\(\int \frac {1-b x^2}{(1-b^2 x^4)^{5/2}} \, dx\) [238]

Optimal result

Integrand size = 23, antiderivative size = 93 \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{5/2}} \, dx=\frac {x \left (1-b x^2\right )}{6 \left (1-b^2 x^4\right )^{3/2}}+\frac {x \left (5-3 b x^2\right )}{12 \sqrt {1-b^2 x^4}}+\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{4 \sqrt {b}}+\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{6 \sqrt {b}} \] 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*EllipticE(b^(1/2)*x,I)/b^(1/2)+1/6*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 = 10.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{5/2}} \, dx=\frac {1}{12} x \left (\frac {7-5 b^2 x^4}{\left (1-b^2 x^4\right )^{3/2}}+5 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )-4 b x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},b^2 x^4\right )\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {1388, 316, 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.

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

\(\Big \downarrow \) 1388

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 399

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

\(\Big \downarrow \) 284

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

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (2 \int \frac {1}{\sqrt {1-b^2 x^4}}dx+\frac {3 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}\right )+\frac {3 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}\right )-\frac {x \sqrt {1-b x^2}}{3 \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b}}+\frac {3 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}\right )+\frac {3 x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}\right )-\frac {x \sqrt {1-b x^2}}{3 \left (b x^2+1\right )^{3/2}}\right )+\frac {x}{2 \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}\)

Input:

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

Output:

x/(2*Sqrt[1 - b*x^2]*(1 + b*x^2)^(3/2)) + (-1/3*(x*Sqrt[1 - b*x^2])/(1 + b 
*x^2)^(3/2) + ((3*x*Sqrt[1 - b*x^2])/(2*Sqrt[1 + b*x^2]) + ((3*EllipticE[A 
rcSin[Sqrt[b]*x], -1])/Sqrt[b] + (2*EllipticF[ArcSin[Sqrt[b]*x], -1])/Sqrt 
[b])/2)/3)/2
 

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 = 1.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.39

method	result	size
meijerg	\(x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {5}{2}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )-\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {5}{2}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3}\)	\(36\)
default	\(-b \left (\frac {x^{3} \sqrt {-b^{2} x^{4}+1}}{6 b^{4} \left (x^{4}-\frac {1}{b^{2}}\right )^{2}}+\frac {x^{3}}{4 \sqrt {-\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{4 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )+\frac {x \sqrt {-b^{2} x^{4}+1}}{6 b^{4} \left (x^{4}-\frac {1}{b^{2}}\right )^{2}}+\frac {5 x}{12 \sqrt {-\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {5 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{12 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\)	\(205\)
elliptic	\(\frac {x \sqrt {-b^{2} x^{4}+1}}{12 b^{2} \left (x^{2}+\frac {1}{b}\right )^{2}}+\frac {3 \left (-b^{2} x^{2}+b \right ) x}{8 b \sqrt {\left (x^{2}+\frac {1}{b}\right ) \left (-b^{2} x^{2}+b \right )}}-\frac {\left (-b^{2} x^{2}-b \right ) x}{8 b \sqrt {\left (x^{2}-\frac {1}{b}\right ) \left (-b^{2} x^{2}-b \right )}}+\frac {5 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{12 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{4 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\)	\(206\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (70) = 140\).

Time = 0.07 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.78 \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{5/2}} \, dx=\frac {3 \, {\left (b^{4} x^{6} + b^{3} x^{4} - b^{2} x^{2} - b\right )} \sqrt {b} E(\arcsin \left (\sqrt {b} x\right )\,|\,-1) - {\left ({\left (3 \, b^{4} - 5 \, b^{3}\right )} x^{6} + {\left (3 \, b^{3} - 5 \, b^{2}\right )} x^{4} - {\left (3 \, b^{2} - 5 \, b\right )} x^{2} - 3 \, b + 5\right )} \sqrt {b} F(\arcsin \left (\sqrt {b} x\right )\,|\,-1) + {\left (3 \, b^{3} x^{5} - 2 \, b^{2} x^{3} - 7 \, b x\right )} \sqrt {-b^{2} x^{4} + 1}}{12 \, {\left (b^{4} x^{6} + b^{3} x^{4} - b^{2} x^{2} - b\right )}} \] Input:

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

Output:

1/12*(3*(b^4*x^6 + b^3*x^4 - b^2*x^2 - b)*sqrt(b)*elliptic_e(arcsin(sqrt(b 
)*x), -1) - ((3*b^4 - 5*b^3)*x^6 + (3*b^3 - 5*b^2)*x^4 - (3*b^2 - 5*b)*x^2 
 - 3*b + 5)*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 = 4.00 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.75 \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{5/2}} \, dx=- \frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

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

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

Output:

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