Integrand size = 29, antiderivative size = 41 \[ \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{4},\frac {5+m}{4},b^2 x^4\right )}{e (1+m)} \] Output:
(e*x)^(1+m)*hypergeom([1/2, 1/4+1/4*m],[5/4+1/4*m],b^2*x^4)/e/(1+m)
Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {x (e x)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{4},1+\frac {1+m}{4},b^2 x^4\right )}{1+m} \] Input:
Integrate[(e*x)^m/(Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
Output:
(x*(e*x)^m*Hypergeometric2F1[1/2, (1 + m)/4, 1 + (1 + m)/4, b^2*x^4])/(1 + m)
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {335, 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.
\(\displaystyle \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {b x^2+1}} \, dx\) |
\(\Big \downarrow \) 335 |
\(\displaystyle \int \frac {(e x)^m}{\sqrt {1-b^2 x^4}}dx\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{4},\frac {m+5}{4},b^2 x^4\right )}{e (m+1)}\) |
Input:
Int[(e*x)^m/(Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
Output:
((e*x)^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/4, (5 + m)/4, b^2*x^4])/(e*( 1 + m))
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] ))
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])
\[\int \frac {\left (e x \right )^{m}}{\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}}d x\]
Input:
int((e*x)^m/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
Output:
int((e*x)^m/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
\[ \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {\left (e x\right )^{m}}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}} \,d x } \] Input:
integrate((e*x)^m/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="fricas")
Output:
integral(-sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*(e*x)^m/(b^2*x^4 - 1), x)
Result contains complex when optimal does not.
Time = 2.37 (sec) , antiderivative size = 177, normalized size of antiderivative = 4.32 \[ \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {i b^{- \frac {m}{2} - \frac {1}{2}} e^{m} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {1}{2} - \frac {m}{4}, 1 - \frac {m}{4}, 1 & \frac {3}{4} - \frac {m}{4}, \frac {3}{4} - \frac {m}{4}, \frac {5}{4} - \frac {m}{4} \\\frac {1}{4} - \frac {m}{4}, \frac {1}{2} - \frac {m}{4}, \frac {3}{4} - \frac {m}{4}, 1 - \frac {m}{4}, \frac {5}{4} - \frac {m}{4} & 0 \end {matrix} \middle | {\frac {1}{b^{2} x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}}} + \frac {i b^{- \frac {m}{2} - \frac {1}{2}} e^{m} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {m}{4} - \frac {1}{4}, - \frac {m}{4}, \frac {1}{4} - \frac {m}{4}, \frac {1}{2} - \frac {m}{4}, \frac {3}{4} - \frac {m}{4}, 1 & \\- \frac {m}{4}, \frac {1}{2} - \frac {m}{4} & - \frac {m}{4} - \frac {1}{4}, \frac {1}{4} - \frac {m}{4}, \frac {1}{4} - \frac {m}{4}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{- \frac {i \pi m}{2}}}{8 \pi ^{\frac {3}{2}}} \] Input:
integrate((e*x)**m/(-b*x**2+1)**(1/2)/(b*x**2+1)**(1/2),x)
Output:
I*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,)), 1/(b**2*x**4))/(8*pi**(3/2)) + I*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)), exp_polar(-2*I*pi)/(b**2*x**4) )*exp(-I*pi*m/2)/(8*pi**(3/2))
\[ \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {\left (e x\right )^{m}}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}} \,d x } \] Input:
integrate((e*x)^m/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="maxima")
Output:
integrate((e*x)^m/(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)), x)
\[ \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {\left (e x\right )^{m}}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}} \,d x } \] Input:
integrate((e*x)^m/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="giac")
Output:
integrate((e*x)^m/(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)), x)
Timed out. \[ \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int \frac {{\left (e\,x\right )}^m}{\sqrt {1-b\,x^2}\,\sqrt {b\,x^2+1}} \,d x \] Input:
int((e*x)^m/((1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2)),x)
Output:
int((e*x)^m/((1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2)), x)
\[ \int \frac {(e x)^m}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=-e^{m} \left (\int \frac {x^{m} \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} x^{4}-1}d x \right ) \] Input:
int((e*x)^m/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
Output:
- e**m*int((x**m*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1))/(b**2*x**4 - 1),x)