Integrand size = 28, antiderivative size = 68 \[ \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=-\frac {\arctan \left (\frac {1-\frac {3 \sqrt {-1-x}}{\sqrt {3+x}}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {\arctan \left (\frac {1+\frac {3 \sqrt {-1-x}}{\sqrt {3+x}}}{\sqrt {2}}\right )}{\sqrt {2}} \] Output:
-1/2*arctan(1/2*(1-3*(-1-x)^(1/2)/(3+x)^(1/2))*2^(1/2))*2^(1/2)+1/2*arctan (1/2*(1+3*(-1-x)^(1/2)/(3+x)^(1/2))*2^(1/2))*2^(1/2)
Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.49 \[ \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=-\frac {\arctan \left (\frac {3+2 x}{\sqrt {2} \sqrt {-3-4 x-x^2}}\right )}{\sqrt {2}} \] Input:
Integrate[x/(Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]
Output:
-(ArcTan[(3 + 2*x)/(Sqrt[2]*Sqrt[-3 - 4*x - x^2])]/Sqrt[2])
Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.50, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1359, 27, 1475, 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.
\(\displaystyle \int \frac {x}{\sqrt {-x^2-4 x-3} \left (2 x^2+4 x+3\right )} \, dx\) |
\(\Big \downarrow \) 1359 |
\(\displaystyle 8 \int -\frac {\frac {(x+3)^2}{3 \left (-x^2-4 x-3\right )}+1}{4 \left (\frac {(x+3)^4}{9 \left (-x^2-4 x-3\right )^2}+\frac {2 (x+3)^2}{9 \left (-x^2-4 x-3\right )}+1\right )}d\frac {x+3}{3 \sqrt {-x^2-4 x-3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -2 \int \frac {\frac {(x+3)^2}{3 \left (-x^2-4 x-3\right )}+1}{\frac {(x+3)^4}{9 \left (-x^2-4 x-3\right )^2}+\frac {2 (x+3)^2}{9 \left (-x^2-4 x-3\right )}+1}d\frac {x+3}{3 \sqrt {-x^2-4 x-3}}\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle 2 \left (-\frac {1}{6} \int \frac {1}{\frac {(x+3)^2}{9 \left (-x^2-4 x-3\right )}-\frac {2 (x+3)}{9 \sqrt {-x^2-4 x-3}}+\frac {1}{3}}d\frac {x+3}{3 \sqrt {-x^2-4 x-3}}-\frac {1}{6} \int \frac {1}{\frac {(x+3)^2}{9 \left (-x^2-4 x-3\right )}+\frac {2 (x+3)}{9 \sqrt {-x^2-4 x-3}}+\frac {1}{3}}d\frac {x+3}{3 \sqrt {-x^2-4 x-3}}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle 2 \left (\frac {1}{3} \int \frac {1}{-\frac {(x+3)^2}{9 \left (-x^2-4 x-3\right )}-\frac {8}{9}}d\left (\frac {2 (x+3)}{3 \sqrt {-x^2-4 x-3}}-\frac {2}{3}\right )+\frac {1}{3} \int \frac {1}{-\frac {(x+3)^2}{9 \left (-x^2-4 x-3\right )}-\frac {8}{9}}d\left (\frac {2 (x+3)}{3 \sqrt {-x^2-4 x-3}}+\frac {2}{3}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\sqrt {2} \arctan \left (\frac {x+3}{2 \sqrt {2} \sqrt {-x^2-4 x-3}}\right )\) |
Input:
Int[x/(Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]
Output:
-(Sqrt[2]*ArcTan[(3 + x)/(2*Sqrt[2]*Sqrt[-3 - 4*x - x^2])])
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[(x_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.) *(x_)^2]), x_Symbol] :> Simp[-2*e Subst[Int[(1 - d*x^2)/(c*e - b*f - e*(2 *c*d - b*e + 2*a*f)*x^2 + d^2*(c*e - b*f)*x^4), x], x, (1 + (e + Sqrt[e^2 - 4*d*f])*(x/(2*d)))/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^ 2, x], x], x] + Simp[e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] , 0]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.69 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.19
method | result | size |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}+20 x \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )+8 x \sqrt {-x^{2}-4 x -3}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )+12 \sqrt {-x^{2}-4 x -3}}{2 x^{2}+4 x +3}\right )}{4}\) | \(81\) |
default | \(\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}\, \sqrt {2}}{6}\right )}{12 \sqrt {\frac {\frac {x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-4}{\left (1+\frac {x}{-\frac {3}{2}-x}\right )^{2}}}\, \left (1+\frac {x}{-\frac {3}{2}-x}\right )}\) | \(92\) |
Input:
int(x/(-x^2-4*x-3)^(1/2)/(2*x^2+4*x+3),x,method=_RETURNVERBOSE)
Output:
1/4*RootOf(_Z^2+2)*ln((6*RootOf(_Z^2+2)*x^2+20*x*RootOf(_Z^2+2)+8*x*(-x^2- 4*x-3)^(1/2)+15*RootOf(_Z^2+2)+12*(-x^2-4*x-3)^(1/2))/(2*x^2+4*x+3))
Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (6 \, x^{2} + 20 \, x + 15\right )} \sqrt {-x^{2} - 4 \, x - 3}}{4 \, {\left (2 \, x^{3} + 11 \, x^{2} + 18 \, x + 9\right )}}\right ) \] Input:
integrate(x/(-x^2-4*x-3)^(1/2)/(2*x^2+4*x+3),x, algorithm="fricas")
Output:
1/4*sqrt(2)*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))
\[ \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\int \frac {x}{\sqrt {- \left (x + 1\right ) \left (x + 3\right )} \left (2 x^{2} + 4 x + 3\right )}\, dx \] Input:
integrate(x/(-x**2-4*x-3)**(1/2)/(2*x**2+4*x+3),x)
Output:
Integral(x/(sqrt(-(x + 1)*(x + 3))*(2*x**2 + 4*x + 3)), x)
\[ \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\int { \frac {x}{{\left (2 \, x^{2} + 4 \, x + 3\right )} \sqrt {-x^{2} - 4 \, x - 3}} \,d x } \] Input:
integrate(x/(-x^2-4*x-3)^(1/2)/(2*x^2+4*x+3),x, algorithm="maxima")
Output:
integrate(x/((2*x^2 + 4*x + 3)*sqrt(-x^2 - 4*x - 3)), x)
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {3 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + 1\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) \] Input:
integrate(x/(-x^2-4*x-3)^(1/2)/(2*x^2+4*x+3),x, algorithm="giac")
Output:
1/2*sqrt(2)*arctan(1/2*sqrt(2)*(3*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 1/2*sqrt(2)*arctan(1/2*sqrt(2)*((sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1) )
Timed out. \[ \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\int \frac {x}{\sqrt {-x^2-4\,x-3}\,\left (2\,x^2+4\,x+3\right )} \,d x \] Input:
int(x/((- 4*x - x^2 - 3)^(1/2)*(4*x + 2*x^2 + 3)),x)
Output:
int(x/((- 4*x - x^2 - 3)^(1/2)*(4*x + 2*x^2 + 3)), x)
Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.54 \[ \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=-\frac {\sqrt {2}\, \left (\mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (x +2\right )}{2}\right )-1}{\sqrt {2}}\right )+\mathit {atan} \left (\frac {3 \tan \left (\frac {\mathit {asin} \left (x +2\right )}{2}\right )-1}{\sqrt {2}}\right )\right )}{2} \] Input:
int(x/(-x^2-4*x-3)^(1/2)/(2*x^2+4*x+3),x)
Output:
( - sqrt(2)*(atan((tan(asin(x + 2)/2) - 1)/sqrt(2)) + atan((3*tan(asin(x + 2)/2) - 1)/sqrt(2))))/2