\(\int (1+b x^2) (1+b^2 x^4)^{3/2} \, dx\) [260]

Optimal result

Integrand size = 21, antiderivative size = 212 \[ \int \left (1+b x^2\right ) \left (1+b^2 x^4\right )^{3/2} \, dx=\frac {4 x \sqrt {1+b^2 x^4}}{15 \left (1+b x^2\right )}+\frac {2}{105} x \left (15+7 b x^2\right ) \sqrt {1+b^2 x^4}+\frac {1}{63} x \left (9+7 b x^2\right ) \left (1+b^2 x^4\right )^{3/2}-\frac {4 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{15 \sqrt {b} \sqrt {1+b^2 x^4}}+\frac {44 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{105 \sqrt {b} \sqrt {1+b^2 x^4}} \] Output:

4*x*(b^2*x^4+1)^(1/2)/(15*b*x^2+15)+2/105*x*(7*b*x^2+15)*(b^2*x^4+1)^(1/2) 
+1/63*x*(7*b*x^2+9)*(b^2*x^4+1)^(3/2)-4/15*(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)+44/105*(b*x^2+1)*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2)*InverseJacobiA 
M(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 = 7.32 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.22 \[ \int \left (1+b x^2\right ) \left (1+b^2 x^4\right )^{3/2} \, dx=x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-b^2 x^4\right )+\frac {1}{3} b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-b^2 x^4\right ) \] Input:

Integrate[(1 + b*x^2)*(1 + b^2*x^4)^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1491, 27, 1491, 27, 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.

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

\(\Big \downarrow \) 1491

\(\displaystyle \frac {1}{21} \int 2 \left (7 b x^2+9\right ) \sqrt {b^2 x^4+1}dx+\frac {1}{63} x \left (7 b x^2+9\right ) \left (b^2 x^4+1\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{21} \int \left (7 b x^2+9\right ) \sqrt {b^2 x^4+1}dx+\frac {1}{63} x \left (7 b x^2+9\right ) \left (b^2 x^4+1\right )^{3/2}\)

\(\Big \downarrow \) 1491

\(\displaystyle \frac {2}{21} \left (\frac {1}{15} \int \frac {6 \left (7 b x^2+15\right )}{\sqrt {b^2 x^4+1}}dx+\frac {1}{5} x \sqrt {b^2 x^4+1} \left (7 b x^2+15\right )\right )+\frac {1}{63} x \left (7 b x^2+9\right ) \left (b^2 x^4+1\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{21} \left (\frac {2}{5} \int \frac {7 b x^2+15}{\sqrt {b^2 x^4+1}}dx+\frac {1}{5} x \sqrt {b^2 x^4+1} \left (7 b x^2+15\right )\right )+\frac {1}{63} x \left (7 b x^2+9\right ) \left (b^2 x^4+1\right )^{3/2}\)

\(\Big \downarrow \) 1512

\(\displaystyle \frac {2}{21} \left (\frac {2}{5} \left (22 \int \frac {1}{\sqrt {b^2 x^4+1}}dx-7 \int \frac {1-b x^2}{\sqrt {b^2 x^4+1}}dx\right )+\frac {1}{5} x \sqrt {b^2 x^4+1} \left (7 b x^2+15\right )\right )+\frac {1}{63} x \left (7 b x^2+9\right ) \left (b^2 x^4+1\right )^{3/2}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{21} \left (\frac {2}{5} \left (\frac {11 \left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}-7 \int \frac {1-b x^2}{\sqrt {b^2 x^4+1}}dx\right )+\frac {1}{5} x \sqrt {b^2 x^4+1} \left (7 b x^2+15\right )\right )+\frac {1}{63} x \left (7 b x^2+9\right ) \left (b^2 x^4+1\right )^{3/2}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2}{21} \left (\frac {2}{5} \left (\frac {11 \left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}-7 \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}-\frac {x \sqrt {b^2 x^4+1}}{b x^2+1}\right )\right )+\frac {1}{5} x \sqrt {b^2 x^4+1} \left (7 b x^2+15\right )\right )+\frac {1}{63} x \left (7 b x^2+9\right ) \left (b^2 x^4+1\right )^{3/2}\)

Input:

Int[(1 + b*x^2)*(1 + b^2*x^4)^(3/2),x]
 

Output:

(x*(9 + 7*b*x^2)*(1 + b^2*x^4)^(3/2))/63 + (2*((x*(15 + 7*b*x^2)*Sqrt[1 + 
b^2*x^4])/5 + (2*(-7*(-((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])/(Sq 
rt[b]*Sqrt[1 + b^2*x^4])) + (11*(1 + b*x^2)*Sqrt[(1 + b^2*x^4)/(1 + b*x^2) 
^2]*EllipticF[2*ArcTan[Sqrt[b]*x], 1/2])/(Sqrt[b]*Sqrt[1 + b^2*x^4])))/5)) 
/21
 

Defintions of rubi rules used

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 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 1491

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*( 
d*(4*p + 3) + e*(4*p + 1)*x^2)*((a + c*x^4)^p/((4*p + 1)*(4*p + 3))), x] + 
Simp[2*(p/((4*p + 1)*(4*p + 3)))   Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4*p + 
 1))*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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[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 = 1.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.18

method	result	size
meijerg	\(x \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )+\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3}\)	\(38\)
risch	\(\frac {x \left (35 b^{3} x^{6}+45 b^{2} x^{4}+77 b \,x^{2}+135\right ) \sqrt {b^{2} x^{4}+1}}{315}+\frac {4 \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{7 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}+\frac {4 i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{15 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}\)	\(159\)
elliptic	\(\frac {b^{3} x^{7} \sqrt {b^{2} x^{4}+1}}{9}+\frac {b^{2} x^{5} \sqrt {b^{2} x^{4}+1}}{7}+\frac {11 b \,x^{3} \sqrt {b^{2} x^{4}+1}}{45}+\frac {3 x \sqrt {b^{2} x^{4}+1}}{7}+\frac {4 \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{7 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}+\frac {4 i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{15 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}\)	\(190\)
default	\(\frac {b^{2} x^{5} \sqrt {b^{2} x^{4}+1}}{7}+\frac {3 x \sqrt {b^{2} x^{4}+1}}{7}+\frac {4 \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{7 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}+b \left (\frac {b^{2} x^{7} \sqrt {b^{2} x^{4}+1}}{9}+\frac {11 x^{3} \sqrt {b^{2} x^{4}+1}}{45}+\frac {4 i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{15 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}\, b}\right )\)	\(195\)

Input:

int((b*x^2+1)*(b^2*x^4+1)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.52 \[ \int \left (1+b x^2\right ) \left (1+b^2 x^4\right )^{3/2} \, dx=\frac {84 \, b x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 12 \, {\left (15 \, b^{2} - 7 \, b\right )} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (35 \, b^{4} x^{8} + 45 \, b^{3} x^{6} + 77 \, b^{2} x^{4} + 135 \, b x^{2} + 84\right )} \sqrt {b^{2} x^{4} + 1}}{315 \, b x} \] Input:

integrate((b*x^2+1)*(b^2*x^4+1)^(3/2),x, algorithm="fricas")
 

Output:

1/315*(84*b*x*(-1/b^2)^(3/4)*elliptic_e(arcsin((-1/b^2)^(1/4)/x), -1) + 12 
*(15*b^2 - 7*b)*x*(-1/b^2)^(3/4)*elliptic_f(arcsin((-1/b^2)^(1/4)/x), -1) 
+ (35*b^4*x^8 + 45*b^3*x^6 + 77*b^2*x^4 + 135*b*x^2 + 84)*sqrt(b^2*x^4 + 1 
))/(b*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.57 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.70 \[ \int \left (1+b x^2\right ) \left (1+b^2 x^4\right )^{3/2} \, dx=\frac {b^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + \frac {b^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((b*x**2+1)*(b**2*x**4+1)**(3/2),x)
 

Output:

b**3*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b**2*x**4*exp_polar(I*pi) 
)/(4*gamma(11/4)) + b**2*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b**2*x 
**4*exp_polar(I*pi))/(4*gamma(9/4)) + b*x**3*gamma(3/4)*hyper((-1/2, 3/4), 
 (7/4,), b**2*x**4*exp_polar(I*pi))/(4*gamma(7/4)) + x*gamma(1/4)*hyper((- 
1/2, 1/4), (5/4,), b**2*x**4*exp_polar(I*pi))/(4*gamma(5/4))
 

Maxima [F]

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

integrate((b*x^2+1)*(b^2*x^4+1)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b^2*x^4 + 1)^(3/2)*(b*x^2 + 1), x)
 

Giac [F]

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

integrate((b*x^2+1)*(b^2*x^4+1)^(3/2),x, algorithm="giac")
 

Output:

integrate((b^2*x^4 + 1)^(3/2)*(b*x^2 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (1+b x^2\right ) \left (1+b^2 x^4\right )^{3/2} \, dx=\int {\left (b^2\,x^4+1\right )}^{3/2}\,\left (b\,x^2+1\right ) \,d x \] Input:

int((b^2*x^4 + 1)^(3/2)*(b*x^2 + 1),x)
 

Output:

int((b^2*x^4 + 1)^(3/2)*(b*x^2 + 1), x)
 

Reduce [F]

\[ \int \left (1+b x^2\right ) \left (1+b^2 x^4\right )^{3/2} \, dx=\frac {\sqrt {b^{2} x^{4}+1}\, b^{3} x^{7}}{9}+\frac {\sqrt {b^{2} x^{4}+1}\, b^{2} x^{5}}{7}+\frac {11 \sqrt {b^{2} x^{4}+1}\, b \,x^{3}}{45}+\frac {3 \sqrt {b^{2} x^{4}+1}\, x}{7}+\frac {4 \left (\int \frac {\sqrt {b^{2} x^{4}+1}}{b^{2} x^{4}+1}d x \right )}{7}+\frac {4 \left (\int \frac {\sqrt {b^{2} x^{4}+1}\, x^{2}}{b^{2} x^{4}+1}d x \right ) b}{15} \] Input:

int((b*x^2+1)*(b^2*x^4+1)^(3/2),x)
 

Output:

(35*sqrt(b**2*x**4 + 1)*b**3*x**7 + 45*sqrt(b**2*x**4 + 1)*b**2*x**5 + 77* 
sqrt(b**2*x**4 + 1)*b*x**3 + 135*sqrt(b**2*x**4 + 1)*x + 180*int(sqrt(b**2 
*x**4 + 1)/(b**2*x**4 + 1),x) + 84*int((sqrt(b**2*x**4 + 1)*x**2)/(b**2*x* 
*4 + 1),x)*b)/315