\(\int \frac {1}{(e x)^{5/2} \sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx\) [1385]

Optimal result

Integrand size = 31, antiderivative size = 74 \[ \int \frac {1}{(e x)^{5/2} \sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=-\frac {2 \sqrt {1-\frac {b^2 x^4}{a^2}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},\frac {1}{2},\frac {5}{8},\frac {b^2 x^4}{a^2}\right )}{3 e (e x)^{3/2} \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 3.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(e x)^{5/2} \sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=-\frac {2 x \sqrt {1-\frac {b^2 x^4}{a^2}} \, _2F_1\left (-\frac {3}{8},\frac {1}{2};\frac {5}{8};\frac {b^2 x^4}{a^2}\right )}{3 (e x)^{5/2} \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Input:

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

Output:

(-2*x*Sqrt[1 - (b^2*x^4)/a^2]*HypergeometricPFQ[{-3/8, 1/2}, {5/8}, (b^2*x 
^4)/a^2])/(3*(e*x)^(5/2)*Sqrt[a - b*x^2]*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.64, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {343, 344, 851, 889, 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.

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

\(\Big \downarrow \) 343

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

\(\Big \downarrow \) 344

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 889

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

(-2*Sqrt[a - b*x^2]*Sqrt[a + b*x^2])/(3*a^2*e*(e*x)^(3/2)) - (2*b^2*(e*x)^ 
(5/2)*Sqrt[1 - (b^2*x^4)/a^2]*Hypergeometric2F1[1/2, 5/8, 13/8, (b^2*x^4)/ 
a^2])/(15*a^2*e^5*Sqrt[a - b*x^2]*Sqrt[a + b*x^2])
 

Defintions of rubi rules used

rule 343

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(p_) 
, x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(p + 1)/ 
(a*c*e*(m + 1))), x] - Simp[b*d*((m + 4*p + 5)/(a*c*e^4*(m + 1)))   Int[(e* 
x)^(m + 4)*(a + b*x^2)^p*(c + d*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, 
 x] && EqQ[b*c + a*d, 0] && LtQ[m, -1]
 

rule 344

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

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])
 

rule 889

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(1/(e*x)^(5/2)/(-b*x^2+a)^(1/2)/(b*x^2+a)^(1/2),x, algorithm="fri 
cas")
 

Output:

integral(-sqrt(b*x^2 + a)*sqrt(-b*x^2 + a)*sqrt(e*x)/(b^2*e^3*x^7 - a^2*e^ 
3*x^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 39.84 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(e x)^{5/2} \sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\frac {i b^{\frac {3}{4}} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {9}{8}, \frac {13}{8}, 1 & \frac {11}{8}, \frac {11}{8}, \frac {15}{8} \\\frac {7}{8}, \frac {9}{8}, \frac {11}{8}, \frac {13}{8}, \frac {15}{8} & 0 \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}} a^{\frac {7}{4}} e^{\frac {5}{2}}} + \frac {b^{\frac {3}{4}} {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {3}{8}, \frac {5}{8}, \frac {7}{8}, \frac {9}{8}, \frac {11}{8}, 1 & \\\frac {5}{8}, \frac {9}{8} & \frac {3}{8}, \frac {7}{8}, \frac {7}{8}, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{- \frac {i \pi }{4}}}{8 \pi ^{\frac {3}{2}} a^{\frac {7}{4}} e^{\frac {5}{2}}} \] Input:

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

Maxima [F]

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

integrate(1/(e*x)^(5/2)/(-b*x^2+a)^(1/2)/(b*x^2+a)^(1/2),x, algorithm="max 
ima")
 

Output:

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

Giac [F]

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

integrate(1/(e*x)^(5/2)/(-b*x^2+a)^(1/2)/(b*x^2+a)^(1/2),x, algorithm="gia 
c")
                                                                                    
                                                                                    
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*int((sqrt(x)*sqrt(a + b*x**2)*sqrt(a - b*x**2))/(a**2*x**3 - b**2 
*x**7),x))/e**3