\(\int \frac {(e x)^{5/2}}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx\) [1380]

Optimal result

Integrand size = 31, antiderivative size = 74 \[ \int \frac {(e x)^{5/2}}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\frac {2 (e x)^{7/2} \sqrt {1-\frac {b^2 x^4}{a^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},\frac {b^2 x^4}{a^2}\right )}{7 e \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Output:

2/7*(e*x)^(7/2)*(1-b^2*x^4/a^2)^(1/2)*hypergeom([1/2, 7/8],[15/8],b^2*x^4/ 
a^2)/e/(-b*x^2+a)^(1/2)/(b*x^2+a)^(1/2)
 

Mathematica [A] (verified)

Time = 2.74 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {(e x)^{5/2}}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\frac {2 x (e x)^{5/2} \sqrt {1-\frac {b^2 x^4}{a^2}} \, _2F_1\left (\frac {1}{2},\frac {7}{8};\frac {15}{8};\frac {b^2 x^4}{a^2}\right )}{7 \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Input:

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

Output:

(2*x*(e*x)^(5/2)*Sqrt[1 - (b^2*x^4)/a^2]*HypergeometricPFQ[{1/2, 7/8}, {15 
/8}, (b^2*x^4)/a^2])/(7*Sqrt[a - b*x^2]*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {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.

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

\(\Big \downarrow \) 344

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 889

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

\(\Big \downarrow \) 888

\(\displaystyle \frac {2 (e x)^{7/2} \sqrt {1-\frac {b^2 x^4}{a^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},\frac {b^2 x^4}{a^2}\right )}{7 e \sqrt {a-b x^2} \sqrt {a+b x^2}}\)

Input:

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

Output:

(2*(e*x)^(7/2)*Sqrt[1 - (b^2*x^4)/a^2]*Hypergeometric2F1[1/2, 7/8, 15/8, ( 
b^2*x^4)/a^2])/(7*e*Sqrt[a - b*x^2]*Sqrt[a + b*x^2])
 

Defintions of rubi rules used

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-sqrt(b*x^2 + a)*sqrt(-b*x^2 + a)*sqrt(e*x)*e^2*x^2/(b^2*x^4 - a^ 
2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 58.50 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.88 \[ \int \frac {(e x)^{5/2}}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\frac {i a^{\frac {3}{4}} e^{\frac {5}{2}} {G_{6, 6}^{5, 3}\left (\begin {matrix} - \frac {1}{8}, \frac {3}{8}, 1 & \frac {1}{8}, \frac {1}{8}, \frac {5}{8} \\- \frac {3}{8}, - \frac {1}{8}, \frac {1}{8}, \frac {3}{8}, \frac {5}{8} & 0 \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}} b^{\frac {7}{4}}} + \frac {a^{\frac {3}{4}} e^{\frac {5}{2}} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {7}{8}, - \frac {5}{8}, - \frac {3}{8}, - \frac {1}{8}, \frac {1}{8}, 1 & \\- \frac {5}{8}, - \frac {1}{8} & - \frac {7}{8}, - \frac {3}{8}, - \frac {3}{8}, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{- \frac {3 i \pi }{4}}}{8 \pi ^{\frac {3}{2}} b^{\frac {7}{4}}} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((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)*x**2)/(a**2 - b**2* 
x**4),x)*e**2