\(\int \frac {1}{(-3-5 x^2-2 x^4)^{3/2}} \, dx\) [275]

Optimal result

Integrand size = 16, antiderivative size = 104 \[ \int \frac {1}{\left (-3-5 x^2-2 x^4\right )^{3/2}} \, dx=-\frac {x}{\sqrt {-3-5 x^2-2 x^4}}+\frac {5 \sqrt {2} \sqrt {-1-x^2} E\left (\arctan \left (\sqrt {\frac {2}{3}} x\right )|-\frac {1}{2}\right )}{3 \sqrt {1+x^2}}+\frac {2 \sqrt {2} \sqrt {1+x^2} \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {2}{3}} x\right ),-\frac {1}{2}\right )}{\sqrt {-1-x^2}} \] Output:

-x/(-2*x^4-5*x^2-3)^(1/2)+5/3*2^(1/2)*(-x^2-1)^(1/2)*EllipticE(x*6^(1/2)/( 
6*x^2+9)^(1/2),1/2*I*2^(1/2))/(x^2+1)^(1/2)+2*(x^2+1)^(1/2)*InverseJacobiA 
M(arctan(1/3*x*6^(1/2)),1/2*I*2^(1/2))*2^(1/2)/(-x^2-1)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (-3-5 x^2-2 x^4\right )^{3/2}} \, dx=-\frac {13 x+10 x^3+5 i \sqrt {2} \sqrt {1+x^2} \sqrt {3+2 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {2}{3}} x\right )|\frac {3}{2}\right )+i \sqrt {2} \sqrt {1+x^2} \sqrt {3+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{3}} x\right ),\frac {3}{2}\right )}{3 \sqrt {-3-5 x^2-2 x^4}} \] Input:

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

Output:

-1/3*(13*x + 10*x^3 + (5*I)*Sqrt[2]*Sqrt[1 + x^2]*Sqrt[3 + 2*x^2]*Elliptic 
E[I*ArcSinh[Sqrt[2/3]*x], 3/2] + I*Sqrt[2]*Sqrt[1 + x^2]*Sqrt[3 + 2*x^2]*E 
llipticF[I*ArcSinh[Sqrt[2/3]*x], 3/2])/Sqrt[-3 - 5*x^2 - 2*x^4]
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.67, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1405, 27, 1494, 27, 406, 320, 388, 313}

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

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1494

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 406

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

\(\Big \downarrow \) 320

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

\(\Big \downarrow \) 388

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

\(\Big \downarrow \) 313

\(\displaystyle \frac {2}{3} \left (\frac {2 \sqrt {3} \sqrt {2 x^2+3} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{3}\right )}{\sqrt {-x^2-1} \sqrt {\frac {2 x^2+3}{x^2+1}}}+5 \left (\frac {x \sqrt {2 x^2+3}}{2 \sqrt {-x^2-1}}-\frac {\sqrt {3} \sqrt {2 x^2+3} E\left (\arctan (x)\left |\frac {1}{3}\right .\right )}{2 \sqrt {-x^2-1} \sqrt {\frac {2 x^2+3}{x^2+1}}}\right )\right )-\frac {x \left (10 x^2+13\right )}{3 \sqrt {-2 x^4-5 x^2-3}}\)

Input:

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

Output:

-1/3*(x*(13 + 10*x^2))/Sqrt[-3 - 5*x^2 - 2*x^4] + (2*(5*((x*Sqrt[3 + 2*x^2 
])/(2*Sqrt[-1 - x^2]) - (Sqrt[3]*Sqrt[3 + 2*x^2]*EllipticE[ArcTan[x], 1/3] 
)/(2*Sqrt[-1 - x^2]*Sqrt[(3 + 2*x^2)/(1 + x^2)])) + (2*Sqrt[3]*Sqrt[3 + 2* 
x^2]*EllipticF[ArcTan[x], 1/3])/(Sqrt[-1 - x^2]*Sqrt[(3 + 2*x^2)/(1 + x^2) 
])))/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 313

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Sim 
p[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c 
+ d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; FreeQ 
[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]
 

rule 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( 
c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 388

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] 
 :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt[c + d*x^2])), x] - Simp[c/b   Int[Sqrt[ 
a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 406

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[e   Int[(a + b*x^2)^p*(c + d*x^2)^q, x], x] + Sim 
p[f   Int[x^2*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, 
f, p, q}, 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 1494

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

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.20

method	result	size
risch	\(-\frac {x \left (10 x^{2}+13\right )}{3 \sqrt {-2 x^{4}-5 x^{2}-3}}-\frac {4 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+9}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{3}\right )}{3 \sqrt {-2 x^{4}-5 x^{2}-3}}+\frac {5 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+9}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{3}\right )\right )}{3 \sqrt {-2 x^{4}-5 x^{2}-3}}\)	\(125\)
default	\(\frac {-\frac {10}{3} x^{3}-\frac {13}{3} x}{\sqrt {-2 x^{4}-5 x^{2}-3}}-\frac {4 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+9}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{3}\right )}{3 \sqrt {-2 x^{4}-5 x^{2}-3}}+\frac {5 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+9}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{3}\right )\right )}{3 \sqrt {-2 x^{4}-5 x^{2}-3}}\)	\(126\)
elliptic	\(\frac {-\frac {10}{3} x^{3}-\frac {13}{3} x}{\sqrt {-2 x^{4}-5 x^{2}-3}}-\frac {4 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+9}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{3}\right )}{3 \sqrt {-2 x^{4}-5 x^{2}-3}}+\frac {5 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+9}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{3}\right )\right )}{3 \sqrt {-2 x^{4}-5 x^{2}-3}}\)	\(126\)

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*I*(x^2+1)^(1/2)*(6*x^2+9)^(1 
/2)/(-2*x^4-5*x^2-3)^(1/2)*EllipticF(I*x,1/3*6^(1/2))+5/3*I*(x^2+1)^(1/2)* 
(6*x^2+9)^(1/2)/(-2*x^4-5*x^2-3)^(1/2)*(EllipticF(I*x,1/3*6^(1/2))-Ellipti 
cE(I*x,1/3*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (-3-5 x^2-2 x^4\right )^{3/2}} \, dx=-\frac {10 \, \sqrt {-\frac {2}{3}} \sqrt {-3} {\left (2 \, x^{4} + 5 \, x^{2} + 3\right )} E(\arcsin \left (\sqrt {-\frac {2}{3}} x\right )\,|\,\frac {3}{2}) - 28 \, \sqrt {-\frac {2}{3}} \sqrt {-3} {\left (2 \, x^{4} + 5 \, x^{2} + 3\right )} F(\arcsin \left (\sqrt {-\frac {2}{3}} x\right )\,|\,\frac {3}{2}) - 3 \, \sqrt {-2 \, x^{4} - 5 \, x^{2} - 3} {\left (10 \, x^{3} + 13 \, x\right )}}{9 \, {\left (2 \, x^{4} + 5 \, x^{2} + 3\right )}} \] Input:

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

Output:

-1/9*(10*sqrt(-2/3)*sqrt(-3)*(2*x^4 + 5*x^2 + 3)*elliptic_e(arcsin(sqrt(-2 
/3)*x), 3/2) - 28*sqrt(-2/3)*sqrt(-3)*(2*x^4 + 5*x^2 + 3)*elliptic_f(arcsi 
n(sqrt(-2/3)*x), 3/2) - 3*sqrt(-2*x^4 - 5*x^2 - 3)*(10*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/(- 5*x^2 - 2*x^4 - 3)^(3/2),x)
 

Output:

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