3.16.3 \(\int \frac {x^2}{1+x^8} \, dx\) [1503]

3.16.3.1 Optimal result
3.16.3.2 Mathematica [A] (verified)
3.16.3.3 Rubi [A] (verified)
3.16.3.4 Maple [C] (verified)
3.16.3.5 Fricas [C] (verification not implemented)
3.16.3.6 Sympy [A] (verification not implemented)
3.16.3.7 Maxima [F]
3.16.3.8 Giac [A] (verification not implemented)
3.16.3.9 Mupad [B] (verification not implemented)

3.16.3.1 Optimal result

Integrand size = 11, antiderivative size = 339 \[ \int \frac {x^2}{1+x^8} \, dx=\frac {1}{8} \sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{8} \sqrt {2+\sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{8} \sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {2-\sqrt {2}}+2 x}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{8} \sqrt {2+\sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}}+2 x}{\sqrt {2-\sqrt {2}}}\right )+\frac {\log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}} \]

output
1/8*arctan((-2*x+(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)-1 
/8*arctan((2*x+(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)+1/8 
*ln(1+x^2-x*(2-2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)-1/8*ln(1+x^2+x*(2-2^(1/ 
2))^(1/2))/(4-2*2^(1/2))^(1/2)-1/8*arctan((-2*x+(2+2^(1/2))^(1/2))/(2-2^(1 
/2))^(1/2))*(2+2^(1/2))^(1/2)+1/8*arctan((2*x+(2+2^(1/2))^(1/2))/(2-2^(1/2 
))^(1/2))*(2+2^(1/2))^(1/2)-1/8*ln(1+x^2-x*(2+2^(1/2))^(1/2))/(4+2*2^(1/2) 
)^(1/2)+1/8*ln(1+x^2+x*(2+2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)
 
3.16.3.2 Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.62 \[ \int \frac {x^2}{1+x^8} \, dx=\frac {1}{4} \arctan \left (\left (x-\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right ) \cos \left (\frac {\pi }{8}\right )+\frac {1}{4} \arctan \left (\left (x+\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right ) \cos \left (\frac {\pi }{8}\right )+\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (1+x^2-2 x \sin \left (\frac {\pi }{8}\right )\right )-\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (1+x^2+2 x \sin \left (\frac {\pi }{8}\right )\right )-\frac {1}{4} \arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x-\sin \left (\frac {\pi }{8}\right )\right )\right ) \sin \left (\frac {\pi }{8}\right )-\frac {1}{4} \arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right ) \sin \left (\frac {\pi }{8}\right )-\frac {1}{8} \log \left (1+x^2-2 x \cos \left (\frac {\pi }{8}\right )\right ) \sin \left (\frac {\pi }{8}\right )+\frac {1}{8} \log \left (1+x^2+2 x \cos \left (\frac {\pi }{8}\right )\right ) \sin \left (\frac {\pi }{8}\right ) \]

input
Integrate[x^2/(1 + x^8),x]
 
output
(ArcTan[(x - Cos[Pi/8])*Csc[Pi/8]]*Cos[Pi/8])/4 + (ArcTan[(x + Cos[Pi/8])* 
Csc[Pi/8]]*Cos[Pi/8])/4 + (Cos[Pi/8]*Log[1 + x^2 - 2*x*Sin[Pi/8]])/8 - (Co 
s[Pi/8]*Log[1 + x^2 + 2*x*Sin[Pi/8]])/8 - (ArcTan[Sec[Pi/8]*(x - Sin[Pi/8] 
)]*Sin[Pi/8])/4 - (ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])]*Sin[Pi/8])/4 - (Log[1 
 + x^2 - 2*x*Cos[Pi/8]]*Sin[Pi/8])/8 + (Log[1 + x^2 + 2*x*Cos[Pi/8]]*Sin[P 
i/8])/8
 
3.16.3.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.20, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {828, 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.

\(\displaystyle \int \frac {x^2}{x^8+1} \, dx\)

\(\Big \downarrow \) 828

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

\(\Big \downarrow \) 1407

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

input
Int[x^2/(1 + x^8),x]
 
output
-1/2*((Sqrt[(2 - Sqrt[2])/(2 + Sqrt[2])]*ArcTan[(-Sqrt[2 - Sqrt[2]] + 2*x) 
/Sqrt[2 + Sqrt[2]]] - Log[1 - Sqrt[2 - Sqrt[2]]*x + x^2]/2)/(2*Sqrt[2 - Sq 
rt[2]]) + (Sqrt[(2 - Sqrt[2])/(2 + Sqrt[2])]*ArcTan[(Sqrt[2 - Sqrt[2]] + 2 
*x)/Sqrt[2 + Sqrt[2]]] + Log[1 + Sqrt[2 - Sqrt[2]]*x + x^2]/2)/(2*Sqrt[2 - 
 Sqrt[2]]))/Sqrt[2] + ((Sqrt[(2 + Sqrt[2])/(2 - Sqrt[2])]*ArcTan[(-Sqrt[2 
+ Sqrt[2]] + 2*x)/Sqrt[2 - Sqrt[2]]] - Log[1 - Sqrt[2 + Sqrt[2]]*x + x^2]/ 
2)/(2*Sqrt[2 + Sqrt[2]]) + (Sqrt[(2 + Sqrt[2])/(2 - Sqrt[2])]*ArcTan[(Sqrt 
[2 + Sqrt[2]] + 2*x)/Sqrt[2 - Sqrt[2]]] + Log[1 + Sqrt[2 + Sqrt[2]]*x + x^ 
2]/2)/(2*Sqrt[2 + Sqrt[2]]))/(2*Sqrt[2])
 

3.16.3.3.1 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 828
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[R 
t[a/b, 4]], s = Denominator[Rt[a/b, 4]]}, Simp[s^3/(2*Sqrt[2]*b*r)   Int[x^ 
(m - n/4)/(r^2 - Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x] - Simp[s^3/(2*S 
qrt[2]*b*r)   Int[x^(m - n/4)/(r^2 + Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x] 
, x]] /; FreeQ[{a, b}, x] && IGtQ[n/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && 
 GtQ[a/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]
 
3.16.3.4 Maple [C] (verified)

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

Time = 3.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.06

method result size
default \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{8}\) \(22\)
risch \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{8}\) \(22\)
meijerg \(-\frac {x^{3} \cos \left (\frac {3 \pi }{8}\right ) \ln \left (1-2 \cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{8 \left (x^{8}\right )^{\frac {3}{8}}}+\frac {x^{3} \sin \left (\frac {3 \pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1-\cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{4 \left (x^{8}\right )^{\frac {3}{8}}}+\frac {x^{3} \cos \left (\frac {\pi }{8}\right ) \ln \left (1-2 \cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{8 \left (x^{8}\right )^{\frac {3}{8}}}-\frac {x^{3} \sin \left (\frac {\pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1-\cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{4 \left (x^{8}\right )^{\frac {3}{8}}}-\frac {x^{3} \cos \left (\frac {\pi }{8}\right ) \ln \left (1+2 \cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{8 \left (x^{8}\right )^{\frac {3}{8}}}-\frac {x^{3} \sin \left (\frac {\pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1+\cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{4 \left (x^{8}\right )^{\frac {3}{8}}}+\frac {x^{3} \cos \left (\frac {3 \pi }{8}\right ) \ln \left (1+2 \cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{8 \left (x^{8}\right )^{\frac {3}{8}}}+\frac {x^{3} \sin \left (\frac {3 \pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1+\cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{4 \left (x^{8}\right )^{\frac {3}{8}}}\) \(292\)

input
int(x^2/(x^8+1),x,method=_RETURNVERBOSE)
 
output
1/8*sum(1/_R^5*ln(x-_R),_R=RootOf(_Z^8+1))
 
3.16.3.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.40 \[ \int \frac {x^2}{1+x^8} \, dx=-\left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {3}{8}}\right ) + \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {3}{8}}\right ) - \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {3}{8}}\right ) + \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {3}{8}}\right ) - \frac {1}{8} \, \left (-1\right )^{\frac {1}{8}} \log \left (x + \left (-1\right )^{\frac {3}{8}}\right ) + \frac {1}{8} i \, \left (-1\right )^{\frac {1}{8}} \log \left (x + i \, \left (-1\right )^{\frac {3}{8}}\right ) - \frac {1}{8} i \, \left (-1\right )^{\frac {1}{8}} \log \left (x - i \, \left (-1\right )^{\frac {3}{8}}\right ) + \frac {1}{8} \, \left (-1\right )^{\frac {1}{8}} \log \left (x - \left (-1\right )^{\frac {3}{8}}\right ) \]

input
integrate(x^2/(x^8+1),x, algorithm="fricas")
 
output
-(1/16*I - 1/16)*sqrt(2)*(-1)^(1/8)*log(2*x + (I + 1)*sqrt(2)*(-1)^(3/8)) 
+ (1/16*I + 1/16)*sqrt(2)*(-1)^(1/8)*log(2*x - (I - 1)*sqrt(2)*(-1)^(3/8)) 
 - (1/16*I + 1/16)*sqrt(2)*(-1)^(1/8)*log(2*x + (I - 1)*sqrt(2)*(-1)^(3/8) 
) + (1/16*I - 1/16)*sqrt(2)*(-1)^(1/8)*log(2*x - (I + 1)*sqrt(2)*(-1)^(3/8 
)) - 1/8*(-1)^(1/8)*log(x + (-1)^(3/8)) + 1/8*I*(-1)^(1/8)*log(x + I*(-1)^ 
(3/8)) - 1/8*I*(-1)^(1/8)*log(x - I*(-1)^(3/8)) + 1/8*(-1)^(1/8)*log(x - ( 
-1)^(3/8))
 
3.16.3.6 Sympy [A] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.04 \[ \int \frac {x^2}{1+x^8} \, dx=\operatorname {RootSum} {\left (16777216 t^{8} + 1, \left ( t \mapsto t \log {\left (- 512 t^{3} + x \right )} \right )\right )} \]

input
integrate(x**2/(x**8+1),x)
 
output
RootSum(16777216*_t**8 + 1, Lambda(_t, _t*log(-512*_t**3 + x)))
 
3.16.3.7 Maxima [F]

\[ \int \frac {x^2}{1+x^8} \, dx=\int { \frac {x^{2}}{x^{8} + 1} \,d x } \]

input
integrate(x^2/(x^8+1),x, algorithm="maxima")
 
output
integrate(x^2/(x^8 + 1), x)
 
3.16.3.8 Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{1+x^8} \, dx=-\frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {-\sqrt {2} + 2} + 1\right ) + \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {-\sqrt {2} + 2} + 1\right ) \]

input
integrate(x^2/(x^8+1),x, algorithm="giac")
 
output
-1/8*sqrt(-sqrt(2) + 2)*arctan((2*x + sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) + 2 
)) - 1/8*sqrt(-sqrt(2) + 2)*arctan((2*x - sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) 
 + 2)) + 1/8*sqrt(sqrt(2) + 2)*arctan((2*x + sqrt(sqrt(2) + 2))/sqrt(-sqrt 
(2) + 2)) + 1/8*sqrt(sqrt(2) + 2)*arctan((2*x - sqrt(sqrt(2) + 2))/sqrt(-s 
qrt(2) + 2)) + 1/16*sqrt(-sqrt(2) + 2)*log(x^2 + x*sqrt(sqrt(2) + 2) + 1) 
- 1/16*sqrt(-sqrt(2) + 2)*log(x^2 - x*sqrt(sqrt(2) + 2) + 1) - 1/16*sqrt(s 
qrt(2) + 2)*log(x^2 + x*sqrt(-sqrt(2) + 2) + 1) + 1/16*sqrt(sqrt(2) + 2)*l 
og(x^2 - x*sqrt(-sqrt(2) + 2) + 1)
 
3.16.3.9 Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.57 \[ \int \frac {x^2}{1+x^8} \, dx=-\mathrm {atan}\left (x\,{\left (\frac {\sqrt {-\sqrt {2}-2}}{16}-\frac {\sqrt {2-\sqrt {2}}}{16}\right )}^5\,32768{}\mathrm {i}\right )\,\left (\frac {\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{8}-\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )-\mathrm {atan}\left (x\,{\left (\frac {\sqrt {\sqrt {2}-2}}{16}+\frac {\sqrt {\sqrt {2}+2}}{16}\right )}^5\,32768{}\mathrm {i}\right )\,\left (\frac {\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )-\mathrm {atan}\left (x\,{\left (\frac {\sqrt {2}}{16}-\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right )}^5\,{\left (\sqrt {2}+2\right )}^{5/2}\,32768{}\mathrm {i}\right )\,\left (\frac {\sqrt {2}}{16}-\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i}-\mathrm {atan}\left (x\,{\left (\frac {\sqrt {2}\,1{}\mathrm {i}}{16}-\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )}^5\,{\left (\sqrt {2}+2\right )}^{5/2}\,32768{}\mathrm {i}\right )\,\left (\frac {\sqrt {2}\,1{}\mathrm {i}}{16}-\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i} \]

input
int(x^2/(x^8 + 1),x)
 
output
- atan(x*((- 2^(1/2) - 2)^(1/2)/16 - (2 - 2^(1/2))^(1/2)/16)^5*32768i)*((( 
- 2^(1/2) - 2)^(1/2)*1i)/8 - ((2 - 2^(1/2))^(1/2)*1i)/8) - atan(x*((2^(1/2 
) - 2)^(1/2)/16 + (2^(1/2) + 2)^(1/2)/16)^5*32768i)*(((2^(1/2) - 2)^(1/2)* 
1i)/8 + ((2^(1/2) + 2)^(1/2)*1i)/8) - atan(x*(2^(1/2)/16 - (1/16 - 1i/16)) 
^5*(2^(1/2) + 2)^(5/2)*32768i)*(2^(1/2)/16 - (1/16 - 1i/16))*(2^(1/2) + 2) 
^(1/2)*2i - atan(x*((2^(1/2)*1i)/16 - (1/16 + 1i/16))^5*(2^(1/2) + 2)^(5/2 
)*32768i)*((2^(1/2)*1i)/16 - (1/16 + 1i/16))*(2^(1/2) + 2)^(1/2)*2i