Integrand size = 18, antiderivative size = 140 \[ \int \frac {1-x^4}{1+x^4+x^8} \, dx=-\frac {1}{4} \sqrt {3} \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )+\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )+\frac {1}{4} \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-\frac {1}{4} \arctan \left (\sqrt {3}+2 x\right )+\frac {1}{8} \log \left (1-x+x^2\right )-\frac {1}{8} \log \left (1+x+x^2\right )-\frac {1}{8} \sqrt {3} \log \left (1-\sqrt {3} x+x^2\right )+\frac {1}{8} \sqrt {3} \log \left (1+\sqrt {3} x+x^2\right ) \]
-1/4*arctan(2*x-3^(1/2))-1/4*arctan(2*x+3^(1/2))+1/8*ln(x^2-x+1)-1/8*ln(x^ 2+x+1)-1/4*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/4*arctan(1/3*(1+2*x)*3^(1 /2))*3^(1/2)-1/8*ln(1+x^2-x*3^(1/2))*3^(1/2)+1/8*ln(1+x^2+x*3^(1/2))*3^(1/ 2)
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.92 \[ \int \frac {1-x^4}{1+x^4+x^8} \, dx=\frac {1}{8} \left (-2 \sqrt {-2-2 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )-2 \sqrt {-2+2 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )+2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+\log \left (1-x+x^2\right )-\log \left (1+x+x^2\right )\right ) \]
(-2*Sqrt[-2 - (2*I)*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x)/2] - 2*Sqrt[-2 + ( 2*I)*Sqrt[3]]*ArcTan[((1 + I*Sqrt[3])*x)/2] + 2*Sqrt[3]*ArcTan[(-1 + 2*x)/ Sqrt[3]] + 2*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] + Log[1 - x + x^2] - Log[1 + x + x^2])/8
Time = 0.37 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1751, 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.
| \(\displaystyle \int \frac {1-x^4}{x^8+x^4+1} \, dx\) |
| \(\Big \downarrow \) 1751 |
| \(\displaystyle -\frac {1}{2} \int -\frac {1-2 x^2}{x^4-x^2+1}dx-\frac {1}{2} \int -\frac {2 x^2+1}{x^4+x^2+1}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx+\frac {1}{2} \int \frac {2 x^2+1}{x^4+x^2+1}dx\) |
| \(\Big \downarrow \) 1483 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {x+1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1-x}{x^2+x+1}dx\right )+\frac {1}{2} \left (\frac {\int \frac {\sqrt {3}-3 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {\int \frac {\sqrt {3} \left (\sqrt {3} x+1\right )}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {x+1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1-x}{x^2+x+1}dx\right )+\frac {1}{2} \left (\frac {\int \frac {\sqrt {3}-3 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {1}{2} \int \frac {\sqrt {3} x+1}{x^2+\sqrt {3} x+1}dx\right )\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {3}{2} \int \frac {1}{x^2-x+1}dx+\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {3}{2} \int \frac {1}{x^2+x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )\right )+\frac {1}{2} \left (\frac {-\frac {1}{2} \sqrt {3} \int \frac {1}{x^2-\sqrt {3} x+1}dx-\frac {3}{2} \int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {1}{2} \left (\frac {1}{2} \sqrt {3} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {3} x+1}dx\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {3}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {3}{2} \int \frac {1}{x^2+x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )\right )+\frac {1}{2} \left (\frac {\frac {3}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx-\frac {1}{2} \sqrt {3} \int \frac {1}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {1}{2} \left (\frac {1}{2} \sqrt {3} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {3} x+1}dx\right )\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-3 \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)\right )+\frac {1}{2} \left (-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx-3 \int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )\right )+\frac {1}{2} \left (\frac {\frac {3}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx+\sqrt {3} \int \frac {1}{-\left (2 x-\sqrt {3}\right )^2-1}d\left (2 x-\sqrt {3}\right )}{2 \sqrt {3}}+\frac {1}{2} \left (\frac {1}{2} \sqrt {3} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx+\int \frac {1}{-\left (2 x+\sqrt {3}\right )^2-1}d\left (2 x+\sqrt {3}\right )\right )\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\sqrt {3} \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )\right )+\frac {1}{2} \left (\frac {\frac {3}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx+\sqrt {3} \arctan \left (\sqrt {3}-2 x\right )}{2 \sqrt {3}}+\frac {1}{2} \left (\frac {1}{2} \sqrt {3} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx-\arctan \left (2 x+\sqrt {3}\right )\right )\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )+\frac {1}{2} \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\sqrt {3} \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (x^2+x+1\right )\right )\right )+\frac {1}{2} \left (\frac {\sqrt {3} \arctan \left (\sqrt {3}-2 x\right )-\frac {3}{2} \log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}+\frac {1}{2} \left (\frac {1}{2} \sqrt {3} \log \left (x^2+\sqrt {3} x+1\right )-\arctan \left (2 x+\sqrt {3}\right )\right )\right )\) |
((Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] + Log[1 - x + x^2]/2)/2 + (Sqrt[3]*Ar cTan[(1 + 2*x)/Sqrt[3]] - Log[1 + x + x^2]/2)/2)/2 + ((Sqrt[3]*ArcTan[Sqrt [3] - 2*x] - (3*Log[1 - Sqrt[3]*x + x^2])/2)/(2*Sqrt[3]) + (-ArcTan[Sqrt[3 ] + 2*x] + (Sqrt[3]*Log[1 + Sqrt[3]*x + x^2])/2)/2)/2
3.1.23.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_) + (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]
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x _Symbol] :> With[{q = Rt[-2*(d/e) - b/c, 2]}, Simp[e/(2*c*q) Int[(q - 2*x ^(n/2))/Simp[d/e + q*x^(n/2) - x^n, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x^(n/2))/Simp[d/e - q*x^(n/2) - x^n, x], x], x]] /; FreeQ[{a, b, c, d, e} , x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && IGt Q[n/2, 0] && !GtQ[b^2 - 4*a*c, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.62
| method | result | size |
| risch | \(-\frac {\ln \left (4 x^{2}+4 x +4\right )}{8}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{4}+\frac {\ln \left (4 x^{2}-4 x +4\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-\textit {\_R}^{3}+\textit {\_R} +x \right )\right )}{4}\) | \(87\) |
| default | \(\frac {\ln \left (x^{2}-x +1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{4}-\frac {\ln \left (x^{2}+x +1\right )}{8}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{4}-\frac {\ln \left (1+x^{2}-x \sqrt {3}\right ) \sqrt {3}}{8}-\frac {\arctan \left (2 x -\sqrt {3}\right )}{4}+\frac {\ln \left (1+x^{2}+x \sqrt {3}\right ) \sqrt {3}}{8}-\frac {\arctan \left (2 x +\sqrt {3}\right )}{4}\) | \(109\) |
-1/8*ln(4*x^2+4*x+4)+1/4*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+1/8*ln(4*x^2- 4*x+4)+1/4*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))+1/4*sum(_R*ln(-_R^3+_R+x),_ R=RootOf(_Z^4-_Z^2+1))
Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.39 \[ \int \frac {1-x^4}{1+x^4+x^8} \, dx=-\frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {-3} + 1} \log \left (\sqrt {2} \sqrt {\sqrt {-3} + 1} {\left (\sqrt {-3} - 1\right )} + 4 \, x\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {-3} + 1} \log \left (-\sqrt {2} \sqrt {\sqrt {-3} + 1} {\left (\sqrt {-3} - 1\right )} + 4 \, x\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {-\sqrt {-3} + 1} \log \left (\sqrt {2} {\left (\sqrt {-3} + 1\right )} \sqrt {-\sqrt {-3} + 1} + 4 \, x\right ) - \frac {1}{8} \, \sqrt {2} \sqrt {-\sqrt {-3} + 1} \log \left (-\sqrt {2} {\left (\sqrt {-3} + 1\right )} \sqrt {-\sqrt {-3} + 1} + 4 \, x\right ) + \frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \]
-1/8*sqrt(2)*sqrt(sqrt(-3) + 1)*log(sqrt(2)*sqrt(sqrt(-3) + 1)*(sqrt(-3) - 1) + 4*x) + 1/8*sqrt(2)*sqrt(sqrt(-3) + 1)*log(-sqrt(2)*sqrt(sqrt(-3) + 1 )*(sqrt(-3) - 1) + 4*x) + 1/8*sqrt(2)*sqrt(-sqrt(-3) + 1)*log(sqrt(2)*(sqr t(-3) + 1)*sqrt(-sqrt(-3) + 1) + 4*x) - 1/8*sqrt(2)*sqrt(-sqrt(-3) + 1)*lo g(-sqrt(2)*(sqrt(-3) + 1)*sqrt(-sqrt(-3) + 1) + 4*x) + 1/4*sqrt(3)*arctan( 1/3*sqrt(3)*(2*x + 1)) + 1/4*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/8*l og(x^2 + x + 1) + 1/8*log(x^2 - x + 1)
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int \frac {1-x^4}{1+x^4+x^8} \, dx=- \left (- \frac {1}{8} - \frac {\sqrt {3} i}{8}\right ) \log {\left (x + 1024 \left (- \frac {1}{8} - \frac {\sqrt {3} i}{8}\right )^{5} \right )} - \left (- \frac {1}{8} + \frac {\sqrt {3} i}{8}\right ) \log {\left (x + 1024 \left (- \frac {1}{8} + \frac {\sqrt {3} i}{8}\right )^{5} \right )} - \left (\frac {1}{8} - \frac {\sqrt {3} i}{8}\right ) \log {\left (x + 1024 \left (\frac {1}{8} - \frac {\sqrt {3} i}{8}\right )^{5} \right )} - \left (\frac {1}{8} + \frac {\sqrt {3} i}{8}\right ) \log {\left (x + 1024 \left (\frac {1}{8} + \frac {\sqrt {3} i}{8}\right )^{5} \right )} - \operatorname {RootSum} {\left (256 t^{4} - 16 t^{2} + 1, \left ( t \mapsto t \log {\left (1024 t^{5} + x \right )} \right )\right )} \]
-(-1/8 - sqrt(3)*I/8)*log(x + 1024*(-1/8 - sqrt(3)*I/8)**5) - (-1/8 + sqrt (3)*I/8)*log(x + 1024*(-1/8 + sqrt(3)*I/8)**5) - (1/8 - sqrt(3)*I/8)*log(x + 1024*(1/8 - sqrt(3)*I/8)**5) - (1/8 + sqrt(3)*I/8)*log(x + 1024*(1/8 + sqrt(3)*I/8)**5) - RootSum(256*_t**4 - 16*_t**2 + 1, Lambda(_t, _t*log(102 4*_t**5 + x)))
\[ \int \frac {1-x^4}{1+x^4+x^8} \, dx=\int { -\frac {x^{4} - 1}{x^{8} + x^{4} + 1} \,d x } \]
1/4*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/4*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) - 1/2*integrate((2*x^2 - 1)/(x^4 - x^2 + 1), x) - 1/8*log(x^2 + x + 1) + 1/8*log(x^2 - x + 1)
Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \frac {1-x^4}{1+x^4+x^8} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \frac {1}{8} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) - \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) - \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) - \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \]
1/4*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/4*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) + 1/8*sqrt(3)*log(x^2 + sqrt(3)*x + 1) - 1/8*sqrt(3)*log(x^2 - sqrt(3)*x + 1) - 1/4*arctan(2*x + sqrt(3)) - 1/4*arctan(2*x - sqrt(3)) - 1/8*log(x^2 + x + 1) + 1/8*log(x^2 - x + 1)
Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.78 \[ \int \frac {1-x^4}{1+x^4+x^8} \, dx=-\mathrm {atan}\left (\frac {54\,\sqrt {3}\,x}{-81+\sqrt {3}\,27{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{4}+\frac {1}{4}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {54\,\sqrt {3}\,x}{81+\sqrt {3}\,27{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{4}-\frac {1}{4}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {\sqrt {3}\,x\,54{}\mathrm {i}}{-81+\sqrt {3}\,27{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\mathrm {atan}\left (\frac {\sqrt {3}\,x\,54{}\mathrm {i}}{81+\sqrt {3}\,27{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right ) \]