Integrand size = 18, antiderivative size = 87 \[ \int \frac {1}{\left (x+\sqrt {-3-4 x-x^2}\right )^2} \, dx=\frac {1-\frac {\sqrt {-1-x}}{\sqrt {3+x}}}{1-\frac {3 (1+x)}{3+x}-\frac {2 \sqrt {-1-x}}{\sqrt {3+x}}}+\frac {\arctan \left (\frac {1-\frac {3 \sqrt {-1-x}}{\sqrt {3+x}}}{\sqrt {2}}\right )}{\sqrt {2}} \]
1/2*arctan(1/2*(1-3*(-1-x)^(1/2)/(3+x)^(1/2))*2^(1/2))*2^(1/2)+(1-(-1-x)^( 1/2)/(3+x)^(1/2))/(1-3*(1+x)/(3+x)-2*(-1-x)^(1/2)/(3+x)^(1/2))
Time = 0.33 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (x+\sqrt {-3-4 x-x^2}\right )^2} \, dx=\frac {3+x+(3+2 x) \sqrt {-3-4 x-x^2}+\sqrt {2} \left (3+4 x+2 x^2\right ) \arctan \left (\frac {\sqrt {2} (1+x)}{1+x+\sqrt {-3-4 x-x^2}}\right )}{2 \left (3+4 x+2 x^2\right )} \]
(3 + x + (3 + 2*x)*Sqrt[-3 - 4*x - x^2] + Sqrt[2]*(3 + 4*x + 2*x^2)*ArcTan [(Sqrt[2]*(1 + x))/(1 + x + Sqrt[-3 - 4*x - x^2])])/(2*(3 + 4*x + 2*x^2))
Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {7287, 27, 1159, 1083, 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.
\(\displaystyle \int \frac {1}{\left (\sqrt {-x^2-4 x-3}+x\right )^2} \, dx\) |
\(\Big \downarrow \) 7287 |
\(\displaystyle 2 \int -\frac {2 \sqrt {-x-1}}{\sqrt {x+3} \left (\frac {3 (-x-1)}{x+3}-\frac {2 \sqrt {-x-1}}{\sqrt {x+3}}+1\right )^2}d\frac {\sqrt {-x-1}}{\sqrt {x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -4 \int \frac {\sqrt {-x-1}}{\sqrt {x+3} \left (\frac {3 (-x-1)}{x+3}-\frac {2 \sqrt {-x-1}}{\sqrt {x+3}}+1\right )^2}d\frac {\sqrt {-x-1}}{\sqrt {x+3}}\) |
\(\Big \downarrow \) 1159 |
\(\displaystyle -4 \left (\frac {1}{4} \int \frac {1}{\frac {3 (-x-1)}{x+3}-\frac {2 \sqrt {-x-1}}{\sqrt {x+3}}+1}d\frac {\sqrt {-x-1}}{\sqrt {x+3}}-\frac {1-\frac {\sqrt {-x-1}}{\sqrt {x+3}}}{4 \left (\frac {3 (-x-1)}{x+3}-\frac {2 \sqrt {-x-1}}{\sqrt {x+3}}+1\right )}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -4 \left (-\frac {1}{2} \int \frac {1}{-\frac {-x-1}{x+3}-8}d\left (\frac {6 \sqrt {-x-1}}{\sqrt {x+3}}-2\right )-\frac {1-\frac {\sqrt {-x-1}}{\sqrt {x+3}}}{4 \left (\frac {3 (-x-1)}{x+3}-\frac {2 \sqrt {-x-1}}{\sqrt {x+3}}+1\right )}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -4 \left (\frac {\arctan \left (\frac {\frac {6 \sqrt {-x-1}}{\sqrt {x+3}}-2}{2 \sqrt {2}}\right )}{4 \sqrt {2}}-\frac {1-\frac {\sqrt {-x-1}}{\sqrt {x+3}}}{4 \left (\frac {3 (-x-1)}{x+3}-\frac {2 \sqrt {-x-1}}{\sqrt {x+3}}+1\right )}\right )\) |
-4*(-1/4*(1 - Sqrt[-1 - x]/Sqrt[3 + x])/(1 + (3*(-1 - x))/(3 + x) - (2*Sqr t[-1 - x])/Sqrt[3 + x]) + ArcTan[(-2 + (6*Sqrt[-1 - x])/Sqrt[3 + x])/(2*Sq rt[2])]/(4*Sqrt[2]))
3.8.60.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* x + c*x^2)^(p + 1), x] - Simp[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 - 4*a* c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] & & LtQ[p, -1] && NeQ[p, -3/2]
Int[u_, x_Symbol] :> With[{lst = FunctionOfSquareRootOfQuadratic[u, x]}, Si mp[2 Subst[Int[lst[[1]], x], x, lst[[2]]], x] /; !FalseQ[lst] && EqQ[lst [[3]], 3]] /; EulerIntegrandQ[u, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.34 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.26
method | result | size |
trager | \(-\frac {\left (2 x +3\right ) x}{2 \left (2 x^{2}+4 x +3\right )}+\frac {\left (2 x +3\right ) \sqrt {-x^{2}-4 x -3}}{4 x^{2}+8 x +6}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +2 \sqrt {-x^{2}-4 x -3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +2 x +3}\right )}{4}\) | \(110\) |
default | \(\text {Expression too large to display}\) | \(2407\) |
-1/2*(2*x+3)*x/(2*x^2+4*x+3)+1/2*(2*x+3)/(2*x^2+4*x+3)*(-x^2-4*x-3)^(1/2)+ 1/4*RootOf(_Z^2+2)*ln((-2*RootOf(_Z^2+2)*x+2*(-x^2-4*x-3)^(1/2)-3*RootOf(_ Z^2+2))/(RootOf(_Z^2+2)*x+2*x+3))
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\left (x+\sqrt {-3-4 x-x^2}\right )^2} \, dx=\frac {2 \, \sqrt {2} {\left (2 \, x^{2} + 4 \, x + 3\right )} \arctan \left (\sqrt {2} {\left (x + 1\right )}\right ) - \sqrt {2} {\left (2 \, x^{2} + 4 \, x + 3\right )} \arctan \left (\frac {\sqrt {2} {\left (6 \, x^{2} + 20 \, x + 15\right )} \sqrt {-x^{2} - 4 \, x - 3}}{4 \, {\left (2 \, x^{3} + 11 \, x^{2} + 18 \, x + 9\right )}}\right ) + 4 \, \sqrt {-x^{2} - 4 \, x - 3} {\left (2 \, x + 3\right )} + 4 \, x + 12}{8 \, {\left (2 \, x^{2} + 4 \, x + 3\right )}} \]
1/8*(2*sqrt(2)*(2*x^2 + 4*x + 3)*arctan(sqrt(2)*(x + 1)) - sqrt(2)*(2*x^2 + 4*x + 3)*arctan(1/4*sqrt(2)*(6*x^2 + 20*x + 15)*sqrt(-x^2 - 4*x - 3)/(2* x^3 + 11*x^2 + 18*x + 9)) + 4*sqrt(-x^2 - 4*x - 3)*(2*x + 3) + 4*x + 12)/( 2*x^2 + 4*x + 3)
\[ \int \frac {1}{\left (x+\sqrt {-3-4 x-x^2}\right )^2} \, dx=\int \frac {1}{\left (x + \sqrt {- x^{2} - 4 x - 3}\right )^{2}}\, dx \]
\[ \int \frac {1}{\left (x+\sqrt {-3-4 x-x^2}\right )^2} \, dx=\int { \frac {1}{{\left (x + \sqrt {-x^{2} - 4 \, x - 3}\right )}^{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (72) = 144\).
Time = 0.34 (sec) , antiderivative size = 263, normalized size of antiderivative = 3.02 \[ \int \frac {1}{\left (x+\sqrt {-3-4 x-x^2}\right )^2} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (x + 1\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {3 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + 1\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) + \frac {x + 3}{2 \, {\left (2 \, x^{2} + 4 \, x + 3\right )}} - \frac {\frac {10 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac {7 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} - \frac {2 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{3}}{{\left (x + 2\right )}^{3}} + 3}{3 \, {\left (\frac {8 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac {14 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + \frac {8 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{3}}{{\left (x + 2\right )}^{3}} + \frac {3 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{4}}{{\left (x + 2\right )}^{4}} + 3\right )}} \]
1/4*sqrt(2)*arctan(sqrt(2)*(x + 1)) - 1/4*sqrt(2)*arctan(1/2*sqrt(2)*(3*(s qrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) - 1/4*sqrt(2)*arctan(1/2*sqrt(2)*(( sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 1/2*(x + 3)/(2*x^2 + 4*x + 3) - 1/3*(10*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 7*(sqrt(-x^2 - 4*x - 3) - 1)^ 2/(x + 2)^2 - 2*(sqrt(-x^2 - 4*x - 3) - 1)^3/(x + 2)^3 + 3)/(8*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 14*(sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 + 8*(s qrt(-x^2 - 4*x - 3) - 1)^3/(x + 2)^3 + 3*(sqrt(-x^2 - 4*x - 3) - 1)^4/(x + 2)^4 + 3)
Timed out. \[ \int \frac {1}{\left (x+\sqrt {-3-4 x-x^2}\right )^2} \, dx=\int \frac {1}{{\left (x+\sqrt {-x^2-4\,x-3}\right )}^2} \,d x \]