Integrand size = 36, antiderivative size = 188 \[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=-\frac {1411 x \left (2+3 x^2\right )}{81 \sqrt {2+5 x^2+3 x^4}}+\frac {x \left (257+365 x^2\right )}{9 \sqrt {2+5 x^2+3 x^4}}+\frac {4}{27} x \sqrt {2+5 x^2+3 x^4}+\frac {1411 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{81 \sqrt {2+5 x^2+3 x^4}}-\frac {725 \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{27 \sqrt {2} \sqrt {2+5 x^2+3 x^4}} \] Output:
-1411/81*x*(3*x^2+2)/(3*x^4+5*x^2+2)^(1/2)+1/9*x*(365*x^2+257)/(3*x^4+5*x^ 2+2)^(1/2)+4/27*x*(3*x^4+5*x^2+2)^(1/2)+1411/81*2^(1/2)*(x^2+1)*((3*x^2+2) /(x^2+1))^(1/2)*EllipticE(x/(x^2+1)^(1/2),1/2*I*2^(1/2))/(3*x^4+5*x^2+2)^( 1/2)-725/54*2^(1/2)*(x^2+1)*((3*x^2+2)/(x^2+1))^(1/2)*InverseJacobiAM(arct an(x),1/2*I*2^(1/2))/(3*x^4+5*x^2+2)^(1/2)
Result contains complex when optimal does not.
Time = 10.15 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.68 \[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\frac {2337 x+3345 x^3+36 x^5+1411 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )-686 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{81 \sqrt {2+5 x^2+3 x^4}} \] Input:
Integrate[((1 + 2*x^2)^2*(4 - 7*x^2 + x^4))/(2 + 5*x^2 + 3*x^4)^(3/2),x]
Output:
(2337*x + 3345*x^3 + 36*x^5 + (1411*I)*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x^ 2]*EllipticE[I*ArcSinh[Sqrt[3/2]*x], 2/3] - (686*I)*Sqrt[3]*Sqrt[1 + x^2]* Sqrt[2 + 3*x^2]*EllipticF[I*ArcSinh[Sqrt[3/2]*x], 2/3])/(81*Sqrt[2 + 5*x^2 + 3*x^4])
Time = 0.44 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2206, 27, 2207, 27, 1503, 1413, 1456}
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.
| \(\displaystyle \int \frac {\left (2 x^2+1\right )^2 \left (x^4-7 x^2+4\right )}{\left (3 x^4+5 x^2+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2206 |
| \(\displaystyle \frac {x \left (365 x^2+257\right )}{9 \sqrt {3 x^4+5 x^2+2}}-\frac {1}{2} \int \frac {2 \left (-12 x^4+457 x^2+239\right )}{9 \sqrt {3 x^4+5 x^2+2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x \left (365 x^2+257\right )}{9 \sqrt {3 x^4+5 x^2+2}}-\frac {1}{9} \int \frac {-12 x^4+457 x^2+239}{\sqrt {3 x^4+5 x^2+2}}dx\) |
| \(\Big \downarrow \) 2207 |
| \(\displaystyle \frac {1}{9} \left (\frac {4}{3} x \sqrt {3 x^4+5 x^2+2}-\frac {1}{9} \int \frac {3 \left (1411 x^2+725\right )}{\sqrt {3 x^4+5 x^2+2}}dx\right )+\frac {x \left (365 x^2+257\right )}{9 \sqrt {3 x^4+5 x^2+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \left (\frac {4}{3} x \sqrt {3 x^4+5 x^2+2}-\frac {1}{3} \int \frac {1411 x^2+725}{\sqrt {3 x^4+5 x^2+2}}dx\right )+\frac {x \left (365 x^2+257\right )}{9 \sqrt {3 x^4+5 x^2+2}}\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{3} \left (-725 \int \frac {1}{\sqrt {3 x^4+5 x^2+2}}dx-1411 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx\right )+\frac {4}{3} \sqrt {3 x^4+5 x^2+2} x\right )+\frac {x \left (365 x^2+257\right )}{9 \sqrt {3 x^4+5 x^2+2}}\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{3} \left (-1411 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx-\frac {725 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )+\frac {4}{3} \sqrt {3 x^4+5 x^2+2} x\right )+\frac {x \left (365 x^2+257\right )}{9 \sqrt {3 x^4+5 x^2+2}}\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{3} \left (-\frac {725 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}-1411 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )\right )+\frac {4}{3} \sqrt {3 x^4+5 x^2+2} x\right )+\frac {x \left (365 x^2+257\right )}{9 \sqrt {3 x^4+5 x^2+2}}\) |
Input:
Int[((1 + 2*x^2)^2*(4 - 7*x^2 + x^4))/(2 + 5*x^2 + 3*x^4)^(3/2),x]
Output:
(x*(257 + 365*x^2))/(9*Sqrt[2 + 5*x^2 + 3*x^4]) + ((4*x*Sqrt[2 + 5*x^2 + 3 *x^4])/3 + (-1411*((x*(2 + 3*x^2))/(3*Sqrt[2 + 5*x^2 + 3*x^4]) - (Sqrt[2]* (1 + x^2)*Sqrt[(2 + 3*x^2)/(1 + x^2)]*EllipticE[ArcTan[x], -1/2])/(3*Sqrt[ 2 + 5*x^2 + 3*x^4])) - (725*(1 + x^2)*Sqrt[(2 + 3*x^2)/(1 + x^2)]*Elliptic F[ArcTan[x], -1/2])/(Sqrt[2]*Sqrt[2 + 5*x^2 + 3*x^4]))/3)/9
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[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]] /; 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]] /; 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[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[Poly nomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^ 4)^(p + 1)*((a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)/(2*a*(p + 1)*(b ^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c *x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[Px, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4* p + 7)*(b*d - 2*a*e)*x^2, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Px, x ^2] && Expon[Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -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]
Time = 12.48 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.69
| method | result | size |
| risch | \(\frac {x \left (12 x^{4}+1115 x^{2}+779\right )}{27 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {725 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{54 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {1411 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{81 \sqrt {3 x^{4}+5 x^{2}+2}}\) | \(130\) |
| elliptic | \(-\frac {6 \left (-\frac {365}{54} x^{3}-\frac {257}{54} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}+\frac {4 x \sqrt {3 x^{4}+5 x^{2}+2}}{27}+\frac {725 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{54 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {1411 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{81 \sqrt {3 x^{4}+5 x^{2}+2}}\) | \(143\) |
| default | \(-\frac {24 \left (-\frac {5}{4} x^{3}-\frac {13}{12} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}+\frac {725 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{54 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {1411 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{81 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {54 \left (x^{3}+\frac {5}{6} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}+\frac {-55 x^{3}-44 x}{\sqrt {3 x^{4}+5 x^{2}+2}}+\frac {104 x^{3}+80 x}{\sqrt {3 x^{4}+5 x^{2}+2}}-\frac {24 \left (-\frac {35}{54} x^{3}-\frac {13}{27} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}+\frac {4 x \sqrt {3 x^{4}+5 x^{2}+2}}{27}\) | \(241\) |
Input:
int((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x,method=_RETURNVERBOS E)
Output:
1/27*x*(12*x^4+1115*x^2+779)/(3*x^4+5*x^2+2)^(1/2)+725/54*I*(x^2+1)^(1/2)* (6*x^2+4)^(1/2)/(3*x^4+5*x^2+2)^(1/2)*EllipticF(I*x,1/2*6^(1/2))-1411/81*I *(x^2+1)^(1/2)*(6*x^2+4)^(1/2)/(3*x^4+5*x^2+2)^(1/2)*(EllipticF(I*x,1/2*6^ (1/2))-EllipticE(I*x,1/2*6^(1/2)))
Time = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.62 \[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\frac {5644 \, \sqrt {3} \sqrt {-\frac {2}{3}} {\left (3 \, x^{5} + 5 \, x^{3} + 2 \, x\right )} E(\arcsin \left (\frac {\sqrt {-\frac {2}{3}}}{x}\right )\,|\,\frac {3}{2}) - 12169 \, \sqrt {3} \sqrt {-\frac {2}{3}} {\left (3 \, x^{5} + 5 \, x^{3} + 2 \, x\right )} F(\arcsin \left (\frac {\sqrt {-\frac {2}{3}}}{x}\right )\,|\,\frac {3}{2}) + 12 \, {\left (18 \, x^{6} - 444 \, x^{4} - 2359 \, x^{2} - 1411\right )} \sqrt {3 \, x^{4} + 5 \, x^{2} + 2}}{486 \, {\left (3 \, x^{5} + 5 \, x^{3} + 2 \, x\right )}} \] Input:
integrate((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x, algorithm="fr icas")
Output:
1/486*(5644*sqrt(3)*sqrt(-2/3)*(3*x^5 + 5*x^3 + 2*x)*elliptic_e(arcsin(sqr t(-2/3)/x), 3/2) - 12169*sqrt(3)*sqrt(-2/3)*(3*x^5 + 5*x^3 + 2*x)*elliptic _f(arcsin(sqrt(-2/3)/x), 3/2) + 12*(18*x^6 - 444*x^4 - 2359*x^2 - 1411)*sq rt(3*x^4 + 5*x^2 + 2))/(3*x^5 + 5*x^3 + 2*x)
\[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {\left (2 x^{2} + 1\right )^{2} \left (x^{4} - 7 x^{2} + 4\right )}{\left (\left (x^{2} + 1\right ) \left (3 x^{2} + 2\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((2*x**2+1)**2*(x**4-7*x**2+4)/(3*x**4+5*x**2+2)**(3/2),x)
Output:
Integral((2*x**2 + 1)**2*(x**4 - 7*x**2 + 4)/((x**2 + 1)*(3*x**2 + 2))**(3 /2), x)
\[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int { \frac {{\left (x^{4} - 7 \, x^{2} + 4\right )} {\left (2 \, x^{2} + 1\right )}^{2}}{{\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x, algorithm="ma xima")
Output:
integrate((x^4 - 7*x^2 + 4)*(2*x^2 + 1)^2/(3*x^4 + 5*x^2 + 2)^(3/2), x)
\[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int { \frac {{\left (x^{4} - 7 \, x^{2} + 4\right )} {\left (2 \, x^{2} + 1\right )}^{2}}{{\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x, algorithm="gi ac")
Output:
integrate((x^4 - 7*x^2 + 4)*(2*x^2 + 1)^2/(3*x^4 + 5*x^2 + 2)^(3/2), x)
Timed out. \[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {{\left (2\,x^2+1\right )}^2\,\left (x^4-7\,x^2+4\right )}{{\left (3\,x^4+5\,x^2+2\right )}^{3/2}} \,d x \] Input:
int(((2*x^2 + 1)^2*(x^4 - 7*x^2 + 4))/(5*x^2 + 3*x^4 + 2)^(3/2),x)
Output:
int(((2*x^2 + 1)^2*(x^4 - 7*x^2 + 4))/(5*x^2 + 3*x^4 + 2)^(3/2), x)
\[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\frac {36 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{5}-888 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{3}-2543 \sqrt {3 x^{4}+5 x^{2}+2}\, x +16230 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right ) x^{4}+27050 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right ) x^{2}+10820 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right )+18171 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right ) x^{4}+30285 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right ) x^{2}+12114 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right )}{243 x^{4}+405 x^{2}+162} \] Input:
int((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x)
Output:
(36*sqrt(3*x**4 + 5*x**2 + 2)*x**5 - 888*sqrt(3*x**4 + 5*x**2 + 2)*x**3 - 2543*sqrt(3*x**4 + 5*x**2 + 2)*x + 16230*int(sqrt(3*x**4 + 5*x**2 + 2)/(9* x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x)*x**4 + 27050*int(sqrt(3*x**4 + 5*x**2 + 2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x)*x**2 + 10820*int (sqrt(3*x**4 + 5*x**2 + 2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x) + 18171*int((sqrt(3*x**4 + 5*x**2 + 2)*x**2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x)*x**4 + 30285*int((sqrt(3*x**4 + 5*x**2 + 2)*x**2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x)*x**2 + 12114*int((sqrt(3*x**4 + 5*x* *2 + 2)*x**2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x))/(81*(3*x**4 + 5*x**2 + 2))