\(\int \frac {1+b x^2}{(-1+b^2 x^4)^{5/2}} \, dx\) [271]

Optimal result

Integrand size = 21, antiderivative size = 144 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{5/2}} \, dx=-\frac {x \left (1+b x^2\right )}{6 \left (-1+b^2 x^4\right )^{3/2}}+\frac {x \left (5+3 b x^2\right )}{12 \sqrt {-1+b^2 x^4}}-\frac {\sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{4 \sqrt {b} \sqrt {-1+b^2 x^4}}+\frac {2 \sqrt {1-b^2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{3 \sqrt {b} \sqrt {-1+b^2 x^4}} \] Output:

-1/6*x*(b*x^2+1)/(b^2*x^4-1)^(3/2)+1/12*x*(3*b*x^2+5)/(b^2*x^4-1)^(1/2)-1/ 
4*(-b^2*x^4+1)^(1/2)*EllipticE(b^(1/2)*x,I)/b^(1/2)/(b^2*x^4-1)^(1/2)+2/3* 
(-b^2*x^4+1)^(1/2)*EllipticF(b^(1/2)*x,I)/b^(1/2)/(b^2*x^4-1)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.68 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{5/2}} \, dx=\frac {x \left (-7+5 b^2 x^4-5 \left (1-b^2 x^4\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )-4 b x^2 \left (1-b^2 x^4\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},b^2 x^4\right )\right )}{12 \left (-1+b^2 x^4\right )^{3/2}} \] Input:

Integrate[(1 + b*x^2)/(-1 + b^2*x^4)^(5/2),x]
 

Output:

(x*(-7 + 5*b^2*x^4 - 5*(1 - b^2*x^4)^(3/2)*Hypergeometric2F1[1/4, 1/2, 5/4 
, b^2*x^4] - 4*b*x^2*(1 - b^2*x^4)^(3/2)*Hypergeometric2F1[3/4, 5/2, 7/4, 
b^2*x^4]))/(12*(-1 + b^2*x^4)^(3/2))
 

Rubi [F]

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.

\(\displaystyle \int \frac {b x^2+1}{\left (b^2 x^4-1\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1571

\(\displaystyle \int \frac {b x^2+1}{\left (b^2 x^4-1\right )^{5/2}}dx\)

Input:

Int[(1 + b*x^2)/(-1 + b^2*x^4)^(5/2),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 1571

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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.61

method	result	size
meijerg	\(\frac {{\left (-\operatorname {signum}\left (b^{2} x^{4}-1\right )\right )}^{\frac {5}{2}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {5}{2}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )}{\operatorname {signum}\left (b^{2} x^{4}-1\right )^{\frac {5}{2}}}+\frac {b {\left (-\operatorname {signum}\left (b^{2} x^{4}-1\right )\right )}^{\frac {5}{2}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {5}{2}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3 \operatorname {signum}\left (b^{2} x^{4}-1\right )^{\frac {5}{2}}}\)	\(88\)
default	\(-\frac {x \sqrt {b^{2} x^{4}-1}}{6 b^{4} \left (x^{4}-\frac {1}{b^{2}}\right )^{2}}+\frac {5 x}{12 \sqrt {\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {5 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{12 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+b \left (-\frac {x^{3} \sqrt {b^{2} x^{4}-1}}{6 b^{4} \left (x^{4}-\frac {1}{b^{2}}\right )^{2}}+\frac {x^{3}}{4 \sqrt {\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{4 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}\, b}\right )\)	\(211\)
elliptic	\(-\frac {x \sqrt {b^{2} x^{4}-1}}{12 b^{2} \left (x^{2}-\frac {1}{b}\right )^{2}}+\frac {3 \left (b^{2} x^{2}+b \right ) x}{8 b \sqrt {\left (x^{2}-\frac {1}{b}\right ) \left (b^{2} x^{2}+b \right )}}-\frac {\left (b^{2} x^{2}-b \right ) x}{8 b \sqrt {\left (x^{2}+\frac {1}{b}\right ) \left (b^{2} x^{2}-b \right )}}+\frac {5 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{12 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{4 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\)	\(211\)

Input:

int((b*x^2+1)/(b^2*x^4-1)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/signum(b^2*x^4-1)^(5/2)*(-signum(b^2*x^4-1))^(5/2)*x*hypergeom([1/4,5/2] 
,[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([3/4,5/2],[7/4],b^2*x^4)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.17 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{5/2}} \, dx=-\frac {3 \, {\left (-i \, b^{4} x^{6} + i \, b^{3} x^{4} + i \, b^{2} x^{2} - i \, b\right )} \sqrt {b} E(\arcsin \left (\sqrt {b} x\right )\,|\,-1) - {\left (-i \, {\left (3 \, b^{4} + 5 \, b^{3}\right )} x^{6} + i \, {\left (3 \, b^{3} + 5 \, b^{2}\right )} x^{4} + i \, {\left (3 \, b^{2} + 5 \, b\right )} x^{2} - 3 i \, b - 5 i\right )} \sqrt {b} F(\arcsin \left (\sqrt {b} x\right )\,|\,-1) - {\left (3 \, b^{3} x^{5} + 2 \, b^{2} x^{3} - 7 \, b x\right )} \sqrt {b^{2} x^{4} - 1}}{12 \, {\left (b^{4} x^{6} - b^{3} x^{4} - b^{2} x^{2} + b\right )}} \] Input:

integrate((b*x^2+1)/(b^2*x^4-1)^(5/2),x, algorithm="fricas")
 

Output:

-1/12*(3*(-I*b^4*x^6 + I*b^3*x^4 + I*b^2*x^2 - I*b)*sqrt(b)*elliptic_e(arc 
sin(sqrt(b)*x), -1) - (-I*(3*b^4 + 5*b^3)*x^6 + I*(3*b^3 + 5*b^2)*x^4 + I* 
(3*b^2 + 5*b)*x^2 - 3*I*b - 5*I)*sqrt(b)*elliptic_f(arcsin(sqrt(b)*x), -1) 
 - (3*b^3*x^5 + 2*b^2*x^3 - 7*b*x)*sqrt(b^2*x^4 - 1))/(b^4*x^6 - b^3*x^4 - 
 b^2*x^2 + b)
 

Sympy [A] (verification not implemented)

Time = 5.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.42 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{5/2}} \, dx=- \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((b*x**2+1)/(b**2*x**4-1)**(5/2),x)
 

Output:

-I*b*x**3*gamma(3/4)*hyper((3/4, 5/2), (7/4,), b**2*x**4)/(4*gamma(7/4)) - 
 I*x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), b**2*x**4)/(4*gamma(5/4))
 

Maxima [F]

\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{5/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (b^{2} x^{4} - 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*x^2+1)/(b^2*x^4-1)^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + 1)/(b^2*x^4 - 1)^(5/2), x)
 

Giac [F]

\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{5/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (b^{2} x^{4} - 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*x^2+1)/(b^2*x^4-1)^(5/2),x, algorithm="giac")
 

Output:

integrate((b*x^2 + 1)/(b^2*x^4 - 1)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{5/2}} \, dx=\int \frac {b\,x^2+1}{{\left (b^2\,x^4-1\right )}^{5/2}} \,d x \] Input:

int((b*x^2 + 1)/(b^2*x^4 - 1)^(5/2),x)
 

Output:

int((b*x^2 + 1)/(b^2*x^4 - 1)^(5/2), x)
 

Reduce [F]

\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{5/2}} \, dx=\int \frac {\sqrt {b^{2} x^{4}-1}}{b^{5} x^{10}-b^{4} x^{8}-2 b^{3} x^{6}+2 b^{2} x^{4}+b \,x^{2}-1}d x \] Input:

int((b*x^2+1)/(b^2*x^4-1)^(5/2),x)
 

Output:

int(sqrt(b**2*x**4 - 1)/(b**5*x**10 - b**4*x**8 - 2*b**3*x**6 + 2*b**2*x** 
4 + b*x**2 - 1),x)