\(\int \frac {x (1+2 x^2)}{\sqrt {1+x^2} (1+x^2+x^4)} \, dx\) [90]

Optimal result

Integrand size = 28, antiderivative size = 77 \[ \int \frac {x \left (1+2 x^2\right )}{\sqrt {1+x^2} \left (1+x^2+x^4\right )} \, dx=-\frac {1}{2} \arctan \left (\sqrt {3}-2 \sqrt {1+x^2}\right )+\frac {1}{2} \arctan \left (\sqrt {3}+2 \sqrt {1+x^2}\right )-\frac {1}{2} \sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt {1+x^2}}{2+x^2}\right ) \] Output:

1/2*arctan(-3^(1/2)+2*(x^2+1)^(1/2))+1/2*arctan(3^(1/2)+2*(x^2+1)^(1/2))-1 
/2*3^(1/2)*arctanh(3^(1/2)*(x^2+1)^(1/2)/(x^2+2))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05 \[ \int \frac {x \left (1+2 x^2\right )}{\sqrt {1+x^2} \left (1+x^2+x^4\right )} \, dx=\frac {1}{2} \left (1-i \sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) \sqrt {1+x^2}\right )+\frac {1}{2} \left (1+i \sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) \sqrt {1+x^2}\right ) \] Input:

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

Output:

((1 - I*Sqrt[3])*ArcTan[((1 - I*Sqrt[3])*Sqrt[1 + x^2])/2])/2 + ((1 + I*Sq 
rt[3])*ArcTan[((1 + I*Sqrt[3])*Sqrt[1 + x^2])/2])/2
 

Rubi [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.48, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2238, 1197, 25, 1483, 27, 1142, 25, 1083, 217, 1103}

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.

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

\(\Big \downarrow \) 2238

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

\(\Big \downarrow \) 1197

\(\displaystyle \int -\frac {1-2 x^4}{x^8-x^4+1}d\sqrt {x^2+1}\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {1-2 x^4}{x^8-x^4+1}d\sqrt {x^2+1}\)

\(\Big \downarrow \) 1483

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

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sqrt {3}-3 \sqrt {x^2+1}}{x^4-\sqrt {3} \sqrt {x^2+1}+1}d\sqrt {x^2+1}}{2 \sqrt {3}}-\frac {1}{2} \int \frac {\sqrt {3} \sqrt {x^2+1}+1}{x^4+\sqrt {3} \sqrt {x^2+1}+1}d\sqrt {x^2+1}\)

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\arctan \left (2 \sqrt {x^2+1}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt {x^2+1}+\sqrt {3}}{x^4+\sqrt {3} \sqrt {x^2+1}+1}d\sqrt {x^2+1}\right )-\frac {\frac {3}{2} \int \frac {\sqrt {3}-2 \sqrt {x^2+1}}{x^4-\sqrt {3} \sqrt {x^2+1}+1}d\sqrt {x^2+1}+\sqrt {3} \arctan \left (\sqrt {3}-2 \sqrt {x^2+1}\right )}{2 \sqrt {3}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (\arctan \left (2 \sqrt {x^2+1}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (x^4+\sqrt {3} \sqrt {x^2+1}+1\right )\right )-\frac {\sqrt {3} \arctan \left (\sqrt {3}-2 \sqrt {x^2+1}\right )-\frac {3}{2} \log \left (x^4-\sqrt {3} \sqrt {x^2+1}+1\right )}{2 \sqrt {3}}\)

Input:

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

Output:

-1/2*(Sqrt[3]*ArcTan[Sqrt[3] - 2*Sqrt[1 + x^2]] - (3*Log[1 + x^4 - Sqrt[3] 
*Sqrt[1 + x^2]])/2)/Sqrt[3] + (ArcTan[Sqrt[3] + 2*Sqrt[1 + x^2]] - (Sqrt[3 
]*Log[1 + x^4 + Sqrt[3]*Sqrt[1 + x^2]])/2)/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 217

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

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1483

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

rule 2238

Int[(Px_)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[(Px /. x -> Sqrt[x])*(d + e*x 
)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] 
 && PolyQ[Px, x^2]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(-\frac {\sqrt {3}\, \ln \left (x^{2}+2+\sqrt {x^{2}+1}\, \sqrt {3}\right )}{4}+\frac {\arctan \left (\sqrt {3}+2 \sqrt {x^{2}+1}\right )}{2}+\frac {\sqrt {3}\, \ln \left (x^{2}+2-\sqrt {x^{2}+1}\, \sqrt {3}\right )}{4}+\frac {\arctan \left (-\sqrt {3}+2 \sqrt {x^{2}+1}\right )}{2}\)	\(81\)
default	\(-\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-1-x \right )^{2}}+2}\, \left (\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-1-x \right )^{2}}+2}\, \sqrt {3}}{2}\right )+\arctan \left (\frac {\sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-1-x \right )^{2}}+2}\, \left (x -1\right )}{\left (\frac {\left (x -1\right )^{2}}{\left (-1-x \right )^{2}}+1\right ) \left (-1-x \right )}\right )\right )}{4 \sqrt {\frac {\frac {\left (x -1\right )^{2}}{\left (-1-x \right )^{2}}+1}{\left (\frac {x -1}{-1-x}+1\right )^{2}}}\, \left (\frac {x -1}{-1-x}+1\right )}-\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x +1\right )^{2}}{\left (1-x \right )^{2}}+2}\, \left (\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {\frac {2 \left (x +1\right )^{2}}{\left (1-x \right )^{2}}+2}\, \sqrt {3}}{2}\right )+\arctan \left (\frac {\sqrt {\frac {2 \left (x +1\right )^{2}}{\left (1-x \right )^{2}}+2}\, \left (x +1\right )}{\left (\frac {\left (x +1\right )^{2}}{\left (1-x \right )^{2}}+1\right ) \left (1-x \right )}\right )\right )}{4 \sqrt {\frac {\frac {\left (x +1\right )^{2}}{\left (1-x \right )^{2}}+1}{\left (\frac {x +1}{1-x}+1\right )^{2}}}\, \left (\frac {x +1}{1-x}+1\right )}\)	\(296\)
trager	\(-4 \ln \left (\frac {-16 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{5} x^{2}+16 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{3} x^{2}+12 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{3}-4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right ) x^{2}+3 \sqrt {x^{2}+1}-6 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )}{\left (4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{2} x -x -1\right ) \left (4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{2} x -x +1\right )}\right ) \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{3}+\ln \left (\frac {-16 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{5} x^{2}+16 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{3} x^{2}+12 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{3}-4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right ) x^{2}+3 \sqrt {x^{2}+1}-6 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )}{\left (4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{2} x -x -1\right ) \left (4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{2} x -x +1\right )}\right ) \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )+\operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right ) \ln \left (-\frac {16 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{5} x^{2}-4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{3} x^{2}+12 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{3}-2 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right ) x^{2}+3 \sqrt {x^{2}+1}-6 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )}{\left (4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{2} x +1\right ) \left (4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )^{2} x -1\right )}\right )\)	\(498\)

Input:

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

Output:

-1/4*3^(1/2)*ln(x^2+2+(x^2+1)^(1/2)*3^(1/2))+1/2*arctan(3^(1/2)+2*(x^2+1)^ 
(1/2))+1/4*3^(1/2)*ln(x^2+2-(x^2+1)^(1/2)*3^(1/2))+1/2*arctan(-3^(1/2)+2*( 
x^2+1)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (59) = 118\).

Time = 0.07 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.95 \[ \int \frac {x \left (1+2 x^2\right )}{\sqrt {1+x^2} \left (1+x^2+x^4\right )} \, dx=\frac {1}{4} \, \sqrt {3} \log \left (2 \, x^{4} + 5 \, x^{2} + 2 \, \sqrt {3} {\left (x^{3} + x\right )} - {\left (2 \, x^{3} + \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 2\right ) - \frac {1}{4} \, \sqrt {3} \log \left (2 \, x^{4} + 5 \, x^{2} - 2 \, \sqrt {3} {\left (x^{3} + x\right )} - {\left (2 \, x^{3} - \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 2\right ) + \frac {1}{2} \, \arctan \left (\sqrt {3} + 2 \, \sqrt {x^{2} + 1}\right ) + \frac {1}{2} \, \arctan \left (-\sqrt {3} + 2 \, \sqrt {x^{2} + 1}\right ) \] Input:

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

Output:

1/4*sqrt(3)*log(2*x^4 + 5*x^2 + 2*sqrt(3)*(x^3 + x) - (2*x^3 + sqrt(3)*(2* 
x^2 + 1) + 4*x)*sqrt(x^2 + 1) + 2) - 1/4*sqrt(3)*log(2*x^4 + 5*x^2 - 2*sqr 
t(3)*(x^3 + x) - (2*x^3 - sqrt(3)*(2*x^2 + 1) + 4*x)*sqrt(x^2 + 1) + 2) + 
1/2*arctan(sqrt(3) + 2*sqrt(x^2 + 1)) + 1/2*arctan(-sqrt(3) + 2*sqrt(x^2 + 
 1))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04 \[ \int \frac {x \left (1+2 x^2\right )}{\sqrt {1+x^2} \left (1+x^2+x^4\right )} \, dx=-\frac {1}{4} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} \sqrt {x^{2} + 1} + 2\right ) + \frac {1}{4} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} \sqrt {x^{2} + 1} + 2\right ) + \frac {1}{2} \, \arctan \left (\sqrt {3} + 2 \, \sqrt {x^{2} + 1}\right ) + \frac {1}{2} \, \arctan \left (-\sqrt {3} + 2 \, \sqrt {x^{2} + 1}\right ) \] Input:

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

Output:

-1/4*sqrt(3)*log(x^2 + sqrt(3)*sqrt(x^2 + 1) + 2) + 1/4*sqrt(3)*log(x^2 - 
sqrt(3)*sqrt(x^2 + 1) + 2) + 1/2*arctan(sqrt(3) + 2*sqrt(x^2 + 1)) + 1/2*a 
rctan(-sqrt(3) + 2*sqrt(x^2 + 1))
 

Mupad [B] (verification not implemented)

Time = 0.79 (sec) , antiderivative size = 397, normalized size of antiderivative = 5.16 \[ \int \frac {x \left (1+2 x^2\right )}{\sqrt {1+x^2} \left (1+x^2+x^4\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

((log(x - (3^(1/2)*1i)/2 - 1/2) - log(x/2 + (3^(1/2)/2 + 1i/2)*(x^2 + 1)^( 
1/2) + (3^(1/2)*x*1i)/2 + 1))*((3^(1/2)*1i)/2 + 2*((3^(1/2)*1i)/2 + 1/2)^3 
 + 1/2))/((((3^(1/2)*1i)/2 + 1/2)^2 + 1)^(1/2)*(3^(1/2)*1i + 4*((3^(1/2)*1 
i)/2 + 1/2)^3 + 1)) + ((log(x - (3^(1/2)*1i)/2 + 1/2) - log((3^(1/2)/2 - 1 
i/2)*(x^2 + 1)^(1/2) - x/2 + (3^(1/2)*x*1i)/2 + 1))*((3^(1/2)*1i)/2 + 2*(( 
3^(1/2)*1i)/2 - 1/2)^3 - 1/2))/((((3^(1/2)*1i)/2 - 1/2)^2 + 1)^(1/2)*(3^(1 
/2)*1i + 4*((3^(1/2)*1i)/2 - 1/2)^3 - 1)) + ((log(x + (3^(1/2)*1i)/2 - 1/2 
) - log(x/2 + (3^(1/2)/2 - 1i/2)*(x^2 + 1)^(1/2) - (3^(1/2)*x*1i)/2 + 1))* 
((3^(1/2)*1i)/2 + 2*((3^(1/2)*1i)/2 - 1/2)^3 - 1/2))/((((3^(1/2)*1i)/2 - 1 
/2)^2 + 1)^(1/2)*(3^(1/2)*1i + 4*((3^(1/2)*1i)/2 - 1/2)^3 - 1)) + ((log(x 
+ (3^(1/2)*1i)/2 + 1/2) - log((3^(1/2)/2 + 1i/2)*(x^2 + 1)^(1/2) - x/2 - ( 
3^(1/2)*x*1i)/2 + 1))*((3^(1/2)*1i)/2 + 2*((3^(1/2)*1i)/2 + 1/2)^3 + 1/2)) 
/((((3^(1/2)*1i)/2 + 1/2)^2 + 1)^(1/2)*(3^(1/2)*1i + 4*((3^(1/2)*1i)/2 + 1 
/2)^3 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.48 \[ \int \frac {x \left (1+2 x^2\right )}{\sqrt {1+x^2} \left (1+x^2+x^4\right )} \, dx=-\frac {\mathit {atan} \left (\frac {\sqrt {x^{2}+1}\, \sqrt {3}-2 \sqrt {x^{2}+1}\, x +\sqrt {3}\, x -2 x^{2}-2}{\sqrt {x^{2}+1}+x}\right )}{2}+\frac {\mathit {atan} \left (\frac {\sqrt {x^{2}+1}\, \sqrt {3}+2 \sqrt {x^{2}+1}\, x +\sqrt {3}\, x +2 x^{2}+2}{\sqrt {x^{2}+1}+x}\right )}{2}+\frac {\sqrt {3}\, \mathrm {log}\left (\frac {-2 \sqrt {x^{2}+1}\, \sqrt {3}\, x^{2}-\sqrt {x^{2}+1}\, \sqrt {3}+2 \sqrt {x^{2}+1}\, x^{3}+4 \sqrt {x^{2}+1}\, x -2 \sqrt {3}\, x^{3}-2 \sqrt {3}\, x +2 x^{4}+5 x^{2}+2}{2 \sqrt {x^{2}+1}\, x +2 x^{2}+1}\right )}{4}-\frac {\sqrt {3}\, \mathrm {log}\left (\frac {2 \sqrt {x^{2}+1}\, \sqrt {3}\, x^{2}+\sqrt {x^{2}+1}\, \sqrt {3}+2 \sqrt {x^{2}+1}\, x^{3}+4 \sqrt {x^{2}+1}\, x +2 \sqrt {3}\, x^{3}+2 \sqrt {3}\, x +2 x^{4}+5 x^{2}+2}{2 \sqrt {x^{2}+1}\, x +2 x^{2}+1}\right )}{4} \] Input:

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

Output:

( - 2*atan((sqrt(x**2 + 1)*sqrt(3) - 2*sqrt(x**2 + 1)*x + sqrt(3)*x - 2*x* 
*2 - 2)/(sqrt(x**2 + 1) + x)) + 2*atan((sqrt(x**2 + 1)*sqrt(3) + 2*sqrt(x* 
*2 + 1)*x + sqrt(3)*x + 2*x**2 + 2)/(sqrt(x**2 + 1) + x)) + sqrt(3)*log(( 
- 2*sqrt(x**2 + 1)*sqrt(3)*x**2 - sqrt(x**2 + 1)*sqrt(3) + 2*sqrt(x**2 + 1 
)*x**3 + 4*sqrt(x**2 + 1)*x - 2*sqrt(3)*x**3 - 2*sqrt(3)*x + 2*x**4 + 5*x* 
*2 + 2)/(2*sqrt(x**2 + 1)*x + 2*x**2 + 1)) - sqrt(3)*log((2*sqrt(x**2 + 1) 
*sqrt(3)*x**2 + sqrt(x**2 + 1)*sqrt(3) + 2*sqrt(x**2 + 1)*x**3 + 4*sqrt(x* 
*2 + 1)*x + 2*sqrt(3)*x**3 + 2*sqrt(3)*x + 2*x**4 + 5*x**2 + 2)/(2*sqrt(x* 
*2 + 1)*x + 2*x**2 + 1)))/4