\(\int \frac {1-b x^2}{(1+b^2 x^4)^{3/2}} \, dx\) [257]

Optimal result

Integrand size = 22, antiderivative size = 186 \[ \int \frac {1-b x^2}{\left (1+b^2 x^4\right )^{3/2}} \, dx=\frac {x \left (1-b x^2\right )}{2 \sqrt {1+b^2 x^4}}+\frac {x \sqrt {1+b^2 x^4}}{2 \left (1+b x^2\right )}-\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{2 \sqrt {b} \sqrt {1+b^2 x^4}}+\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{2 \sqrt {b} \sqrt {1+b^2 x^4}} \] 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 = 10.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.35 \[ \int \frac {1-b x^2}{\left (1+b^2 x^4\right )^{3/2}} \, dx=\frac {1}{6} x \left (\frac {3}{\sqrt {1+b^2 x^4}}+3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-b^2 x^4\right )-2 b x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-b^2 x^4\right )\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, 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

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

\(\Big \downarrow \) 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

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 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:

(x*(1 - b*x^2))/(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[Sqrt[ 
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] (verified)

Time = 0.90 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.20

method	result	size
meijerg	\(x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )-\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{2}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3}\)	\(38\)
elliptic	\(-\frac {2 b^{2} \left (\frac {x^{3}}{4 b}-\frac {x}{4 b^{2}}\right )}{\sqrt {\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}+\frac {\sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{2 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}+\frac {i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{2 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}\)	\(154\)
default	\(-b \left (\frac {x^{3}}{2 \sqrt {\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}-\frac {i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{2 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}\, b}\right )+\frac {x}{2 \sqrt {\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}+\frac {\sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{2 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}\)	\(162\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.58 \[ \int \frac {1-b x^2}{\left (1+b^2 x^4\right )^{3/2}} \, dx=-\frac {{\left (b^{3} x^{4} + b\right )} \left (-b^{2}\right )^{\frac {3}{4}} E(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) - {\left ({\left (b^{3} - b^{2}\right )} x^{4} + b - 1\right )} \left (-b^{2}\right )^{\frac {3}{4}} F(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) + {\left (b^{3} x^{3} - b^{2} x\right )} \sqrt {b^{2} x^{4} + 1}}{2 \, {\left (b^{4} x^{4} + b^{2}\right )}} \] Input:

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

Output:

-1/2*((b^3*x^4 + b)*(-b^2)^(3/4)*elliptic_e(arcsin((-b^2)^(1/4)*x), -1) - 
((b^3 - b^2)*x^4 + b - 1)*(-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 = 2.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.35 \[ \int \frac {1-b x^2}{\left (1+b^2 x^4\right )^{3/2}} \, dx=- \frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

-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)) + x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), b**2*x**4*exp_polar( 
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]

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

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

Output:

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