3.2.9 \(\int \sqrt {2+5 x-3 x^2} \, dx\) [109]

Optimal. Leaf size=43 \[ -\frac {1}{12} (5-6 x) \sqrt {2+5 x-3 x^2}-\frac {49 \sin ^{-1}\left (\frac {1}{7} (5-6 x)\right )}{24 \sqrt {3}} \]

[Out]

49/72*arcsin(-5/7+6/7*x)*3^(1/2)-1/12*(5-6*x)*(-3*x^2+5*x+2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 43, 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 {49 \text {ArcSin}\left (\frac {1}{7} (5-6 x)\right )}{24 \sqrt {3}}-\frac {1}{12} \sqrt {-3 x^2+5 x+2} (5-6 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + 5*x - 3*x^2],x]

[Out]

-1/12*((5 - 6*x)*Sqrt[2 + 5*x - 3*x^2]) - (49*ArcSin[(5 - 6*x)/7])/(24*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+5 x-3 x^2} \, dx &=-\frac {1}{12} (5-6 x) \sqrt {2+5 x-3 x^2}+\frac {49}{24} \int \frac {1}{\sqrt {2+5 x-3 x^2}} \, dx\\ &=-\frac {1}{12} (5-6 x) \sqrt {2+5 x-3 x^2}-\frac {7 \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{49}}} \, dx,x,5-6 x\right )}{24 \sqrt {3}}\\ &=-\frac {1}{12} (5-6 x) \sqrt {2+5 x-3 x^2}-\frac {49 \sin ^{-1}\left (\frac {1}{7} (5-6 x)\right )}{24 \sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 56, normalized size = 1.30 \begin {gather*} \frac {1}{36} \left (3 (-5+6 x) \sqrt {2+5 x-3 x^2}-49 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {6+15 x-9 x^2}}{1+3 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + 5*x - 3*x^2],x]

[Out]

(3*(-5 + 6*x)*Sqrt[2 + 5*x - 3*x^2] - 49*Sqrt[3]*ArcTan[Sqrt[6 + 15*x - 9*x^2]/(1 + 3*x)])/36

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Maple [A]
time = 0.51, size = 32, normalized size = 0.74

method	result	size
default	\(\frac {49 \arcsin \left (-\frac {5}{7}+\frac {6 x}{7}\right ) \sqrt {3}}{72}-\frac {\left (5-6 x \right ) \sqrt {-3 x^{2}+5 x +2}}{12}\)	\(32\)
risch	\(-\frac {\left (3 x^{2}-5 x -2\right ) \left (-5+6 x \right )}{12 \sqrt {-3 x^{2}+5 x +2}}+\frac {49 \arcsin \left (-\frac {5}{7}+\frac {6 x}{7}\right ) \sqrt {3}}{72}\)	\(42\)
trager	\(\left (-\frac {5}{12}+\frac {x}{2}\right ) \sqrt {-3 x^{2}+5 x +2}+\frac {49 \RootOf \left (\textit {\_Z}^{2}+3\right ) \ln \left (-6 x \RootOf \left (\textit {\_Z}^{2}+3\right )+6 \sqrt {-3 x^{2}+5 x +2}+5 \RootOf \left (\textit {\_Z}^{2}+3\right )\right )}{72}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^2+5*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

49/72*arcsin(-5/7+6/7*x)*3^(1/2)-1/12*(5-6*x)*(-3*x^2+5*x+2)^(1/2)

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Maxima [A]
time = 0.53, size = 41, normalized size = 0.95 \begin {gather*} \frac {1}{2} \, \sqrt {-3 \, x^{2} + 5 \, x + 2} x - \frac {49}{72} \, \sqrt {3} \arcsin \left (-\frac {6}{7} \, x + \frac {5}{7}\right ) - \frac {5}{12} \, \sqrt {-3 \, x^{2} + 5 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+5*x+2)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(-3*x^2 + 5*x + 2)*x - 49/72*sqrt(3)*arcsin(-6/7*x + 5/7) - 5/12*sqrt(-3*x^2 + 5*x + 2)

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Fricas [A]
time = 1.02, size = 60, normalized size = 1.40 \begin {gather*} \frac {1}{12} \, \sqrt {-3 \, x^{2} + 5 \, x + 2} {\left (6 \, x - 5\right )} - \frac {49}{72} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {-3 \, x^{2} + 5 \, x + 2} {\left (6 \, x - 5\right )}}{6 \, {\left (3 \, x^{2} - 5 \, x - 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+5*x+2)^(1/2),x, algorithm="fricas")

[Out]

1/12*sqrt(-3*x^2 + 5*x + 2)*(6*x - 5) - 49/72*sqrt(3)*arctan(1/6*sqrt(3)*sqrt(-3*x^2 + 5*x + 2)*(6*x - 5)/(3*x
^2 - 5*x - 2))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {- 3 x^{2} + 5 x + 2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**2+5*x+2)**(1/2),x)

[Out]

Integral(sqrt(-3*x**2 + 5*x + 2), x)

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Giac [A]
time = 0.52, size = 31, normalized size = 0.72 \begin {gather*} \frac {1}{12} \, \sqrt {-3 \, x^{2} + 5 \, x + 2} {\left (6 \, x - 5\right )} + \frac {49}{72} \, \sqrt {3} \arcsin \left (\frac {6}{7} \, x - \frac {5}{7}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+5*x+2)^(1/2),x, algorithm="giac")

[Out]

1/12*sqrt(-3*x^2 + 5*x + 2)*(6*x - 5) + 49/72*sqrt(3)*arcsin(6/7*x - 5/7)

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Mupad [B]
time = 0.15, size = 30, normalized size = 0.70 \begin {gather*} \frac {49\,\sqrt {3}\,\mathrm {asin}\left (\frac {6\,x}{7}-\frac {5}{7}\right )}{72}+\left (\frac {x}{2}-\frac {5}{12}\right )\,\sqrt {-3\,x^2+5\,x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x - 3*x^2 + 2)^(1/2),x)

[Out]

(49*3^(1/2)*asin((6*x)/7 - 5/7))/72 + (x/2 - 5/12)*(5*x - 3*x^2 + 2)^(1/2)

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