Integrand size = 25, antiderivative size = 286 \[ \int \frac {x^2 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=-\frac {11 x \left (5+\sqrt {13}+2 x^2\right )}{26 \sqrt {3+5 x^2+x^4}}+\frac {x \left (8+11 x^2\right )}{13 \sqrt {3+5 x^2+x^4}}+\frac {11 \sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} \sqrt {\frac {6+\left (5-\sqrt {13}\right ) x^2}{6+\left (5+\sqrt {13}\right ) x^2}} \left (6+\left (5+\sqrt {13}\right ) x^2\right ) E\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{26 \sqrt {3+5 x^2+x^4}}-\frac {4 \sqrt {\frac {2}{3 \left (5+\sqrt {13}\right )}} \sqrt {\frac {6+\left (5-\sqrt {13}\right ) x^2}{6+\left (5+\sqrt {13}\right ) x^2}} \left (6+\left (5+\sqrt {13}\right ) x^2\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right ),\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{13 \sqrt {3+5 x^2+x^4}} \]
1/13*x*(11*x^2+8)/(x^4+5*x^2+3)^(1/2)-11/26*x*(5+2*x^2+13^(1/2))/(x^4+5*x^ 2+3)^(1/2)-4/39*(1/(36+x^2*(30+6*13^(1/2))))^(1/2)*(36+x^2*(30+6*13^(1/2)) )^(1/2)*EllipticF(x*(30+6*13^(1/2))^(1/2)/(36+x^2*(30+6*13^(1/2)))^(1/2),1 /6*(-78+30*13^(1/2))^(1/2))*(6+x^2*(5+13^(1/2)))*6^(1/2)/(5+13^(1/2))^(1/2 )*((6+x^2*(5-13^(1/2)))/(6+x^2*(5+13^(1/2))))^(1/2)/(x^4+5*x^2+3)^(1/2)+11 /156*(1/(36+x^2*(30+6*13^(1/2))))^(1/2)*(36+x^2*(30+6*13^(1/2)))^(1/2)*Ell ipticE(x*(30+6*13^(1/2))^(1/2)/(36+x^2*(30+6*13^(1/2)))^(1/2),1/6*(-78+30* 13^(1/2))^(1/2))*(6+x^2*(5+13^(1/2)))*(30+6*13^(1/2))^(1/2)*((6+x^2*(5-13^ (1/2)))/(6+x^2*(5+13^(1/2))))^(1/2)/(x^4+5*x^2+3)^(1/2)
Result contains complex when optimal does not.
Time = 10.25 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\frac {4 x \left (8+11 x^2\right )-11 i \sqrt {2} \left (-5+\sqrt {13}\right ) \sqrt {\frac {-5+\sqrt {13}-2 x^2}{-5+\sqrt {13}}} \sqrt {5+\sqrt {13}+2 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )+i \sqrt {2} \left (-39+11 \sqrt {13}\right ) \sqrt {\frac {-5+\sqrt {13}-2 x^2}{-5+\sqrt {13}}} \sqrt {5+\sqrt {13}+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right ),\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )}{52 \sqrt {3+5 x^2+x^4}} \]
(4*x*(8 + 11*x^2) - (11*I)*Sqrt[2]*(-5 + Sqrt[13])*Sqrt[(-5 + Sqrt[13] - 2 *x^2)/(-5 + Sqrt[13])]*Sqrt[5 + Sqrt[13] + 2*x^2]*EllipticE[I*ArcSinh[Sqrt [2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6] + I*Sqrt[2]*(-39 + 11*Sqrt[1 3])*Sqrt[(-5 + Sqrt[13] - 2*x^2)/(-5 + Sqrt[13])]*Sqrt[5 + Sqrt[13] + 2*x^ 2]*EllipticF[I*ArcSinh[Sqrt[2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6])/ (52*Sqrt[3 + 5*x^2 + x^4])
Time = 0.37 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1598, 25, 1503, 1412, 1455}
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.
| \(\displaystyle \int \frac {x^2 \left (3 x^2+2\right )}{\left (x^4+5 x^2+3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1598 |
| \(\displaystyle \frac {1}{13} \int -\frac {11 x^2+8}{\sqrt {x^4+5 x^2+3}}dx+\frac {x \left (11 x^2+8\right )}{13 \sqrt {x^4+5 x^2+3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {x \left (11 x^2+8\right )}{13 \sqrt {x^4+5 x^2+3}}-\frac {1}{13} \int \frac {11 x^2+8}{\sqrt {x^4+5 x^2+3}}dx\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle \frac {1}{13} \left (-8 \int \frac {1}{\sqrt {x^4+5 x^2+3}}dx-11 \int \frac {x^2}{\sqrt {x^4+5 x^2+3}}dx\right )+\frac {x \left (11 x^2+8\right )}{13 \sqrt {x^4+5 x^2+3}}\) |
| \(\Big \downarrow \) 1412 |
| \(\displaystyle \frac {1}{13} \left (-11 \int \frac {x^2}{\sqrt {x^4+5 x^2+3}}dx-\frac {4 \sqrt {\frac {2}{3 \left (5+\sqrt {13}\right )}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right ),\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {x^4+5 x^2+3}}\right )+\frac {x \left (11 x^2+8\right )}{13 \sqrt {x^4+5 x^2+3}}\) |
| \(\Big \downarrow \) 1455 |
| \(\displaystyle \frac {1}{13} \left (-\frac {4 \sqrt {\frac {2}{3 \left (5+\sqrt {13}\right )}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right ),\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {x^4+5 x^2+3}}-11 \left (\frac {x \left (2 x^2+\sqrt {13}+5\right )}{2 \sqrt {x^4+5 x^2+3}}-\frac {\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{2 \sqrt {x^4+5 x^2+3}}\right )\right )+\frac {x \left (11 x^2+8\right )}{13 \sqrt {x^4+5 x^2+3}}\) |
(x*(8 + 11*x^2))/(13*Sqrt[3 + 5*x^2 + x^4]) + (-11*((x*(5 + Sqrt[13] + 2*x ^2))/(2*Sqrt[3 + 5*x^2 + x^4]) - (Sqrt[(5 + Sqrt[13])/6]*Sqrt[(6 + (5 - Sq rt[13])*x^2)/(6 + (5 + Sqrt[13])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticE[ ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(2*Sqrt[3 + 5*x^2 + x^4])) - (4*Sqrt[2/(3*(5 + Sqrt[13]))]*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[13])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticF[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/Sqrt[3 + 5*x^2 + x^4])/13
3.2.100.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b ^2 - 4*a*c, 2]}, Simp[(2*a + (b + q)*x^2)*(Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF [ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] && !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[ {a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[x*((b + q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4 ])), x] - Simp[Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*(Sqrt[(2*a + (b - q )*x^2)/(2*a + (b + q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan [Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] && !(PosQ[ (b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d Int[1/Sqrt[a + b*x^2 + c*x^4] , x], x] + Simp[e Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b + q) /a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_.), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1) *((b*d - 2*a*e - (b*e - 2*c*d)*x^2)/(2*(p + 1)*(b^2 - 4*a*c))), x] - Simp[f ^2/(2*(p + 1)*(b^2 - 4*a*c)) Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1 )*Simp[(m - 1)*(b*d - 2*a*e) - (4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Time = 2.62 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.76
| method | result | size |
| risch | \(\frac {x \left (11 x^{2}+8\right )}{13 \sqrt {x^{4}+5 x^{2}+3}}-\frac {48 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{13 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}+\frac {396 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-E\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{13 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}\) | \(216\) |
| elliptic | \(-\frac {2 \left (-\frac {11}{26} x^{3}-\frac {4}{13} x \right )}{\sqrt {x^{4}+5 x^{2}+3}}-\frac {48 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{13 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}+\frac {396 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-E\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{13 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}\) | \(217\) |
| default | \(-\frac {6 \left (-\frac {5}{26} x^{3}-\frac {3}{13} x \right )}{\sqrt {x^{4}+5 x^{2}+3}}-\frac {48 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{13 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}+\frac {396 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-E\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{13 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}-\frac {4 \left (\frac {1}{13} x^{3}+\frac {5}{26} x \right )}{\sqrt {x^{4}+5 x^{2}+3}}\) | \(240\) |
1/13*x*(11*x^2+8)/(x^4+5*x^2+3)^(1/2)-48/13/(-30+6*13^(1/2))^(1/2)*(1-(-5/ 6+1/6*13^(1/2))*x^2)^(1/2)*(1-(-5/6-1/6*13^(1/2))*x^2)^(1/2)/(x^4+5*x^2+3) ^(1/2)*EllipticF(1/6*x*(-30+6*13^(1/2))^(1/2),5/6*3^(1/2)+1/6*39^(1/2))+39 6/13/(-30+6*13^(1/2))^(1/2)*(1-(-5/6+1/6*13^(1/2))*x^2)^(1/2)*(1-(-5/6-1/6 *13^(1/2))*x^2)^(1/2)/(x^4+5*x^2+3)^(1/2)/(5+13^(1/2))*(EllipticF(1/6*x*(- 30+6*13^(1/2))^(1/2),5/6*3^(1/2)+1/6*39^(1/2))-EllipticE(1/6*x*(-30+6*13^( 1/2))^(1/2),5/6*3^(1/2)+1/6*39^(1/2)))
Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.63 \[ \int \frac {x^2 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\frac {11 \, {\left (\sqrt {13} \sqrt {6} \sqrt {3} {\left (x^{4} + 5 \, x^{2} + 3\right )} - 5 \, \sqrt {6} \sqrt {3} {\left (x^{4} + 5 \, x^{2} + 3\right )}\right )} \sqrt {\sqrt {13} - 5} E(\arcsin \left (\frac {1}{6} \, \sqrt {6} x \sqrt {\sqrt {13} - 5}\right )\,|\,\frac {5}{6} \, \sqrt {13} + \frac {19}{6}) - {\left (3 \, \sqrt {13} \sqrt {6} \sqrt {3} {\left (x^{4} + 5 \, x^{2} + 3\right )} - 95 \, \sqrt {6} \sqrt {3} {\left (x^{4} + 5 \, x^{2} + 3\right )}\right )} \sqrt {\sqrt {13} - 5} F(\arcsin \left (\frac {1}{6} \, \sqrt {6} x \sqrt {\sqrt {13} - 5}\right )\,|\,\frac {5}{6} \, \sqrt {13} + \frac {19}{6}) + 36 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (11 \, x^{3} + 8 \, x\right )}}{468 \, {\left (x^{4} + 5 \, x^{2} + 3\right )}} \]
1/468*(11*(sqrt(13)*sqrt(6)*sqrt(3)*(x^4 + 5*x^2 + 3) - 5*sqrt(6)*sqrt(3)* (x^4 + 5*x^2 + 3))*sqrt(sqrt(13) - 5)*elliptic_e(arcsin(1/6*sqrt(6)*x*sqrt (sqrt(13) - 5)), 5/6*sqrt(13) + 19/6) - (3*sqrt(13)*sqrt(6)*sqrt(3)*(x^4 + 5*x^2 + 3) - 95*sqrt(6)*sqrt(3)*(x^4 + 5*x^2 + 3))*sqrt(sqrt(13) - 5)*ell iptic_f(arcsin(1/6*sqrt(6)*x*sqrt(sqrt(13) - 5)), 5/6*sqrt(13) + 19/6) + 3 6*sqrt(x^4 + 5*x^2 + 3)*(11*x^3 + 8*x))/(x^4 + 5*x^2 + 3)
\[ \int \frac {x^2 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\int \frac {x^{2} \cdot \left (3 x^{2} + 2\right )}{\left (x^{4} + 5 x^{2} + 3\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {x^2 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\int { \frac {{\left (3 \, x^{2} + 2\right )} x^{2}}{{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^2 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\int { \frac {{\left (3 \, x^{2} + 2\right )} x^{2}}{{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\int \frac {x^2\,\left (3\,x^2+2\right )}{{\left (x^4+5\,x^2+3\right )}^{3/2}} \,d x \]