\(\int \frac {1}{\sqrt {e x} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx\) [1372]

Optimal result

Integrand size = 31, antiderivative size = 29 \[ \int \frac {1}{\sqrt {e x} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {2 \sqrt {e x} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},b^2 x^4\right )}{e} \] Output:

2*(e*x)^(1/2)*hypergeom([1/8, 1/2],[9/8],b^2*x^4)/e
 

Mathematica [A] (verified)

Time = 10.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\sqrt {e x} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {2 x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},b^2 x^4\right )}{\sqrt {e x}} \] Input:

Integrate[1/(Sqrt[e*x]*Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
 

Output:

(2*x*Hypergeometric2F1[1/8, 1/2, 9/8, b^2*x^4])/Sqrt[e*x]
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(442\) vs. \(2(29)=58\).

Time = 0.55 (sec) , antiderivative size = 442, normalized size of antiderivative = 15.24, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {335, 851, 767, 27, 2422}

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.

\(\displaystyle \int \frac {1}{\sqrt {1-b x^2} \sqrt {b x^2+1} \sqrt {e x}} \, dx\)

\(\Big \downarrow \) 335

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

\(\Big \downarrow \) 851

\(\displaystyle \frac {2 \int \frac {1}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{e}\)

\(\Big \downarrow \) 767

\(\displaystyle \frac {2 \left (\frac {1}{2} \int \frac {e-\sqrt [4]{-b^2} e x}{e \sqrt {1-b^2 x^4}}d\sqrt {e x}+\frac {1}{2} \int \frac {\sqrt [4]{-b^2} x e+e}{e \sqrt {1-b^2 x^4}}d\sqrt {e x}\right )}{e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {\int \frac {e-\sqrt [4]{-b^2} e x}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{2 e}+\frac {\int \frac {\sqrt [4]{-b^2} x e+e}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{2 e}\right )}{e}\)

\(\Big \downarrow \) 2422

\(\displaystyle \frac {2 \left (\frac {\sqrt [4]{-b^2} (e x)^{3/2} \sqrt {\frac {\left (\sqrt [4]{-b^2} e x+e\right )^2}{\sqrt [4]{-b^2} e^2 x}} \sqrt {-\frac {e^4-b^2 e^4 x^4}{\sqrt {-b^2} e^4 x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} \sqrt {-b^2} x^2 e^2-2 \sqrt [4]{-b^2} x e^2+\sqrt {2} e^2}{\sqrt [4]{-b^2} e^2 x}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \sqrt {1-b^2 x^4} \left (\sqrt [4]{-b^2} e x+e\right )}-\frac {\sqrt [4]{-b^2} (e x)^{3/2} \sqrt {-\frac {\left (e-\sqrt [4]{-b^2} e x\right )^2}{\sqrt [4]{-b^2} e^2 x}} \sqrt {-\frac {e^4-b^2 e^4 x^4}{\sqrt {-b^2} e^4 x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt {2} \sqrt {-b^2} x^2 e^2+2 \sqrt [4]{-b^2} x e^2+\sqrt {2} e^2}{\sqrt [4]{-b^2} e^2 x}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \sqrt {1-b^2 x^4} \left (e-\sqrt [4]{-b^2} e x\right )}\right )}{e}\)

Input:

Int[1/(Sqrt[e*x]*Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
 

Output:

(2*(((-b^2)^(1/4)*(e*x)^(3/2)*Sqrt[(e + (-b^2)^(1/4)*e*x)^2/((-b^2)^(1/4)* 
e^2*x)]*Sqrt[-((e^4 - b^2*e^4*x^4)/(Sqrt[-b^2]*e^4*x^2))]*EllipticF[ArcSin 
[Sqrt[-((Sqrt[2]*e^2 - 2*(-b^2)^(1/4)*e^2*x + Sqrt[2]*Sqrt[-b^2]*e^2*x^2)/ 
((-b^2)^(1/4)*e^2*x))]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]*(e + (- 
b^2)^(1/4)*e*x)*Sqrt[1 - b^2*x^4]) - ((-b^2)^(1/4)*(e*x)^(3/2)*Sqrt[-((e - 
 (-b^2)^(1/4)*e*x)^2/((-b^2)^(1/4)*e^2*x))]*Sqrt[-((e^4 - b^2*e^4*x^4)/(Sq 
rt[-b^2]*e^4*x^2))]*EllipticF[ArcSin[Sqrt[(Sqrt[2]*e^2 + 2*(-b^2)^(1/4)*e^ 
2*x + Sqrt[2]*Sqrt[-b^2]*e^2*x^2)/((-b^2)^(1/4)*e^2*x)]/2], -2*(1 - Sqrt[2 
])])/(2*Sqrt[2 + Sqrt[2]]*(e - (-b^2)^(1/4)*e*x)*Sqrt[1 - b^2*x^4])))/e
 

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 335

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

rule 767

Int[1/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[1/2   Int[(1 - Rt[b/a, 4 
]*x^2)/Sqrt[a + b*x^8], x], x] + Simp[1/2   Int[(1 + Rt[b/a, 4]*x^2)/Sqrt[a 
 + b*x^8], x], x] /; FreeQ[{a, b}, x]
 

rule 851

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ 
n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && 
FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 2422

Int[((c_) + (d_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[(-c) 
*d*x^3*Sqrt[-(c - d*x^2)^2/(c*d*x^2)]*(Sqrt[(-d^2)*((a + b*x^8)/(b*c^2*x^4) 
)]/(Sqrt[2 + Sqrt[2]]*(c - d*x^2)*Sqrt[a + b*x^8]))*EllipticF[ArcSin[(1/2)* 
Sqrt[(Sqrt[2]*c^2 + 2*c*d*x^2 + Sqrt[2]*d^2*x^4)/(c*d*x^2)]], -2*(1 - Sqrt[ 
2])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^4 - a*d^4, 0]
 

Maple [F]

\[\int \frac {1}{\sqrt {e x}\, \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {1}{\sqrt {e x} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1} \sqrt {e x}} \,d x } \] Input:

integrate(1/(e*x)^(1/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="fri 
cas")
 

Output:

integral(-sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x)/(b^2*e*x^5 - e*x), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.83 (sec) , antiderivative size = 107, normalized size of antiderivative = 3.69 \[ \int \frac {1}{\sqrt {e x} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {i {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{8}, \frac {9}{8}, 1 & \frac {7}{8}, \frac {7}{8}, \frac {11}{8} \\\frac {3}{8}, \frac {5}{8}, \frac {7}{8}, \frac {9}{8}, \frac {11}{8} & 0 \end {matrix} \middle | {\frac {1}{b^{2} x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}} \sqrt [4]{b} \sqrt {e}} + \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{8}, \frac {1}{8}, \frac {3}{8}, \frac {5}{8}, \frac {7}{8}, 1 & \\\frac {1}{8}, \frac {5}{8} & - \frac {1}{8}, \frac {3}{8}, \frac {3}{8}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{\frac {3 i \pi }{4}}}{8 \pi ^{\frac {3}{2}} \sqrt [4]{b} \sqrt {e}} \] Input:

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

Output:

I*meijerg(((5/8, 9/8, 1), (7/8, 7/8, 11/8)), ((3/8, 5/8, 7/8, 9/8, 11/8), 
(0,)), 1/(b**2*x**4))/(8*pi**(3/2)*b**(1/4)*sqrt(e)) + meijerg(((-1/8, 1/8 
, 3/8, 5/8, 7/8, 1), ()), ((1/8, 5/8), (-1/8, 3/8, 3/8, 0)), exp_polar(-2* 
I*pi)/(b**2*x**4))*exp(3*I*pi/4)/(8*pi**(3/2)*b**(1/4)*sqrt(e))
 

Maxima [F]

\[ \int \frac {1}{\sqrt {e x} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1} \sqrt {e x}} \,d x } \] Input:

integrate(1/(e*x)^(1/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="max 
ima")
 

Output:

integrate(1/(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {e x} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1} \sqrt {e x}} \,d x } \] Input:

integrate(1/(e*x)^(1/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="gia 
c")
                                                                                    
                                                                                    
 

Output:

integrate(1/(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {e x} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int \frac {1}{\sqrt {e\,x}\,\sqrt {1-b\,x^2}\,\sqrt {b\,x^2+1}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

( - sqrt(e)*int((sqrt(x)*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1))/(b**2*x**5 
- x),x))/e