\(\int \frac {1+b x^2}{(1-b^2 x^4)^{7/2}} \, dx\) [232]

Optimal result

Integrand size = 22, antiderivative size = 119 \[ \int \frac {1+b x^2}{\left (1-b^2 x^4\right )^{7/2}} \, dx=\frac {x \left (1+b x^2\right )}{10 \left (1-b^2 x^4\right )^{5/2}}+\frac {x \left (9+7 b x^2\right )}{60 \left (1-b^2 x^4\right )^{3/2}}+\frac {x \left (15+7 b x^2\right )}{40 \sqrt {1-b^2 x^4}}-\frac {7 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{40 \sqrt {b}}+\frac {11 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{20 \sqrt {b}} \] 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*EllipticE(b^(1/2)*x,I)/b^(1/2) 
+11/20*EllipticF(b^(1/2)*x,I)/b^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.68 \[ \int \frac {1+b x^2}{\left (1-b^2 x^4\right )^{7/2}} \, dx=\frac {1}{120} x \left (\frac {75-108 b^2 x^4+45 b^4 x^8}{\left (1-b^2 x^4\right )^{5/2}}+45 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )+40 b x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},b^2 x^4\right )\right ) \] Input:

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

Output:

(x*((75 - 108*b^2*x^4 + 45*b^4*x^8)/(1 - b^2*x^4)^(5/2) + 45*Hypergeometri 
c2F1[1/4, 1/2, 5/4, b^2*x^4] + 40*b*x^2*Hypergeometric2F1[3/4, 7/2, 7/4, b 
^2*x^4]))/120
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.62, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {1388, 316, 27, 402, 27, 402, 27, 402, 27, 402, 25, 27, 399, 284, 327, 762}

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.

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

\(\Big \downarrow \) 1388

\(\displaystyle \int \frac {1}{\left (1-b x^2\right )^{7/2} \left (b x^2+1\right )^{5/2}}dx\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {\int \frac {b \left (7 b x^2+9\right )}{\left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{5/2}}dx}{10 b}+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \int \frac {7 b x^2+9}{\left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{5/2}}dx+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {1}{10} \left (\frac {\int \frac {2 b \left (40 b x^2+19\right )}{\left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{5/2}}dx}{6 b}+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \int \frac {40 b x^2+19}{\left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{5/2}}dx+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {\int -\frac {3 b \left (7-59 b x^2\right )}{\sqrt {1-b x^2} \left (b x^2+1\right )^{5/2}}dx}{2 b}+\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \int \frac {7-59 b x^2}{\sqrt {1-b x^2} \left (b x^2+1\right )^{5/2}}dx\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}-\frac {\int \frac {6 b \left (11 b x^2+4\right )}{\sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}dx}{6 b}\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}-\int \frac {11 b x^2+4}{\sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}dx\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (\frac {\int -\frac {b \left (15-7 b x^2\right )}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx}{2 b}+\frac {7 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}+\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (-\frac {\int \frac {b \left (15-7 b x^2\right )}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx}{2 b}+\frac {7 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}+\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (-\frac {1}{2} \int \frac {15-7 b x^2}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx+\frac {7 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}+\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (\frac {1}{2} \left (7 \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx-22 \int \frac {1}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx\right )+\frac {7 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}+\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 284

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (\frac {1}{2} \left (7 \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx-22 \int \frac {1}{\sqrt {1-b^2 x^4}}dx\right )+\frac {7 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}+\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (\frac {1}{2} \left (\frac {7 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}-22 \int \frac {1}{\sqrt {1-b^2 x^4}}dx\right )+\frac {7 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}+\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {1}{10} \left (\frac {1}{3} \left (\frac {59 x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}-\frac {3}{2} \left (\frac {1}{2} \left (\frac {7 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}-\frac {22 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b}}\right )+\frac {7 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}+\frac {11 x \sqrt {1-b x^2}}{\left (b x^2+1\right )^{3/2}}\right )\right )+\frac {8 x}{3 \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{10 \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}}\)

Input:

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

Output:

x/(10*(1 - b*x^2)^(5/2)*(1 + b*x^2)^(3/2)) + ((8*x)/(3*(1 - b*x^2)^(3/2)*( 
1 + b*x^2)^(3/2)) + ((59*x)/(2*Sqrt[1 - b*x^2]*(1 + b*x^2)^(3/2)) - (3*((1 
1*x*Sqrt[1 - b*x^2])/(1 + b*x^2)^(3/2) + (7*x*Sqrt[1 - b*x^2])/(2*Sqrt[1 + 
 b*x^2]) + ((7*EllipticE[ArcSin[Sqrt[b]*x], -1])/Sqrt[b] - (22*EllipticF[A 
rcSin[Sqrt[b]*x], -1])/Sqrt[b])/2))/2)/3)/10
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 284

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> I 
nt[(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c + a*d, 0] 
&& (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0]))
 

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 1388

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

Maple [C] (verified)

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

Time = 0.93 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.30

method	result	size
meijerg	\(x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {7}{2}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )+\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {7}{2}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3}\)	\(36\)
default	\(-\frac {x \sqrt {-b^{2} x^{4}+1}}{10 b^{6} \left (x^{4}-\frac {1}{b^{2}}\right )^{3}}+\frac {3 x \sqrt {-b^{2} x^{4}+1}}{20 b^{4} \left (x^{4}-\frac {1}{b^{2}}\right )^{2}}+\frac {3 x}{8 \sqrt {-\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {3 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{8 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}+b \left (-\frac {x^{3} \sqrt {-b^{2} x^{4}+1}}{10 b^{6} \left (x^{4}-\frac {1}{b^{2}}\right )^{3}}+\frac {7 x^{3} \sqrt {-b^{2} x^{4}+1}}{60 b^{4} \left (x^{4}-\frac {1}{b^{2}}\right )^{2}}+\frac {7 x^{3}}{40 \sqrt {-\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {7 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{40 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )\)	\(264\)
elliptic	\(-\frac {x \sqrt {-b^{2} x^{4}+1}}{40 b^{3} \left (x^{2}-\frac {1}{b}\right )^{3}}+\frac {11 x \sqrt {-b^{2} x^{4}+1}}{120 b^{2} \left (x^{2}-\frac {1}{b}\right )^{2}}-\frac {53 \left (-b^{2} x^{2}-b \right ) x}{160 b \sqrt {\left (x^{2}-\frac {1}{b}\right ) \left (-b^{2} x^{2}-b \right )}}+\frac {x \sqrt {-b^{2} x^{4}+1}}{48 b^{2} \left (x^{2}+\frac {1}{b}\right )^{2}}+\frac {5 \left (-b^{2} x^{2}+b \right ) x}{32 b \sqrt {\left (x^{2}+\frac {1}{b}\right ) \left (-b^{2} x^{2}+b \right )}}+\frac {3 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{8 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}+\frac {7 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{40 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\)	\(264\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (93) = 186\).

Time = 0.08 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.07 \[ \int \frac {1+b x^2}{\left (1-b^2 x^4\right )^{7/2}} \, dx=-\frac {21 \, {\left (b^{6} x^{10} - b^{5} x^{8} - 2 \, b^{4} x^{6} + 2 \, b^{3} x^{4} + b^{2} x^{2} - b\right )} \sqrt {b} E(\arcsin \left (\sqrt {b} x\right )\,|\,-1) - 3 \, {\left ({\left (7 \, b^{6} + 15 \, b^{5}\right )} x^{10} - {\left (7 \, b^{5} + 15 \, b^{4}\right )} x^{8} - 2 \, {\left (7 \, b^{4} + 15 \, b^{3}\right )} x^{6} + 2 \, {\left (7 \, b^{3} + 15 \, b^{2}\right )} x^{4} + {\left (7 \, b^{2} + 15 \, b\right )} x^{2} - 7 \, b - 15\right )} \sqrt {b} F(\arcsin \left (\sqrt {b} x\right )\,|\,-1) + {\left (21 \, b^{5} x^{9} + 24 \, b^{4} x^{7} - 80 \, b^{3} x^{5} - 28 \, b^{2} x^{3} + 75 \, b x\right )} \sqrt {-b^{2} x^{4} + 1}}{120 \, {\left (b^{6} x^{10} - b^{5} x^{8} - 2 \, b^{4} x^{6} + 2 \, b^{3} x^{4} + b^{2} x^{2} - b\right )}} \] Input:

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

Output:

-1/120*(21*(b^6*x^10 - b^5*x^8 - 2*b^4*x^6 + 2*b^3*x^4 + b^2*x^2 - b)*sqrt 
(b)*elliptic_e(arcsin(sqrt(b)*x), -1) - 3*((7*b^6 + 15*b^5)*x^10 - (7*b^5 
+ 15*b^4)*x^8 - 2*(7*b^4 + 15*b^3)*x^6 + 2*(7*b^3 + 15*b^2)*x^4 + (7*b^2 + 
 15*b)*x^2 - 7*b - 15)*sqrt(b)*elliptic_f(arcsin(sqrt(b)*x), -1) + (21*b^5 
*x^9 + 24*b^4*x^7 - 80*b^3*x^5 - 28*b^2*x^3 + 75*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 = 12.94 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.59 \[ \int \frac {1+b x^2}{\left (1-b^2 x^4\right )^{7/2}} \, dx=\frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

b*x**3*gamma(3/4)*hyper((3/4, 7/2), (7/4,), b**2*x**4*exp_polar(2*I*pi))/( 
4*gamma(7/4)) + x*gamma(1/4)*hyper((1/4, 7/2), (5/4,), b**2*x**4*exp_polar 
(2*I*pi))/(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. \[ \int \frac {1+b x^2}{\left (1-b^2 x^4\right )^{7/2}} \, dx=\int \frac {b\,x^2+1}{{\left (1-b^2\,x^4\right )}^{7/2}} \,d x \] Input:

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

Output:

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