\(\int \frac {1-b x^2}{(-1-b^2 x^4)^{3/2}} \, dx\) [284]

Optimal result

Integrand size = 23, antiderivative size = 190 \[ \int \frac {1-b x^2}{\left (-1-b^2 x^4\right )^{3/2}} \, dx=-\frac {x \left (1-b x^2\right )}{2 \sqrt {-1-b^2 x^4}}+\frac {x \sqrt {-1-b^2 x^4}}{2 \left (1+b x^2\right )}+\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{2 \sqrt {b} \sqrt {-1-b^2 x^4}}-\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{2 \sqrt {b} \sqrt {-1-b^2 x^4}} \] Output:

-1/2*x*(-b*x^2+1)/(-b^2*x^4-1)^(1/2)+x*(-b^2*x^4-1)^(1/2)/(2*b*x^2+2)+1/2* 
(b*x^2+1)*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/2)*x 
)),1/2*2^(1/2))/b^(1/2)/(-b^2*x^4-1)^(1/2)-1/2*(b*x^2+1)*((b^2*x^4+1)/(b*x 
^2+1)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/2)*x),1/2*2^(1/2))/b^(1/2)/(- 
b^2*x^4-1)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.48 \[ \int \frac {1-b x^2}{\left (-1-b^2 x^4\right )^{3/2}} \, dx=\frac {x \left (-3-3 \sqrt {1+b^2 x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-b^2 x^4\right )+2 b x^2 \sqrt {1+b^2 x^4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-b^2 x^4\right )\right )}{6 \sqrt {-1-b^2 x^4}} \] Input:

Integrate[(1 - b*x^2)/(-1 - b^2*x^4)^(3/2),x]
 

Output:

(x*(-3 - 3*Sqrt[1 + b^2*x^4]*Hypergeometric2F1[1/4, 1/2, 5/4, -(b^2*x^4)] 
+ 2*b*x^2*Sqrt[1 + b^2*x^4]*Hypergeometric2F1[3/4, 3/2, 7/4, -(b^2*x^4)])) 
/(6*Sqrt[-1 - b^2*x^4])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1493, 25, 1512, 761, 1510}

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.

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

\(\Big \downarrow \) 1493

\(\displaystyle \frac {1}{2} \int -\frac {b x^2+1}{\sqrt {-b^2 x^4-1}}dx-\frac {x \left (1-b x^2\right )}{2 \sqrt {-b^2 x^4-1}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{2} \int \frac {b x^2+1}{\sqrt {-b^2 x^4-1}}dx-\frac {x \left (1-b x^2\right )}{2 \sqrt {-b^2 x^4-1}}\)

\(\Big \downarrow \) 1512

\(\displaystyle \frac {1}{2} \left (\int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx-2 \int \frac {1}{\sqrt {-b^2 x^4-1}}dx\right )-\frac {x \left (1-b x^2\right )}{2 \sqrt {-b^2 x^4-1}}\)

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

\(\displaystyle \frac {1}{2} \left (-\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}+\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}+\frac {x \sqrt {-b^2 x^4-1}}{b x^2+1}\right )-\frac {x \left (1-b x^2\right )}{2 \sqrt {-b^2 x^4-1}}\)

Input:

Int[(1 - b*x^2)/(-1 - b^2*x^4)^(3/2),x]
 

Output:

-1/2*(x*(1 - b*x^2))/Sqrt[-1 - b^2*x^4] + ((x*Sqrt[-1 - b^2*x^4])/(1 + b*x 
^2) + ((1 + b*x^2)*Sqrt[(1 + b^2*x^4)/(1 + b*x^2)^2]*EllipticE[2*ArcTan[Sq 
rt[b]*x], 1/2])/(Sqrt[b]*Sqrt[-1 - b^2*x^4]) - ((1 + b*x^2)*Sqrt[(1 + b^2* 
x^4)/(1 + b*x^2)^2]*EllipticF[2*ArcTan[Sqrt[b]*x], 1/2])/(Sqrt[b]*Sqrt[-1 
- b^2*x^4]))/2
 

Defintions of rubi rules used

rule 25

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

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1493

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x 
)*(d + e*x^2)*((a + c*x^4)^(p + 1)/(4*a*(p + 1))), x] + Simp[1/(4*a*(p + 1) 
)   Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x] 
/; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && Integer 
Q[2*p]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 1512

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q 
 Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c 
, d, e}, x] && PosQ[c/a]
 

Maple [A] (warning: unable to verify)

Time = 1.36 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.47

method	result	size
meijerg	\(\frac {\operatorname {signum}\left (b^{2} x^{4}+1\right )^{\frac {3}{2}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )}{{\left (-\operatorname {signum}\left (b^{2} x^{4}+1\right )\right )}^{\frac {3}{2}}}-\frac {b \operatorname {signum}\left (b^{2} x^{4}+1\right )^{\frac {3}{2}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{2}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3 {\left (-\operatorname {signum}\left (b^{2} x^{4}+1\right )\right )}^{\frac {3}{2}}}\)	\(90\)
elliptic	\(\frac {2 b^{2} \left (\frac {x^{3}}{4 b}-\frac {x}{4 b^{2}}\right )}{\sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}-\frac {\sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{2 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}+\frac {i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{2 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\)	\(157\)
default	\(-b \left (-\frac {x^{3}}{2 \sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}-\frac {i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{2 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}\, b}\right )-\frac {x}{2 \sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}-\frac {\sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{2 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\)	\(166\)

Input:

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

Output:

1/(-signum(b^2*x^4+1))^(3/2)*signum(b^2*x^4+1)^(3/2)*x*hypergeom([1/4,3/2] 
,[5/4],-b^2*x^4)-1/3*b/(-signum(b^2*x^4+1))^(3/2)*signum(b^2*x^4+1)^(3/2)* 
x^3*hypergeom([3/4,3/2],[7/4],-b^2*x^4)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.61 \[ \int \frac {1-b x^2}{\left (-1-b^2 x^4\right )^{3/2}} \, dx=\frac {{\left (-i \, b^{3} x^{4} - i \, b\right )} \left (-b^{2}\right )^{\frac {3}{4}} E(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) + {\left (i \, {\left (b^{3} - b^{2}\right )} x^{4} + i \, b - i\right )} \left (-b^{2}\right )^{\frac {3}{4}} F(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) - {\left (b^{3} x^{3} - b^{2} x\right )} \sqrt {-b^{2} x^{4} - 1}}{2 \, {\left (b^{4} x^{4} + b^{2}\right )}} \] Input:

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

Output:

1/2*((-I*b^3*x^4 - I*b)*(-b^2)^(3/4)*elliptic_e(arcsin((-b^2)^(1/4)*x), -1 
) + (I*(b^3 - b^2)*x^4 + I*b - I)*(-b^2)^(3/4)*elliptic_f(arcsin((-b^2)^(1 
/4)*x), -1) - (b^3*x^3 - b^2*x)*sqrt(-b^2*x^4 - 1))/(b^4*x^4 + b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.89 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.37 \[ \int \frac {1-b x^2}{\left (-1-b^2 x^4\right )^{3/2}} \, dx=- \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

-I*b*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), b**2*x**4*exp_polar(I*pi))/ 
(4*gamma(7/4)) + I*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), b**2*x**4*exp_po 
lar(I*pi))/(4*gamma(5/4))
 

Maxima [F]

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

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

Output:

-integrate((b*x^2 - 1)/(-b^2*x^4 - 1)^(3/2), x)
 

Giac [F]

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

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

Output:

integrate(-(b*x^2 - 1)/(-b^2*x^4 - 1)^(3/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

i*(int(sqrt(b**2*x**4 + 1)/(b**4*x**8 + 2*b**2*x**4 + 1),x) - int((sqrt(b* 
*2*x**4 + 1)*x**2)/(b**4*x**8 + 2*b**2*x**4 + 1),x)*b)