\(\int \frac {1}{(3-3 x^2+2 x^4)^{3/2}} \, dx\) [257]

Optimal result

Integrand size = 16, antiderivative size = 257 \[ \int \frac {1}{\left (3-3 x^2+2 x^4\right )^{3/2}} \, dx=\frac {x \left (1+2 x^2\right )}{15 \sqrt {3-3 x^2+2 x^4}}-\frac {2 x \sqrt {3-3 x^2+2 x^4}}{15 \left (\sqrt {6}+2 x^2\right )}+\frac {\sqrt [4]{2} \left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-3 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{5\ 3^{3/4} \sqrt {3-3 x^2+2 x^4}}-\frac {\left (3-2 \sqrt {6}\right ) \left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-3 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{15\ 6^{3/4} \sqrt {3-3 x^2+2 x^4}} \] Output:

1/15*x*(2*x^2+1)/(2*x^4-3*x^2+3)^(1/2)-2*x*(2*x^4-3*x^2+3)^(1/2)/(15*6^(1/ 
2)+30*x^2)+1/15*2^(1/4)*(3+6^(1/2)*x^2)*((2*x^4-3*x^2+3)/(3+6^(1/2)*x^2)^2 
)^(1/2)*EllipticE(sin(2*arctan(1/3*2^(1/4)*3^(3/4)*x)),1/4*(8+2*6^(1/2))^( 
1/2))*3^(1/4)/(2*x^4-3*x^2+3)^(1/2)-1/90*(3-2*6^(1/2))*(3+6^(1/2)*x^2)*((2 
*x^4-3*x^2+3)/(3+6^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(1/3*2^(1/4 
)*3^(3/4)*x),1/4*(8+2*6^(1/2))^(1/2))*6^(1/4)/(2*x^4-3*x^2+3)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.15 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\left (3-3 x^2+2 x^4\right )^{3/2}} \, dx=\frac {4 \sqrt {-\frac {i}{3 i+\sqrt {15}}} x \left (1+2 x^2\right )+\left (-3 i+\sqrt {15}\right ) \sqrt {\frac {3 i+\sqrt {15}-4 i x^2}{3 i+\sqrt {15}}} \sqrt {\frac {-3 i+\sqrt {15}+4 i x^2}{-3 i+\sqrt {15}}} E\left (i \text {arcsinh}\left (2 \sqrt {-\frac {i}{3 i+\sqrt {15}}} x\right )|\frac {3 i+\sqrt {15}}{3 i-\sqrt {15}}\right )-\left (5 i+\sqrt {15}\right ) \sqrt {\frac {3 i+\sqrt {15}-4 i x^2}{3 i+\sqrt {15}}} \sqrt {\frac {-3 i+\sqrt {15}+4 i x^2}{-3 i+\sqrt {15}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (2 \sqrt {-\frac {i}{3 i+\sqrt {15}}} x\right ),\frac {3 i+\sqrt {15}}{3 i-\sqrt {15}}\right )}{60 \sqrt {-\frac {i}{3 i+\sqrt {15}}} \sqrt {3-3 x^2+2 x^4}} \] Input:

Integrate[(3 - 3*x^2 + 2*x^4)^(-3/2),x]
 

Output:

(4*Sqrt[(-I)/(3*I + Sqrt[15])]*x*(1 + 2*x^2) + (-3*I + Sqrt[15])*Sqrt[(3*I 
 + Sqrt[15] - (4*I)*x^2)/(3*I + Sqrt[15])]*Sqrt[(-3*I + Sqrt[15] + (4*I)*x 
^2)/(-3*I + Sqrt[15])]*EllipticE[I*ArcSinh[2*Sqrt[(-I)/(3*I + Sqrt[15])]*x 
], (3*I + Sqrt[15])/(3*I - Sqrt[15])] - (5*I + Sqrt[15])*Sqrt[(3*I + Sqrt[ 
15] - (4*I)*x^2)/(3*I + Sqrt[15])]*Sqrt[(-3*I + Sqrt[15] + (4*I)*x^2)/(-3* 
I + Sqrt[15])]*EllipticF[I*ArcSinh[2*Sqrt[(-I)/(3*I + Sqrt[15])]*x], (3*I 
+ Sqrt[15])/(3*I - Sqrt[15])])/(60*Sqrt[(-I)/(3*I + Sqrt[15])]*Sqrt[3 - 3* 
x^2 + 2*x^4])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 27, 1511, 27, 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.

\(\displaystyle \int \frac {1}{\left (2 x^4-3 x^2+3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{45} \int \frac {6 \left (2-x^2\right )}{\sqrt {2 x^4-3 x^2+3}}dx+\frac {x \left (2 x^2+1\right )}{15 \sqrt {2 x^4-3 x^2+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{15} \int \frac {2-x^2}{\sqrt {2 x^4-3 x^2+3}}dx+\frac {x \left (2 x^2+1\right )}{15 \sqrt {2 x^4-3 x^2+3}}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {2}{15} \left (\frac {1}{2} \left (4-\sqrt {6}\right ) \int \frac {1}{\sqrt {2 x^4-3 x^2+3}}dx+\sqrt {\frac {3}{2}} \int \frac {3-\sqrt {6} x^2}{3 \sqrt {2 x^4-3 x^2+3}}dx\right )+\frac {x \left (2 x^2+1\right )}{15 \sqrt {2 x^4-3 x^2+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{15} \left (\frac {1}{2} \left (4-\sqrt {6}\right ) \int \frac {1}{\sqrt {2 x^4-3 x^2+3}}dx+\frac {\int \frac {3-\sqrt {6} x^2}{\sqrt {2 x^4-3 x^2+3}}dx}{\sqrt {6}}\right )+\frac {x \left (2 x^2+1\right )}{15 \sqrt {2 x^4-3 x^2+3}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {2}{15} \left (\frac {\int \frac {3-\sqrt {6} x^2}{\sqrt {2 x^4-3 x^2+3}}dx}{\sqrt {6}}+\frac {\left (4-\sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {2 x^4-3 x^2+3}}\right )+\frac {x \left (2 x^2+1\right )}{15 \sqrt {2 x^4-3 x^2+3}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {2}{15} \left (\frac {\left (4-\sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {2 x^4-3 x^2+3}}+\frac {\frac {3^{3/4} \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{\sqrt [4]{2} \sqrt {2 x^4-3 x^2+3}}-\frac {3 x \sqrt {2 x^4-3 x^2+3}}{\sqrt {6} x^2+3}}{\sqrt {6}}\right )+\frac {x \left (2 x^2+1\right )}{15 \sqrt {2 x^4-3 x^2+3}}\)

Input:

Int[(3 - 3*x^2 + 2*x^4)^(-3/2),x]
 

Output:

(x*(1 + 2*x^2))/(15*Sqrt[3 - 3*x^2 + 2*x^4]) + (2*(((-3*x*Sqrt[3 - 3*x^2 + 
 2*x^4])/(3 + Sqrt[6]*x^2) + (3^(3/4)*(3 + Sqrt[6]*x^2)*Sqrt[(3 - 3*x^2 + 
2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticE[2*ArcTan[(2/3)^(1/4)*x], (4 + Sqrt[6 
])/8])/(2^(1/4)*Sqrt[3 - 3*x^2 + 2*x^4]))/Sqrt[6] + ((4 - Sqrt[6])*(3 + Sq 
rt[6]*x^2)*Sqrt[(3 - 3*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTa 
n[(2/3)^(1/4)*x], (4 + Sqrt[6])/8])/(4*6^(1/4)*Sqrt[3 - 3*x^2 + 2*x^4])))/ 
15
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1405

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x)*(b^2 
- 2*a*c + b*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c)) 
), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[(b^2 - 2*a*c + 2*(p + 1)*( 
b^2 - 4*a*c) + b*c*(4*p + 7)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; Fr 
eeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]
 

rule 1416

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]
 

rule 1509

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]
 

rule 1511

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.30 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {x \left (2 x^{2}+1\right )}{15 \sqrt {2 x^{4}-3 x^{2}+3}}+\frac {8 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )}{5 \sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}}+\frac {24 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}\, \left (-3+i \sqrt {15}\right )}\)	\(237\)
default	\(-\frac {4 \left (-\frac {1}{60} x -\frac {1}{30} x^{3}\right )}{\sqrt {2 x^{4}-3 x^{2}+3}}+\frac {8 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )}{5 \sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}}+\frac {24 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}\, \left (-3+i \sqrt {15}\right )}\)	\(238\)
elliptic	\(-\frac {4 \left (-\frac {1}{60} x -\frac {1}{30} x^{3}\right )}{\sqrt {2 x^{4}-3 x^{2}+3}}+\frac {8 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )}{5 \sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}}+\frac {24 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}\, \left (-3+i \sqrt {15}\right )}\)	\(238\)

Input:

int(1/(2*x^4-3*x^2+3)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/15*x*(2*x^2+1)/(2*x^4-3*x^2+3)^(1/2)+8/5/(18+6*I*15^(1/2))^(1/2)*(1-(1/2 
+1/6*I*15^(1/2))*x^2)^(1/2)*(1-(1/2-1/6*I*15^(1/2))*x^2)^(1/2)/(2*x^4-3*x^ 
2+3)^(1/2)*EllipticF(1/6*x*(18+6*I*15^(1/2))^(1/2),1/2*(-1-I*15^(1/2))^(1/ 
2))+24/5/(18+6*I*15^(1/2))^(1/2)*(1-(1/2+1/6*I*15^(1/2))*x^2)^(1/2)*(1-(1/ 
2-1/6*I*15^(1/2))*x^2)^(1/2)/(2*x^4-3*x^2+3)^(1/2)/(-3+I*15^(1/2))*(Ellipt 
icF(1/6*x*(18+6*I*15^(1/2))^(1/2),1/2*(-1-I*15^(1/2))^(1/2))-EllipticE(1/6 
*x*(18+6*I*15^(1/2))^(1/2),1/2*(-1-I*15^(1/2))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\left (3-3 x^2+2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {3} {\left (2 \, x^{4} - 3 \, x^{2} + \sqrt {-\frac {5}{3}} {\left (2 \, x^{4} - 3 \, x^{2} + 3\right )} + 3\right )} \sqrt {\frac {1}{2} \, \sqrt {-\frac {5}{3}} + \frac {1}{2}} E(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {-\frac {5}{3}} + \frac {1}{2}}\right )\,|\,-\frac {3}{4} \, \sqrt {-\frac {5}{3}} - \frac {1}{4}) + \sqrt {3} {\left (2 \, x^{4} - 3 \, x^{2} - 3 \, \sqrt {-\frac {5}{3}} {\left (2 \, x^{4} - 3 \, x^{2} + 3\right )} + 3\right )} \sqrt {\frac {1}{2} \, \sqrt {-\frac {5}{3}} + \frac {1}{2}} F(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {-\frac {5}{3}} + \frac {1}{2}}\right )\,|\,-\frac {3}{4} \, \sqrt {-\frac {5}{3}} - \frac {1}{4}) + 2 \, \sqrt {2 \, x^{4} - 3 \, x^{2} + 3} {\left (2 \, x^{3} + x\right )}}{30 \, {\left (2 \, x^{4} - 3 \, x^{2} + 3\right )}} \] Input:

integrate(1/(2*x^4-3*x^2+3)^(3/2),x, algorithm="fricas")
 

Output:

1/30*(sqrt(3)*(2*x^4 - 3*x^2 + sqrt(-5/3)*(2*x^4 - 3*x^2 + 3) + 3)*sqrt(1/ 
2*sqrt(-5/3) + 1/2)*elliptic_e(arcsin(x*sqrt(1/2*sqrt(-5/3) + 1/2)), -3/4* 
sqrt(-5/3) - 1/4) + sqrt(3)*(2*x^4 - 3*x^2 - 3*sqrt(-5/3)*(2*x^4 - 3*x^2 + 
 3) + 3)*sqrt(1/2*sqrt(-5/3) + 1/2)*elliptic_f(arcsin(x*sqrt(1/2*sqrt(-5/3 
) + 1/2)), -3/4*sqrt(-5/3) - 1/4) + 2*sqrt(2*x^4 - 3*x^2 + 3)*(2*x^3 + x)) 
/(2*x^4 - 3*x^2 + 3)
 

Sympy [F]

\[ \int \frac {1}{\left (3-3 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (2 x^{4} - 3 x^{2} + 3\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(2*x**4-3*x**2+3)**(3/2),x)
 

Output:

Integral((2*x**4 - 3*x**2 + 3)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (3-3 x^2+2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, x^{4} - 3 \, x^{2} + 3\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(2*x^4-3*x^2+3)^(3/2),x, algorithm="maxima")
 

Output:

integrate((2*x^4 - 3*x^2 + 3)^(-3/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (3-3 x^2+2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, x^{4} - 3 \, x^{2} + 3\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(2*x^4-3*x^2+3)^(3/2),x, algorithm="giac")
 

Output:

integrate((2*x^4 - 3*x^2 + 3)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (3-3 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (2\,x^4-3\,x^2+3\right )}^{3/2}} \,d x \] Input:

int(1/(2*x^4 - 3*x^2 + 3)^(3/2),x)
 

Output:

int(1/(2*x^4 - 3*x^2 + 3)^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (3-3 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {2 x^{4}-3 x^{2}+3}}{4 x^{8}-12 x^{6}+21 x^{4}-18 x^{2}+9}d x \] Input:

int(1/(2*x^4-3*x^2+3)^(3/2),x)
 

Output:

int(sqrt(2*x**4 - 3*x**2 + 3)/(4*x**8 - 12*x**6 + 21*x**4 - 18*x**2 + 9),x 
)