Integrand size = 29, antiderivative size = 418 \[ \int \frac {x^4}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\frac {x \sqrt {1+2 x^2+2 x^4}}{2 \sqrt {2} \left (1+\sqrt {2} x^2\right )}-\frac {3 \sqrt {\frac {3}{10}} \left (3-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )}{4 \left (2-3 \sqrt {2}\right )}-\frac {\left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{2\ 2^{3/4} \sqrt {1+2 x^2+2 x^4}}+\frac {\left (1-3 \sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{2\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {1+2 x^2+2 x^4}}+\frac {3 \left (3+\sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{24} \left (12-11 \sqrt {2}\right ),2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{8\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {1+2 x^2+2 x^4}} \]
-3/40*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*30^(1/2)*(3-2^(1/2))/(2 -3*2^(1/2))+1/4*x*(2*x^4+2*x^2+1)^(1/2)*2^(1/2)/(1+x^2*2^(1/2))-1/4*(cos(2 *arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticE(sin(2*arct an(2^(1/4)*x)),1/2*(2-2^(1/2))^(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)+1/4*(cos(2*arctan(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))*(1-3*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)/(2-3*2^(1/2))/(2*x^4+2*x^2+1)^(1/2)+3/16*( cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticPi(sin( 2*arctan(2^(1/4)*x)),1/2-11/24*2^(1/2),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)/(2-3*2^( 1/2))/(2*x^4+2*x^2+1)^(1/2)
Result contains complex when optimal does not.
Time = 10.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.30 \[ \int \frac {x^4}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=-\frac {\sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} \left ((1+i) E\left (\left .i \text {arcsinh}\left (\sqrt {1-i} x\right )\right |i\right )-(1+4 i) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-i} x\right ),i\right )+3 i \operatorname {EllipticPi}\left (\frac {1}{3}+\frac {i}{3},i \text {arcsinh}\left (\sqrt {1-i} x\right ),i\right )\right )}{4 \sqrt {1-i} \sqrt {1+2 x^2+2 x^4}} \]
-1/4*(Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*((1 + I)*EllipticE[I*Arc Sinh[Sqrt[1 - I]*x], I] - (1 + 4*I)*EllipticF[I*ArcSinh[Sqrt[1 - I]*x], I] + (3*I)*EllipticPi[1/3 + I/3, I*ArcSinh[Sqrt[1 - I]*x], I]))/(Sqrt[1 - I] *Sqrt[1 + 2*x^2 + 2*x^4])
Time = 0.60 (sec) , antiderivative size = 410, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1662, 1416, 1509, 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.
| \(\displaystyle \int \frac {x^4}{\left (2 x^2+3\right ) \sqrt {2 x^4+2 x^2+1}} \, dx\) |
| \(\Big \downarrow \) 1662 |
| \(\displaystyle -\frac {\left (6-\sqrt {2}\right ) \int \frac {1}{\sqrt {2 x^4+2 x^2+1}}dx}{2 \left (2-3 \sqrt {2}\right )}-\frac {\int \frac {1-\sqrt {2} x^2}{\sqrt {2 x^4+2 x^2+1}}dx}{2 \sqrt {2}}+\frac {9 \int \frac {\sqrt {2} x^2+1}{\left (2 x^2+3\right ) \sqrt {2 x^4+2 x^2+1}}dx}{2 \left (2-3 \sqrt {2}\right )}\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle -\frac {\int \frac {1-\sqrt {2} x^2}{\sqrt {2 x^4+2 x^2+1}}dx}{2 \sqrt {2}}+\frac {9 \int \frac {\sqrt {2} x^2+1}{\left (2 x^2+3\right ) \sqrt {2 x^4+2 x^2+1}}dx}{2 \left (2-3 \sqrt {2}\right )}-\frac {\left (6-\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{4 \sqrt [4]{2} \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}}\) |
| \(\Big \downarrow \) 1509 |
| \(\displaystyle \frac {9 \int \frac {\sqrt {2} x^2+1}{\left (2 x^2+3\right ) \sqrt {2 x^4+2 x^2+1}}dx}{2 \left (2-3 \sqrt {2}\right )}-\frac {\left (6-\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{4 \sqrt [4]{2} \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}}-\frac {\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{\sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {x \sqrt {2 x^4+2 x^2+1}}{\sqrt {2} x^2+1}}{2 \sqrt {2}}\) |
| \(\Big \downarrow \) 2220 |
| \(\displaystyle -\frac {\left (6-\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{4 \sqrt [4]{2} \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}}-\frac {\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{\sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {x \sqrt {2 x^4+2 x^2+1}}{\sqrt {2} x^2+1}}{2 \sqrt {2}}+\frac {9 \left (\frac {\left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{24} \left (12-11 \sqrt {2}\right ),2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{12\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}-\frac {\left (3-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )}{2 \sqrt {30}}\right )}{2 \left (2-3 \sqrt {2}\right )}\) |
-1/2*(-((x*Sqrt[1 + 2*x^2 + 2*x^4])/(1 + Sqrt[2]*x^2)) + ((1 + Sqrt[2]*x^2 )*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticE[2*ArcTan[2^(1/4) *x], (2 - Sqrt[2])/4])/(2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4]))/Sqrt[2] - ((6 - Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*E llipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(4*2^(1/4)*(2 - 3*Sqrt[2]) *Sqrt[1 + 2*x^2 + 2*x^4]) + (9*(-1/2*((3 - Sqrt[2])*ArcTan[(Sqrt[5/3]*x)/S qrt[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*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(12*2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4] )))/(2*(2 - 3*Sqrt[2]))
3.4.37.3.1 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[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q ^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 /(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[(x_)^4/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]) , x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[-(2*c*d - a*e*q)/(c*e*(e - d*q)) Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + (-Simp[1/(e*q) Int[(1 - q*x^2) /Sqrt[a + b*x^2 + c*x^4], x], x] + Simp[d^2/(e*(e - d*q)) Int[(1 + q*x^2) /((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x])] /; FreeQ[{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 = 1.89 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.53
| method | result | size |
| default | \(-\frac {3 \sqrt {1+\left (1-i\right ) x^{2}}\, \sqrt {1+\left (1+i\right ) x^{2}}\, F\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{4 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {\left (-\frac {1}{4}+\frac {i}{4}\right ) \sqrt {1+\left (1-i\right ) x^{2}}\, \sqrt {1+\left (1+i\right ) x^{2}}\, \left (F\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )-E\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {3 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \Pi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{4 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\) | \(222\) |
| elliptic | \(-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, F\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, F\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{4 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, E\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{4 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, E\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{4 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {3 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \Pi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{4 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\) | \(341\) |
-3/4/(-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))+(-1/4+1/4*I)/(-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)*(E llipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-EllipticE(x*(-1+I)^(1/2 ),1/2*2^(1/2)+1/2*I*2^(1/2)))+3/4/(-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))
\[ \int \frac {x^4}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int { \frac {x^{4}}{\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} {\left (2 \, x^{2} + 3\right )}} \,d x } \]
\[ \int \frac {x^4}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int \frac {x^{4}}{\left (2 x^{2} + 3\right ) \sqrt {2 x^{4} + 2 x^{2} + 1}}\, dx \]
\[ \int \frac {x^4}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int { \frac {x^{4}}{\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} {\left (2 \, x^{2} + 3\right )}} \,d x } \]
\[ \int \frac {x^4}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int { \frac {x^{4}}{\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} {\left (2 \, x^{2} + 3\right )}} \,d x } \]
Timed out. \[ \int \frac {x^4}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int \frac {x^4}{\left (2\,x^2+3\right )\,\sqrt {2\,x^4+2\,x^2+1}} \,d x \]