\(\int (1-b x^2) \sqrt {-1-b^2 x^4} \, dx\) [282]

Optimal result

Integrand size = 23, antiderivative size = 190 \[ \int \left (1-b x^2\right ) \sqrt {-1-b^2 x^4} \, dx=\frac {1}{15} x \left (5-3 b x^2\right ) \sqrt {-1-b^2 x^4}-\frac {2 x \sqrt {-1-b^2 x^4}}{5 \left (1+b x^2\right )}-\frac {2 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{5 \sqrt {b} \sqrt {-1-b^2 x^4}}-\frac {2 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{15 \sqrt {b} \sqrt {-1-b^2 x^4}} \] Output:

1/15*x*(-3*b*x^2+5)*(-b^2*x^4-1)^(1/2)-2*x*(-b^2*x^4-1)^(1/2)/(5*b*x^2+5)- 
2/5*(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)-2/15*(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 = 6.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.40 \[ \int \left (1-b x^2\right ) \sqrt {-1-b^2 x^4} \, dx=-\frac {\sqrt {-1-b^2 x^4} \left (-3 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-b^2 x^4\right )\right )}{3 \sqrt {1+b^2 x^4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1491, 27, 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.

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

\(\Big \downarrow \) 1491

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1512

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

(x*(5 - 3*b*x^2)*Sqrt[-1 - b^2*x^4])/15 - (2*(3*((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*Arc 
Tan[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]*S 
qrt[-1 - b^2*x^4])))/15
 

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 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 1491

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*( 
d*(4*p + 3) + e*(4*p + 1)*x^2)*((a + c*x^4)^p/((4*p + 1)*(4*p + 3))), x] + 
Simp[2*(p/((4*p + 1)*(4*p + 3)))   Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4*p + 
 1))*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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[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 = 2.43 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.47

method	result	size
meijerg	\(\frac {\sqrt {-\operatorname {signum}\left (b^{2} x^{4}+1\right )}\, x \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )}{\sqrt {\operatorname {signum}\left (b^{2} x^{4}+1\right )}}-\frac {b \sqrt {-\operatorname {signum}\left (b^{2} x^{4}+1\right )}\, x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3 \sqrt {\operatorname {signum}\left (b^{2} x^{4}+1\right )}}\)	\(90\)
risch	\(\frac {x \left (3 b \,x^{2}-5\right ) \left (b^{2} x^{4}+1\right )}{15 \sqrt {-b^{2} x^{4}-1}}-\frac {2 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{3 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {2 i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{5 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\)	\(155\)
elliptic	\(-\frac {b \,x^{3} \sqrt {-b^{2} x^{4}-1}}{5}+\frac {x \sqrt {-b^{2} x^{4}-1}}{3}-\frac {2 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{3 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {2 i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{5 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\)	\(156\)
default	\(-b \left (\frac {x^{3} \sqrt {-b^{2} x^{4}-1}}{5}+\frac {2 i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{5 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}\, b}\right )+\frac {x \sqrt {-b^{2} x^{4}-1}}{3}-\frac {2 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{3 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\)	\(162\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.56 \[ \int \left (1-b x^2\right ) \sqrt {-1-b^2 x^4} \, dx=\frac {2 \, \sqrt {-b^{2}} {\left (5 \, b + 3\right )} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 6 \, \sqrt {-b^{2}} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (3 \, b^{2} x^{4} - 5 \, b x^{2} + 6\right )} \sqrt {-b^{2} x^{4} - 1}}{15 \, b x} \] Input:

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

Output:

1/15*(2*sqrt(-b^2)*(5*b + 3)*x*(-1/b^2)^(3/4)*elliptic_f(arcsin((-1/b^2)^( 
1/4)/x), -1) - 6*sqrt(-b^2)*x*(-1/b^2)^(3/4)*elliptic_e(arcsin((-1/b^2)^(1 
/4)/x), -1) - (3*b^2*x^4 - 5*b*x^2 + 6)*sqrt(-b^2*x^4 - 1))/(b*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.99 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.38 \[ \int \left (1-b x^2\right ) \sqrt {-1-b^2 x^4} \, dx=- \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (1-b x^2\right ) \sqrt {-1-b^2 x^4} \, dx=\frac {i \left (-3 \sqrt {b^{2} x^{4}+1}\, b \,x^{3}+5 \sqrt {b^{2} x^{4}+1}\, x +10 \left (\int \frac {\sqrt {b^{2} x^{4}+1}}{b^{2} x^{4}+1}d x \right )-6 \left (\int \frac {\sqrt {b^{2} x^{4}+1}\, x^{2}}{b^{2} x^{4}+1}d x \right ) b \right )}{15} \] Input:

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

Output:

(i*( - 3*sqrt(b**2*x**4 + 1)*b*x**3 + 5*sqrt(b**2*x**4 + 1)*x + 10*int(sqr 
t(b**2*x**4 + 1)/(b**2*x**4 + 1),x) - 6*int((sqrt(b**2*x**4 + 1)*x**2)/(b* 
*2*x**4 + 1),x)*b))/15