Integrand size = 12, antiderivative size = 63 \[ \int \frac {x}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \text {arctanh}\left (\frac {x^2}{1+x^4}\right ) \] Output:
-1/12*arctan(1/3*(-2*x^2+1)*3^(1/2))*3^(1/2)+1/12*arctan(1/3*(2*x^2+1)*3^( 1/2))*3^(1/2)+1/4*arctanh(x^2/(x^4+1))
Result contains complex when optimal does not.
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25 \[ \int \frac {x}{1+x^4+x^8} \, dx=\frac {i \left (\sqrt {1-i \sqrt {3}} \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x^2\right )-\sqrt {1+i \sqrt {3}} \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x^2\right )\right )}{2 \sqrt {6}} \] Input:
Integrate[x/(1 + x^4 + x^8),x]
Output:
((I/2)*(Sqrt[1 - I*Sqrt[3]]*ArcTan[((-I + Sqrt[3])*x^2)/2] - Sqrt[1 + I*Sq rt[3]]*ArcTan[((I + Sqrt[3])*x^2)/2]))/Sqrt[6]
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, 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 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx^2+\frac {1}{2} \int \frac {x^2+1}{x^4+x^2+1}dx^2\right )\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^4-x^2+1}dx^2-\frac {1}{2} \int -\frac {1-2 x^2}{x^4-x^2+1}dx^2\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^4+x^2+1}dx^2+\frac {1}{2} \int \frac {2 x^2+1}{x^4+x^2+1}dx^2\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^4-x^2+1}dx^2+\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^4+x^2+1}dx^2+\frac {1}{2} \int \frac {2 x^2+1}{x^4+x^2+1}dx^2\right )\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2-\int \frac {1}{-x^4-3}d\left (2 x^2-1\right )\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {2 x^2+1}{x^4+x^2+1}dx^2-\int \frac {1}{-x^4-3}d\left (2 x^2+1\right )\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2+\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {2 x^2+1}{x^4+x^2+1}dx^2+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^4-x^2+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (x^4+x^2+1\right )\right )\right )\) |
Input:
Int[x/(1 + x^4 + x^8),x]
Output:
((ArcTan[(-1 + 2*x^2)/Sqrt[3]]/Sqrt[3] - Log[1 - x^2 + x^4]/2)/2 + (ArcTan [(1 + 2*x^2)/Sqrt[3]]/Sqrt[3] + Log[1 + x^2 + x^4]/2)/2)/2
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]
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98
method | result | size |
default | \(-\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{4}+x^{2}+1\right )}{8}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}\) | \(62\) |
risch | \(\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (4 x^{4}-4 x^{2}+4\right )}{8}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}+\frac {\ln \left (4 x^{4}+4 x^{2}+4\right )}{8}\) | \(68\) |
Input:
int(x/(x^8+x^4+1),x,method=_RETURNVERBOSE)
Output:
-1/8*ln(x^4-x^2+1)+1/12*3^(1/2)*arctan(1/3*(2*x^2-1)*3^(1/2))+1/8*ln(x^4+x ^2+1)+1/12*arctan(1/3*(2*x^2+1)*3^(1/2))*3^(1/2)
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {x}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \] Input:
integrate(x/(x^8+x^4+1),x, algorithm="fricas")
Output:
1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqr t(3)*(2*x^2 - 1)) + 1/8*log(x^4 + x^2 + 1) - 1/8*log(x^4 - x^2 + 1)
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.21 \[ \int \frac {x}{1+x^4+x^8} \, dx=- \frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{8} + \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{12} \] Input:
integrate(x/(x**8+x**4+1),x)
Output:
-log(x**4 - x**2 + 1)/8 + log(x**4 + x**2 + 1)/8 + sqrt(3)*atan(2*sqrt(3)* x**2/3 - sqrt(3)/3)/12 + sqrt(3)*atan(2*sqrt(3)*x**2/3 + sqrt(3)/3)/12
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {x}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \] Input:
integrate(x/(x^8+x^4+1),x, algorithm="maxima")
Output:
1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqr t(3)*(2*x^2 - 1)) + 1/8*log(x^4 + x^2 + 1) - 1/8*log(x^4 - x^2 + 1)
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {x}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \] Input:
integrate(x/(x^8+x^4+1),x, algorithm="giac")
Output:
1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqr t(3)*(2*x^2 - 1)) + 1/8*log(x^4 + x^2 + 1) - 1/8*log(x^4 - x^2 + 1)
Time = 19.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int \frac {x}{1+x^4+x^8} \, dx=\mathrm {atan}\left (\frac {\sqrt {3}\,x^2}{2}-\frac {x^2\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{12}+\frac {1}{4}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {\sqrt {3}\,x^2}{2}+\frac {x^2\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{12}-\frac {1}{4}{}\mathrm {i}\right ) \] Input:
int(x/(x^4 + x^8 + 1),x)
Output:
atan((3^(1/2)*x^2)/2 - (x^2*1i)/2)*(3^(1/2)/12 + 1i/4) + atan((3^(1/2)*x^2 )/2 + (x^2*1i)/2)*(3^(1/2)/12 - 1i/4)
Time = 0.15 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.56 \[ \int \frac {x}{1+x^4+x^8} \, dx=-\frac {\sqrt {3}\, \mathit {atan} \left (\sqrt {3}-2 x \right )}{12}-\frac {\sqrt {3}\, \mathit {atan} \left (\sqrt {3}+2 x \right )}{12}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{12}-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )}{12}+\frac {\mathrm {log}\left (x^{2}-x +1\right )}{8}+\frac {\mathrm {log}\left (x^{2}+x +1\right )}{8}-\frac {\mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )}{8}-\frac {\mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )}{8} \] Input:
int(x/(x^8+x^4+1),x)
Output:
( - 2*sqrt(3)*atan(sqrt(3) - 2*x) - 2*sqrt(3)*atan(sqrt(3) + 2*x) + 2*sqrt (3)*atan((2*x - 1)/sqrt(3)) - 2*sqrt(3)*atan((2*x + 1)/sqrt(3)) + 3*log(x* *2 - x + 1) + 3*log(x**2 + x + 1) - 3*log( - sqrt(3)*x + x**2 + 1) - 3*log (sqrt(3)*x + x**2 + 1))/24