\(\int \frac {1}{((-1-x^2) (2+3 x^2))^{3/2}} \, dx\) [296]

Optimal result

Integrand size = 19, antiderivative size = 96 \[ \int \frac {1}{\left (\left (-1-x^2\right ) \left (2+3 x^2\right )\right )^{3/2}} \, dx=-\frac {3 x}{2 \sqrt {-2-3 x^2} \sqrt {1+x^2}}-\frac {5 \sqrt {2+3 x^2} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{\sqrt {2} \sqrt {-2-3 x^2}}+\frac {3 \sqrt {2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {-2-3 x^2}} \] Output:

-3/2*x/(-3*x^2-2)^(1/2)/(x^2+1)^(1/2)-5/2*(3*x^2+2)^(1/2)*EllipticE(x/(x^2 
+1)^(1/2),1/2*I*2^(1/2))*2^(1/2)/(-3*x^2-2)^(1/2)+3*(3*x^2+2)^(1/2)*Invers 
eJacobiAM(arctan(x),1/2*I*2^(1/2))*2^(1/2)/(-3*x^2-2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.01 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (\left (-1-x^2\right ) \left (2+3 x^2\right )\right )^{3/2}} \, dx=-\frac {13 x+15 x^3+5 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )-i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{2 \sqrt {-2-5 x^2-3 x^4}} \] Input:

Integrate[((-1 - x^2)*(2 + 3*x^2))^(-3/2),x]
 

Output:

-1/2*(13*x + 15*x^3 + (5*I)*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*Elliptic 
E[I*ArcSinh[Sqrt[3/2]*x], 2/3] - I*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*E 
llipticF[I*ArcSinh[Sqrt[3/2]*x], 2/3])/Sqrt[-2 - 5*x^2 - 3*x^4]
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.75, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {2048, 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.

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

\(\Big \downarrow \) 2048

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{2} \int \frac {3 \left (5 x^2+4\right )}{\sqrt {-3 x^4-5 x^2-2}}dx-\frac {x \left (15 x^2+13\right )}{2 \sqrt {-3 x^4-5 x^2-2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{2} \int \frac {5 x^2+4}{\sqrt {-3 x^4-5 x^2-2}}dx-\frac {x \left (15 x^2+13\right )}{2 \sqrt {-3 x^4-5 x^2-2}}\)

\(\Big \downarrow \) 1494

\(\displaystyle 3 \sqrt {3} \int \frac {5 x^2+4}{2 \sqrt {3} \sqrt {-3 x^2-2} \sqrt {x^2+1}}dx-\frac {x \left (15 x^2+13\right )}{2 \sqrt {-3 x^4-5 x^2-2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{2} \int \frac {5 x^2+4}{\sqrt {-3 x^2-2} \sqrt {x^2+1}}dx-\frac {x \left (15 x^2+13\right )}{2 \sqrt {-3 x^4-5 x^2-2}}\)

\(\Big \downarrow \) 406

\(\displaystyle \frac {3}{2} \left (4 \int \frac {1}{\sqrt {-3 x^2-2} \sqrt {x^2+1}}dx+5 \int \frac {x^2}{\sqrt {-3 x^2-2} \sqrt {x^2+1}}dx\right )-\frac {x \left (15 x^2+13\right )}{2 \sqrt {-3 x^4-5 x^2-2}}\)

\(\Big \downarrow \) 320

\(\displaystyle \frac {3}{2} \left (5 \int \frac {x^2}{\sqrt {-3 x^2-2} \sqrt {x^2+1}}dx-\frac {2 \sqrt {2} \sqrt {-3 x^2-2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}\right )-\frac {x \left (15 x^2+13\right )}{2 \sqrt {-3 x^4-5 x^2-2}}\)

\(\Big \downarrow \) 388

\(\displaystyle \frac {3}{2} \left (5 \left (\frac {1}{3} \int \frac {\sqrt {-3 x^2-2}}{\left (x^2+1\right )^{3/2}}dx-\frac {x \sqrt {-3 x^2-2}}{3 \sqrt {x^2+1}}\right )-\frac {2 \sqrt {2} \sqrt {-3 x^2-2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}\right )-\frac {x \left (15 x^2+13\right )}{2 \sqrt {-3 x^4-5 x^2-2}}\)

\(\Big \downarrow \) 313

\(\displaystyle \frac {3}{2} \left (5 \left (\frac {\sqrt {2} \sqrt {-3 x^2-2} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}-\frac {x \sqrt {-3 x^2-2}}{3 \sqrt {x^2+1}}\right )-\frac {2 \sqrt {2} \sqrt {-3 x^2-2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}\right )-\frac {x \left (15 x^2+13\right )}{2 \sqrt {-3 x^4-5 x^2-2}}\)

Input:

Int[((-1 - x^2)*(2 + 3*x^2))^(-3/2),x]
 

Output:

-1/2*(x*(13 + 15*x^2))/Sqrt[-2 - 5*x^2 - 3*x^4] + (3*(5*(-1/3*(x*Sqrt[-2 - 
 3*x^2])/Sqrt[1 + x^2] + (Sqrt[2]*Sqrt[-2 - 3*x^2]*EllipticE[ArcTan[x], -1 
/2])/(3*Sqrt[1 + x^2]*Sqrt[(2 + 3*x^2)/(1 + x^2)])) - (2*Sqrt[2]*Sqrt[-2 - 
 3*x^2]*EllipticF[ArcTan[x], -1/2])/(Sqrt[1 + x^2]*Sqrt[(2 + 3*x^2)/(1 + x 
^2)])))/2
 

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]
 

rule 2048

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_) 
, x_Symbol] :> Int[u*(a*c*e + (b*c + a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; F 
reeQ[{a, b, c, d, e, n, p}, x]
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.47

method	result	size
default	\(\frac {-\frac {15}{2} x^{3}-\frac {13}{2} x}{\sqrt {-3 x^{4}-5 x^{2}-2}}-\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )}{\sqrt {-3 x^{4}-5 x^{2}-2}}+\frac {5 i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )\right )}{4 \sqrt {-3 x^{4}-5 x^{2}-2}}\)	\(141\)
elliptic	\(\frac {-\frac {15}{2} x^{3}-\frac {13}{2} x}{\sqrt {-3 x^{4}-5 x^{2}-2}}-\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )}{\sqrt {-3 x^{4}-5 x^{2}-2}}+\frac {5 i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )\right )}{4 \sqrt {-3 x^{4}-5 x^{2}-2}}\)	\(141\)

Input:

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

Output:

6*(-5/4*x^3-13/12*x)/(-3*x^4-5*x^2-2)^(1/2)-I*6^(1/2)*(6*x^2+4)^(1/2)*(x^2 
+1)^(1/2)/(-3*x^4-5*x^2-2)^(1/2)*EllipticF(1/2*I*x*6^(1/2),1/3*6^(1/2))+5/ 
4*I*6^(1/2)*(6*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-3*x^4-5*x^2-2)^(1/2)*(Elliptic 
F(1/2*I*x*6^(1/2),1/3*6^(1/2))-EllipticE(1/2*I*x*6^(1/2),1/3*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (\left (-1-x^2\right ) \left (2+3 x^2\right )\right )^{3/2}} \, dx=\frac {-5 i \, \sqrt {-2} {\left (3 \, x^{4} + 5 \, x^{2} + 2\right )} E(\arcsin \left (i \, x\right )\,|\,\frac {3}{2}) + 11 i \, \sqrt {-2} {\left (3 \, x^{4} + 5 \, x^{2} + 2\right )} F(\arcsin \left (i \, x\right )\,|\,\frac {3}{2}) + \sqrt {-3 \, x^{4} - 5 \, x^{2} - 2} {\left (15 \, x^{3} + 13 \, x\right )}}{2 \, {\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}} \] Input:

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

Output:

1/2*(-5*I*sqrt(-2)*(3*x^4 + 5*x^2 + 2)*elliptic_e(arcsin(I*x), 3/2) + 11*I 
*sqrt(-2)*(3*x^4 + 5*x^2 + 2)*elliptic_f(arcsin(I*x), 3/2) + sqrt(-3*x^4 - 
 5*x^2 - 2)*(15*x^3 + 13*x))/(3*x^4 + 5*x^2 + 2)
 

Sympy [F]

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

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

Output:

Integral(((-x**2 - 1)*(3*x**2 + 2))**(-3/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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