Integrand size = 21, antiderivative size = 119 \[ \int \left (1+b x^2\right ) \sqrt {-1+b^2 x^4} \, dx=\frac {1}{15} x \left (5+3 b x^2\right ) \sqrt {-1+b^2 x^4}-\frac {2 \sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{5 \sqrt {b} \sqrt {-1+b^2 x^4}}-\frac {4 \sqrt {1-b^2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{15 \sqrt {b} \sqrt {-1+b^2 x^4}} \] Output:
1/15*x*(3*b*x^2+5)*(b^2*x^4-1)^(1/2)-2/5*(-b^2*x^4+1)^(1/2)*EllipticE(b^(1 /2)*x,I)/b^(1/2)/(b^2*x^4-1)^(1/2)-4/15*(-b^2*x^4+1)^(1/2)*EllipticF(b^(1/ 2)*x,I)/b^(1/2)/(b^2*x^4-1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.95 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.62 \[ \int \left (1+b x^2\right ) \sqrt {-1+b^2 x^4} \, dx=\frac {\sqrt {-1+b^2 x^4} \left (3 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right )\right )}{3 \sqrt {1-b^2 x^4}} \] Input:
Integrate[(1 + b*x^2)*Sqrt[-1 + b^2*x^4],x]
Output:
(Sqrt[-1 + b^2*x^4]*(3*x*Hypergeometric2F1[-1/2, 1/4, 5/4, b^2*x^4] + b*x^ 3*Hypergeometric2F1[-1/2, 3/4, 7/4, b^2*x^4]))/(3*Sqrt[1 - b^2*x^4])
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.
\(\displaystyle \int \left (b x^2+1\right ) \sqrt {b^2 x^4-1} \, dx\) |
\(\Big \downarrow \) 1571 |
\(\displaystyle \int \left (b x^2+1\right ) \sqrt {b^2 x^4-1}dx\) |
Input:
Int[(1 + b*x^2)*Sqrt[-1 + b^2*x^4],x]
Output:
$Aborted
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> U nintegrable[(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, p, q}, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.99 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.74
method | result | size |
meijerg | \(\frac {\sqrt {\operatorname {signum}\left (b^{2} x^{4}-1\right )}\, x \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )}{\sqrt {-\operatorname {signum}\left (b^{2} x^{4}-1\right )}}+\frac {b \sqrt {\operatorname {signum}\left (b^{2} x^{4}-1\right )}\, x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3 \sqrt {-\operatorname {signum}\left (b^{2} x^{4}-1\right )}}\) | \(88\) |
risch | \(\frac {x \left (3 b \,x^{2}+5\right ) \sqrt {b^{2} x^{4}-1}}{15}-\frac {2 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{3 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {2 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{5 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(131\) |
elliptic | \(\frac {b \,x^{3} \sqrt {b^{2} x^{4}-1}}{5}+\frac {x \sqrt {b^{2} x^{4}-1}}{3}-\frac {2 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{3 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {2 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{5 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(140\) |
default | \(\frac {x \sqrt {b^{2} x^{4}-1}}{3}-\frac {2 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{3 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+b \left (\frac {x^{3} \sqrt {b^{2} x^{4}-1}}{5}-\frac {2 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{5 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}\, b}\right )\) | \(145\) |
Input:
int((b*x^2+1)*(b^2*x^4-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
signum(b^2*x^4-1)^(1/2)/(-signum(b^2*x^4-1))^(1/2)*x*hypergeom([-1/2,1/4], [5/4],b^2*x^4)+1/3*b*signum(b^2*x^4-1)^(1/2)/(-signum(b^2*x^4-1))^(1/2)*x^ 3*hypergeom([-1/2,3/4],[7/4],b^2*x^4)
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.62 \[ \int \left (1+b x^2\right ) \sqrt {-1+b^2 x^4} \, dx=\frac {\frac {2 \, {\left (5 \, b + 3\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + {\left (3 \, b^{2} x^{4} + 5 \, b x^{2} - 6\right )} \sqrt {b^{2} x^{4} - 1} - \frac {6 \, x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}}}{15 \, b x} \] Input:
integrate((b*x^2+1)*(b^2*x^4-1)^(1/2),x, algorithm="fricas")
Output:
1/15*(2*(5*b + 3)*x*elliptic_f(arcsin(1/(sqrt(b)*x)), -1)/sqrt(b) + (3*b^2 *x^4 + 5*b*x^2 - 6)*sqrt(b^2*x^4 - 1) - 6*x*elliptic_e(arcsin(1/(sqrt(b)*x )), -1)/sqrt(b))/(b*x)
Time = 0.96 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.53 \[ \int \left (1+b x^2\right ) \sqrt {-1+b^2 x^4} \, dx=\frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**2+1)*(b**2*x**4-1)**(1/2),x)
Output:
I*b*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b**2*x**4)/(4*gamma(7/4)) + I*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b**2*x**4)/(4*gamma(5/4))
\[ \int \left (1+b x^2\right ) \sqrt {-1+b^2 x^4} \, dx=\int { \sqrt {b^{2} x^{4} - 1} {\left (b x^{2} + 1\right )} \,d x } \] Input:
integrate((b*x^2+1)*(b^2*x^4-1)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b^2*x^4 - 1)*(b*x^2 + 1), x)
\[ \int \left (1+b x^2\right ) \sqrt {-1+b^2 x^4} \, dx=\int { \sqrt {b^{2} x^{4} - 1} {\left (b x^{2} + 1\right )} \,d x } \] Input:
integrate((b*x^2+1)*(b^2*x^4-1)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(b^2*x^4 - 1)*(b*x^2 + 1), x)
Timed out. \[ \int \left (1+b x^2\right ) \sqrt {-1+b^2 x^4} \, dx=\int \sqrt {b^2\,x^4-1}\,\left (b\,x^2+1\right ) \,d x \] Input:
int((b^2*x^4 - 1)^(1/2)*(b*x^2 + 1),x)
Output:
int((b^2*x^4 - 1)^(1/2)*(b*x^2 + 1), x)
\[ \int \left (1+b x^2\right ) \sqrt {-1+b^2 x^4} \, dx=\frac {\sqrt {b^{2} x^{4}-1}\, b \,x^{3}}{5}+\frac {\sqrt {b^{2} x^{4}-1}\, x}{3}-\frac {2 \left (\int \frac {\sqrt {b^{2} x^{4}-1}}{b^{2} x^{4}-1}d x \right )}{3}-\frac {2 \left (\int \frac {\sqrt {b^{2} x^{4}-1}\, x^{2}}{b^{2} x^{4}-1}d x \right ) b}{5} \] Input:
int((b*x^2+1)*(b^2*x^4-1)^(1/2),x)
Output:
(3*sqrt(b**2*x**4 - 1)*b*x**3 + 5*sqrt(b**2*x**4 - 1)*x - 10*int(sqrt(b**2 *x**4 - 1)/(b**2*x**4 - 1),x) - 6*int((sqrt(b**2*x**4 - 1)*x**2)/(b**2*x** 4 - 1),x)*b)/15