3.2.7 \(\int \sqrt {2+4 x-3 x^2} \, dx\) [107]

Optimal. Leaf size=45 \[ -\frac {1}{6} (2-3 x) \sqrt {2+4 x-3 x^2}-\frac {5 \sin ^{-1}\left (\frac {2-3 x}{\sqrt {10}}\right )}{3 \sqrt {3}} \]

[Out]

-5/9*arcsin(1/10*(2-3*x)*10^(1/2))*3^(1/2)-1/6*(2-3*x)*(-3*x^2+4*x+2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {626, 633, 222} \begin {gather*} -\frac {5 \text {ArcSin}\left (\frac {2-3 x}{\sqrt {10}}\right )}{3 \sqrt {3}}-\frac {1}{6} \sqrt {-3 x^2+4 x+2} (2-3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + 4*x - 3*x^2],x]

[Out]

-1/6*((2 - 3*x)*Sqrt[2 + 4*x - 3*x^2]) - (5*ArcSin[(2 - 3*x)/Sqrt[10]])/(3*Sqrt[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 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 {2+4 x-3 x^2} \, dx &=-\frac {1}{6} (2-3 x) \sqrt {2+4 x-3 x^2}+\frac {5}{3} \int \frac {1}{\sqrt {2+4 x-3 x^2}} \, dx\\ &=-\frac {1}{6} (2-3 x) \sqrt {2+4 x-3 x^2}-\frac {1}{6} \sqrt {\frac {5}{6}} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{40}}} \, dx,x,4-6 x\right )\\ &=-\frac {1}{6} (2-3 x) \sqrt {2+4 x-3 x^2}-\frac {5 \sin ^{-1}\left (\frac {2-3 x}{\sqrt {10}}\right )}{3 \sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 61, normalized size = 1.36 \begin {gather*} \frac {1}{6} (-2+3 x) \sqrt {2+4 x-3 x^2}+\frac {10 \tan ^{-1}\left (\frac {-2-\sqrt {10}+3 x}{\sqrt {6+12 x-9 x^2}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + 4*x - 3*x^2],x]

[Out]

((-2 + 3*x)*Sqrt[2 + 4*x - 3*x^2])/6 + (10*ArcTan[(-2 - Sqrt[10] + 3*x)/Sqrt[6 + 12*x - 9*x^2]])/(3*Sqrt[3])

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Maple [A]
time = 0.53, size = 35, normalized size = 0.78

method	result	size
default	\(-\frac {\left (-6 x +4\right ) \sqrt {-3 x^{2}+4 x +2}}{12}+\frac {5 \sqrt {3}\, \arcsin \left (\frac {3 \sqrt {10}\, \left (x -\frac {2}{3}\right )}{10}\right )}{9}\)	\(35\)
risch	\(-\frac {\left (3 x^{2}-4 x -2\right ) \left (-2+3 x \right )}{6 \sqrt {-3 x^{2}+4 x +2}}+\frac {5 \sqrt {3}\, \arcsin \left (\frac {3 \sqrt {10}\, \left (x -\frac {2}{3}\right )}{10}\right )}{9}\)	\(45\)
trager	\(\left (-\frac {1}{3}+\frac {x}{2}\right ) \sqrt {-3 x^{2}+4 x +2}+\frac {5 \RootOf \left (\textit {\_Z}^{2}+3\right ) \ln \left (-3 x \RootOf \left (\textit {\_Z}^{2}+3\right )+3 \sqrt {-3 x^{2}+4 x +2}+2 \RootOf \left (\textit {\_Z}^{2}+3\right )\right )}{9}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^2+4*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(-6*x+4)*(-3*x^2+4*x+2)^(1/2)+5/9*3^(1/2)*arcsin(3/10*10^(1/2)*(x-2/3))

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Maxima [A]
time = 0.54, size = 46, normalized size = 1.02 \begin {gather*} \frac {1}{2} \, \sqrt {-3 \, x^{2} + 4 \, x + 2} x - \frac {5}{9} \, \sqrt {3} \arcsin \left (-\frac {1}{10} \, \sqrt {10} {\left (3 \, x - 2\right )}\right ) - \frac {1}{3} \, \sqrt {-3 \, x^{2} + 4 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+4*x+2)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(-3*x^2 + 4*x + 2)*x - 5/9*sqrt(3)*arcsin(-1/10*sqrt(10)*(3*x - 2)) - 1/3*sqrt(-3*x^2 + 4*x + 2)

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Fricas [A]
time = 1.17, size = 60, normalized size = 1.33 \begin {gather*} \frac {1}{6} \, \sqrt {-3 \, x^{2} + 4 \, x + 2} {\left (3 \, x - 2\right )} - \frac {5}{9} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {-3 \, x^{2} + 4 \, x + 2} {\left (3 \, x - 2\right )}}{3 \, {\left (3 \, x^{2} - 4 \, x - 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+4*x+2)^(1/2),x, algorithm="fricas")

[Out]

1/6*sqrt(-3*x^2 + 4*x + 2)*(3*x - 2) - 5/9*sqrt(3)*arctan(1/3*sqrt(3)*sqrt(-3*x^2 + 4*x + 2)*(3*x - 2)/(3*x^2
- 4*x - 2))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {- 3 x^{2} + 4 x + 2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**2+4*x+2)**(1/2),x)

[Out]

Integral(sqrt(-3*x**2 + 4*x + 2), x)

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Giac [A]
time = 0.75, size = 36, normalized size = 0.80 \begin {gather*} \frac {1}{6} \, \sqrt {-3 \, x^{2} + 4 \, x + 2} {\left (3 \, x - 2\right )} + \frac {5}{9} \, \sqrt {3} \arcsin \left (\frac {1}{10} \, \sqrt {10} {\left (3 \, x - 2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+4*x+2)^(1/2),x, algorithm="giac")

[Out]

1/6*sqrt(-3*x^2 + 4*x + 2)*(3*x - 2) + 5/9*sqrt(3)*arcsin(1/10*sqrt(10)*(3*x - 2))

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Mupad [B]
time = 0.05, size = 35, normalized size = 0.78 \begin {gather*} \frac {5\,\sqrt {3}\,\mathrm {asin}\left (\frac {\sqrt {10}\,\left (3\,x-2\right )}{10}\right )}{9}+\left (\frac {x}{2}-\frac {1}{3}\right )\,\sqrt {-3\,x^2+4\,x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - 3*x^2 + 2)^(1/2),x)

[Out]

(5*3^(1/2)*asin((10^(1/2)*(3*x - 2))/10))/9 + (x/2 - 1/3)*(4*x - 3*x^2 + 2)^(1/2)

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