\(\int \sqrt {5 x-9 x^2} \, dx\) [11]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \sqrt {5 x-9 x^2} \, dx=-\frac {1}{36} (5-18 x) \sqrt {5 x-9 x^2}-\frac {25}{216} \arcsin \left (1-\frac {18 x}{5}\right ) \]

[Out]

25/216*arcsin(-1+18/5*x)-1/36*(5-18*x)*(-9*x^2+5*x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {626, 633, 222} \[ \int \sqrt {5 x-9 x^2} \, dx=-\frac {25}{216} \arcsin \left (1-\frac {18 x}{5}\right )-\frac {1}{36} \sqrt {5 x-9 x^2} (5-18 x) \]

[In]

Int[Sqrt[5*x - 9*x^2],x]

[Out]

-1/36*((5 - 18*x)*Sqrt[5*x - 9*x^2]) - (25*ArcSin[1 - (18*x)/5])/216

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*} \text {integral}& = -\frac {1}{36} (5-18 x) \sqrt {5 x-9 x^2}+\frac {25}{72} \int \frac {1}{\sqrt {5 x-9 x^2}} \, dx \\ & = -\frac {1}{36} (5-18 x) \sqrt {5 x-9 x^2}-\frac {5}{216} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{25}}} \, dx,x,5-18 x\right ) \\ & = -\frac {1}{36} (5-18 x) \sqrt {5 x-9 x^2}-\frac {25}{216} \sin ^{-1}\left (1-\frac {18 x}{5}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {1}{108} \sqrt {-x (-5+9 x)} \left (-15+54 x+\frac {25 \log \left (-3 \sqrt {x}+\sqrt {-5+9 x}\right )}{\sqrt {x} \sqrt {-5+9 x}}\right ) \]

[In]

Integrate[Sqrt[5*x - 9*x^2],x]

[Out]

(Sqrt[-(x*(-5 + 9*x))]*(-15 + 54*x + (25*Log[-3*Sqrt[x] + Sqrt[-5 + 9*x]])/(Sqrt[x]*Sqrt[-5 + 9*x])))/108

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {25 \arcsin \left (-1+\frac {18 x}{5}\right )}{216}-\frac {\left (5-18 x \right ) \sqrt {-9 x^{2}+5 x}}{36}\)	\(28\)
risch	\(-\frac {\left (-5+18 x \right ) x \left (9 x -5\right )}{36 \sqrt {-x \left (9 x -5\right )}}+\frac {25 \arcsin \left (-1+\frac {18 x}{5}\right )}{216}\)	\(33\)
pseudoelliptic	\(-\frac {25 \arctan \left (\frac {\sqrt {-9 x^{2}+5 x}}{3 x}\right )}{108}+\frac {\left (-5+18 x \right ) \sqrt {-9 x^{2}+5 x}}{36}\)	\(39\)
meijerg	\(-\frac {25 i \left (-\frac {i \sqrt {\pi }\, \sqrt {x}\, \sqrt {5}\, \left (-\frac {54 x}{5}+3\right ) \sqrt {-\frac {9 x}{5}+1}}{10}+\frac {i \sqrt {\pi }\, \arcsin \left (\frac {3 \sqrt {5}\, \sqrt {x}}{5}\right )}{2}\right )}{54 \sqrt {\pi }}\)	\(47\)
trager	\(\left (-\frac {5}{36}+\frac {x}{2}\right ) \sqrt {-9 x^{2}+5 x}-\frac {25 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (18 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+6 \sqrt {-9 x^{2}+5 x}\right )}{216}\)	\(59\)

[In]

int((-9*x^2+5*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

25/216*arcsin(-1+18/5*x)-1/36*(5-18*x)*(-9*x^2+5*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {1}{36} \, \sqrt {-9 \, x^{2} + 5 \, x} {\left (18 \, x - 5\right )} - \frac {25}{108} \, \arctan \left (\frac {\sqrt {-9 \, x^{2} + 5 \, x}}{3 \, x}\right ) \]

[In]

integrate((-9*x^2+5*x)^(1/2),x, algorithm="fricas")

[Out]

1/36*sqrt(-9*x^2 + 5*x)*(18*x - 5) - 25/108*arctan(1/3*sqrt(-9*x^2 + 5*x)/x)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \sqrt {5 x-9 x^2} \, dx=\left (\frac {x}{2} - \frac {5}{36}\right ) \sqrt {- 9 x^{2} + 5 x} + \frac {25 \operatorname {asin}{\left (\frac {18 x}{5} - 1 \right )}}{216} \]

[In]

integrate((-9*x**2+5*x)**(1/2),x)

[Out]

(x/2 - 5/36)*sqrt(-9*x**2 + 5*x) + 25*asin(18*x/5 - 1)/216

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {1}{2} \, \sqrt {-9 \, x^{2} + 5 \, x} x - \frac {5}{36} \, \sqrt {-9 \, x^{2} + 5 \, x} - \frac {25}{216} \, \arcsin \left (-\frac {18}{5} \, x + 1\right ) \]

[In]

integrate((-9*x^2+5*x)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(-9*x^2 + 5*x)*x - 5/36*sqrt(-9*x^2 + 5*x) - 25/216*arcsin(-18/5*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {1}{36} \, \sqrt {-9 \, x^{2} + 5 \, x} {\left (18 \, x - 5\right )} + \frac {25}{216} \, \arcsin \left (\frac {18}{5} \, x - 1\right ) \]

[In]

integrate((-9*x^2+5*x)^(1/2),x, algorithm="giac")

[Out]

1/36*sqrt(-9*x^2 + 5*x)*(18*x - 5) + 25/216*arcsin(18/5*x - 1)

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {25\,\mathrm {asin}\left (\frac {18\,x}{5}-1\right )}{216}+\left (\frac {x}{2}-\frac {5}{36}\right )\,\sqrt {5\,x-9\,x^2} \]

[In]

int((5*x - 9*x^2)^(1/2),x)

[Out]

(25*asin((18*x)/5 - 1))/216 + (x/2 - 5/36)*(5*x - 9*x^2)^(1/2)