\(\int \frac {1+b x^2}{(-1+b^2 x^4)^{7/2}} \, dx\) [272]

Optimal result

Integrand size = 21, antiderivative size = 170 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{7/2}} \, dx=-\frac {x \left (1+b x^2\right )}{10 \left (-1+b^2 x^4\right )^{5/2}}+\frac {x \left (9+7 b x^2\right )}{60 \left (-1+b^2 x^4\right )^{3/2}}-\frac {x \left (15+7 b x^2\right )}{40 \sqrt {-1+b^2 x^4}}+\frac {7 \sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{40 \sqrt {b} \sqrt {-1+b^2 x^4}}-\frac {11 \sqrt {1-b^2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{20 \sqrt {b} \sqrt {-1+b^2 x^4}} \] Output:

-1/10*x*(b*x^2+1)/(b^2*x^4-1)^(5/2)+1/60*x*(7*b*x^2+9)/(b^2*x^4-1)^(3/2)-1 
/40*x*(7*b*x^2+15)/(b^2*x^4-1)^(1/2)+7/40*(-b^2*x^4+1)^(1/2)*EllipticE(b^( 
1/2)*x,I)/b^(1/2)/(b^2*x^4-1)^(1/2)-11/20*(-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 = 10.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{7/2}} \, dx=-\frac {x \left (75-108 b^2 x^4+45 b^4 x^8+45 \left (1-b^2 x^4\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )+40 b x^2 \left (1-b^2 x^4\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},b^2 x^4\right )\right )}{120 \left (-1+b^2 x^4\right )^{5/2}} \] Input:

Integrate[(1 + b*x^2)/(-1 + b^2*x^4)^(7/2),x]
 

Output:

-1/120*(x*(75 - 108*b^2*x^4 + 45*b^4*x^8 + 45*(1 - b^2*x^4)^(5/2)*Hypergeo 
metric2F1[1/4, 1/2, 5/4, b^2*x^4] + 40*b*x^2*(1 - b^2*x^4)^(5/2)*Hypergeom 
etric2F1[3/4, 7/2, 7/4, b^2*x^4]))/(-1 + b^2*x^4)^(5/2)
 

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.

\(\displaystyle \int \frac {b x^2+1}{\left (b^2 x^4-1\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 1571

\(\displaystyle \int \frac {b x^2+1}{\left (b^2 x^4-1\right )^{7/2}}dx\)

Input:

Int[(1 + b*x^2)/(-1 + b^2*x^4)^(7/2),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 = 1.00 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.52

method	result	size
meijerg	\(\frac {{\left (-\operatorname {signum}\left (b^{2} x^{4}-1\right )\right )}^{\frac {7}{2}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {7}{2}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )}{\operatorname {signum}\left (b^{2} x^{4}-1\right )^{\frac {7}{2}}}+\frac {b {\left (-\operatorname {signum}\left (b^{2} x^{4}-1\right )\right )}^{\frac {7}{2}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {7}{2}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3 \operatorname {signum}\left (b^{2} x^{4}-1\right )^{\frac {7}{2}}}\)	\(88\)
elliptic	\(-\frac {x \sqrt {b^{2} x^{4}-1}}{40 b^{3} \left (x^{2}-\frac {1}{b}\right )^{3}}+\frac {11 x \sqrt {b^{2} x^{4}-1}}{120 b^{2} \left (x^{2}-\frac {1}{b}\right )^{2}}-\frac {53 \left (b^{2} x^{2}+b \right ) x}{160 b \sqrt {\left (x^{2}-\frac {1}{b}\right ) \left (b^{2} x^{2}+b \right )}}+\frac {x \sqrt {b^{2} x^{4}-1}}{48 b^{2} \left (x^{2}+\frac {1}{b}\right )^{2}}+\frac {5 \left (b^{2} x^{2}-b \right ) x}{32 b \sqrt {\left (x^{2}+\frac {1}{b}\right ) \left (b^{2} x^{2}-b \right )}}-\frac {3 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{8 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+\frac {7 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{40 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\)	\(265\)
default	\(-\frac {x \sqrt {b^{2} x^{4}-1}}{10 b^{6} \left (x^{4}-\frac {1}{b^{2}}\right )^{3}}+\frac {3 x \sqrt {b^{2} x^{4}-1}}{20 b^{4} \left (x^{4}-\frac {1}{b^{2}}\right )^{2}}-\frac {3 x}{8 \sqrt {\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}-\frac {3 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-b}, i\right )}{8 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+b \left (-\frac {x^{3} \sqrt {b^{2} x^{4}-1}}{10 b^{6} \left (x^{4}-\frac {1}{b^{2}}\right )^{3}}+\frac {7 x^{3} \sqrt {b^{2} x^{4}-1}}{60 b^{4} \left (x^{4}-\frac {1}{b^{2}}\right )^{2}}-\frac {7 x^{3}}{40 \sqrt {\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {7 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-b}, i\right )\right )}{40 \sqrt {-b}\, \sqrt {b^{2} x^{4}-1}\, b}\right )\)	\(269\)

Input:

int((b*x^2+1)/(b^2*x^4-1)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.46 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{7/2}} \, dx=-\frac {21 \, {\left (i \, b^{6} x^{10} - i \, b^{5} x^{8} - 2 i \, b^{4} x^{6} + 2 i \, b^{3} x^{4} + i \, b^{2} x^{2} - i \, b\right )} \sqrt {b} E(\arcsin \left (\sqrt {b} x\right )\,|\,-1) + 3 \, {\left (-i \, {\left (7 \, b^{6} + 15 \, b^{5}\right )} x^{10} + i \, {\left (7 \, b^{5} + 15 \, b^{4}\right )} x^{8} + 2 i \, {\left (7 \, b^{4} + 15 \, b^{3}\right )} x^{6} - 2 i \, {\left (7 \, b^{3} + 15 \, b^{2}\right )} x^{4} - i \, {\left (7 \, b^{2} + 15 \, b\right )} x^{2} + 7 i \, b + 15 i\right )} \sqrt {b} F(\arcsin \left (\sqrt {b} x\right )\,|\,-1) + {\left (21 \, b^{5} x^{9} + 24 \, b^{4} x^{7} - 80 \, b^{3} x^{5} - 28 \, b^{2} x^{3} + 75 \, b x\right )} \sqrt {b^{2} x^{4} - 1}}{120 \, {\left (b^{6} x^{10} - b^{5} x^{8} - 2 \, b^{4} x^{6} + 2 \, b^{3} x^{4} + b^{2} x^{2} - b\right )}} \] Input:

integrate((b*x^2+1)/(b^2*x^4-1)^(7/2),x, algorithm="fricas")
 

Output:

-1/120*(21*(I*b^6*x^10 - I*b^5*x^8 - 2*I*b^4*x^6 + 2*I*b^3*x^4 + I*b^2*x^2 
 - I*b)*sqrt(b)*elliptic_e(arcsin(sqrt(b)*x), -1) + 3*(-I*(7*b^6 + 15*b^5) 
*x^10 + I*(7*b^5 + 15*b^4)*x^8 + 2*I*(7*b^4 + 15*b^3)*x^6 - 2*I*(7*b^3 + 1 
5*b^2)*x^4 - I*(7*b^2 + 15*b)*x^2 + 7*I*b + 15*I)*sqrt(b)*elliptic_f(arcsi 
n(sqrt(b)*x), -1) + (21*b^5*x^9 + 24*b^4*x^7 - 80*b^3*x^5 - 28*b^2*x^3 + 7 
5*b*x)*sqrt(b^2*x^4 - 1))/(b^6*x^10 - b^5*x^8 - 2*b^4*x^6 + 2*b^3*x^4 + b^ 
2*x^2 - b)
 

Sympy [A] (verification not implemented)

Time = 13.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.35 \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{7/2}} \, dx=\frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((b*x**2+1)/(b**2*x**4-1)**(7/2),x)
 

Output:

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

Maxima [F]

\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{7/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (b^{2} x^{4} - 1\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((b*x^2+1)/(b^2*x^4-1)^(7/2),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + 1)/(b^2*x^4 - 1)^(7/2), x)
 

Giac [F]

\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{7/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (b^{2} x^{4} - 1\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((b*x^2+1)/(b^2*x^4-1)^(7/2),x, algorithm="giac")
 

Output:

integrate((b*x^2 + 1)/(b^2*x^4 - 1)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{7/2}} \, dx=\int \frac {b\,x^2+1}{{\left (b^2\,x^4-1\right )}^{7/2}} \,d x \] Input:

int((b*x^2 + 1)/(b^2*x^4 - 1)^(7/2),x)
 

Output:

int((b*x^2 + 1)/(b^2*x^4 - 1)^(7/2), x)
 

Reduce [F]

\[ \int \frac {1+b x^2}{\left (-1+b^2 x^4\right )^{7/2}} \, dx=\int \frac {\sqrt {b^{2} x^{4}-1}}{b^{7} x^{14}-b^{6} x^{12}-3 b^{5} x^{10}+3 b^{4} x^{8}+3 b^{3} x^{6}-3 b^{2} x^{4}-b \,x^{2}+1}d x \] Input:

int((b*x^2+1)/(b^2*x^4-1)^(7/2),x)
 

Output:

int(sqrt(b**2*x**4 - 1)/(b**7*x**14 - b**6*x**12 - 3*b**5*x**10 + 3*b**4*x 
**8 + 3*b**3*x**6 - 3*b**2*x**4 - b*x**2 + 1),x)