Integrand size = 21, antiderivative size = 171 \[ \int \left (1+b x^2\right ) \left (-1+b^2 x^4\right )^{5/2} \, dx=\frac {4 x \left (195+77 b x^2\right ) \sqrt {-1+b^2 x^4}}{3003}-\frac {10 x \left (117+77 b x^2\right ) \left (-1+b^2 x^4\right )^{3/2}}{9009}+\frac {1}{143} x \left (13+11 b x^2\right ) \left (-1+b^2 x^4\right )^{5/2}-\frac {8 \sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{39 \sqrt {b} \sqrt {-1+b^2 x^4}}-\frac {944 \sqrt {1-b^2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{3003 \sqrt {b} \sqrt {-1+b^2 x^4}} \] Output:
4/3003*x*(77*b*x^2+195)*(b^2*x^4-1)^(1/2)-10/9009*x*(77*b*x^2+117)*(b^2*x^ 4-1)^(3/2)+1/143*x*(11*b*x^2+13)*(b^2*x^4-1)^(5/2)-8/39*(-b^2*x^4+1)^(1/2) *EllipticE(b^(1/2)*x,I)/b^(1/2)/(b^2*x^4-1)^(1/2)-944/3003*(-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 = 9.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.43 \[ \int \left (1+b x^2\right ) \left (-1+b^2 x^4\right )^{5/2} \, dx=\frac {\sqrt {-1+b^2 x^4} \left (3 x \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{4},\frac {5}{4},b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right )\right )}{3 \sqrt {1-b^2 x^4}} \] Input:
Integrate[(1 + b*x^2)*(-1 + b^2*x^4)^(5/2),x]
Output:
(Sqrt[-1 + b^2*x^4]*(3*x*Hypergeometric2F1[-5/2, 1/4, 5/4, b^2*x^4] + b*x^ 3*Hypergeometric2F1[-5/2, 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.
| \(\displaystyle \int \left (b x^2+1\right ) \left (b^2 x^4-1\right )^{5/2} \, dx\) |
| \(\Big \downarrow \) 1571 |
| \(\displaystyle \int \left (b x^2+1\right ) \left (b^2 x^4-1\right )^{5/2}dx\) |
Input:
Int[(1 + b*x^2)*(-1 + b^2*x^4)^(5/2),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 = 2.00 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.51
| method | result | size |
| meijerg | \(\frac {\operatorname {signum}\left (b^{2} x^{4}-1\right )^{\frac {5}{2}} x \operatorname {hypergeom}\left (\left [-\frac {5}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )}{{\left (-\operatorname {signum}\left (b^{2} x^{4}-1\right )\right )}^{\frac {5}{2}}}+\frac {b \operatorname {signum}\left (b^{2} x^{4}-1\right )^{\frac {5}{2}} x^{3} \operatorname {hypergeom}\left (\left [-\frac {5}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3 {\left (-\operatorname {signum}\left (b^{2} x^{4}-1\right )\right )}^{\frac {5}{2}}}\) | \(88\) |
| risch | \(\frac {x \left (693 x^{10} b^{5}+819 b^{4} x^{8}-2156 b^{3} x^{6}-2808 b^{2} x^{4}+2387 b \,x^{2}+4329\right ) \sqrt {b^{2} x^{4}-1}}{9009}-\frac {40 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{77 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {8 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{39 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(163\) |
| elliptic | \(\frac {b^{5} x^{11} \sqrt {b^{2} x^{4}-1}}{13}+\frac {b^{4} x^{9} \sqrt {b^{2} x^{4}-1}}{11}-\frac {28 b^{3} x^{7} \sqrt {b^{2} x^{4}-1}}{117}-\frac {24 b^{2} x^{5} \sqrt {b^{2} x^{4}-1}}{77}+\frac {31 b \,x^{3} \sqrt {b^{2} x^{4}-1}}{117}+\frac {37 x \sqrt {b^{2} x^{4}-1}}{77}-\frac {40 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{77 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {8 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{39 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(216\) |
| default | \(\frac {b^{4} x^{9} \sqrt {b^{2} x^{4}-1}}{11}-\frac {24 b^{2} x^{5} \sqrt {b^{2} x^{4}-1}}{77}+\frac {37 x \sqrt {b^{2} x^{4}-1}}{77}-\frac {40 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{77 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+b \left (\frac {b^{4} x^{11} \sqrt {b^{2} x^{4}-1}}{13}-\frac {28 b^{2} x^{7} \sqrt {b^{2} x^{4}-1}}{117}+\frac {31 x^{3} \sqrt {b^{2} x^{4}-1}}{117}-\frac {8 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{39 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}\, b}\right )\) | \(221\) |
Input:
int((b*x^2+1)*(b^2*x^4-1)^(5/2),x,method=_RETURNVERBOSE)
Output:
signum(b^2*x^4-1)^(5/2)/(-signum(b^2*x^4-1))^(5/2)*x*hypergeom([-5/2,1/4], [5/4],b^2*x^4)+1/3*b*signum(b^2*x^4-1)^(5/2)/(-signum(b^2*x^4-1))^(5/2)*x^ 3*hypergeom([-5/2,3/4],[7/4],b^2*x^4)
Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62 \[ \int \left (1+b x^2\right ) \left (-1+b^2 x^4\right )^{5/2} \, dx=\frac {\frac {24 \, {\left (195 \, b + 77\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + {\left (693 \, b^{6} x^{12} + 819 \, b^{5} x^{10} - 2156 \, b^{4} x^{8} - 2808 \, b^{3} x^{6} + 2387 \, b^{2} x^{4} + 4329 \, b x^{2} - 1848\right )} \sqrt {b^{2} x^{4} - 1} - \frac {1848 \, x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}}}{9009 \, b x} \] Input:
integrate((b*x^2+1)*(b^2*x^4-1)^(5/2),x, algorithm="fricas")
Output:
1/9009*(24*(195*b + 77)*x*elliptic_f(arcsin(1/(sqrt(b)*x)), -1)/sqrt(b) + (693*b^6*x^12 + 819*b^5*x^10 - 2156*b^4*x^8 - 2808*b^3*x^6 + 2387*b^2*x^4 + 4329*b*x^2 - 1848)*sqrt(b^2*x^4 - 1) - 1848*x*elliptic_e(arcsin(1/(sqrt( b)*x)), -1)/sqrt(b))/(b*x)
Time = 2.47 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.20 \[ \int \left (1+b x^2\right ) \left (-1+b^2 x^4\right )^{5/2} \, dx=\frac {i b^{5} x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} + \frac {i b^{4} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} - \frac {i b^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} - \frac {i b^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**2+1)*(b**2*x**4-1)**(5/2),x)
Output:
I*b**5*x**11*gamma(11/4)*hyper((-1/2, 11/4), (15/4,), b**2*x**4)/(4*gamma( 15/4)) + I*b**4*x**9*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), b**2*x**4)/(4* gamma(13/4)) - I*b**3*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b**2*x** 4)/(2*gamma(11/4)) - I*b**2*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b** 2*x**4)/(2*gamma(9/4)) + 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))
\[ \int \left (1+b x^2\right ) \left (-1+b^2 x^4\right )^{5/2} \, dx=\int { {\left (b^{2} x^{4} - 1\right )}^{\frac {5}{2}} {\left (b x^{2} + 1\right )} \,d x } \] Input:
integrate((b*x^2+1)*(b^2*x^4-1)^(5/2),x, algorithm="maxima")
Output:
integrate((b^2*x^4 - 1)^(5/2)*(b*x^2 + 1), x)
\[ \int \left (1+b x^2\right ) \left (-1+b^2 x^4\right )^{5/2} \, dx=\int { {\left (b^{2} x^{4} - 1\right )}^{\frac {5}{2}} {\left (b x^{2} + 1\right )} \,d x } \] Input:
integrate((b*x^2+1)*(b^2*x^4-1)^(5/2),x, algorithm="giac")
Output:
integrate((b^2*x^4 - 1)^(5/2)*(b*x^2 + 1), x)
Timed out. \[ \int \left (1+b x^2\right ) \left (-1+b^2 x^4\right )^{5/2} \, dx=\int {\left (b^2\,x^4-1\right )}^{5/2}\,\left (b\,x^2+1\right ) \,d x \] Input:
int((b^2*x^4 - 1)^(5/2)*(b*x^2 + 1),x)
Output:
int((b^2*x^4 - 1)^(5/2)*(b*x^2 + 1), x)
\[ \int \left (1+b x^2\right ) \left (-1+b^2 x^4\right )^{5/2} \, dx=\frac {\sqrt {b^{2} x^{4}-1}\, b^{5} x^{11}}{13}+\frac {\sqrt {b^{2} x^{4}-1}\, b^{4} x^{9}}{11}-\frac {28 \sqrt {b^{2} x^{4}-1}\, b^{3} x^{7}}{117}-\frac {24 \sqrt {b^{2} x^{4}-1}\, b^{2} x^{5}}{77}+\frac {31 \sqrt {b^{2} x^{4}-1}\, b \,x^{3}}{117}+\frac {37 \sqrt {b^{2} x^{4}-1}\, x}{77}-\frac {40 \left (\int \frac {\sqrt {b^{2} x^{4}-1}}{b^{2} x^{4}-1}d x \right )}{77}-\frac {8 \left (\int \frac {\sqrt {b^{2} x^{4}-1}\, x^{2}}{b^{2} x^{4}-1}d x \right ) b}{39} \] Input:
int((b*x^2+1)*(b^2*x^4-1)^(5/2),x)
Output:
(693*sqrt(b**2*x**4 - 1)*b**5*x**11 + 819*sqrt(b**2*x**4 - 1)*b**4*x**9 - 2156*sqrt(b**2*x**4 - 1)*b**3*x**7 - 2808*sqrt(b**2*x**4 - 1)*b**2*x**5 + 2387*sqrt(b**2*x**4 - 1)*b*x**3 + 4329*sqrt(b**2*x**4 - 1)*x - 4680*int(sq rt(b**2*x**4 - 1)/(b**2*x**4 - 1),x) - 1848*int((sqrt(b**2*x**4 - 1)*x**2) /(b**2*x**4 - 1),x)*b)/9009