}{}\mathrm {i}\right )\,1{}\mathrm {i} \]

[In]

int((2*x + 1)/(x - x^2 + 2)^(1/2),x)

[Out]

asin((2*x)/3 - 1/3) - log(x*1i + (x - x^2 + 2)^(1/2) - 1i/2)*1i - 2*(x - x^2 + 2)^(1/2)