\(\int \frac {1}{(e x)^{5/2} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx\) [1374]

Optimal result

Integrand size = 31, antiderivative size = 31 \[ \int \frac {1}{(e x)^{5/2} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=-\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},\frac {1}{2},\frac {5}{8},b^2 x^4\right )}{3 e (e x)^{3/2}} \] Output:

-2/3*hypergeom([-3/8, 1/2],[5/8],b^2*x^4)/e/(e*x)^(3/2)
 

Mathematica [A] (verified)

Time = 10.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(e x)^{5/2} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=-\frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},\frac {1}{2},\frac {5}{8},b^2 x^4\right )}{3 (e x)^{5/2}} \] Input:

Integrate[1/((e*x)^(5/2)*Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
 

Output:

(-2*x*Hypergeometric2F1[-3/8, 1/2, 5/8, b^2*x^4])/(3*(e*x)^(5/2))
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(31)=62\).

Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {335, 847, 851, 888}

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.

\(\displaystyle \int \frac {1}{\sqrt {1-b x^2} \sqrt {b x^2+1} (e x)^{5/2}} \, dx\)

\(\Big \downarrow \) 335

\(\displaystyle \int \frac {1}{\sqrt {1-b^2 x^4} (e x)^{5/2}}dx\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {b^2 \int \frac {(e x)^{3/2}}{\sqrt {1-b^2 x^4}}dx}{3 e^4}-\frac {2 \sqrt {1-b^2 x^4}}{3 e (e x)^{3/2}}\)

\(\Big \downarrow \) 851

\(\displaystyle -\frac {2 b^2 \int \frac {e^2 x^2}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{3 e^5}-\frac {2 \sqrt {1-b^2 x^4}}{3 e (e x)^{3/2}}\)

\(\Big \downarrow \) 888

\(\displaystyle -\frac {2 b^2 (e x)^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right )}{15 e^5}-\frac {2 \sqrt {1-b^2 x^4}}{3 e (e x)^{3/2}}\)

Input:

Int[1/((e*x)^(5/2)*Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
 

Output:

(-2*Sqrt[1 - b^2*x^4])/(3*e*(e*x)^(3/2)) - (2*b^2*(e*x)^(5/2)*Hypergeometr 
ic2F1[1/2, 5/8, 13/8, b^2*x^4])/(15*e^5)
 

Defintions of rubi rules used

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 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, 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 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(1/(e*x)^(5/2)/(-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^3*x^7 - e^3*x^ 
3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 39.95 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.29 \[ \int \frac {1}{(e x)^{5/2} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {i b^{\frac {3}{4}} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {9}{8}, \frac {13}{8}, 1 & \frac {11}{8}, \frac {11}{8}, \frac {15}{8} \\\frac {7}{8}, \frac {9}{8}, \frac {11}{8}, \frac {13}{8}, \frac {15}{8} & 0 \end {matrix} \middle | {\frac {1}{b^{2} x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}} e^{\frac {5}{2}}} + \frac {b^{\frac {3}{4}} {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {3}{8}, \frac {5}{8}, \frac {7}{8}, \frac {9}{8}, \frac {11}{8}, 1 & \\\frac {5}{8}, \frac {9}{8} & \frac {3}{8}, \frac {7}{8}, \frac {7}{8}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{- \frac {i \pi }{4}}}{8 \pi ^{\frac {3}{2}} e^{\frac {5}{2}}} \] Input:

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

Output:

I*b**(3/4)*meijerg(((9/8, 13/8, 1), (11/8, 11/8, 15/8)), ((7/8, 9/8, 11/8, 
 13/8, 15/8), (0,)), 1/(b**2*x**4))/(8*pi**(3/2)*e**(5/2)) + b**(3/4)*meij 
erg(((3/8, 5/8, 7/8, 9/8, 11/8, 1), ()), ((5/8, 9/8), (3/8, 7/8, 7/8, 0)), 
 exp_polar(-2*I*pi)/(b**2*x**4))*exp(-I*pi/4)/(8*pi**(3/2)*e**(5/2))
 

Maxima [F]

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

integrate(1/(e*x)^(5/2)/(-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)*(e*x)^(5/2)), x)
 

Giac [F]

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

integrate(1/(e*x)^(5/2)/(-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)*(e*x)^(5/2)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(e x)^{5/2} \sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=-\frac {\sqrt {e}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} x^{7}-x^{3}}d x \right )}{e^{3}} \] Input:

int(1/(e*x)^(5/2)/(-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**7 
- x**3),x))/e**3