Integrand size = 20, antiderivative size = 159 \[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\frac {11}{15} x \sqrt {1+x^2+x^4}+\frac {1}{5} x^3 \sqrt {1+x^2+x^4}+\frac {14 x \sqrt {1+x^2+x^4}}{15 \left (1+x^2\right )}-\frac {14 \left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{15 \sqrt {1+x^2+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{5 \sqrt {1+x^2+x^4}} \]
11/15*x*(x^4+x^2+1)^(1/2)+1/5*x^3*(x^4+x^2+1)^(1/2)+14/15*x*(x^4+x^2+1)^(1 /2)/(x^2+1)-14/15*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*Elli pticE(sin(2*arctan(x)),1/2)*((x^4+x^2+1)/(x^2+1)^2)^(1/2)/(x^4+x^2+1)^(1/2 )+3/5*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticF(sin(2* arctan(x)),1/2)*((x^4+x^2+1)/(x^2+1)^2)^(1/2)/(x^4+x^2+1)^(1/2)
Result contains complex when optimal does not.
Time = 10.15 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99 \[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\frac {x \left (11+14 x^2+14 x^4+3 x^6\right )+14 \sqrt [3]{-1} \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} E\left (i \text {arcsinh}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+2 \sqrt [3]{-1} \left (-7+2 \sqrt [3]{-1}\right ) \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )}{15 \sqrt {1+x^2+x^4}} \]
(x*(11 + 14*x^2 + 14*x^4 + 3*x^6) + 14*(-1)^(1/3)*Sqrt[1 + (-1)^(1/3)*x^2] *Sqrt[1 - (-1)^(2/3)*x^2]*EllipticE[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] + 2*(-1)^(1/3)*(-7 + 2*(-1)^(1/3))*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^( 2/3)*x^2]*EllipticF[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)])/(15*Sqrt[1 + x^2 + x^4])
Time = 0.33 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1518, 2207, 27, 1511, 1416, 1509}
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.
| \(\displaystyle \int \frac {\left (x^2+1\right )^3}{\sqrt {x^4+x^2+1}} \, dx\) |
| \(\Big \downarrow \) 1518 |
| \(\displaystyle \frac {1}{5} \int \frac {11 x^4+12 x^2+5}{\sqrt {x^4+x^2+1}}dx+\frac {1}{5} \sqrt {x^4+x^2+1} x^3\) |
| \(\Big \downarrow \) 2207 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {2 \left (7 x^2+2\right )}{\sqrt {x^4+x^2+1}}dx+\frac {11}{3} \sqrt {x^4+x^2+1} x\right )+\frac {1}{5} \sqrt {x^4+x^2+1} x^3\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {7 x^2+2}{\sqrt {x^4+x^2+1}}dx+\frac {11}{3} \sqrt {x^4+x^2+1} x\right )+\frac {1}{5} \sqrt {x^4+x^2+1} x^3\) |
| \(\Big \downarrow \) 1511 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (9 \int \frac {1}{\sqrt {x^4+x^2+1}}dx-7 \int \frac {1-x^2}{\sqrt {x^4+x^2+1}}dx\right )+\frac {11}{3} \sqrt {x^4+x^2+1} x\right )+\frac {1}{5} \sqrt {x^4+x^2+1} x^3\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\frac {9 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{2 \sqrt {x^4+x^2+1}}-7 \int \frac {1-x^2}{\sqrt {x^4+x^2+1}}dx\right )+\frac {11}{3} \sqrt {x^4+x^2+1} x\right )+\frac {1}{5} \sqrt {x^4+x^2+1} x^3\) |
| \(\Big \downarrow \) 1509 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\frac {9 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{2 \sqrt {x^4+x^2+1}}-7 \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {x^4+x^2+1}}-\frac {x \sqrt {x^4+x^2+1}}{x^2+1}\right )\right )+\frac {11}{3} \sqrt {x^4+x^2+1} x\right )+\frac {1}{5} \sqrt {x^4+x^2+1} x^3\) |
(x^3*Sqrt[1 + x^2 + x^4])/5 + ((11*x*Sqrt[1 + x^2 + x^4])/3 + (2*(-7*(-((x *Sqrt[1 + x^2 + x^4])/(1 + x^2)) + ((1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^ 2)^2]*EllipticE[2*ArcTan[x], 1/4])/Sqrt[1 + x^2 + x^4]) + (9*(1 + x^2)*Sqr t[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/4])/(2*Sqrt[1 + x^ 2 + x^4])))/3)/5
3.3.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + b*x^2 + c*x^ 4], x], x] - Simp[e/q Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos Q[c/a]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x _Symbol] :> Simp[e^q*x^(2*q - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Simp[1/(c*(4*p + 2*q + 1)) Int[(a + b*x^2 + c*x^4)^p*Expand ToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2* p + 2*q - 1)*e^q*x^(2*q - 2) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1]
Int[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{n = Expon[Px, x^2], e = Coeff[Px, x^2, Expon[Px, x^2]]}, Simp[e*x^(2*n - 3)*(( a + b*x^2 + c*x^4)^(p + 1)/(c*(2*n + 4*p + 1))), x] + Simp[1/(c*(2*n + 4*p + 1)) Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*n + 4*p + 1)*Px - a*e*(2 *n - 3)*x^(2*n - 4) - b*e*(2*n + 2*p - 1)*x^(2*n - 2) - c*e*(2*n + 4*p + 1) *x^(2*n), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Px, x^2] && Expon[ Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && !LtQ[p, -1]
Result contains complex when optimal does not.
Time = 1.12 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.42
| method | result | size |
| risch | \(\frac {x \left (3 x^{2}+11\right ) \sqrt {x^{4}+x^{2}+1}}{15}+\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}-\frac {56 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\) | \(225\) |
| default | \(\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {x^{3} \sqrt {x^{4}+x^{2}+1}}{5}+\frac {11 x \sqrt {x^{4}+x^{2}+1}}{15}-\frac {56 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\) | \(233\) |
| elliptic | \(\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {x^{3} \sqrt {x^{4}+x^{2}+1}}{5}+\frac {11 x \sqrt {x^{4}+x^{2}+1}}{15}-\frac {56 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\) | \(233\) |
1/15*x*(3*x^2+11)*(x^4+x^2+1)^(1/2)+8/15/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1 /2*I*3^(1/2))*x^2)^(1/2)*(1-(-1/2-1/2*I*3^(1/2))*x^2)^(1/2)/(x^4+x^2+1)^(1 /2)*EllipticF(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2))-56/ 15/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1-(-1/2-1/2* I*3^(1/2))*x^2)^(1/2)/(x^4+x^2+1)^(1/2)/(1+I*3^(1/2))*(EllipticF(1/2*x*(-2 +2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2))-EllipticE(1/2*x*(-2+2*I*3^ (1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2)))
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.76 \[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\frac {7 \, \sqrt {2} {\left (\sqrt {-3} x - x\right )} \sqrt {\sqrt {-3} - 1} E(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) - \sqrt {2} {\left (5 \, \sqrt {-3} x - 9 \, x\right )} \sqrt {\sqrt {-3} - 1} F(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) + 2 \, {\left (3 \, x^{4} + 11 \, x^{2} + 14\right )} \sqrt {x^{4} + x^{2} + 1}}{30 \, x} \]
1/30*(7*sqrt(2)*(sqrt(-3)*x - x)*sqrt(sqrt(-3) - 1)*elliptic_e(arcsin(1/2* sqrt(2)*sqrt(sqrt(-3) - 1)/x), 1/2*sqrt(-3) - 1/2) - sqrt(2)*(5*sqrt(-3)*x - 9*x)*sqrt(sqrt(-3) - 1)*elliptic_f(arcsin(1/2*sqrt(2)*sqrt(sqrt(-3) - 1 )/x), 1/2*sqrt(-3) - 1/2) + 2*(3*x^4 + 11*x^2 + 14)*sqrt(x^4 + x^2 + 1))/x
\[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\int \frac {\left (x^{2} + 1\right )^{3}}{\sqrt {\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )}}\, dx \]
\[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\int { \frac {{\left (x^{2} + 1\right )}^{3}}{\sqrt {x^{4} + x^{2} + 1}} \,d x } \]
\[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\int { \frac {{\left (x^{2} + 1\right )}^{3}}{\sqrt {x^{4} + x^{2} + 1}} \,d x } \]
Timed out. \[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\int \frac {{\left (x^2+1\right )}^3}{\sqrt {x^4+x^2+1}} \,d x \]