Integrand size = 21, antiderivative size = 116 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{3/2}} \, dx=-\frac {x \left (1+b x^2\right )}{2 \sqrt {-1+b^2 x^4}}+\frac {\sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{2 \sqrt {b} \sqrt {-1+b^2 x^4}}-\frac {\sqrt {1-b^2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}} \] Output:
-1/2*x*(b*x^2+1)/(b^2*x^4-1)^(1/2)+1/2*(-b^2*x^4+1)^(1/2)*EllipticE(b^(1/2 )*x,I)/b^(1/2)/(b^2*x^4-1)^(1/2)-(-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 = 10.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{3/2}} \, dx=-\frac {x \left (3+3 \sqrt {1-b^2 x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )+2 b x^2 \sqrt {1-b^2 x^4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},b^2 x^4\right )\right )}{6 \sqrt {-1+b^2 x^4}} \] Input:
Integrate[(1 + b*x^2)/(-1 + b^2*x^4)^(3/2),x]
Output:
-1/6*(x*(3 + 3*Sqrt[1 - b^2*x^4]*Hypergeometric2F1[1/4, 1/2, 5/4, b^2*x^4] + 2*b*x^2*Sqrt[1 - b^2*x^4]*Hypergeometric2F1[3/4, 3/2, 7/4, b^2*x^4]))/S qrt[-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.
\(\displaystyle \int \frac {b x^2+1}{\left (b^2 x^4-1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1571 |
\(\displaystyle \int \frac {b x^2+1}{\left (b^2 x^4-1\right )^{3/2}}dx\) |
Input:
Int[(1 + b*x^2)/(-1 + b^2*x^4)^(3/2),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 = 1.00 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.76
method | result | size |
meijerg | \(\frac {{\left (-\operatorname {signum}\left (b^{2} x^{4}-1\right )\right )}^{\frac {3}{2}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )}{\operatorname {signum}\left (b^{2} x^{4}-1\right )^{\frac {3}{2}}}+\frac {b {\left (-\operatorname {signum}\left (b^{2} x^{4}-1\right )\right )}^{\frac {3}{2}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{2}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3 \operatorname {signum}\left (b^{2} x^{4}-1\right )^{\frac {3}{2}}}\) | \(88\) |
elliptic | \(-\frac {\left (b^{2} x^{2}+b \right ) x}{2 b \sqrt {\left (x^{2}-\frac {1}{b}\right ) \left (b^{2} x^{2}+b \right )}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{2 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{2 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(145\) |
default | \(-\frac {x}{2 \sqrt {\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{2 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+b \left (-\frac {x^{3}}{2 \sqrt {\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{2 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}\, b}\right )\) | \(153\) |
Input:
int((b*x^2+1)/(b^2*x^4-1)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/signum(b^2*x^4-1)^(3/2)*(-signum(b^2*x^4-1))^(3/2)*x*hypergeom([1/4,3/2] ,[5/4],b^2*x^4)+1/3*b/signum(b^2*x^4-1)^(3/2)*(-signum(b^2*x^4-1))^(3/2)*x ^3*hypergeom([3/4,3/2],[7/4],b^2*x^4)
Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{3/2}} \, dx=-\frac {\sqrt {b^{2} x^{4} - 1} b x - {\left (-i \, b^{2} x^{2} + i \, b\right )} \sqrt {b} E(\arcsin \left (\sqrt {b} x\right )\,|\,-1) - {\left (i \, {\left (b^{2} + b\right )} x^{2} - i \, b - i\right )} \sqrt {b} F(\arcsin \left (\sqrt {b} x\right )\,|\,-1)}{2 \, {\left (b^{2} x^{2} - b\right )}} \] Input:
integrate((b*x^2+1)/(b^2*x^4-1)^(3/2),x, algorithm="fricas")
Output:
-1/2*(sqrt(b^2*x^4 - 1)*b*x - (-I*b^2*x^2 + I*b)*sqrt(b)*elliptic_e(arcsin (sqrt(b)*x), -1) - (I*(b^2 + b)*x^2 - I*b - I)*sqrt(b)*elliptic_f(arcsin(s qrt(b)*x), -1))/(b^2*x^2 - b)
Time = 1.92 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.52 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{3/2}} \, dx=\frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**2+1)/(b**2*x**4-1)**(3/2),x)
Output:
I*b*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), b**2*x**4)/(4*gamma(7/4)) + I*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), b**2*x**4)/(4*gamma(5/4))
\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{3/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (b^{2} x^{4} - 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((b*x^2+1)/(b^2*x^4-1)^(3/2),x, algorithm="maxima")
Output:
integrate((b*x^2 + 1)/(b^2*x^4 - 1)^(3/2), x)
\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{3/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (b^{2} x^{4} - 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((b*x^2+1)/(b^2*x^4-1)^(3/2),x, algorithm="giac")
Output:
integrate((b*x^2 + 1)/(b^2*x^4 - 1)^(3/2), x)
Timed out. \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{3/2}} \, dx=\int \frac {b\,x^2+1}{{\left (b^2\,x^4-1\right )}^{3/2}} \,d x \] Input:
int((b*x^2 + 1)/(b^2*x^4 - 1)^(3/2),x)
Output:
int((b*x^2 + 1)/(b^2*x^4 - 1)^(3/2), x)
\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {b^{2} x^{4}-1}}{b^{3} x^{6}-b^{2} x^{4}-b \,x^{2}+1}d x \] Input:
int((b*x^2+1)/(b^2*x^4-1)^(3/2),x)
Output:
int(sqrt(b**2*x**4 - 1)/(b**3*x**6 - b**2*x**4 - b*x**2 + 1),x)