\(\int \frac {1}{(-2-3 x^2)^{3/2} (1+x^2)^{3/2}} \, dx\) [295]

Optimal result

Integrand size = 21, antiderivative size = 96 \[ \int \frac {1}{\left (-2-3 x^2\right )^{3/2} \left (1+x^2\right )^{3/2}} \, dx=-\frac {3 x}{2 \sqrt {-2-3 x^2} \sqrt {1+x^2}}-\frac {5 \sqrt {2+3 x^2} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{\sqrt {2} \sqrt {-2-3 x^2}}+\frac {3 \sqrt {2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {-2-3 x^2}} \] 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 = 2.63 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.23 \[ \int \frac {1}{\left (-2-3 x^2\right )^{3/2} \left (1+x^2\right )^{3/2}} \, dx=\frac {-x \sqrt {1+x^2} \left (13+15 x^2\right )-5 i \left (1+x^2\right ) \sqrt {6+9 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )+i \left (1+x^2\right ) \sqrt {6+9 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{2 \sqrt {-2-3 x^2} \left (1+x^2\right )} \] Input:

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

Output:

(-(x*Sqrt[1 + x^2]*(13 + 15*x^2)) - (5*I)*(1 + x^2)*Sqrt[6 + 9*x^2]*Ellipt 
icE[I*ArcSinh[Sqrt[3/2]*x], 2/3] + I*(1 + x^2)*Sqrt[6 + 9*x^2]*EllipticF[I 
*ArcSinh[Sqrt[3/2]*x], 2/3])/(2*Sqrt[-2 - 3*x^2]*(1 + x^2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.41, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {316, 25, 400, 313, 320}

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.

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 400

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

\(\Big \downarrow \) 313

\(\displaystyle \frac {1}{2} \left (12 \int \frac {1}{\sqrt {-3 x^2-2} \sqrt {x^2+1}}dx+\frac {5 \sqrt {2} \sqrt {-3 x^2-2} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}\right )-\frac {3 x}{2 \sqrt {-3 x^2-2} \sqrt {x^2+1}}\)

\(\Big \downarrow \) 320

\(\displaystyle \frac {1}{2} \left (\frac {5 \sqrt {2} \sqrt {-3 x^2-2} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}-\frac {6 \sqrt {2} \sqrt {-3 x^2-2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}\right )-\frac {3 x}{2 \sqrt {-3 x^2-2} \sqrt {x^2+1}}\)

Input:

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

Output:

(-3*x)/(2*Sqrt[-2 - 3*x^2]*Sqrt[1 + x^2]) + ((5*Sqrt[2]*Sqrt[-2 - 3*x^2]*E 
llipticE[ArcTan[x], -1/2])/(Sqrt[1 + x^2]*Sqrt[(2 + 3*x^2)/(1 + x^2)]) - ( 
6*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 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

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 400

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^ 
(3/2)), x_Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(Sqrt[a + b*x^2]* 
Sqrt[c + d*x^2]), x], x] - Simp[(d*e - c*f)/(b*c - a*d)   Int[Sqrt[a + b*x^ 
2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] & 
& PosQ[d/c]
 

Maple [A] (verified)

Time = 5.99 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17

method	result	size
default	\(-\frac {\sqrt {-3 x^{2}-2}\, \sqrt {x^{2}+1}\, \left (i \sqrt {3}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right ) \sqrt {x^{2}+1}\, \sqrt {3 x^{2}+2}-5 i \sqrt {3}\, \operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right ) \sqrt {x^{2}+1}\, \sqrt {3 x^{2}+2}-15 x^{3}-13 x \right )}{2 \left (3 x^{4}+5 x^{2}+2\right )}\)	\(112\)
elliptic	\(\frac {\sqrt {-\left (3 x^{2}+2\right ) \left (x^{2}+1\right )}\, \left (\frac {-\frac {15}{2} x^{3}-\frac {13}{2} x}{\sqrt {-3 x^{4}-5 x^{2}-2}}-\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )}{\sqrt {-3 x^{4}-5 x^{2}-2}}+\frac {5 i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{2}, \frac {\sqrt {6}}{3}\right )\right )}{4 \sqrt {-3 x^{4}-5 x^{2}-2}}\right )}{\sqrt {x^{2}+1}\, \sqrt {-3 x^{2}-2}}\)	\(174\)

Input:

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

Output:

-1/2*(-3*x^2-2)^(1/2)*(x^2+1)^(1/2)*(I*3^(1/2)*EllipticF(1/2*I*x*6^(1/2),1 
/3*6^(1/2))*(x^2+1)^(1/2)*(3*x^2+2)^(1/2)-5*I*3^(1/2)*EllipticE(1/2*I*x*6^ 
(1/2),1/3*6^(1/2))*(x^2+1)^(1/2)*(3*x^2+2)^(1/2)-15*x^3-13*x)/(3*x^4+5*x^2 
+2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (-2-3 x^2\right )^{3/2} \left (1+x^2\right )^{3/2}} \, dx=\frac {-5 i \, \sqrt {-2} {\left (3 \, x^{4} + 5 \, x^{2} + 2\right )} E(\arcsin \left (i \, x\right )\,|\,\frac {3}{2}) + 11 i \, \sqrt {-2} {\left (3 \, x^{4} + 5 \, x^{2} + 2\right )} F(\arcsin \left (i \, x\right )\,|\,\frac {3}{2}) + {\left (15 \, x^{3} + 13 \, x\right )} \sqrt {x^{2} + 1} \sqrt {-3 \, x^{2} - 2}}{2 \, {\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}} \] Input:

integrate(1/(-3*x^2-2)^(3/2)/(x^2+1)^(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) + (15*x^3 + 13* 
x)*sqrt(x^2 + 1)*sqrt(-3*x^2 - 2))/(3*x^4 + 5*x^2 + 2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt( - 3*x**2 - 2)*sqrt(x**2 + 1))/(9*x**8 + 30*x**6 + 37*x**4 + 20* 
x**2 + 4),x)