\(\int \frac {1}{(-3-5 x^2+2 x^4)^{3/2}} \, dx\) [229]

Optimal result

Integrand size = 16, antiderivative size = 133 \[ \int \frac {1}{\left (-3-5 x^2+2 x^4\right )^{3/2}} \, dx=-\frac {x \left (37-10 x^2\right )}{147 \sqrt {-3-5 x^2+2 x^4}}-\frac {5 \sqrt {3-x^2} \sqrt {1+2 x^2} E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-6\right )}{147 \sqrt {-3-5 x^2+2 x^4}}-\frac {\sqrt {3-x^2} \sqrt {1+2 x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3}}\right ),-6\right )}{21 \sqrt {-3-5 x^2+2 x^4}} \] Output:

-1/147*x*(-10*x^2+37)/(2*x^4-5*x^2-3)^(1/2)-5/147*(-x^2+3)^(1/2)*(2*x^2+1) 
^(1/2)*EllipticE(1/3*x*3^(1/2),I*6^(1/2))/(2*x^4-5*x^2-3)^(1/2)-1/21*(-x^2 
+3)^(1/2)*(2*x^2+1)^(1/2)*EllipticF(1/3*x*3^(1/2),I*6^(1/2))/(2*x^4-5*x^2- 
3)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (-3-5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {-37 x+10 x^3-5 i \sqrt {6} \sqrt {3-x^2} \sqrt {1+2 x^2} E\left (i \text {arcsinh}\left (\sqrt {2} x\right )|-\frac {1}{6}\right )+7 i \sqrt {6} \sqrt {3-x^2} \sqrt {1+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} x\right ),-\frac {1}{6}\right )}{147 \sqrt {-3-5 x^2+2 x^4}} \] Input:

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

Output:

(-37*x + 10*x^3 - (5*I)*Sqrt[6]*Sqrt[3 - x^2]*Sqrt[1 + 2*x^2]*EllipticE[I* 
ArcSinh[Sqrt[2]*x], -1/6] + (7*I)*Sqrt[6]*Sqrt[3 - x^2]*Sqrt[1 + 2*x^2]*El 
lipticF[I*ArcSinh[Sqrt[2]*x], -1/6])/(147*Sqrt[-3 - 5*x^2 + 2*x^4])
 

Rubi [A] (warning: unable to verify)

Time = 0.54 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.64, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 27, 1501, 27, 1410, 1498}

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

\(\displaystyle \frac {1}{147} \int -\frac {2 \left (5 x^2+6\right )}{\sqrt {2 x^4-5 x^2-3}}dx-\frac {x \left (37-10 x^2\right )}{147 \sqrt {2 x^4-5 x^2-3}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2}{147} \int \frac {5 x^2+6}{\sqrt {2 x^4-5 x^2-3}}dx-\frac {x \left (37-10 x^2\right )}{147 \sqrt {2 x^4-5 x^2-3}}\)

\(\Big \downarrow \) 1501

\(\displaystyle -\frac {2}{147} \left (21 \int \frac {1}{\sqrt {2 x^4-5 x^2-3}}dx+\frac {5}{4} \int -\frac {4 \left (3-x^2\right )}{\sqrt {2 x^4-5 x^2-3}}dx\right )-\frac {x \left (37-10 x^2\right )}{147 \sqrt {2 x^4-5 x^2-3}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2}{147} \left (21 \int \frac {1}{\sqrt {2 x^4-5 x^2-3}}dx-5 \int \frac {3-x^2}{\sqrt {2 x^4-5 x^2-3}}dx\right )-\frac {x \left (37-10 x^2\right )}{147 \sqrt {2 x^4-5 x^2-3}}\)

\(\Big \downarrow \) 1410

\(\displaystyle -\frac {2}{147} \left (\frac {3 \sqrt {7} \sqrt {x^2-3} \sqrt {2 x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {7} x}{\sqrt {x^2-3}}\right ),\frac {1}{7}\right )}{\sqrt {2 x^4-5 x^2-3}}-5 \int \frac {3-x^2}{\sqrt {2 x^4-5 x^2-3}}dx\right )-\frac {x \left (37-10 x^2\right )}{147 \sqrt {2 x^4-5 x^2-3}}\)

\(\Big \downarrow \) 1498

\(\displaystyle -\frac {2}{147} \left (\frac {3 \sqrt {7} \sqrt {x^2-3} \sqrt {2 x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {7} x}{\sqrt {x^2-3}}\right ),\frac {1}{7}\right )}{\sqrt {2 x^4-5 x^2-3}}-5 \left (\frac {\sqrt {7} \sqrt {x^2-3} \sqrt {\frac {2 x^2+1}{3-x^2}} E\left (\arcsin \left (\frac {\sqrt {7} x}{\sqrt {x^2-3}}\right )|\frac {1}{7}\right )}{2 \sqrt {\frac {1}{3-x^2}} \sqrt {2 x^4-5 x^2-3}}-\frac {x \left (2 x^2+1\right )}{2 \sqrt {2 x^4-5 x^2-3}}\right )\right )-\frac {x \left (37-10 x^2\right )}{147 \sqrt {2 x^4-5 x^2-3}}\)

Input:

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

Output:

-1/147*(x*(37 - 10*x^2))/Sqrt[-3 - 5*x^2 + 2*x^4] - (2*(-5*(-1/2*(x*(1 + 2 
*x^2))/Sqrt[-3 - 5*x^2 + 2*x^4] + (Sqrt[7]*Sqrt[-3 + x^2]*Sqrt[(1 + 2*x^2) 
/(3 - x^2)]*EllipticE[ArcSin[(Sqrt[7]*x)/Sqrt[-3 + x^2]], 1/7])/(2*Sqrt[(3 
 - x^2)^(-1)]*Sqrt[-3 - 5*x^2 + 2*x^4])) + (3*Sqrt[7]*Sqrt[-3 + x^2]*Sqrt[ 
1 + 2*x^2]*EllipticF[ArcSin[(Sqrt[7]*x)/Sqrt[-3 + x^2]], 1/7])/Sqrt[-3 - 5 
*x^2 + 2*x^4]))/147
 

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 1410

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b 
^2 - 4*a*c, 2]}, Simp[Sqrt[-2*a - (b - q)*x^2]*(Sqrt[(2*a + (b + q)*x^2)/q] 
/(2*Sqrt[-a]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + 
q)*x^2)/(2*q)]], (b + q)/(2*q)], x] /; IntegerQ[q]] /; FreeQ[{a, b, c}, x] 
&& GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

rule 1498

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[e*x*((b + q + 2*c*x^2)/(2*c*Sqrt[ 
a + b*x^2 + c*x^4])), x] - Simp[e*q*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q) 
*x^2)]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*c*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2* 
a + (b + q)*x^2)]))*EllipticE[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b 
 + q)/(2*q)], x] /; EqQ[2*c*d - e*(b - q), 0]] /; FreeQ[{a, b, c, d, e}, x] 
 && GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

rule 1501

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[(2*c*d - e*(b - q))/(2*c)   Int[1 
/Sqrt[a + b*x^2 + c*x^4], x], x] + Simp[e/(2*c)   Int[(b - q + 2*c*x^2)/Sqr 
t[a + b*x^2 + c*x^4], x], x] /; NeQ[2*c*d - e*(b - q), 0]] /; FreeQ[{a, b, 
c, d, e}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

Maple [A] (verified)

Time = 2.77 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.11

method	result	size
risch	\(\frac {x \left (10 x^{2}-37\right )}{147 \sqrt {2 x^{4}-5 x^{2}-3}}+\frac {2 i \sqrt {2}\, \sqrt {2 x^{2}+1}\, \sqrt {-3 x^{2}+9}\, \operatorname {EllipticF}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )}{147 \sqrt {2 x^{4}-5 x^{2}-3}}+\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+1}\, \sqrt {-3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )-\operatorname {EllipticE}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )\right )}{147 \sqrt {2 x^{4}-5 x^{2}-3}}\)	\(147\)
default	\(-\frac {4 \left (\frac {37}{588} x -\frac {5}{294} x^{3}\right )}{\sqrt {2 x^{4}-5 x^{2}-3}}+\frac {2 i \sqrt {2}\, \sqrt {2 x^{2}+1}\, \sqrt {-3 x^{2}+9}\, \operatorname {EllipticF}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )}{147 \sqrt {2 x^{4}-5 x^{2}-3}}+\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+1}\, \sqrt {-3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )-\operatorname {EllipticE}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )\right )}{147 \sqrt {2 x^{4}-5 x^{2}-3}}\)	\(148\)
elliptic	\(-\frac {4 \left (\frac {37}{588} x -\frac {5}{294} x^{3}\right )}{\sqrt {2 x^{4}-5 x^{2}-3}}+\frac {2 i \sqrt {2}\, \sqrt {2 x^{2}+1}\, \sqrt {-3 x^{2}+9}\, \operatorname {EllipticF}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )}{147 \sqrt {2 x^{4}-5 x^{2}-3}}+\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+1}\, \sqrt {-3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )-\operatorname {EllipticE}\left (i x \sqrt {2}, \frac {i \sqrt {6}}{6}\right )\right )}{147 \sqrt {2 x^{4}-5 x^{2}-3}}\)	\(148\)

Input:

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

Output:

1/147*x*(10*x^2-37)/(2*x^4-5*x^2-3)^(1/2)+2/147*I*2^(1/2)*(2*x^2+1)^(1/2)* 
(-3*x^2+9)^(1/2)/(2*x^4-5*x^2-3)^(1/2)*EllipticF(I*2^(1/2)*x,1/6*I*6^(1/2) 
)+5/147*I*2^(1/2)*(2*x^2+1)^(1/2)*(-3*x^2+9)^(1/2)/(2*x^4-5*x^2-3)^(1/2)*( 
EllipticF(I*2^(1/2)*x,1/6*I*6^(1/2))-EllipticE(I*2^(1/2)*x,1/6*I*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (-3-5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {5 \, \sqrt {3} \sqrt {-3} {\left (2 \, x^{4} - 5 \, x^{2} - 3\right )} E(\arcsin \left (\frac {1}{3} \, \sqrt {3} x\right )\,|\,-6) + 31 \, \sqrt {3} \sqrt {-3} {\left (2 \, x^{4} - 5 \, x^{2} - 3\right )} F(\arcsin \left (\frac {1}{3} \, \sqrt {3} x\right )\,|\,-6) + 9 \, \sqrt {2 \, x^{4} - 5 \, x^{2} - 3} {\left (10 \, x^{3} - 37 \, x\right )}}{1323 \, {\left (2 \, x^{4} - 5 \, x^{2} - 3\right )}} \] Input:

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

Output:

1/1323*(5*sqrt(3)*sqrt(-3)*(2*x^4 - 5*x^2 - 3)*elliptic_e(arcsin(1/3*sqrt( 
3)*x), -6) + 31*sqrt(3)*sqrt(-3)*(2*x^4 - 5*x^2 - 3)*elliptic_f(arcsin(1/3 
*sqrt(3)*x), -6) + 9*sqrt(2*x^4 - 5*x^2 - 3)*(10*x^3 - 37*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]

\[ \int \frac {1}{\left (-3-5 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {2 x^{4}-5 x^{2}-3}}{4 x^{8}-20 x^{6}+13 x^{4}+30 x^{2}+9}d x \] 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 + 13*x**4 + 30*x**2 + 9),x 
)