\(\int \frac {2+3 x^2}{(1+x^2) \sqrt {1+x^2+x^4}} \, dx\) [101]

Optimal result

Integrand size = 27, antiderivative size = 69 \[ \int \frac {2+3 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=-\frac {1}{2} \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {5 \left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {1+x^2+x^4}} \] Output:

-1/2*arctan(x/(x^4+x^2+1)^(1/2))+5/4*(x^2+1)*((x^4+x^2+1)/(x^2+1)^2)^(1/2) 
*InverseJacobiAM(2*arctan(x),1/2)/(x^4+x^2+1)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \frac {2+3 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\frac {(-1)^{2/3} \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} \left (3 \operatorname {EllipticF}\left (i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )-\operatorname {EllipticPi}\left (\sqrt [3]{-1},i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )\right )}{\sqrt {1+x^2+x^4}} \] Input:

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

Output:

((-1)^(2/3)*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^(2/3)*x^2]*(3*EllipticF 
[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] - EllipticPi[(-1)^(1/3), I*ArcSinh[( 
-1)^(5/6)*x], (-1)^(2/3)]))/Sqrt[1 + x^2 + x^4]
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2214, 1416, 2212, 216}

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.

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

\(\Big \downarrow \) 2214

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

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 2212

\(\displaystyle \frac {5 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}-\frac {1}{2} \int \frac {1}{\frac {x^2}{x^4+x^2+1}+1}d\frac {x}{\sqrt {x^4+x^2+1}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {5 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}-\frac {1}{2} \arctan \left (\frac {x}{\sqrt {x^4+x^2+1}}\right )\)

Input:

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

Output:

-1/2*ArcTan[x/Sqrt[1 + x^2 + x^4]] + (5*(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 
+ x^2)^2]*EllipticF[2*ArcTan[x], 1/4])/(4*Sqrt[1 + x^2 + x^4])
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 1416

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

rule 2212

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4]), x_Symbol] :> Simp[A   Subst[Int[1/(d - (b*d - 2*a*e)*x^2), 
 x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B}, x] & 
& EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]
 

rule 2214

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4]), x_Symbol] :> Simp[(B*d + A*e)/(2*d*e)   Int[1/Sqrt[a + b*x 
^2 + c*x^4], x], x] - Simp[(B*d - A*e)/(2*d*e)   Int[(d - e*x^2)/((d + e*x^ 
2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && E 
qQ[c*d^2 - a*e^2, 0] && NeQ[B*d + A*e, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.55 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.72

method	result	size
default	\(\frac {6 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}-\frac {\sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \operatorname {EllipticPi}\left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(188\)
elliptic	\(\frac {6 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}-\frac {\sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \operatorname {EllipticPi}\left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(190\)

Input:

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

Output:

6/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1-(-1/2-1/2*I 
*3^(1/2))*x^2)^(1/2)/(x^4+x^2+1)^(1/2)*EllipticF(1/2*x*(-2+2*I*3^(1/2))^(1 
/2),1/2*(-2+2*I*3^(1/2))^(1/2))-1/(-1/2+1/2*I*3^(1/2))^(1/2)*(1+1/2*x^2-1/ 
2*I*x^2*3^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(1/2)/(x^4+x^2+1)^(1/ 
2)*EllipticPi((-1/2+1/2*I*3^(1/2))^(1/2)*x,-1/(-1/2+1/2*I*3^(1/2)),(-1/2-1 
/2*I*3^(1/2))^(1/2)/(-1/2+1/2*I*3^(1/2))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75 \[ \int \frac {2+3 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=-\frac {5}{4} \, {\left (\sqrt {-3} + 1\right )} \sqrt {\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}} F(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) - \frac {1}{2} \, \arctan \left (\frac {x}{\sqrt {x^{4} + x^{2} + 1}}\right ) \] Input:

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

Output:

-5/4*(sqrt(-3) + 1)*sqrt(1/2*sqrt(-3) - 1/2)*elliptic_f(arcsin(x*sqrt(1/2* 
sqrt(-3) - 1/2)), 1/2*sqrt(-3) - 1/2) - 1/2*arctan(x/sqrt(x^4 + x^2 + 1))
 

Sympy [F]

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

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

Output:

Integral((3*x**2 + 2)/(sqrt((x**2 - x + 1)*(x**2 + x + 1))*(x**2 + 1)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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