\(\int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx\) [2368]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 56 \[ \int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx=-\frac {3}{2} (2-x) \sqrt {9+16 x-4 x^2}+\frac {1}{6} \left (9+16 x-4 x^2\right )^{3/2}-\frac {75}{4} \arcsin \left (\frac {2 (2-x)}{5}\right ) \]

[Out]

1/6*(-4*x^2+16*x+9)^(3/2)+75/4*arcsin(-4/5+2/5*x)-3/2*(2-x)*(-4*x^2+16*x+9)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {654, 626, 633, 222} \[ \int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx=-\frac {75}{4} \arcsin \left (\frac {2 (2-x)}{5}\right )+\frac {1}{6} \left (-4 x^2+16 x+9\right )^{3/2}-\frac {3}{2} (2-x) \sqrt {-4 x^2+16 x+9} \]

[In]

Int[(7 - 2*x)*Sqrt[9 + 16*x - 4*x^2],x]

[Out]

(-3*(2 - x)*Sqrt[9 + 16*x - 4*x^2])/2 + (9 + 16*x - 4*x^2)^(3/2)/6 - (75*ArcSin[(2*(2 - x))/5])/4

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]

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}& = \frac {1}{6} \left (9+16 x-4 x^2\right )^{3/2}+3 \int \sqrt {9+16 x-4 x^2} \, dx \\ & = -\frac {3}{2} (2-x) \sqrt {9+16 x-4 x^2}+\frac {1}{6} \left (9+16 x-4 x^2\right )^{3/2}+\frac {75}{2} \int \frac {1}{\sqrt {9+16 x-4 x^2}} \, dx \\ & = -\frac {3}{2} (2-x) \sqrt {9+16 x-4 x^2}+\frac {1}{6} \left (9+16 x-4 x^2\right )^{3/2}-\frac {15}{16} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{400}}} \, dx,x,16-8 x\right ) \\ & = -\frac {3}{2} (2-x) \sqrt {9+16 x-4 x^2}+\frac {1}{6} \left (9+16 x-4 x^2\right )^{3/2}-\frac {75}{4} \sin ^{-1}\left (\frac {2 (2-x)}{5}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx=\frac {1}{6} \sqrt {9+16 x-4 x^2} \left (-9+25 x-4 x^2\right )-\frac {75}{2} \arctan \left (\frac {\sqrt {9+16 x-4 x^2}}{1+2 x}\right ) \]

[In]

Integrate[(7 - 2*x)*Sqrt[9 + 16*x - 4*x^2],x]

[Out]

(Sqrt[9 + 16*x - 4*x^2]*(-9 + 25*x - 4*x^2))/6 - (75*ArcTan[Sqrt[9 + 16*x - 4*x^2]/(1 + 2*x)])/2

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.77

method result size
default \(-\frac {3 \left (-8 x +16\right ) \sqrt {-4 x^{2}+16 x +9}}{16}+\frac {75 \arcsin \left (-\frac {4}{5}+\frac {2 x}{5}\right )}{4}+\frac {\left (-4 x^{2}+16 x +9\right )^{\frac {3}{2}}}{6}\) \(43\)
risch \(\frac {\left (4 x^{2}-25 x +9\right ) \left (4 x^{2}-16 x -9\right )}{6 \sqrt {-4 x^{2}+16 x +9}}+\frac {75 \arcsin \left (-\frac {4}{5}+\frac {2 x}{5}\right )}{4}\) \(44\)
trager \(\left (-\frac {2}{3} x^{2}+\frac {25}{6} x -\frac {3}{2}\right ) \sqrt {-4 x^{2}+16 x +9}+\frac {75 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\sqrt {-4 x^{2}+16 x +9}\right )}{4}\) \(64\)

[In]

int((7-2*x)*(-4*x^2+16*x+9)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-3/16*(-8*x+16)*(-4*x^2+16*x+9)^(1/2)+75/4*arcsin(-4/5+2/5*x)+1/6*(-4*x^2+16*x+9)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.84 \[ \int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx=-\frac {1}{6} \, {\left (4 \, x^{2} - 25 \, x + 9\right )} \sqrt {-4 \, x^{2} + 16 \, x + 9} - \frac {75}{2} \, \arctan \left (\frac {\sqrt {-4 \, x^{2} + 16 \, x + 9} - 3}{2 \, x}\right ) \]

[In]

integrate((7-2*x)*(-4*x^2+16*x+9)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(4*x^2 - 25*x + 9)*sqrt(-4*x^2 + 16*x + 9) - 75/2*arctan(1/2*(sqrt(-4*x^2 + 16*x + 9) - 3)/x)

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.73 \[ \int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx=\sqrt {- 4 x^{2} + 16 x + 9} \left (- \frac {2 x^{2}}{3} + \frac {25 x}{6} - \frac {3}{2}\right ) + \frac {75 \operatorname {asin}{\left (\frac {2 x}{5} - \frac {4}{5} \right )}}{4} \]

[In]

integrate((7-2*x)*(-4*x**2+16*x+9)**(1/2),x)

[Out]

sqrt(-4*x**2 + 16*x + 9)*(-2*x**2/3 + 25*x/6 - 3/2) + 75*asin(2*x/5 - 4/5)/4

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx=\frac {1}{6} \, {\left (-4 \, x^{2} + 16 \, x + 9\right )}^{\frac {3}{2}} + \frac {3}{2} \, \sqrt {-4 \, x^{2} + 16 \, x + 9} x - 3 \, \sqrt {-4 \, x^{2} + 16 \, x + 9} - \frac {75}{4} \, \arcsin \left (-\frac {2}{5} \, x + \frac {4}{5}\right ) \]

[In]

integrate((7-2*x)*(-4*x^2+16*x+9)^(1/2),x, algorithm="maxima")

[Out]

1/6*(-4*x^2 + 16*x + 9)^(3/2) + 3/2*sqrt(-4*x^2 + 16*x + 9)*x - 3*sqrt(-4*x^2 + 16*x + 9) - 75/4*arcsin(-2/5*x
 + 4/5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.57 \[ \int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx=-\frac {1}{6} \, {\left ({\left (4 \, x - 25\right )} x + 9\right )} \sqrt {-4 \, x^{2} + 16 \, x + 9} + \frac {75}{4} \, \arcsin \left (\frac {2}{5} \, x - \frac {4}{5}\right ) \]

[In]

integrate((7-2*x)*(-4*x^2+16*x+9)^(1/2),x, algorithm="giac")

[Out]

-1/6*((4*x - 25)*x + 9)*sqrt(-4*x^2 + 16*x + 9) + 75/4*arcsin(2/5*x - 4/5)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.32 \[ \int (7-2 x) \sqrt {9+16 x-4 x^2} \, dx=\frac {175\,\mathrm {asin}\left (\frac {2\,x}{5}-\frac {4}{5}\right )}{4}+7\,\left (\frac {x}{2}-1\right )\,\sqrt {-4\,x^2+16\,x+9}+\frac {\sqrt {-4\,x^2+16\,x+9}\,\left (-128\,x^2+128\,x+1056\right )}{192}+\ln \left (x-2-\frac {\sqrt {-4\,x^2+16\,x+9}\,1{}\mathrm {i}}{2}\right )\,25{}\mathrm {i} \]

[In]

int(-(2*x - 7)*(16*x - 4*x^2 + 9)^(1/2),x)

[Out]

log(x - ((16*x - 4*x^2 + 9)^(1/2)*1i)/2 - 2)*25i + (175*asin((2*x)/5 - 4/5))/4 + 7*(x/2 - 1)*(16*x - 4*x^2 + 9
)^(1/2) + ((16*x - 4*x^2 + 9)^(1/2)*(128*x - 128*x^2 + 1056))/192