\(\int \frac {1}{(3-5 x^2+2 x^4)^{3/2}} \, dx\) [259]

Optimal result

Integrand size = 16, antiderivative size = 130 \[ \int \frac {1}{\left (3-5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {x \left (13-10 x^2\right )}{3 \sqrt {3-5 x^2+2 x^4}}-\frac {5 \sqrt {3-2 x^2} \sqrt {1-x^2} E\left (\arcsin (x)\left |\frac {2}{3}\right .\right )}{\sqrt {3} \sqrt {3-5 x^2+2 x^4}}+\frac {\sqrt {3-2 x^2} \sqrt {1-x^2} \operatorname {EllipticF}\left (\arcsin (x),\frac {2}{3}\right )}{\sqrt {3} \sqrt {3-5 x^2+2 x^4}} \] Output:

1/3*x*(-10*x^2+13)/(2*x^4-5*x^2+3)^(1/2)-5/3*(-2*x^2+3)^(1/2)*(-x^2+1)^(1/ 
2)*EllipticE(x,1/3*6^(1/2))*3^(1/2)/(2*x^4-5*x^2+3)^(1/2)+1/3*(-2*x^2+3)^( 
1/2)*(-x^2+1)^(1/2)*EllipticF(x,1/3*6^(1/2))*3^(1/2)/(2*x^4-5*x^2+3)^(1/2)
 

Mathematica [A] (verified)

Time = 5.82 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (3-5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {13 x-10 x^3-5 \sqrt {6-4 x^2} \sqrt {1-x^2} E\left (\arcsin \left (\sqrt {\frac {2}{3}} x\right )|\frac {3}{2}\right )-\sqrt {6-4 x^2} \sqrt {1-x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{3}} x\right ),\frac {3}{2}\right )}{3 \sqrt {3-5 x^2+2 x^4}} \] Input:

Integrate[(3 - 5*x^2 + 2*x^4)^(-3/2),x]
 

Output:

(13*x - 10*x^3 - 5*Sqrt[6 - 4*x^2]*Sqrt[1 - x^2]*EllipticE[ArcSin[Sqrt[2/3 
]*x], 3/2] - Sqrt[6 - 4*x^2]*Sqrt[1 - x^2]*EllipticF[ArcSin[Sqrt[2/3]*x], 
3/2])/(3*Sqrt[3 - 5*x^2 + 2*x^4])
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(269\) vs. \(2(130)=260\).

Time = 0.63 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 27, 1497, 27, 1409, 1496}

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.

\(\displaystyle \int \frac {1}{\left (2 x^4-5 x^2+3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (13-10 x^2\right )}{3 \sqrt {2 x^4-5 x^2+3}}-\frac {1}{3} \int \frac {2 \left (6-5 x^2\right )}{\sqrt {2 x^4-5 x^2+3}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (13-10 x^2\right )}{3 \sqrt {2 x^4-5 x^2+3}}-\frac {2}{3} \int \frac {6-5 x^2}{\sqrt {2 x^4-5 x^2+3}}dx\)

\(\Big \downarrow \) 1497

\(\displaystyle \frac {x \left (13-10 x^2\right )}{3 \sqrt {2 x^4-5 x^2+3}}-\frac {2}{3} \left (\frac {1}{2} \left (12-5 \sqrt {6}\right ) \int \frac {1}{\sqrt {2 x^4-5 x^2+3}}dx+5 \sqrt {\frac {3}{2}} \int \frac {3-\sqrt {6} x^2}{3 \sqrt {2 x^4-5 x^2+3}}dx\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (13-10 x^2\right )}{3 \sqrt {2 x^4-5 x^2+3}}-\frac {2}{3} \left (\frac {1}{2} \left (12-5 \sqrt {6}\right ) \int \frac {1}{\sqrt {2 x^4-5 x^2+3}}dx+\frac {5 \int \frac {3-\sqrt {6} x^2}{\sqrt {2 x^4-5 x^2+3}}dx}{\sqrt {6}}\right )\)

\(\Big \downarrow \) 1409

\(\displaystyle \frac {x \left (13-10 x^2\right )}{3 \sqrt {2 x^4-5 x^2+3}}-\frac {2}{3} \left (\frac {5 \int \frac {3-\sqrt {6} x^2}{\sqrt {2 x^4-5 x^2+3}}dx}{\sqrt {6}}+\frac {\left (12-5 \sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-5 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{24} \left (12+5 \sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {2 x^4-5 x^2+3}}\right )\)

\(\Big \downarrow \) 1496

\(\displaystyle \frac {x \left (13-10 x^2\right )}{3 \sqrt {2 x^4-5 x^2+3}}-\frac {2}{3} \left (\frac {\left (12-5 \sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-5 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{24} \left (12+5 \sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {2 x^4-5 x^2+3}}+\frac {5 \left (\frac {3^{3/4} \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-5 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{24} \left (12+5 \sqrt {6}\right )\right )}{\sqrt [4]{2} \sqrt {2 x^4-5 x^2+3}}-\frac {3 x \sqrt {2 x^4-5 x^2+3}}{\sqrt {6} x^2+3}\right )}{\sqrt {6}}\right )\)

Input:

Int[(3 - 5*x^2 + 2*x^4)^(-3/2),x]
 

Output:

(x*(13 - 10*x^2))/(3*Sqrt[3 - 5*x^2 + 2*x^4]) - (2*((5*((-3*x*Sqrt[3 - 5*x 
^2 + 2*x^4])/(3 + Sqrt[6]*x^2) + (3^(3/4)*(3 + Sqrt[6]*x^2)*Sqrt[(3 - 5*x^ 
2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticE[2*ArcTan[(2/3)^(1/4)*x], (12 + 5 
*Sqrt[6])/24])/(2^(1/4)*Sqrt[3 - 5*x^2 + 2*x^4])))/Sqrt[6] + ((12 - 5*Sqrt 
[6])*(3 + Sqrt[6]*x^2)*Sqrt[(3 - 5*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*Ellip 
ticF[2*ArcTan[(2/3)^(1/4)*x], (12 + 5*Sqrt[6])/24])/(4*6^(1/4)*Sqrt[3 - 5* 
x^2 + 2*x^4])))/3
 

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 1409

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] && GtQ[b^2 - 4*a*c, 0] && GtQ[c/a, 0] && LtQ[ 
b/a, 0]
 

rule 1496

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] && GtQ[b^2 
- 4*a*c, 0] && GtQ[c/a, 0] && LtQ[b/a, 0]
 

rule 1497

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] && GtQ[b^2 - 4*a*c, 0] && GtQ 
[c/a, 0] && LtQ[b/a, 0]
 

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {x \left (10 x^{2}-13\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}-\frac {4 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+9}\, \operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{3}\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+9}\, \left (\operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (x , \frac {\sqrt {6}}{3}\right )\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}\)	\(118\)
default	\(-\frac {4 \left (\frac {5}{6} x^{3}-\frac {13}{12} x \right )}{\sqrt {2 x^{4}-5 x^{2}+3}}-\frac {4 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+9}\, \operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{3}\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+9}\, \left (\operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (x , \frac {\sqrt {6}}{3}\right )\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}\)	\(119\)
elliptic	\(-\frac {4 \left (\frac {5}{6} x^{3}-\frac {13}{12} x \right )}{\sqrt {2 x^{4}-5 x^{2}+3}}-\frac {4 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+9}\, \operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{3}\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+9}\, \left (\operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (x , \frac {\sqrt {6}}{3}\right )\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}\)	\(119\)

Input:

int(1/(2*x^4-5*x^2+3)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3*x*(10*x^2-13)/(2*x^4-5*x^2+3)^(1/2)-4/3*(-x^2+1)^(1/2)*(-6*x^2+9)^(1/ 
2)/(2*x^4-5*x^2+3)^(1/2)*EllipticF(x,1/3*6^(1/2))+5/3*(-x^2+1)^(1/2)*(-6*x 
^2+9)^(1/2)/(2*x^4-5*x^2+3)^(1/2)*(EllipticF(x,1/3*6^(1/2))-EllipticE(x,1/ 
3*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\left (3-5 x^2+2 x^4\right )^{3/2}} \, dx=-\frac {5 \, \sqrt {3} {\left (2 \, x^{4} - 5 \, x^{2} + 3\right )} E(\arcsin \left (x\right )\,|\,\frac {2}{3}) - \sqrt {3} {\left (2 \, x^{4} - 5 \, x^{2} + 3\right )} F(\arcsin \left (x\right )\,|\,\frac {2}{3}) + \sqrt {2 \, x^{4} - 5 \, x^{2} + 3} {\left (10 \, x^{3} - 13 \, x\right )}}{3 \, {\left (2 \, x^{4} - 5 \, x^{2} + 3\right )}} \] Input:

integrate(1/(2*x^4-5*x^2+3)^(3/2),x, algorithm="fricas")
 

Output:

-1/3*(5*sqrt(3)*(2*x^4 - 5*x^2 + 3)*elliptic_e(arcsin(x), 2/3) - sqrt(3)*( 
2*x^4 - 5*x^2 + 3)*elliptic_f(arcsin(x), 2/3) + sqrt(2*x^4 - 5*x^2 + 3)*(1 
0*x^3 - 13*x))/(2*x^4 - 5*x^2 + 3)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{\left (3-5 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (2 x^{4} - 5 x^{2} + 3\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(2*x**4-5*x**2+3)**(3/2),x)
 

Output:

Integral((2*x**4 - 5*x**2 + 3)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (3-5 x^2+2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, x^{4} - 5 \, x^{2} + 3\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(2*x^4-5*x^2+3)^(3/2),x, algorithm="maxima")
 

Output:

integrate((2*x^4 - 5*x^2 + 3)^(-3/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (3-5 x^2+2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, x^{4} - 5 \, x^{2} + 3\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(2*x^4-5*x^2+3)^(3/2),x, algorithm="giac")
 

Output:

integrate((2*x^4 - 5*x^2 + 3)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (3-5 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (2\,x^4-5\,x^2+3\right )}^{3/2}} \,d x \] Input:

int(1/(2*x^4 - 5*x^2 + 3)^(3/2),x)
 

Output:

int(1/(2*x^4 - 5*x^2 + 3)^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (3-5 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {2 x^{4}-5 x^{2}+3}}{4 x^{8}-20 x^{6}+37 x^{4}-30 x^{2}+9}d x \] Input:

int(1/(2*x^4-5*x^2+3)^(3/2),x)
 

Output:

int(sqrt(2*x**4 - 5*x**2 + 3)/(4*x**8 - 20*x**6 + 37*x**4 - 30*x**2 + 9),x 
)