\(\int (1-b x^2) \sqrt {1-b^2 x^4} \, dx\) [235]

Optimal result

Integrand size = 23, antiderivative size = 66 \[ \int \left (1-b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\frac {1}{15} x \left (5-3 b x^2\right ) \sqrt {1-b^2 x^4}-\frac {2 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{5 \sqrt {b}}+\frac {16 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{15 \sqrt {b}} \] Output:

1/15*x*(-3*b*x^2+5)*(-b^2*x^4+1)^(1/2)-2/5*EllipticE(b^(1/2)*x,I)/b^(1/2)+ 
16/15*EllipticF(b^(1/2)*x,I)/b^(1/2)
 

Mathematica [C] (verified)

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

Time = 6.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.68 \[ \int \left (1-b x^2\right ) \sqrt {1-b^2 x^4} \, dx=x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},b^2 x^4\right )-\frac {1}{3} b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right ) \] Input:

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

Output:

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])/3
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.53, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {1388, 318, 27, 403, 25, 27, 399, 284, 327, 762}

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.

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

\(\Big \downarrow \) 1388

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 403

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 399

\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (8 \int \frac {1}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx-3 \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx\right )+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}\)

\(\Big \downarrow \) 284

\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (8 \int \frac {1}{\sqrt {1-b^2 x^4}}dx-3 \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx\right )+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (8 \int \frac {1}{\sqrt {1-b^2 x^4}}dx-\frac {3 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}\right )+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (\frac {8 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b}}-\frac {3 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}\right )+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}\)

Input:

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

Output:

-1/5*(x*Sqrt[1 - b*x^2]*(1 + b*x^2)^(3/2)) + (2*((4*x*Sqrt[1 - b*x^2]*Sqrt 
[1 + b*x^2])/3 + ((-3*EllipticE[ArcSin[Sqrt[b]*x], -1])/Sqrt[b] + (8*Ellip 
ticF[ArcSin[Sqrt[b]*x], -1])/Sqrt[b])/3))/5
 

Defintions of rubi rules used

rule 25

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

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 284

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> I 
nt[(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c + a*d, 0] 
&& (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0]))
 

rule 318

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

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 1388

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

Maple [C] (verified)

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

Time = 1.82 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.55

method	result	size
meijerg	\(x \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )-\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3}\)	\(36\)
risch	\(\frac {x \left (3 b \,x^{2}-5\right ) \left (b^{2} x^{4}-1\right )}{15 \sqrt {-b^{2} x^{4}+1}}+\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{3 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}+\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{5 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\)	\(133\)
elliptic	\(-\frac {b \,x^{3} \sqrt {-b^{2} x^{4}+1}}{5}+\frac {x \sqrt {-b^{2} x^{4}+1}}{3}+\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{3 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}+\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{5 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\)	\(134\)
default	\(-b \left (\frac {x^{3} \sqrt {-b^{2} x^{4}+1}}{5}-\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )+\frac {x \sqrt {-b^{2} x^{4}+1}}{3}+\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{3 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\)	\(137\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.42 \[ \int \left (1-b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\frac {\frac {2 \, \sqrt {-b^{2}} {\left (5 \, b - 3\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + \frac {6 \, \sqrt {-b^{2}} x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - {\left (3 \, b^{3} x^{4} - 5 \, b^{2} x^{2} - 6 \, b\right )} \sqrt {-b^{2} x^{4} + 1}}{15 \, b^{2} x} \] 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*elliptic_f(arcsin(1/(sqrt(b)*x)), -1)/sqrt( 
b) + 6*sqrt(-b^2)*x*elliptic_e(arcsin(1/(sqrt(b)*x)), -1)/sqrt(b) - (3*b^3 
*x^4 - 5*b^2*x^2 - 6*b)*sqrt(-b^2*x^4 + 1))/(b^2*x)
 

Sympy [A] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11 \[ \int \left (1-b x^2\right ) \sqrt {1-b^2 x^4} \, dx=- \frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

-b*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b**2*x**4*exp_polar(2*I*pi)) 
/(4*gamma(7/4)) + x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b**2*x**4*exp_po 
lar(2*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. \[ \int \left (1-b x^2\right ) \sqrt {1-b^2 x^4} \, dx=-\int \sqrt {1-b^2\,x^4}\,\left (b\,x^2-1\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

( - 3*sqrt( - b**2*x**4 + 1)*b*x**3 + 5*sqrt( - b**2*x**4 + 1)*x - 10*int( 
sqrt( - 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