Integrand size = 25, antiderivative size = 298 \[ \int \frac {x^4 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx=\frac {419 x \left (5+\sqrt {13}+2 x^2\right )}{30 \sqrt {3+5 x^2+x^4}}-\frac {10}{3} x \sqrt {3+5 x^2+x^4}+\frac {3}{5} x^3 \sqrt {3+5 x^2+x^4}-\frac {419 \sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} \sqrt {\frac {6+\left (5-\sqrt {13}\right ) x^2}{6+\left (5+\sqrt {13}\right ) x^2}} \left (6+\left (5+\sqrt {13}\right ) x^2\right ) E\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{30 \sqrt {3+5 x^2+x^4}}+\frac {5 \sqrt {\frac {2}{3 \left (5+\sqrt {13}\right )}} \sqrt {\frac {6+\left (5-\sqrt {13}\right ) x^2}{6+\left (5+\sqrt {13}\right ) x^2}} \left (6+\left (5+\sqrt {13}\right ) x^2\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right ),\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {3+5 x^2+x^4}} \]
419/30*x*(5+2*x^2+13^(1/2))/(x^4+5*x^2+3)^(1/2)-10/3*x*(x^4+5*x^2+3)^(1/2) +3/5*x^3*(x^4+5*x^2+3)^(1/2)+5/3*(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*1 3^(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)-419/180*(1/(36+x^2*(30+6*13^(1/2))))^(1/2)*(36+x^2*(30+6*13 ^(1/2)))^(1/2)*EllipticE(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 = 10.23 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.77 \[ \int \frac {x^4 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx=\frac {4 x \left (-150-223 x^2-5 x^4+9 x^6\right )+419 i \sqrt {2} \left (-5+\sqrt {13}\right ) \sqrt {\frac {-5+\sqrt {13}-2 x^2}{-5+\sqrt {13}}} \sqrt {5+\sqrt {13}+2 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )-i \sqrt {2} \left (-1795+419 \sqrt {13}\right ) \sqrt {\frac {-5+\sqrt {13}-2 x^2}{-5+\sqrt {13}}} \sqrt {5+\sqrt {13}+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right ),\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )}{60 \sqrt {3+5 x^2+x^4}} \]
(4*x*(-150 - 223*x^2 - 5*x^4 + 9*x^6) + (419*I)*Sqrt[2]*(-5 + Sqrt[13])*Sq rt[(-5 + Sqrt[13] - 2*x^2)/(-5 + Sqrt[13])]*Sqrt[5 + Sqrt[13] + 2*x^2]*Ell ipticE[I*ArcSinh[Sqrt[2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6] - I*Sqr t[2]*(-1795 + 419*Sqrt[13])*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])/(60*Sqrt[3 + 5*x^2 + x^4])
Time = 0.39 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1602, 1602, 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.
| \(\displaystyle \int \frac {x^4 \left (3 x^2+2\right )}{\sqrt {x^4+5 x^2+3}} \, dx\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle \frac {3}{5} x^3 \sqrt {x^4+5 x^2+3}-\frac {1}{5} \int \frac {x^2 \left (50 x^2+27\right )}{\sqrt {x^4+5 x^2+3}}dx\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {419 x^2+150}{\sqrt {x^4+5 x^2+3}}dx-\frac {50}{3} x \sqrt {x^4+5 x^2+3}\right )+\frac {3}{5} \sqrt {x^4+5 x^2+3} x^3\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (150 \int \frac {1}{\sqrt {x^4+5 x^2+3}}dx+419 \int \frac {x^2}{\sqrt {x^4+5 x^2+3}}dx\right )-\frac {50}{3} x \sqrt {x^4+5 x^2+3}\right )+\frac {3}{5} \sqrt {x^4+5 x^2+3} x^3\) |
| \(\Big \downarrow \) 1412 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (419 \int \frac {x^2}{\sqrt {x^4+5 x^2+3}}dx+\frac {25 \sqrt {\frac {6}{5+\sqrt {13}}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right ),\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {x^4+5 x^2+3}}\right )-\frac {50}{3} x \sqrt {x^4+5 x^2+3}\right )+\frac {3}{5} \sqrt {x^4+5 x^2+3} x^3\) |
| \(\Big \downarrow \) 1455 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {25 \sqrt {\frac {6}{5+\sqrt {13}}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right ),\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {x^4+5 x^2+3}}+419 \left (\frac {x \left (2 x^2+\sqrt {13}+5\right )}{2 \sqrt {x^4+5 x^2+3}}-\frac {\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\arctan \left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{2 \sqrt {x^4+5 x^2+3}}\right )\right )-\frac {50}{3} x \sqrt {x^4+5 x^2+3}\right )+\frac {3}{5} \sqrt {x^4+5 x^2+3} x^3\) |
(3*x^3*Sqrt[3 + 5*x^2 + x^4])/5 + ((-50*x*Sqrt[3 + 5*x^2 + x^4])/3 + (419* ((x*(5 + Sqrt[13] + 2*x^2))/(2*Sqrt[3 + 5*x^2 + x^4]) - (Sqrt[(5 + Sqrt[13 ])/6]*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[13])*x^2)]*(6 + (5 + Sq rt[13])*x^2)*EllipticE[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13] )/6])/(2*Sqrt[3 + 5*x^2 + x^4])) + (25*Sqrt[6/(5 + Sqrt[13])]*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[13])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*Ellip ticF[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/Sqrt[3 + 5*x ^2 + x^4])/3)/5
3.2.89.3.1 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[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
Time = 3.89 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.72
| method | result | size |
| risch | \(\frac {x \left (9 x^{2}-50\right ) \sqrt {x^{4}+5 x^{2}+3}}{15}+\frac {60 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}-\frac {5028 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-E\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{5 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}\) | \(216\) |
| default | \(\frac {3 x^{3} \sqrt {x^{4}+5 x^{2}+3}}{5}-\frac {10 x \sqrt {x^{4}+5 x^{2}+3}}{3}+\frac {60 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}-\frac {5028 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-E\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{5 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}\) | \(226\) |
| elliptic | \(\frac {3 x^{3} \sqrt {x^{4}+5 x^{2}+3}}{5}-\frac {10 x \sqrt {x^{4}+5 x^{2}+3}}{3}+\frac {60 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}-\frac {5028 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-E\left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{5 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}\) | \(226\) |
1/15*x*(9*x^2-50)*(x^4+5*x^2+3)^(1/2)+60/(-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))-5028/ 5/(-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 = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.43 \[ \int \frac {x^4 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx=\frac {419 \, {\left (\sqrt {13} \sqrt {2} x - 5 \, \sqrt {2} x\right )} \sqrt {\sqrt {13} - 5} E(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {13} - 5}}{2 \, x}\right )\,|\,\frac {5}{6} \, \sqrt {13} + \frac {19}{6}) - {\left (369 \, \sqrt {13} \sqrt {2} x - 2345 \, \sqrt {2} x\right )} \sqrt {\sqrt {13} - 5} F(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {13} - 5}}{2 \, x}\right )\,|\,\frac {5}{6} \, \sqrt {13} + \frac {19}{6}) + 4 \, {\left (9 \, x^{4} - 50 \, x^{2} + 419\right )} \sqrt {x^{4} + 5 \, x^{2} + 3}}{60 \, x} \]
1/60*(419*(sqrt(13)*sqrt(2)*x - 5*sqrt(2)*x)*sqrt(sqrt(13) - 5)*elliptic_e (arcsin(1/2*sqrt(2)*sqrt(sqrt(13) - 5)/x), 5/6*sqrt(13) + 19/6) - (369*sqr t(13)*sqrt(2)*x - 2345*sqrt(2)*x)*sqrt(sqrt(13) - 5)*elliptic_f(arcsin(1/2 *sqrt(2)*sqrt(sqrt(13) - 5)/x), 5/6*sqrt(13) + 19/6) + 4*(9*x^4 - 50*x^2 + 419)*sqrt(x^4 + 5*x^2 + 3))/x
\[ \int \frac {x^4 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx=\int \frac {x^{4} \cdot \left (3 x^{2} + 2\right )}{\sqrt {x^{4} + 5 x^{2} + 3}}\, dx \]
\[ \int \frac {x^4 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx=\int { \frac {{\left (3 \, x^{2} + 2\right )} x^{4}}{\sqrt {x^{4} + 5 \, x^{2} + 3}} \,d x } \]
\[ \int \frac {x^4 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx=\int { \frac {{\left (3 \, x^{2} + 2\right )} x^{4}}{\sqrt {x^{4} + 5 \, x^{2} + 3}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx=\int \frac {x^4\,\left (3\,x^2+2\right )}{\sqrt {x^4+5\,x^2+3}} \,d x \]