Integrand size = 7, antiderivative size = 339 \[ \int \frac {1}{1+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\arctan \left (\frac {\sqrt {2-\sqrt {2}}+2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2}}+2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}-\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )+\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right ) \]
-1/16*ln(1+x^2-x*(2-2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)+1/16*ln(1+x^2+x*(2-2 ^(1/2))^(1/2))*(2-2^(1/2))^(1/2)-1/4*arctan((-2*x+(2-2^(1/2))^(1/2))/(2+2^ (1/2))^(1/2))/(4-2*2^(1/2))^(1/2)+1/4*arctan((2*x+(2-2^(1/2))^(1/2))/(2+2^ (1/2))^(1/2))/(4-2*2^(1/2))^(1/2)-1/16*ln(1+x^2-x*(2+2^(1/2))^(1/2))*(2+2^ (1/2))^(1/2)+1/16*ln(1+x^2+x*(2+2^(1/2))^(1/2))*(2+2^(1/2))^(1/2)-1/4*arct an((-2*x+(2+2^(1/2))^(1/2))/(2-2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)+1/4*arc tan((2*x+(2+2^(1/2))^(1/2))/(2-2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)
Time = 0.01 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.62 \[ \int \frac {1}{1+x^8} \, dx=\frac {1}{4} \arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x-\sin \left (\frac {\pi }{8}\right )\right )\right ) \cos \left (\frac {\pi }{8}\right )+\frac {1}{4} \arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right ) \cos \left (\frac {\pi }{8}\right )-\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (1+x^2-2 x \cos \left (\frac {\pi }{8}\right )\right )+\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (1+x^2+2 x \cos \left (\frac {\pi }{8}\right )\right )+\frac {1}{4} \arctan \left (\left (x-\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right ) \sin \left (\frac {\pi }{8}\right )+\frac {1}{4} \arctan \left (\left (x+\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right ) \sin \left (\frac {\pi }{8}\right )-\frac {1}{8} \log \left (1+x^2-2 x \sin \left (\frac {\pi }{8}\right )\right ) \sin \left (\frac {\pi }{8}\right )+\frac {1}{8} \log \left (1+x^2+2 x \sin \left (\frac {\pi }{8}\right )\right ) \sin \left (\frac {\pi }{8}\right ) \]
(ArcTan[Sec[Pi/8]*(x - Sin[Pi/8])]*Cos[Pi/8])/4 + (ArcTan[Sec[Pi/8]*(x + S in[Pi/8])]*Cos[Pi/8])/4 - (Cos[Pi/8]*Log[1 + x^2 - 2*x*Cos[Pi/8]])/8 + (Co s[Pi/8]*Log[1 + x^2 + 2*x*Cos[Pi/8]])/8 + (ArcTan[(x - Cos[Pi/8])*Csc[Pi/8 ]]*Sin[Pi/8])/4 + (ArcTan[(x + Cos[Pi/8])*Csc[Pi/8]]*Sin[Pi/8])/4 - (Log[1 + x^2 - 2*x*Sin[Pi/8]]*Sin[Pi/8])/8 + (Log[1 + x^2 + 2*x*Sin[Pi/8]]*Sin[P i/8])/8
Time = 0.58 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {757, 1483, 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.
| \(\displaystyle \int \frac {1}{x^8+1} \, dx\) |
| \(\Big \downarrow \) 757 |
| \(\displaystyle \frac {\int \frac {\sqrt {2}-x^2}{x^4-\sqrt {2} x^2+1}dx}{2 \sqrt {2}}+\frac {\int \frac {x^2+\sqrt {2}}{x^4+\sqrt {2} x^2+1}dx}{2 \sqrt {2}}\) |
| \(\Big \downarrow \) 1483 |
| \(\displaystyle \frac {\frac {\int \frac {\left (1-\sqrt {2}\right ) x+\sqrt {2 \left (2-\sqrt {2}\right )}}{x^2-\sqrt {2-\sqrt {2}} x+1}dx}{2 \sqrt {2-\sqrt {2}}}+\frac {\int \frac {\sqrt {2 \left (2-\sqrt {2}\right )}-\left (1-\sqrt {2}\right ) x}{x^2+\sqrt {2-\sqrt {2}} x+1}dx}{2 \sqrt {2-\sqrt {2}}}}{2 \sqrt {2}}+\frac {\frac {\int \frac {\sqrt {2 \left (2+\sqrt {2}\right )}-\left (1+\sqrt {2}\right ) x}{x^2-\sqrt {2+\sqrt {2}} x+1}dx}{2 \sqrt {2+\sqrt {2}}}+\frac {\int \frac {\left (1+\sqrt {2}\right ) x+\sqrt {2 \left (2+\sqrt {2}\right )}}{x^2+\sqrt {2+\sqrt {2}} x+1}dx}{2 \sqrt {2+\sqrt {2}}}}{2 \sqrt {2}}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {\frac {\frac {1}{2} \sqrt {2+\sqrt {2}} \int \frac {1}{x^2-\sqrt {2-\sqrt {2}} x+1}dx+\frac {1}{2} \left (1-\sqrt {2}\right ) \int -\frac {\sqrt {2-\sqrt {2}}-2 x}{x^2-\sqrt {2-\sqrt {2}} x+1}dx}{2 \sqrt {2-\sqrt {2}}}+\frac {\frac {1}{2} \sqrt {2+\sqrt {2}} \int \frac {1}{x^2+\sqrt {2-\sqrt {2}} x+1}dx-\frac {1}{2} \left (1-\sqrt {2}\right ) \int \frac {2 x+\sqrt {2-\sqrt {2}}}{x^2+\sqrt {2-\sqrt {2}} x+1}dx}{2 \sqrt {2-\sqrt {2}}}}{2 \sqrt {2}}+\frac {\frac {\frac {1}{2} \sqrt {2-\sqrt {2}} \int \frac {1}{x^2-\sqrt {2+\sqrt {2}} x+1}dx-\frac {1}{2} \left (1+\sqrt {2}\right ) \int -\frac {\sqrt {2+\sqrt {2}}-2 x}{x^2-\sqrt {2+\sqrt {2}} x+1}dx}{2 \sqrt {2+\sqrt {2}}}+\frac {\frac {1}{2} \sqrt {2-\sqrt {2}} \int \frac {1}{x^2+\sqrt {2+\sqrt {2}} x+1}dx+\frac {1}{2} \left (1+\sqrt {2}\right ) \int \frac {2 x+\sqrt {2+\sqrt {2}}}{x^2+\sqrt {2+\sqrt {2}} x+1}dx}{2 \sqrt {2+\sqrt {2}}}}{2 \sqrt {2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {1}{2} \sqrt {2+\sqrt {2}} \int \frac {1}{x^2-\sqrt {2-\sqrt {2}} x+1}dx-\frac {1}{2} \left (1-\sqrt {2}\right ) \int \frac {\sqrt {2-\sqrt {2}}-2 x}{x^2-\sqrt {2-\sqrt {2}} x+1}dx}{2 \sqrt {2-\sqrt {2}}}+\frac {\frac {1}{2} \sqrt {2+\sqrt {2}} \int \frac {1}{x^2+\sqrt {2-\sqrt {2}} x+1}dx-\frac {1}{2} \left (1-\sqrt {2}\right ) \int \frac {2 x+\sqrt {2-\sqrt {2}}}{x^2+\sqrt {2-\sqrt {2}} x+1}dx}{2 \sqrt {2-\sqrt {2}}}}{2 \sqrt {2}}+\frac {\frac {\frac {1}{2} \sqrt {2-\sqrt {2}} \int \frac {1}{x^2-\sqrt {2+\sqrt {2}} x+1}dx+\frac {1}{2} \left (1+\sqrt {2}\right ) \int \frac {\sqrt {2+\sqrt {2}}-2 x}{x^2-\sqrt {2+\sqrt {2}} x+1}dx}{2 \sqrt {2+\sqrt {2}}}+\frac {\frac {1}{2} \sqrt {2-\sqrt {2}} \int \frac {1}{x^2+\sqrt {2+\sqrt {2}} x+1}dx+\frac {1}{2} \left (1+\sqrt {2}\right ) \int \frac {2 x+\sqrt {2+\sqrt {2}}}{x^2+\sqrt {2+\sqrt {2}} x+1}dx}{2 \sqrt {2+\sqrt {2}}}}{2 \sqrt {2}}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {-\frac {1}{2} \left (1-\sqrt {2}\right ) \int \frac {\sqrt {2-\sqrt {2}}-2 x}{x^2-\sqrt {2-\sqrt {2}} x+1}dx-\sqrt {2+\sqrt {2}} \int \frac {1}{-\left (2 x-\sqrt {2-\sqrt {2}}\right )^2-\sqrt {2}-2}d\left (2 x-\sqrt {2-\sqrt {2}}\right )}{2 \sqrt {2-\sqrt {2}}}+\frac {-\frac {1}{2} \left (1-\sqrt {2}\right ) \int \frac {2 x+\sqrt {2-\sqrt {2}}}{x^2+\sqrt {2-\sqrt {2}} x+1}dx-\sqrt {2+\sqrt {2}} \int \frac {1}{-\left (2 x+\sqrt {2-\sqrt {2}}\right )^2-\sqrt {2}-2}d\left (2 x+\sqrt {2-\sqrt {2}}\right )}{2 \sqrt {2-\sqrt {2}}}}{2 \sqrt {2}}+\frac {\frac {\frac {1}{2} \left (1+\sqrt {2}\right ) \int \frac {\sqrt {2+\sqrt {2}}-2 x}{x^2-\sqrt {2+\sqrt {2}} x+1}dx-\sqrt {2-\sqrt {2}} \int \frac {1}{-\left (2 x-\sqrt {2+\sqrt {2}}\right )^2+\sqrt {2}-2}d\left (2 x-\sqrt {2+\sqrt {2}}\right )}{2 \sqrt {2+\sqrt {2}}}+\frac {\frac {1}{2} \left (1+\sqrt {2}\right ) \int \frac {2 x+\sqrt {2+\sqrt {2}}}{x^2+\sqrt {2+\sqrt {2}} x+1}dx-\sqrt {2-\sqrt {2}} \int \frac {1}{-\left (2 x+\sqrt {2+\sqrt {2}}\right )^2+\sqrt {2}-2}d\left (2 x+\sqrt {2+\sqrt {2}}\right )}{2 \sqrt {2+\sqrt {2}}}}{2 \sqrt {2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {\arctan \left (\frac {2 x-\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{2} \left (1-\sqrt {2}\right ) \int \frac {\sqrt {2-\sqrt {2}}-2 x}{x^2-\sqrt {2-\sqrt {2}} x+1}dx}{2 \sqrt {2-\sqrt {2}}}+\frac {\arctan \left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{2} \left (1-\sqrt {2}\right ) \int \frac {2 x+\sqrt {2-\sqrt {2}}}{x^2+\sqrt {2-\sqrt {2}} x+1}dx}{2 \sqrt {2-\sqrt {2}}}}{2 \sqrt {2}}+\frac {\frac {\frac {1}{2} \left (1+\sqrt {2}\right ) \int \frac {\sqrt {2+\sqrt {2}}-2 x}{x^2-\sqrt {2+\sqrt {2}} x+1}dx+\arctan \left (\frac {2 x-\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{2 \sqrt {2+\sqrt {2}}}+\frac {\frac {1}{2} \left (1+\sqrt {2}\right ) \int \frac {2 x+\sqrt {2+\sqrt {2}}}{x^2+\sqrt {2+\sqrt {2}} x+1}dx+\arctan \left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{2 \sqrt {2+\sqrt {2}}}}{2 \sqrt {2}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {\arctan \left (\frac {2 x-\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{2} \left (1-\sqrt {2}\right ) \log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )}{2 \sqrt {2-\sqrt {2}}}+\frac {\arctan \left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{2} \left (1-\sqrt {2}\right ) \log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )}{2 \sqrt {2-\sqrt {2}}}}{2 \sqrt {2}}+\frac {\frac {\arctan \left (\frac {2 x-\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{2} \left (1+\sqrt {2}\right ) \log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )}{2 \sqrt {2+\sqrt {2}}}+\frac {\arctan \left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{2} \left (1+\sqrt {2}\right ) \log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )}{2 \sqrt {2+\sqrt {2}}}}{2 \sqrt {2}}\) |
((ArcTan[(-Sqrt[2 - Sqrt[2]] + 2*x)/Sqrt[2 + Sqrt[2]]] + ((1 - Sqrt[2])*Lo g[1 - Sqrt[2 - Sqrt[2]]*x + x^2])/2)/(2*Sqrt[2 - Sqrt[2]]) + (ArcTan[(Sqrt [2 - Sqrt[2]] + 2*x)/Sqrt[2 + Sqrt[2]]] - ((1 - Sqrt[2])*Log[1 + Sqrt[2 - Sqrt[2]]*x + x^2])/2)/(2*Sqrt[2 - Sqrt[2]]))/(2*Sqrt[2]) + ((ArcTan[(-Sqrt [2 + Sqrt[2]] + 2*x)/Sqrt[2 - Sqrt[2]]] - ((1 + Sqrt[2])*Log[1 - Sqrt[2 + Sqrt[2]]*x + x^2])/2)/(2*Sqrt[2 + Sqrt[2]]) + (ArcTan[(Sqrt[2 + Sqrt[2]] + 2*x)/Sqrt[2 - Sqrt[2]]] + ((1 + Sqrt[2])*Log[1 + Sqrt[2 + Sqrt[2]]*x + x^ 2])/2)/(2*Sqrt[2 + Sqrt[2]]))/(2*Sqrt[2])
3.1.49.3.1 Defintions of rubi rules used
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])
Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 4]], s = Denominator[Rt[a/b, 4]]}, Simp[r/(2*Sqrt[2]*a) Int[(Sqrt[2]*r - s*x^(n/4))/(r^2 - Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x] + Simp[r/(2*S qrt[2]*a) Int[(Sqrt[2]*r + s*x^(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, 1] && GtQ[a/b, 0]
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]
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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.06
| method | result | size |
| default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{7}}\right )}{8}\) | \(22\) |
| risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{7}}\right )}{8}\) | \(22\) |
| meijerg | \(-\frac {x \cos \left (\frac {\pi }{8}\right ) \ln \left (1-2 \cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{8 \left (x^{8}\right )^{\frac {1}{8}}}+\frac {x \sin \left (\frac {\pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1-\cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{4 \left (x^{8}\right )^{\frac {1}{8}}}-\frac {x \cos \left (\frac {3 \pi }{8}\right ) \ln \left (1-2 \cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{8 \left (x^{8}\right )^{\frac {1}{8}}}+\frac {x \sin \left (\frac {3 \pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1-\cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{4 \left (x^{8}\right )^{\frac {1}{8}}}+\frac {x \cos \left (\frac {3 \pi }{8}\right ) \ln \left (1+2 \cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{8 \left (x^{8}\right )^{\frac {1}{8}}}+\frac {x \sin \left (\frac {3 \pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1+\cos \left (\frac {3 \pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{4 \left (x^{8}\right )^{\frac {1}{8}}}+\frac {x \cos \left (\frac {\pi }{8}\right ) \ln \left (1+2 \cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{8 \left (x^{8}\right )^{\frac {1}{8}}}+\frac {x \sin \left (\frac {\pi }{8}\right ) \arctan \left (\frac {\sin \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}{1+\cos \left (\frac {\pi }{8}\right ) \left (x^{8}\right )^{\frac {1}{8}}}\right )}{4 \left (x^{8}\right )^{\frac {1}{8}}}\) | \(276\) |
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.40 \[ \int \frac {1}{1+x^8} \, dx=\left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}}\right ) + \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}}\right ) - \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2} \left (-1\right )^{\frac {1}{8}}\right ) + \frac {1}{8} \, \left (-1\right )^{\frac {1}{8}} \log \left (x + \left (-1\right )^{\frac {1}{8}}\right ) + \frac {1}{8} i \, \left (-1\right )^{\frac {1}{8}} \log \left (x + i \, \left (-1\right )^{\frac {1}{8}}\right ) - \frac {1}{8} i \, \left (-1\right )^{\frac {1}{8}} \log \left (x - i \, \left (-1\right )^{\frac {1}{8}}\right ) - \frac {1}{8} \, \left (-1\right )^{\frac {1}{8}} \log \left (x - \left (-1\right )^{\frac {1}{8}}\right ) \]
(1/16*I + 1/16)*sqrt(2)*(-1)^(1/8)*log(2*x + (I + 1)*sqrt(2)*(-1)^(1/8)) - (1/16*I - 1/16)*sqrt(2)*(-1)^(1/8)*log(2*x - (I - 1)*sqrt(2)*(-1)^(1/8)) + (1/16*I - 1/16)*sqrt(2)*(-1)^(1/8)*log(2*x + (I - 1)*sqrt(2)*(-1)^(1/8)) - (1/16*I + 1/16)*sqrt(2)*(-1)^(1/8)*log(2*x - (I + 1)*sqrt(2)*(-1)^(1/8) ) + 1/8*(-1)^(1/8)*log(x + (-1)^(1/8)) + 1/8*I*(-1)^(1/8)*log(x + I*(-1)^( 1/8)) - 1/8*I*(-1)^(1/8)*log(x - I*(-1)^(1/8)) - 1/8*(-1)^(1/8)*log(x - (- 1)^(1/8))
Time = 1.34 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.04 \[ \int \frac {1}{1+x^8} \, dx=\operatorname {RootSum} {\left (16777216 t^{8} + 1, \left ( t \mapsto t \log {\left (8 t + x \right )} \right )\right )} \]
\[ \int \frac {1}{1+x^8} \, dx=\int { \frac {1}{x^{8} + 1} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.71 \[ \int \frac {1}{1+x^8} \, dx=\frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} + 1\right ) + \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {-\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {-\sqrt {2} + 2} + 1\right ) \]
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(-sq rt(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(-sqr t(2) + 2)*log(x^2 + x*sqrt(-sqrt(2) + 2) + 1) - 1/16*sqrt(-sqrt(2) + 2)*lo g(x^2 - x*sqrt(-sqrt(2) + 2) + 1)
Time = 0.31 (sec) , antiderivative size = 288, normalized size of antiderivative = 0.85 \[ \int \frac {1}{1+x^8} \, dx=\mathrm {atan}\left (\frac {x\,\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{\sqrt {2-\sqrt {2}}\,\sqrt {-\sqrt {2}-2}+\sqrt {2}}-\frac {x\,\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{\sqrt {2-\sqrt {2}}\,\sqrt {-\sqrt {2}-2}+\sqrt {2}}\right )\,\left (\frac {\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{8}-\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )-\mathrm {atan}\left (\frac {x\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{\sqrt {2}+\sqrt {\sqrt {2}-2}\,\sqrt {\sqrt {2}+2}}+\frac {x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{\sqrt {2}+\sqrt {\sqrt {2}-2}\,\sqrt {\sqrt {2}+2}}\right )\,\left (\frac {\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )+\mathrm {atan}\left (-\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}}{2}+x\,\sqrt {\sqrt {2}+2}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {2}\,1{}\mathrm {i}}{16}-\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i}-\mathrm {atan}\left (x\,\sqrt {\sqrt {2}+2}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {2}}{16}-\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i} \]
atan((x*(- 2^(1/2) - 2)^(1/2)*1i)/((2 - 2^(1/2))^(1/2)*(- 2^(1/2) - 2)^(1/ 2) + 2^(1/2)) - (x*(2 - 2^(1/2))^(1/2)*1i)/((2 - 2^(1/2))^(1/2)*(- 2^(1/2) - 2)^(1/2) + 2^(1/2)))*(((- 2^(1/2) - 2)^(1/2)*1i)/8 - ((2 - 2^(1/2))^(1/ 2)*1i)/8) - atan((x*(2^(1/2) - 2)^(1/2)*1i)/(2^(1/2) + (2^(1/2) - 2)^(1/2) *(2^(1/2) + 2)^(1/2)) + (x*(2^(1/2) + 2)^(1/2)*1i)/(2^(1/2) + (2^(1/2) - 2 )^(1/2)*(2^(1/2) + 2)^(1/2)))*(((2^(1/2) - 2)^(1/2)*1i)/8 + ((2^(1/2) + 2) ^(1/2)*1i)/8) + atan(x*(2^(1/2) + 2)^(1/2)*(1/2 + 1i/2) - (2^(1/2)*x*(2^(1 /2) + 2)^(1/2))/2)*((2^(1/2)*1i)/16 - (1/16 + 1i/16))*(2^(1/2) + 2)^(1/2)* 2i - atan(x*(2^(1/2) + 2)^(1/2)*(1/2 - 1i/2) + (2^(1/2)*x*(2^(1/2) + 2)^(1 /2)*1i)/2)*(2^(1/2)/16 - (1/16 - 1i/16))*(2^(1/2) + 2)^(1/2)*2i
Time = 0.00 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.60 \[ \int \frac {1}{1+x^8} \, dx=-\frac {\sqrt {\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {-\sqrt {2}+2}-2 x}{\sqrt {\sqrt {2}+2}}\right )}{8}+\frac {\sqrt {\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {-\sqrt {2}+2}+2 x}{\sqrt {\sqrt {2}+2}}\right )}{8}-\frac {\sqrt {-\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}+2}-2 x}{\sqrt {-\sqrt {2}+2}}\right )}{8}+\frac {\sqrt {-\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}+2}+2 x}{\sqrt {-\sqrt {2}+2}}\right )}{8}-\frac {\sqrt {-\sqrt {2}+2}\, \mathrm {log}\left (-\sqrt {-\sqrt {2}+2}\, x +x^{2}+1\right )}{16}+\frac {\sqrt {-\sqrt {2}+2}\, \mathrm {log}\left (\sqrt {-\sqrt {2}+2}\, x +x^{2}+1\right )}{16}-\frac {\sqrt {\sqrt {2}+2}\, \mathrm {log}\left (-\sqrt {\sqrt {2}+2}\, x +x^{2}+1\right )}{16}+\frac {\sqrt {\sqrt {2}+2}\, \mathrm {log}\left (\sqrt {\sqrt {2}+2}\, x +x^{2}+1\right )}{16} \]
( - 2*sqrt(sqrt(2) + 2)*atan((sqrt( - sqrt(2) + 2) - 2*x)/sqrt(sqrt(2) + 2 )) + 2*sqrt(sqrt(2) + 2)*atan((sqrt( - sqrt(2) + 2) + 2*x)/sqrt(sqrt(2) + 2)) - 2*sqrt( - sqrt(2) + 2)*atan((sqrt(sqrt(2) + 2) - 2*x)/sqrt( - sqrt(2 ) + 2)) + 2*sqrt( - sqrt(2) + 2)*atan((sqrt(sqrt(2) + 2) + 2*x)/sqrt( - sq rt(2) + 2)) - sqrt( - sqrt(2) + 2)*log( - sqrt( - sqrt(2) + 2)*x + x**2 + 1) + sqrt( - sqrt(2) + 2)*log(sqrt( - sqrt(2) + 2)*x + x**2 + 1) - sqrt(sq rt(2) + 2)*log( - sqrt(sqrt(2) + 2)*x + x**2 + 1) + sqrt(sqrt(2) + 2)*log( sqrt(sqrt(2) + 2)*x + x**2 + 1))/16