Integrand size = 29, antiderivative size = 247 \[ \int \frac {x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=-\frac {1}{4} \sqrt {\frac {3}{5}} \arctan \left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )-\frac {\left (3+\sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{14\ 2^{3/4} \sqrt {1+2 x^2+2 x^4}}+\frac {\left (3+\sqrt {2}\right )^2 \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{24} \left (12-11 \sqrt {2}\right ),2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{56 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}} \]
-1/20*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)-1/28*(cos(2*ar ctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticF(sin(2*arctan( 2^(1/4)*x)),1/2*(2-2^(1/2))^(1/2))*(3+2^(1/2))*(1+x^2*2^(1/2))*((2*x^4+2*x ^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(1/4)/(2*x^4+2*x^2+1)^(1/2)+1/112*(cos(2* arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticPi(sin(2*arct an(2^(1/4)*x)),1/2-11/24*2^(1/2),1/2*(2-2^(1/2))^(1/2))*(3+2^(1/2))^2*(1+x ^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(3/4)/(2*x^4+2*x^2 +1)^(1/2)
Result contains complex when optimal does not.
Time = 10.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.40 \[ \int \frac {x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\frac {(1-i)^{3/2} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-i} x\right ),i\right )-\operatorname {EllipticPi}\left (\frac {1}{3}+\frac {i}{3},i \text {arcsinh}\left (\sqrt {1-i} x\right ),i\right )\right )}{4 \sqrt {1+2 x^2+2 x^4}} \]
((1 - I)^(3/2)*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*(EllipticF[I*Ar cSinh[Sqrt[1 - I]*x], I] - EllipticPi[1/3 + I/3, I*ArcSinh[Sqrt[1 - I]*x], I]))/(4*Sqrt[1 + 2*x^2 + 2*x^4])
Time = 0.40 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1656, 1416, 2220}
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.
| \(\displaystyle \int \frac {x^2}{\left (2 x^2+3\right ) \sqrt {2 x^4+2 x^2+1}} \, dx\) |
| \(\Big \downarrow \) 1656 |
| \(\displaystyle \frac {3}{14} \left (2+3 \sqrt {2}\right ) \int \frac {\sqrt {2} x^2+1}{\left (2 x^2+3\right ) \sqrt {2 x^4+2 x^2+1}}dx-\frac {1}{14} \left (2+3 \sqrt {2}\right ) \int \frac {1}{\sqrt {2 x^4+2 x^2+1}}dx\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle \frac {3}{14} \left (2+3 \sqrt {2}\right ) \int \frac {\sqrt {2} x^2+1}{\left (2 x^2+3\right ) \sqrt {2 x^4+2 x^2+1}}dx-\frac {\left (2+3 \sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{28 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}\) |
| \(\Big \downarrow \) 2220 |
| \(\displaystyle \frac {3}{14} \left (2+3 \sqrt {2}\right ) \left (\frac {\left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{24} \left (12-11 \sqrt {2}\right ),2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{12\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}-\frac {\left (3-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )}{2 \sqrt {30}}\right )-\frac {\left (2+3 \sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{28 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}\) |
-1/28*((2 + 3*Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqr t[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(2^(1/4)*Sqr t[1 + 2*x^2 + 2*x^4]) + (3*(2 + 3*Sqrt[2])*(-1/2*((3 - Sqrt[2])*ArcTan[(Sq rt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/Sqrt[30] + ((3 + Sqrt[2])*(1 + Sqrt[2 ]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticPi[(12 - 11*S qrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(12*2^(3/4)*Sqrt[1 + 2* x^2 + 2*x^4])))/14
3.4.38.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c /a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ (2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) ], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[(x_)^2/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]) , x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(-a)*((e + d*q)/(c*d^2 - a*e^2)) Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Simp[a*d*((e + d*q)/(c*d^2 - a*e ^2)) Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; Fr eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a] && NeQ[c*d^2 - a*e^2, 0]
Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(A rcTan[Rt[-b + c*(d/e) + a*(e/d), 2]*(x/Sqrt[a + b*x^2 + c*x^4])]/(2*d*e*Rt[ -b + c*(d/e) + a*(e/d), 2])), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4*d*e*q*Sqrt[a + b*x^2 + c*x^4]))*El lipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x], 1/2 - b/(4*a*q^2)], x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] & & EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && PosQ[-b + c*(d/e) + a*(e/d)]
Result contains complex when optimal does not.
Time = 0.75 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.54
| method | result | size |
| default | \(\frac {\sqrt {1+\left (1-i\right ) x^{2}}\, \sqrt {1+\left (1+i\right ) x^{2}}\, F\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \Pi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\) | \(134\) |
| elliptic | \(\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, F\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \Pi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\) | \(138\) |
1/2/(-1+I)^(1/2)*(1+(1-I)*x^2)^(1/2)*(1+(1+I)*x^2)^(1/2)/(2*x^4+2*x^2+1)^( 1/2)*EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-1/2/(-1+I)^(1/2)* (1-I*x^2+x^2)^(1/2)*(1+I*x^2+x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticPi(x *(-1+I)^(1/2),1/3+1/3*I,(-1-I)^(1/2)/(-1+I)^(1/2))
\[ \int \frac {x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} {\left (2 \, x^{2} + 3\right )}} \,d x } \]
\[ \int \frac {x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int \frac {x^{2}}{\left (2 x^{2} + 3\right ) \sqrt {2 x^{4} + 2 x^{2} + 1}}\, dx \]
\[ \int \frac {x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} {\left (2 \, x^{2} + 3\right )}} \,d x } \]
\[ \int \frac {x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} {\left (2 \, x^{2} + 3\right )}} \,d x } \]
Timed out. \[ \int \frac {x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int \frac {x^2}{\left (2\,x^2+3\right )\,\sqrt {2\,x^4+2\,x^2+1}} \,d x \]