3.4.25 \(\int \sqrt {1+x-x^2} \, dx\) [325]

Optimal. Leaf size=38 \[ -\frac {1}{4} (1-2 x) \sqrt {1+x-x^2}-\frac {5}{8} \sin ^{-1}\left (\frac {1-2 x}{\sqrt {5}}\right ) \]

[Out]

-5/8*arcsin(1/5*(1-2*x)*5^(1/2))-1/4*(1-2*x)*(-x^2+x+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {626, 633, 222} \begin {gather*} -\frac {5}{8} \text {ArcSin}\left (\frac {1-2 x}{\sqrt {5}}\right )-\frac {1}{4} \sqrt {-x^2+x+1} (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x - x^2],x]

[Out]

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

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 626

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

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*} \int \sqrt {1+x-x^2} \, dx &=-\frac {1}{4} (1-2 x) \sqrt {1+x-x^2}+\frac {5}{8} \int \frac {1}{\sqrt {1+x-x^2}} \, dx\\ &=-\frac {1}{4} (1-2 x) \sqrt {1+x-x^2}-\frac {1}{8} \sqrt {5} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{5}}} \, dx,x,1-2 x\right )\\ &=-\frac {1}{4} (1-2 x) \sqrt {1+x-x^2}-\frac {5}{8} \sin ^{-1}\left (\frac {1-2 x}{\sqrt {5}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 45, normalized size = 1.18 \begin {gather*} \frac {1}{4} (-1+2 x) \sqrt {1+x-x^2}+\frac {5}{4} \tan ^{-1}\left (\frac {x}{-1+\sqrt {1+x-x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x - x^2],x]

[Out]

((-1 + 2*x)*Sqrt[1 + x - x^2])/4 + (5*ArcTan[x/(-1 + Sqrt[1 + x - x^2])])/4

________________________________________________________________________________________

Maple [A]
time = 0.14, size = 30, normalized size = 0.79

method	result	size
default	\(-\frac {\left (1-2 x \right ) \sqrt {-x^{2}+x +1}}{4}+\frac {5 \arcsin \left (\frac {2 \sqrt {5}\, \left (x -\frac {1}{2}\right )}{5}\right )}{8}\)	\(30\)
risch	\(-\frac {\left (2 x -1\right ) \left (x^{2}-x -1\right )}{4 \sqrt {-x^{2}+x +1}}+\frac {5 \arcsin \left (\frac {2 \sqrt {5}\, \left (x -\frac {1}{2}\right )}{5}\right )}{8}\)	\(38\)
trager	\(\left (-\frac {1}{4}+\frac {x}{2}\right ) \sqrt {-x^{2}+x +1}-\frac {5 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {-x^{2}+x +1}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{8}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(1-2*x)*(-x^2+x+1)^(1/2)+5/8*arcsin(2/5*5^(1/2)*(x-1/2))

________________________________________________________________________________________

Maxima [A]
time = 4.78, size = 39, normalized size = 1.03 \begin {gather*} \frac {1}{2} \, \sqrt {-x^{2} + x + 1} x - \frac {1}{4} \, \sqrt {-x^{2} + x + 1} - \frac {5}{8} \, \arcsin \left (-\frac {1}{5} \, \sqrt {5} {\left (2 \, x - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-x^2 + x + 1)*x - 1/4*sqrt(-x^2 + x + 1) - 5/8*arcsin(-1/5*sqrt(5)*(2*x - 1))

________________________________________________________________________________________

Fricas [A]
time = 1.00, size = 37, normalized size = 0.97 \begin {gather*} \frac {1}{4} \, \sqrt {-x^{2} + x + 1} {\left (2 \, x - 1\right )} - \frac {5}{4} \, \arctan \left (\frac {\sqrt {-x^{2} + x + 1} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(-x^2 + x + 1)*(2*x - 1) - 5/4*arctan((sqrt(-x^2 + x + 1) - 1)/x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {- x^{2} + x + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-x**2 + x + 1), x)

________________________________________________________________________________________

Giac [A]
time = 1.36, size = 31, normalized size = 0.82 \begin {gather*} \frac {1}{4} \, \sqrt {-x^{2} + x + 1} {\left (2 \, x - 1\right )} + \frac {5}{8} \, \arcsin \left (\frac {1}{5} \, \sqrt {5} {\left (2 \, x - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(-x^2 + x + 1)*(2*x - 1) + 5/8*arcsin(1/5*sqrt(5)*(2*x - 1))

________________________________________________________________________________________

Mupad [B]
time = 0.05, size = 28, normalized size = 0.74 \begin {gather*} \frac {5\,\mathrm {asin}\left (\frac {2\,\sqrt {5}\,\left (x-\frac {1}{2}\right )}{5}\right )}{8}+\left (\frac {x}{2}-\frac {1}{4}\right )\,\sqrt {-x^2+x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - x^2 + 1)^(1/2),x)

[Out]

(5*asin((2*5^(1/2)*(x - 1/2))/5))/8 + (x/2 - 1/4)*(x - x^2 + 1)^(1/2)

________________________________________________________________________________________