\(\int \frac {1+b x^2}{(1+b^2 x^4)^{5/2}} \, dx\) [264]

Optimal result

Integrand size = 21, antiderivative size = 211 \[ \int \frac {1+b x^2}{\left (1+b^2 x^4\right )^{5/2}} \, dx=\frac {x \left (1+b x^2\right )}{6 \left (1+b^2 x^4\right )^{3/2}}+\frac {x \left (5+3 b x^2\right )}{12 \sqrt {1+b^2 x^4}}-\frac {x \sqrt {1+b^2 x^4}}{4 \left (1+b x^2\right )}+\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{4 \sqrt {b} \sqrt {1+b^2 x^4}}+\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{12 \sqrt {b} \sqrt {1+b^2 x^4}} \] 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)-x*( 
b^2*x^4+1)^(1/2)/(4*b*x^2+4)+1/4*(b*x^2+1)*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2) 
*EllipticE(sin(2*arctan(b^(1/2)*x)),1/2*2^(1/2))/b^(1/2)/(b^2*x^4+1)^(1/2) 
+1/12*(b*x^2+1)*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2)*InverseJacobiAM(2*arctan(b 
^(1/2)*x),1/2*2^(1/2))/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 = 74, normalized size of antiderivative = 0.35 \[ \int \frac {1+b x^2}{\left (1+b^2 x^4\right )^{5/2}} \, dx=\frac {1}{12} x \left (\frac {7+5 b^2 x^4}{\left (1+b^2 x^4\right )^{3/2}}+5 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-b^2 x^4\right )+4 b x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-b^2 x^4\right )\right ) \] 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 = 0.57 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1493, 25, 1493, 25, 1512, 761, 1510}

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.

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

\(\Big \downarrow \) 1493

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

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{6} \int \frac {3 b x^2+5}{\left (b^2 x^4+1\right )^{3/2}}dx+\frac {x \left (b x^2+1\right )}{6 \left (b^2 x^4+1\right )^{3/2}}\)

\(\Big \downarrow \) 1493

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1512

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

\(\Big \downarrow \) 761

\(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (3 \int \frac {1-b x^2}{\sqrt {b^2 x^4+1}}dx+\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}\right )+\frac {x \left (3 b x^2+5\right )}{2 \sqrt {b^2 x^4+1}}\right )+\frac {x \left (b x^2+1\right )}{6 \left (b^2 x^4+1\right )^{3/2}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}+3 \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}-\frac {x \sqrt {b^2 x^4+1}}{b x^2+1}\right )\right )+\frac {x \left (3 b x^2+5\right )}{2 \sqrt {b^2 x^4+1}}\right )+\frac {x \left (b x^2+1\right )}{6 \left (b^2 x^4+1\right )^{3/2}}\)

Input:

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

Output:

(x*(1 + b*x^2))/(6*(1 + b^2*x^4)^(3/2)) + ((x*(5 + 3*b*x^2))/(2*Sqrt[1 + b 
^2*x^4]) + (3*(-((x*Sqrt[1 + b^2*x^4])/(1 + b*x^2)) + ((1 + b*x^2)*Sqrt[(1 
 + b^2*x^4)/(1 + b*x^2)^2]*EllipticE[2*ArcTan[Sqrt[b]*x], 1/2])/(Sqrt[b]*S 
qrt[1 + b^2*x^4])) + ((1 + b*x^2)*Sqrt[(1 + b^2*x^4)/(1 + b*x^2)^2]*Ellipt 
icF[2*ArcTan[Sqrt[b]*x], 1/2])/(Sqrt[b]*Sqrt[1 + b^2*x^4]))/2)/6
 

Defintions of rubi rules used

rule 25

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

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1493

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x 
)*(d + e*x^2)*((a + c*x^4)^(p + 1)/(4*a*(p + 1))), x] + Simp[1/(4*a*(p + 1) 
)   Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x] 
/; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && Integer 
Q[2*p]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 1512

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q 
 Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c 
, d, e}, x] && PosQ[c/a]
 

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.18

method	result	size
meijerg	\(x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {5}{2}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )+\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {5}{2}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3}\)	\(38\)
elliptic	\(\frac {\left (\frac {x^{3}}{6 b^{3}}+\frac {x}{6 b^{4}}\right ) \sqrt {b^{2} x^{4}+1}}{\left (x^{4}+\frac {1}{b^{2}}\right )^{2}}-\frac {2 b^{2} \left (-\frac {x^{3}}{8 b}-\frac {5 x}{24 b^{2}}\right )}{\sqrt {\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}+\frac {5 \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{12 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}-\frac {i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{4 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}\)	\(190\)
default	\(\frac {x \sqrt {b^{2} x^{4}+1}}{6 b^{4} \left (x^{4}+\frac {1}{b^{2}}\right )^{2}}+\frac {5 x}{12 \sqrt {\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}+\frac {5 \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{12 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}+b \left (\frac {x^{3} \sqrt {b^{2} x^{4}+1}}{6 b^{4} \left (x^{4}+\frac {1}{b^{2}}\right )^{2}}+\frac {x^{3}}{4 \sqrt {\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}-\frac {i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{4 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}\, b}\right )\)	\(215\)

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 [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.77 \[ \int \frac {1+b x^2}{\left (1+b^2 x^4\right )^{5/2}} \, dx=\frac {3 \, {\left (b^{5} x^{8} + 2 \, b^{3} x^{4} + b\right )} \left (-b^{2}\right )^{\frac {3}{4}} E(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) - {\left ({\left (3 \, b^{5} + 5 \, b^{4}\right )} x^{8} + 2 \, {\left (3 \, b^{3} + 5 \, b^{2}\right )} x^{4} + 3 \, b + 5\right )} \left (-b^{2}\right )^{\frac {3}{4}} F(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) + {\left (3 \, b^{5} x^{7} + 5 \, b^{4} x^{5} + 5 \, b^{3} x^{3} + 7 \, b^{2} x\right )} \sqrt {b^{2} x^{4} + 1}}{12 \, {\left (b^{6} x^{8} + 2 \, b^{4} x^{4} + b^{2}\right )}} \] Input:

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

Output:

1/12*(3*(b^5*x^8 + 2*b^3*x^4 + b)*(-b^2)^(3/4)*elliptic_e(arcsin((-b^2)^(1 
/4)*x), -1) - ((3*b^5 + 5*b^4)*x^8 + 2*(3*b^3 + 5*b^2)*x^4 + 3*b + 5)*(-b^ 
2)^(3/4)*elliptic_f(arcsin((-b^2)^(1/4)*x), -1) + (3*b^5*x^7 + 5*b^4*x^5 + 
 5*b^3*x^3 + 7*b^2*x)*sqrt(b^2*x^4 + 1))/(b^6*x^8 + 2*b^4*x^4 + b^2)
                                                                                    
                                                                                    
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 8.55 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.31 \[ \int \frac {1+b x^2}{\left (1+b^2 x^4\right )^{5/2}} \, dx=\frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] 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(I*pi))/(4* 
gamma(7/4)) + x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), b**2*x**4*exp_polar(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. \[ \int \frac {1+b x^2}{\left (1+b^2 x^4\right )^{5/2}} \, dx=\int \frac {b\,x^2+1}{{\left (b^2\,x^4+1\right )}^{5/2}} \,d x \] 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]

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

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

Output:

int(sqrt(b**2*x**4 + 1)/(b**6*x**12 + 3*b**4*x**8 + 3*b**2*x**4 + 1),x) + 
int((sqrt(b**2*x**4 + 1)*x**2)/(b**6*x**12 + 3*b**4*x**8 + 3*b**2*x**4 + 1 
),x)*b