Integrand size = 14, antiderivative size = 82 \[ \int \frac {x}{1-x^4+x^8} \, dx=-\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )+\frac {1}{4} \arctan \left (\sqrt {3}+2 x^2\right )-\frac {\log \left (1-\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}} \]
1/4*arctan(2*x^2-3^(1/2))+1/4*arctan(2*x^2+3^(1/2))-1/24*ln(1+x^4-x^2*3^(1 /2))*3^(1/2)+1/24*ln(1+x^4+x^2*3^(1/2))*3^(1/2)
Result contains complex when optimal does not.
Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01 \[ \int \frac {x}{1-x^4+x^8} \, dx=\frac {i \left (\sqrt {-1-i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )-\sqrt {-1+i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )\right )}{2 \sqrt {6}} \]
((I/2)*(Sqrt[-1 - I*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x^2)/2] - Sqrt[-1 + I *Sqrt[3]]*ArcTan[((1 + I*Sqrt[3])*x^2)/2]))/Sqrt[6]
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.23, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1695, 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.
\(\displaystyle \int \frac {x}{x^8-x^4+1} \, dx\) |
\(\Big \downarrow \) 1695 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^8-x^4+1}dx^2\) |
\(\Big \downarrow \) 1407 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {3}-x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}+\frac {\int \frac {x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^4-\sqrt {3} x^2+1}dx^2-\frac {1}{2} \int -\frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}+\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^4+\sqrt {3} x^2+1}dx^2+\frac {1}{2} \int \frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^4-\sqrt {3} x^2+1}dx^2+\frac {1}{2} \int \frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}+\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^4+\sqrt {3} x^2+1}dx^2+\frac {1}{2} \int \frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} \int \frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2-\sqrt {3} \int \frac {1}{-x^4-1}d\left (2 x^2-\sqrt {3}\right )}{2 \sqrt {3}}+\frac {\frac {1}{2} \int \frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2-\sqrt {3} \int \frac {1}{-x^4-1}d\left (2 x^2+\sqrt {3}\right )}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} \int \frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2-\sqrt {3} \arctan \left (\sqrt {3}-2 x^2\right )}{2 \sqrt {3}}+\frac {\frac {1}{2} \int \frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2+\sqrt {3} \arctan \left (2 x^2+\sqrt {3}\right )}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {-\sqrt {3} \arctan \left (\sqrt {3}-2 x^2\right )-\frac {1}{2} \log \left (x^4-\sqrt {3} x^2+1\right )}{2 \sqrt {3}}+\frac {\sqrt {3} \arctan \left (2 x^2+\sqrt {3}\right )+\frac {1}{2} \log \left (x^4+\sqrt {3} x^2+1\right )}{2 \sqrt {3}}\right )\) |
((-(Sqrt[3]*ArcTan[Sqrt[3] - 2*x^2]) - Log[1 - Sqrt[3]*x^2 + x^4]/2)/(2*Sq rt[3]) + (Sqrt[3]*ArcTan[Sqrt[3] + 2*x^2] + Log[1 + Sqrt[3]*x^2 + x^4]/2)/ (2*Sqrt[3]))/2
3.4.53.3.1 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_) + (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[((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]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b *x^(n/k) + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.39
method | result | size |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-3 \textit {\_R}^{3}+x^{2}+\textit {\_R} \right )\right )}{4}\) | \(32\) |
default | \(\frac {\arctan \left (2 x^{2}-\sqrt {3}\right )}{4}+\frac {\arctan \left (2 x^{2}+\sqrt {3}\right )}{4}-\frac {\ln \left (1+x^{4}-x^{2} \sqrt {3}\right ) \sqrt {3}}{24}+\frac {\ln \left (1+x^{4}+x^{2} \sqrt {3}\right ) \sqrt {3}}{24}\) | \(65\) |
Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.01 \[ \int \frac {x}{1-x^4+x^8} \, dx=-\frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (12 \, x^{2} + \sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (i \, \sqrt {3} - 3\right )}\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (12 \, x^{2} + \sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} + 3\right )}\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (12 \, x^{2} + \sqrt {6} {\left (i \, \sqrt {3} + 3\right )} \sqrt {-i \, \sqrt {3} - 1}\right ) - \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (12 \, x^{2} + \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} - 3\right )}\right ) \]
-1/24*sqrt(6)*sqrt(I*sqrt(3) - 1)*log(12*x^2 + sqrt(6)*sqrt(I*sqrt(3) - 1) *(I*sqrt(3) - 3)) + 1/24*sqrt(6)*sqrt(I*sqrt(3) - 1)*log(12*x^2 + sqrt(6)* sqrt(I*sqrt(3) - 1)*(-I*sqrt(3) + 3)) + 1/24*sqrt(6)*sqrt(-I*sqrt(3) - 1)* log(12*x^2 + sqrt(6)*(I*sqrt(3) + 3)*sqrt(-I*sqrt(3) - 1)) - 1/24*sqrt(6)* sqrt(-I*sqrt(3) - 1)*log(12*x^2 + sqrt(6)*sqrt(-I*sqrt(3) - 1)*(-I*sqrt(3) - 3))
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {x}{1-x^4+x^8} \, dx=- \frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\operatorname {atan}{\left (2 x^{2} - \sqrt {3} \right )}}{4} + \frac {\operatorname {atan}{\left (2 x^{2} + \sqrt {3} \right )}}{4} \]
-sqrt(3)*log(x**4 - sqrt(3)*x**2 + 1)/24 + sqrt(3)*log(x**4 + sqrt(3)*x**2 + 1)/24 + atan(2*x**2 - sqrt(3))/4 + atan(2*x**2 + sqrt(3))/4
\[ \int \frac {x}{1-x^4+x^8} \, dx=\int { \frac {x}{x^{8} - x^{4} + 1} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int \frac {x}{1-x^4+x^8} \, dx=\frac {1}{24} \, \sqrt {3} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) + \frac {1}{4} \, \arctan \left (2 \, x^{2} + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (2 \, x^{2} - \sqrt {3}\right ) \]
1/24*sqrt(3)*log(x^4 + sqrt(3)*x^2 + 1) - 1/24*sqrt(3)*log(x^4 - sqrt(3)*x ^2 + 1) + 1/4*arctan(2*x^2 + sqrt(3)) + 1/4*arctan(2*x^2 - sqrt(3))
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65 \[ \int \frac {x}{1-x^4+x^8} \, dx=-\mathrm {atan}\left (-\frac {x^2}{2}+\frac {\sqrt {3}\,x^2\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {x^2}{2}+\frac {\sqrt {3}\,x^2\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]