Integrand size = 24, antiderivative size = 79 \[ \int \frac {-2+3 x}{(1+x)^3 \sqrt {2 x-x^2}} \, dx=-\frac {5 \sqrt {2 x-x^2}}{6 (1+x)^2}-\frac {2 \sqrt {2 x-x^2}}{3 (1+x)}+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3} \sqrt {2 x-x^2}}\right )}{2 \sqrt {3}} \]
1/6*arctan(1/3*(1-2*x)*3^(1/2)/(-x^2+2*x)^(1/2))*3^(1/2)-5/6*(-x^2+2*x)^(1 /2)/(1+x)^2-2/3*(-x^2+2*x)^(1/2)/(1+x)
Time = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99 \[ \int \frac {-2+3 x}{(1+x)^3 \sqrt {2 x-x^2}} \, dx=\frac {x \left (-18+x+4 x^2\right )-2 \sqrt {3} \sqrt {-2+x} \sqrt {x} (1+x)^2 \text {arctanh}\left (\frac {1-\sqrt {-2+x} \sqrt {x}+x}{\sqrt {3}}\right )}{6 \sqrt {-((-2+x) x)} (1+x)^2} \]
(x*(-18 + x + 4*x^2) - 2*Sqrt[3]*Sqrt[-2 + x]*Sqrt[x]*(1 + x)^2*ArcTanh[(1 - Sqrt[-2 + x]*Sqrt[x] + x)/Sqrt[3]])/(6*Sqrt[-((-2 + x)*x)]*(1 + x)^2)
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1237, 25, 1228, 1154, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {3 x-2}{(x+1)^3 \sqrt {2 x-x^2}} \, dx\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle \frac {1}{6} \int -\frac {7-5 x}{(x+1)^2 \sqrt {2 x-x^2}}dx-\frac {5 \sqrt {2 x-x^2}}{6 (x+1)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{6} \int \frac {7-5 x}{(x+1)^2 \sqrt {2 x-x^2}}dx-\frac {5 \sqrt {2 x-x^2}}{6 (x+1)^2}\) |
\(\Big \downarrow \) 1228 |
\(\displaystyle \frac {1}{6} \left (-3 \int \frac {1}{(x+1) \sqrt {2 x-x^2}}dx-\frac {4 \sqrt {2 x-x^2}}{x+1}\right )-\frac {5 \sqrt {2 x-x^2}}{6 (x+1)^2}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{6} \left (6 \int \frac {1}{-\frac {4 (1-2 x)^2}{2 x-x^2}-12}d\left (-\frac {2 (1-2 x)}{\sqrt {2 x-x^2}}\right )-\frac {4 \sqrt {2 x-x^2}}{x+1}\right )-\frac {5 \sqrt {2 x-x^2}}{6 (x+1)^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{6} \left (\sqrt {3} \arctan \left (\frac {1-2 x}{\sqrt {3} \sqrt {2 x-x^2}}\right )-\frac {4 \sqrt {2 x-x^2}}{x+1}\right )-\frac {5 \sqrt {2 x-x^2}}{6 (x+1)^2}\) |
(-5*Sqrt[2*x - x^2])/(6*(1 + x)^2) + ((-4*Sqrt[2*x - x^2])/(1 + x) + Sqrt[ 3]*ArcTan[(1 - 2*x)/(Sqrt[3]*Sqrt[2*x - x^2])])/6
3.3.65.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Time = 0.56 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.63
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {3}\, \left (1+x \right )^{2} \arctan \left (\frac {\sqrt {3}\, \sqrt {-x \left (-2+x \right )}}{3 x}\right )+\left (-4 x -9\right ) \sqrt {-x \left (-2+x \right )}}{6 \left (1+x \right )^{2}}\) | \(50\) |
risch | \(\frac {x \left (-2+x \right ) \left (4 x +9\right )}{6 \left (1+x \right )^{2} \sqrt {-x \left (-2+x \right )}}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (-2+4 x \right ) \sqrt {3}}{6 \sqrt {-\left (1+x \right )^{2}+1+4 x}}\right )}{6}\) | \(56\) |
trager | \(-\frac {\left (4 x +9\right ) \sqrt {-x^{2}+2 x}}{6 \left (1+x \right )^{2}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x +3 \sqrt {-x^{2}+2 x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right )}{1+x}\right )}{6}\) | \(69\) |
default | \(-\frac {5 \sqrt {-\left (1+x \right )^{2}+1+4 x}}{6 \left (1+x \right )^{2}}-\frac {2 \sqrt {-\left (1+x \right )^{2}+1+4 x}}{3 \left (1+x \right )}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (-2+4 x \right ) \sqrt {3}}{6 \sqrt {-\left (1+x \right )^{2}+1+4 x}}\right )}{6}\) | \(74\) |
1/6*(2*3^(1/2)*(1+x)^2*arctan(1/3*3^(1/2)*(-x*(-2+x))^(1/2)/x)+(-4*x-9)*(- x*(-2+x))^(1/2))/(1+x)^2
Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.81 \[ \int \frac {-2+3 x}{(1+x)^3 \sqrt {2 x-x^2}} \, dx=\frac {2 \, \sqrt {3} {\left (x^{2} + 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {3} \sqrt {-x^{2} + 2 \, x}}{3 \, x}\right ) - \sqrt {-x^{2} + 2 \, x} {\left (4 \, x + 9\right )}}{6 \, {\left (x^{2} + 2 \, x + 1\right )}} \]
1/6*(2*sqrt(3)*(x^2 + 2*x + 1)*arctan(1/3*sqrt(3)*sqrt(-x^2 + 2*x)/x) - sq rt(-x^2 + 2*x)*(4*x + 9))/(x^2 + 2*x + 1)
\[ \int \frac {-2+3 x}{(1+x)^3 \sqrt {2 x-x^2}} \, dx=\int \frac {3 x - 2}{\sqrt {- x \left (x - 2\right )} \left (x + 1\right )^{3}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {-2+3 x}{(1+x)^3 \sqrt {2 x-x^2}} \, dx=-\frac {1}{6} \, \sqrt {3} \arcsin \left (\frac {2 \, x}{{\left | x + 1 \right |}} - \frac {1}{{\left | x + 1 \right |}}\right ) - \frac {5 \, \sqrt {-x^{2} + 2 \, x}}{6 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {2 \, \sqrt {-x^{2} + 2 \, x}}{3 \, {\left (x + 1\right )}} \]
-1/6*sqrt(3)*arcsin(2*x/abs(x + 1) - 1/abs(x + 1)) - 5/6*sqrt(-x^2 + 2*x)/ (x^2 + 2*x + 1) - 2/3*sqrt(-x^2 + 2*x)/(x + 1)
Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (64) = 128\).
Time = 0.30 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.86 \[ \int \frac {-2+3 x}{(1+x)^3 \sqrt {2 x-x^2}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}}{x - 1} - 1\right )}\right ) + \frac {\frac {34 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}}{x - 1} - \frac {39 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}^{2}}{{\left (x - 1\right )}^{2}} + \frac {18 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}^{3}}{{\left (x - 1\right )}^{3}} - 26}{24 \, {\left (\frac {\sqrt {-x^{2} + 2 \, x} - 1}{x - 1} - \frac {{\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}^{2}}{{\left (x - 1\right )}^{2}} - 1\right )}^{2}} \]
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(sqrt(-x^2 + 2*x) - 1)/(x - 1) - 1)) + 1 /24*(34*(sqrt(-x^2 + 2*x) - 1)/(x - 1) - 39*(sqrt(-x^2 + 2*x) - 1)^2/(x - 1)^2 + 18*(sqrt(-x^2 + 2*x) - 1)^3/(x - 1)^3 - 26)/((sqrt(-x^2 + 2*x) - 1) /(x - 1) - (sqrt(-x^2 + 2*x) - 1)^2/(x - 1)^2 - 1)^2
Timed out. \[ \int \frac {-2+3 x}{(1+x)^3 \sqrt {2 x-x^2}} \, dx=\int \frac {3\,x-2}{\sqrt {2\,x-x^2}\,{\left (x+1\right )}^3} \,d x \]