\(\int \frac {(e x)^m}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx\) [1396]

Optimal result

Integrand size = 29, antiderivative size = 84 \[ \int \frac {(e x)^m}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\frac {(e x)^{1+m} \sqrt {1-\frac {b^2 x^4}{a^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{4},\frac {5+m}{4},\frac {b^2 x^4}{a^2}\right )}{e (1+m) \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {(e x)^m}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\frac {x (e x)^m \sqrt {1-\frac {b^2 x^4}{a^2}} \, _2F_1\left (\frac {1}{2},\frac {1}{4}+\frac {m}{4};\frac {5}{4}+\frac {m}{4};\frac {b^2 x^4}{a^2}\right )}{(1+m) \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {344, 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.

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

\(\Big \downarrow \) 344

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

\(\Big \downarrow \) 889

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.55 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.37 \[ \int \frac {(e x)^m}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\frac {i a^{\frac {m}{2} - \frac {1}{2}} b^{- \frac {m}{2} - \frac {1}{2}} e^{m} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {1}{2} - \frac {m}{4}, 1 - \frac {m}{4}, 1 & \frac {3}{4} - \frac {m}{4}, \frac {3}{4} - \frac {m}{4}, \frac {5}{4} - \frac {m}{4} \\\frac {1}{4} - \frac {m}{4}, \frac {1}{2} - \frac {m}{4}, \frac {3}{4} - \frac {m}{4}, 1 - \frac {m}{4}, \frac {5}{4} - \frac {m}{4} & 0 \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}}} + \frac {i a^{\frac {m}{2} - \frac {1}{2}} b^{- \frac {m}{2} - \frac {1}{2}} e^{m} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {m}{4} - \frac {1}{4}, - \frac {m}{4}, \frac {1}{4} - \frac {m}{4}, \frac {1}{2} - \frac {m}{4}, \frac {3}{4} - \frac {m}{4}, 1 & \\- \frac {m}{4}, \frac {1}{2} - \frac {m}{4} & - \frac {m}{4} - \frac {1}{4}, \frac {1}{4} - \frac {m}{4}, \frac {1}{4} - \frac {m}{4}, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{- \frac {i \pi m}{2}}}{8 \pi ^{\frac {3}{2}}} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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