Integrand size = 9, antiderivative size = 49 \[ \int \frac {x}{1+x^6} \, dx=-\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{6} \log \left (1+x^2\right )-\frac {1}{12} \log \left (1-x^2+x^4\right ) \] Output:
1/6*ln(x^2+1)-1/12*ln(x^4-x^2+1)-1/6*arctan(1/3*(-2*x^2+1)*3^(1/2))*3^(1/2 )
Time = 0.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.59 \[ \int \frac {x}{1+x^6} \, dx=\frac {1}{12} \left (-2 \sqrt {3} \arctan \left (\sqrt {3}-2 x\right )-2 \sqrt {3} \arctan \left (\sqrt {3}+2 x\right )+2 \log \left (1+x^2\right )-\log \left (1-\sqrt {3} x+x^2\right )-\log \left (1+\sqrt {3} x+x^2\right )\right ) \] Input:
Integrate[x/(1 + x^6),x]
Output:
(-2*Sqrt[3]*ArcTan[Sqrt[3] - 2*x] - 2*Sqrt[3]*ArcTan[Sqrt[3] + 2*x] + 2*Lo g[1 + x^2] - Log[1 - Sqrt[3]*x + x^2] - Log[1 + Sqrt[3]*x + x^2])/12
Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.889, Rules used = {807, 750, 16, 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.
\(\displaystyle \int \frac {x}{x^6+1} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^6+1}dx^2\) |
\(\Big \downarrow \) 750 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {1}{x^2+1}dx^2+\frac {1}{3} \int \frac {2-x^2}{x^4-x^2+1}dx^2\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {2-x^2}{x^4-x^2+1}dx^2+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^4-x^2+1}dx^2-\frac {1}{2} \int -\frac {1-2 x^2}{x^4-x^2+1}dx^2\right )+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^4-x^2+1}dx^2+\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2\right )+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2-3 \int \frac {1}{-x^4-3}d\left (2 x^2-1\right )\right )+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2+\sqrt {3} \arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (x^4-x^2+1\right )\right )+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
Input:
Int[x/(1 + x^6),x]
Output:
(Log[1 + x^2]/3 + (Sqrt[3]*ArcTan[(-1 + 2*x^2)/Sqrt[3]] - Log[1 - x^2 + x^ 4]/2)/3)/2
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_.)*((a_) + (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))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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]
Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80
method | result | size |
risch | \(\frac {\ln \left (x^{2}+1\right )}{6}-\frac {\ln \left (x^{4}-x^{2}+1\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{2}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}\) | \(39\) |
default | \(\frac {\ln \left (x^{2}+1\right )}{6}-\frac {\ln \left (x^{4}-x^{2}+1\right )}{12}+\frac {\arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}\) | \(41\) |
meijerg | \(\frac {x^{2} \ln \left (1+\left (x^{6}\right )^{\frac {1}{3}}\right )}{6 \left (x^{6}\right )^{\frac {1}{3}}}-\frac {x^{2} \ln \left (1-\left (x^{6}\right )^{\frac {1}{3}}+\left (x^{6}\right )^{\frac {2}{3}}\right )}{12 \left (x^{6}\right )^{\frac {1}{3}}}+\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{3}}}{2-\left (x^{6}\right )^{\frac {1}{3}}}\right )}{6 \left (x^{6}\right )^{\frac {1}{3}}}\) | \(80\) |
Input:
int(x/(x^6+1),x,method=_RETURNVERBOSE)
Output:
1/6*ln(x^2+1)-1/12*ln(x^4-x^2+1)+1/6*3^(1/2)*arctan(2/3*(x^2-1/2)*3^(1/2))
Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {x}{1+x^6} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \] Input:
integrate(x/(x^6+1),x, algorithm="fricas")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/12*log(x^4 - x^2 + 1) + 1/ 6*log(x^2 + 1)
Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int \frac {x}{1+x^6} \, dx=\frac {\log {\left (x^{2} + 1 \right )}}{6} - \frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{6} \] Input:
integrate(x/(x**6+1),x)
Output:
log(x**2 + 1)/6 - log(x**4 - x**2 + 1)/12 + sqrt(3)*atan(2*sqrt(3)*x**2/3 - sqrt(3)/3)/6
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {x}{1+x^6} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \] Input:
integrate(x/(x^6+1),x, algorithm="maxima")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/12*log(x^4 - x^2 + 1) + 1/ 6*log(x^2 + 1)
Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {x}{1+x^6} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \] Input:
integrate(x/(x^6+1),x, algorithm="giac")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/12*log(x^4 - x^2 + 1) + 1/ 6*log(x^2 + 1)
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {x}{1+x^6} \, dx=\frac {\ln \left (x^2+1\right )}{6}-\ln \left (x^2-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\ln \left (x^2+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \] Input:
int(x/(x^6 + 1),x)
Output:
log(x^2 + 1)/6 - log(x^2 - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/12 + 1/12) + log((3^(1/2)*1i)/2 + x^2 - 1/2)*((3^(1/2)*1i)/12 - 1/12)
Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.14 \[ \int \frac {x}{1+x^6} \, dx=-\frac {\sqrt {3}\, \mathit {atan} \left (\sqrt {3}-2 x \right )}{6}-\frac {\sqrt {3}\, \mathit {atan} \left (\sqrt {3}+2 x \right )}{6}+\frac {\mathrm {log}\left (x^{2}+1\right )}{6}-\frac {\mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )}{12}-\frac {\mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )}{12} \] Input:
int(x/(x^6+1),x)
Output:
( - 2*sqrt(3)*atan(sqrt(3) - 2*x) - 2*sqrt(3)*atan(sqrt(3) + 2*x) + 2*log( x**2 + 1) - log( - sqrt(3)*x + x**2 + 1) - log(sqrt(3)*x + x**2 + 1))/12