3.1.44 \(\int \frac {1}{2+x^2+x^4} \, dx\) [44]

3.1.44.1 Optimal result

Integrand size = 10, antiderivative size = 196 \[ \int \frac {1}{2+x^2+x^4} \, dx=-\frac {1}{2} \sqrt {\frac {1}{14} \left (-1+2 \sqrt {2}\right )} \arctan \left (\frac {\sqrt {-1+2 \sqrt {2}}-2 x}{\sqrt {1+2 \sqrt {2}}}\right )+\frac {1}{2} \sqrt {\frac {1}{14} \left (-1+2 \sqrt {2}\right )} \arctan \left (\frac {\sqrt {-1+2 \sqrt {2}}+2 x}{\sqrt {1+2 \sqrt {2}}}\right )-\frac {\log \left (\sqrt {2}-\sqrt {-1+2 \sqrt {2}} x+x^2\right )}{4 \sqrt {2 \left (-1+2 \sqrt {2}\right )}}+\frac {\log \left (\sqrt {2}+\sqrt {-1+2 \sqrt {2}} x+x^2\right )}{4 \sqrt {2 \left (-1+2 \sqrt {2}\right )}} \]

output

-1/28*arctan((-2*x+(-1+2*2^(1/2))^(1/2))/(1+2*2^(1/2))^(1/2))*(-14+28*2^(1 
/2))^(1/2)+1/28*arctan((2*x+(-1+2*2^(1/2))^(1/2))/(1+2*2^(1/2))^(1/2))*(-1 
4+28*2^(1/2))^(1/2)-1/4*ln(x^2+2^(1/2)-x*(-1+2*2^(1/2))^(1/2))/(-2+4*2^(1/ 
2))^(1/2)+1/4*ln(x^2+2^(1/2)+x*(-1+2*2^(1/2))^(1/2))/(-2+4*2^(1/2))^(1/2)
 

3.1.44.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.46 \[ \int \frac {1}{2+x^2+x^4} \, dx=-\frac {i \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \left (1-i \sqrt {7}\right )}}\right )}{\sqrt {\frac {7}{2} \left (1-i \sqrt {7}\right )}}+\frac {i \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \left (1+i \sqrt {7}\right )}}\right )}{\sqrt {\frac {7}{2} \left (1+i \sqrt {7}\right )}} \]

input

Integrate[(2 + x^2 + x^4)^(-1),x]
 

output

((-I)*ArcTan[x/Sqrt[(1 - I*Sqrt[7])/2]])/Sqrt[(7*(1 - I*Sqrt[7]))/2] + (I* 
ArcTan[x/Sqrt[(1 + I*Sqrt[7])/2]])/Sqrt[(7*(1 + I*Sqrt[7]))/2]
 

3.1.44.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {1407, 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.

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

\(\Big \downarrow \) 1407

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

input

Int[(2 + x^2 + x^4)^(-1),x]
 

output

(Sqrt[(-1 + 2*Sqrt[2])/(1 + 2*Sqrt[2])]*ArcTan[(-Sqrt[-1 + 2*Sqrt[2]] + 2* 
x)/Sqrt[1 + 2*Sqrt[2]]] - Log[Sqrt[2] - Sqrt[-1 + 2*Sqrt[2]]*x + x^2]/2)/( 
2*Sqrt[2*(-1 + 2*Sqrt[2])]) + (Sqrt[(-1 + 2*Sqrt[2])/(1 + 2*Sqrt[2])]*ArcT 
an[(Sqrt[-1 + 2*Sqrt[2]] + 2*x)/Sqrt[1 + 2*Sqrt[2]]] + Log[Sqrt[2] + Sqrt[ 
-1 + 2*Sqrt[2]]*x + x^2]/2)/(2*Sqrt[2*(-1 + 2*Sqrt[2])])
 

3.1.44.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 1407

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

3.1.44.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.16

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+\textit {\_Z}^{2}+2\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3}+\textit {\_R}}\right )}{2}\)	\(31\)
default	\(\frac {\left (-\sqrt {-1+2 \sqrt {2}}\, \sqrt {2}-4 \sqrt {-1+2 \sqrt {2}}\right ) \ln \left (x^{2}+\sqrt {2}-x \sqrt {-1+2 \sqrt {2}}\right )}{56}+\frac {\left (7 \sqrt {2}+\frac {\left (-\sqrt {-1+2 \sqrt {2}}\, \sqrt {2}-4 \sqrt {-1+2 \sqrt {2}}\right ) \sqrt {-1+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 x -\sqrt {-1+2 \sqrt {2}}}{\sqrt {1+2 \sqrt {2}}}\right )}{14 \sqrt {1+2 \sqrt {2}}}+\frac {\left (\sqrt {-1+2 \sqrt {2}}\, \sqrt {2}+4 \sqrt {-1+2 \sqrt {2}}\right ) \ln \left (x^{2}+\sqrt {2}+x \sqrt {-1+2 \sqrt {2}}\right )}{56}+\frac {\left (7 \sqrt {2}-\frac {\left (\sqrt {-1+2 \sqrt {2}}\, \sqrt {2}+4 \sqrt {-1+2 \sqrt {2}}\right ) \sqrt {-1+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 x +\sqrt {-1+2 \sqrt {2}}}{\sqrt {1+2 \sqrt {2}}}\right )}{14 \sqrt {1+2 \sqrt {2}}}\)	\(253\)

input

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

output

1/2*sum(1/(2*_R^3+_R)*ln(x-_R),_R=RootOf(_Z^4+_Z^2+2))
 

3.1.44.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.71 \[ \int \frac {1}{2+x^2+x^4} \, dx=\frac {1}{28} \, \sqrt {7} \sqrt {i \, \sqrt {7} + 1} \log \left ({\left (\sqrt {7} + i\right )} \sqrt {i \, \sqrt {7} + 1} + 4 \, x\right ) - \frac {1}{28} \, \sqrt {7} \sqrt {i \, \sqrt {7} + 1} \log \left (-{\left (\sqrt {7} + i\right )} \sqrt {i \, \sqrt {7} + 1} + 4 \, x\right ) + \frac {1}{28} \, \sqrt {7} \sqrt {-i \, \sqrt {7} + 1} \log \left ({\left (\sqrt {7} - i\right )} \sqrt {-i \, \sqrt {7} + 1} + 4 \, x\right ) - \frac {1}{28} \, \sqrt {7} \sqrt {-i \, \sqrt {7} + 1} \log \left (-{\left (\sqrt {7} - i\right )} \sqrt {-i \, \sqrt {7} + 1} + 4 \, x\right ) \]

input

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

output

1/28*sqrt(7)*sqrt(I*sqrt(7) + 1)*log((sqrt(7) + I)*sqrt(I*sqrt(7) + 1) + 4 
*x) - 1/28*sqrt(7)*sqrt(I*sqrt(7) + 1)*log(-(sqrt(7) + I)*sqrt(I*sqrt(7) + 
 1) + 4*x) + 1/28*sqrt(7)*sqrt(-I*sqrt(7) + 1)*log((sqrt(7) - I)*sqrt(-I*s 
qrt(7) + 1) + 4*x) - 1/28*sqrt(7)*sqrt(-I*sqrt(7) + 1)*log(-(sqrt(7) - I)* 
sqrt(-I*sqrt(7) + 1) + 4*x)
 

3.1.44.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 994 vs. \(2 (158) = 316\).

Time = 0.68 (sec) , antiderivative size = 994, normalized size of antiderivative = 5.07 \[ \int \frac {1}{2+x^2+x^4} \, dx =\text {Too large to display} \]

input

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

output

sqrt(1/224 + sqrt(2)/112)*log(x**2 + x*(-4*sqrt(7)*sqrt(1 + 2*sqrt(2))/7 + 
 5*sqrt(14)*sqrt(1 + 2*sqrt(2))/28 + 3*sqrt(14)*sqrt(1 + 2*sqrt(2))*sqrt(4 
*sqrt(2) + 9)/28) - 33*sqrt(4*sqrt(2) + 9)/28 - 11/28 + 11*sqrt(2)*sqrt(4* 
sqrt(2) + 9)/28 + 83*sqrt(2)/28) - sqrt(1/224 + sqrt(2)/112)*log(x**2 + x* 
(-3*sqrt(14)*sqrt(1 + 2*sqrt(2))*sqrt(4*sqrt(2) + 9)/28 - 5*sqrt(14)*sqrt( 
1 + 2*sqrt(2))/28 + 4*sqrt(7)*sqrt(1 + 2*sqrt(2))/7) - 33*sqrt(4*sqrt(2) + 
 9)/28 - 11/28 + 11*sqrt(2)*sqrt(4*sqrt(2) + 9)/28 + 83*sqrt(2)/28) + 2*sq 
rt(-sqrt(4*sqrt(2) + 9)/112 + 1/224 + 3*sqrt(2)/112)*atan(4*sqrt(14)*x/(sq 
rt(4*sqrt(2) + 9)*sqrt(-2*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2)) + 7*sqrt(-2 
*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2))) - 8*sqrt(2)*sqrt(1 + 2*sqrt(2))/(sq 
rt(4*sqrt(2) + 9)*sqrt(-2*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2)) + 7*sqrt(-2 
*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2))) + 5*sqrt(1 + 2*sqrt(2))/(sqrt(4*sqr 
t(2) + 9)*sqrt(-2*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2)) + 7*sqrt(-2*sqrt(4* 
sqrt(2) + 9) + 1 + 6*sqrt(2))) + 3*sqrt(1 + 2*sqrt(2))*sqrt(4*sqrt(2) + 9) 
/(sqrt(4*sqrt(2) + 9)*sqrt(-2*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2)) + 7*sqr 
t(-2*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2)))) + 2*sqrt(-sqrt(4*sqrt(2) + 9)/ 
112 + 1/224 + 3*sqrt(2)/112)*atan(4*sqrt(14)*x/(sqrt(4*sqrt(2) + 9)*sqrt(- 
2*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2)) + 7*sqrt(-2*sqrt(4*sqrt(2) + 9) + 1 
 + 6*sqrt(2))) - 3*sqrt(1 + 2*sqrt(2))*sqrt(4*sqrt(2) + 9)/(sqrt(4*sqrt(2) 
 + 9)*sqrt(-2*sqrt(4*sqrt(2) + 9) + 1 + 6*sqrt(2)) + 7*sqrt(-2*sqrt(4*s...
 

3.1.44.7 Maxima [F]

\[ \int \frac {1}{2+x^2+x^4} \, dx=\int { \frac {1}{x^{4} + x^{2} + 2} \,d x } \]

input

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

output

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

3.1.44.8 Giac [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.29 \[ \int \frac {1}{2+x^2+x^4} \, dx=\frac {1}{112} \, \sqrt {7} {\left (\sqrt {7} 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 8} - 2^{\frac {1}{4}} \sqrt {-2 \, \sqrt {2} + 8}\right )} \arctan \left (\frac {2 \cdot 2^{\frac {3}{4}} \sqrt {\frac {1}{2}} {\left (x + 2^{\frac {1}{4}} \sqrt {-\frac {1}{8} \, \sqrt {2} + \frac {1}{2}}\right )}}{\sqrt {\sqrt {2} + 4}}\right ) + \frac {1}{112} \, \sqrt {7} {\left (\sqrt {7} 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 8} - 2^{\frac {1}{4}} \sqrt {-2 \, \sqrt {2} + 8}\right )} \arctan \left (\frac {2 \cdot 2^{\frac {3}{4}} \sqrt {\frac {1}{2}} {\left (x - 2^{\frac {1}{4}} \sqrt {-\frac {1}{8} \, \sqrt {2} + \frac {1}{2}}\right )}}{\sqrt {\sqrt {2} + 4}}\right ) + \frac {1}{224} \, \sqrt {7} {\left (\sqrt {7} 2^{\frac {1}{4}} \sqrt {-2 \, \sqrt {2} + 8} + 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 8}\right )} \log \left (x^{2} + 2 \cdot 2^{\frac {1}{4}} x \sqrt {-\frac {1}{8} \, \sqrt {2} + \frac {1}{2}} + \sqrt {2}\right ) - \frac {1}{224} \, \sqrt {7} {\left (\sqrt {7} 2^{\frac {1}{4}} \sqrt {-2 \, \sqrt {2} + 8} + 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 8}\right )} \log \left (x^{2} - 2 \cdot 2^{\frac {1}{4}} x \sqrt {-\frac {1}{8} \, \sqrt {2} + \frac {1}{2}} + \sqrt {2}\right ) \]

input

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

output

1/112*sqrt(7)*(sqrt(7)*2^(1/4)*sqrt(2*sqrt(2) + 8) - 2^(1/4)*sqrt(-2*sqrt( 
2) + 8))*arctan(2*2^(3/4)*sqrt(1/2)*(x + 2^(1/4)*sqrt(-1/8*sqrt(2) + 1/2)) 
/sqrt(sqrt(2) + 4)) + 1/112*sqrt(7)*(sqrt(7)*2^(1/4)*sqrt(2*sqrt(2) + 8) - 
 2^(1/4)*sqrt(-2*sqrt(2) + 8))*arctan(2*2^(3/4)*sqrt(1/2)*(x - 2^(1/4)*sqr 
t(-1/8*sqrt(2) + 1/2))/sqrt(sqrt(2) + 4)) + 1/224*sqrt(7)*(sqrt(7)*2^(1/4) 
*sqrt(-2*sqrt(2) + 8) + 2^(1/4)*sqrt(2*sqrt(2) + 8))*log(x^2 + 2*2^(1/4)*x 
*sqrt(-1/8*sqrt(2) + 1/2) + sqrt(2)) - 1/224*sqrt(7)*(sqrt(7)*2^(1/4)*sqrt 
(-2*sqrt(2) + 8) + 2^(1/4)*sqrt(2*sqrt(2) + 8))*log(x^2 - 2*2^(1/4)*x*sqrt 
(-1/8*sqrt(2) + 1/2) + sqrt(2))
 

3.1.44.9 Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.31 \[ \int \frac {1}{2+x^2+x^4} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {7}\,x\,\sqrt {7-\sqrt {7}\,7{}\mathrm {i}}}{14}\right )\,\sqrt {7-\sqrt {7}\,7{}\mathrm {i}}\,1{}\mathrm {i}}{14}-\frac {\sqrt {7}\,\mathrm {atan}\left (\frac {x\,\sqrt {1+\sqrt {7}\,1{}\mathrm {i}}}{2}\right )\,\sqrt {1+\sqrt {7}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{14} \]

input

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

output

(atan((7^(1/2)*x*(7 - 7^(1/2)*7i)^(1/2))/14)*(7 - 7^(1/2)*7i)^(1/2)*1i)/14 
 - (7^(1/2)*atan((x*(7^(1/2)*1i + 1)^(1/2))/2)*(7^(1/2)*1i + 1)^(1/2)*1i)/ 
14
 

3.1.44.10 Reduce [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.20 \[ \int \frac {1}{2+x^2+x^4} \, dx=\frac {\sqrt {2 \sqrt {2}+1}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {2}-1}-2 x}{\sqrt {2 \sqrt {2}+1}}\right )}{28}-\frac {\sqrt {2 \sqrt {2}+1}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {2}-1}-2 x}{\sqrt {2 \sqrt {2}+1}}\right )}{7}-\frac {\sqrt {2 \sqrt {2}+1}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {2}-1}+2 x}{\sqrt {2 \sqrt {2}+1}}\right )}{28}+\frac {\sqrt {2 \sqrt {2}+1}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {2}-1}+2 x}{\sqrt {2 \sqrt {2}+1}}\right )}{7}-\frac {\sqrt {2 \sqrt {2}-1}\, \sqrt {2}\, \mathrm {log}\left (-\sqrt {2 \sqrt {2}-1}\, x +\sqrt {2}+x^{2}\right )}{56}+\frac {\sqrt {2 \sqrt {2}-1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {2 \sqrt {2}-1}\, x +\sqrt {2}+x^{2}\right )}{56}-\frac {\sqrt {2 \sqrt {2}-1}\, \mathrm {log}\left (-\sqrt {2 \sqrt {2}-1}\, x +\sqrt {2}+x^{2}\right )}{14}+\frac {\sqrt {2 \sqrt {2}-1}\, \mathrm {log}\left (\sqrt {2 \sqrt {2}-1}\, x +\sqrt {2}+x^{2}\right )}{14} \]

input

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

output

(2*sqrt(2*sqrt(2) + 1)*sqrt(2)*atan((sqrt(2*sqrt(2) - 1) - 2*x)/sqrt(2*sqr 
t(2) + 1)) - 8*sqrt(2*sqrt(2) + 1)*atan((sqrt(2*sqrt(2) - 1) - 2*x)/sqrt(2 
*sqrt(2) + 1)) - 2*sqrt(2*sqrt(2) + 1)*sqrt(2)*atan((sqrt(2*sqrt(2) - 1) + 
 2*x)/sqrt(2*sqrt(2) + 1)) + 8*sqrt(2*sqrt(2) + 1)*atan((sqrt(2*sqrt(2) - 
1) + 2*x)/sqrt(2*sqrt(2) + 1)) - sqrt(2*sqrt(2) - 1)*sqrt(2)*log( - sqrt(2 
*sqrt(2) - 1)*x + sqrt(2) + x**2) + sqrt(2*sqrt(2) - 1)*sqrt(2)*log(sqrt(2 
*sqrt(2) - 1)*x + sqrt(2) + x**2) - 4*sqrt(2*sqrt(2) - 1)*log( - sqrt(2*sq 
rt(2) - 1)*x + sqrt(2) + x**2) + 4*sqrt(2*sqrt(2) - 1)*log(sqrt(2*sqrt(2) 
- 1)*x + sqrt(2) + x**2))/56