Integrand size = 14, antiderivative size = 71 \[ \int \frac {1}{x^6 \left (1+x^2+x^4\right )} \, dx=-\frac {1}{5 x^5}+\frac {1}{3 x^3}-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{2} \text {arctanh}\left (\frac {x}{1+x^2}\right ) \] Output:
-1/5/x^5+1/3/x^3-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/6*arctan(1/3*(1 +2*x)*3^(1/2))*3^(1/2)+1/2*arctanh(x/(x^2+1))
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^6 \left (1+x^2+x^4\right )} \, dx=\frac {-6+10 x^2+5 i \sqrt {6-6 i \sqrt {3}} x^5 \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x\right )-5 i \sqrt {6+6 i \sqrt {3}} x^5 \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x\right )}{30 x^5} \] Input:
Integrate[1/(x^6*(1 + x^2 + x^4)),x]
Output:
(-6 + 10*x^2 + (5*I)*Sqrt[6 - (6*I)*Sqrt[3]]*x^5*ArcTan[((-I + Sqrt[3])*x) /2] - (5*I)*Sqrt[6 + (6*I)*Sqrt[3]]*x^5*ArcTan[((I + Sqrt[3])*x)/2])/(30*x ^5)
Time = 0.47 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {1443, 27, 1604, 27, 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.
\(\displaystyle \int \frac {1}{x^6 \left (x^4+x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 1443 |
\(\displaystyle \frac {1}{5} \int -\frac {5 \left (x^2+1\right )}{x^4 \left (x^4+x^2+1\right )}dx-\frac {1}{5 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {x^2+1}{x^4 \left (x^4+x^2+1\right )}dx-\frac {1}{5 x^5}\) |
\(\Big \downarrow \) 1604 |
\(\displaystyle \frac {1}{3} \int \frac {3}{x^4+x^2+1}dx-\frac {1}{5 x^5}+\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x^4+x^2+1}dx-\frac {1}{5 x^5}+\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 1407 |
\(\displaystyle \frac {1}{2} \int \frac {1-x}{x^2-x+1}dx+\frac {1}{2} \int \frac {x+1}{x^2+x+1}dx-\frac {1}{5 x^5}+\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2+x+1}dx+\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )-\frac {1}{5 x^5}+\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2+x+1}dx+\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )-\frac {1}{5 x^5}+\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx-\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )-\frac {1}{5 x^5}+\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx+\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{5 x^5}+\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (x^2+x+1\right )\right )-\frac {1}{5 x^5}+\frac {1}{3 x^3}\) |
Input:
Int[1/(x^6*(1 + x^2 + x^4)),x]
Output:
-1/5*1/x^5 + 1/(3*x^3) + (ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3] - Log[1 - x + x^2]/2)/2 + (ArcTan[(1 + 2*x)/Sqrt[3]]/Sqrt[3] + Log[1 + x + x^2]/2)/2
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[((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[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1)/(a*d*(m + 1))), x] - Sim p[1/(a*d^2*(m + 1)) Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p + 5)* x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) /(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1)) Int[(f*x)^(m + 2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x ], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {\ln \left (x^{2}+x +1\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (x^{2}-x +1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {1}{5 x^{5}}+\frac {1}{3 x^{3}}\) | \(64\) |
risch | \(\frac {\frac {x^{2}}{3}-\frac {1}{5}}{x^{5}}+\frac {\ln \left (4 x^{2}+4 x +4\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (x^{2}-x +1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}\) | \(67\) |
Input:
int(1/x^6/(x^4+x^2+1),x,method=_RETURNVERBOSE)
Output:
1/4*ln(x^2+x+1)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/4*ln(x^2-x+1)+1/ 6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))-1/5/x^5+1/3/x^3
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^6 \left (1+x^2+x^4\right )} \, dx=\frac {10 \, \sqrt {3} x^{5} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + 10 \, \sqrt {3} x^{5} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 15 \, x^{5} \log \left (x^{2} + x + 1\right ) - 15 \, x^{5} \log \left (x^{2} - x + 1\right ) + 20 \, x^{2} - 12}{60 \, x^{5}} \] Input:
integrate(1/x^6/(x^4+x^2+1),x, algorithm="fricas")
Output:
1/60*(10*sqrt(3)*x^5*arctan(1/3*sqrt(3)*(2*x + 1)) + 10*sqrt(3)*x^5*arctan (1/3*sqrt(3)*(2*x - 1)) + 15*x^5*log(x^2 + x + 1) - 15*x^5*log(x^2 - x + 1 ) + 20*x^2 - 12)/x^5
Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^6 \left (1+x^2+x^4\right )} \, dx=- \frac {\log {\left (x^{2} - x + 1 \right )}}{4} + \frac {\log {\left (x^{2} + x + 1 \right )}}{4} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} + \frac {5 x^{2} - 3}{15 x^{5}} \] Input:
integrate(1/x**6/(x**4+x**2+1),x)
Output:
-log(x**2 - x + 1)/4 + log(x**2 + x + 1)/4 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/6 + sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/6 + (5*x**2 - 3)/(1 5*x**5)
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^6 \left (1+x^2+x^4\right )} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {5 \, x^{2} - 3}{15 \, x^{5}} + \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(1/x^6/(x^4+x^2+1),x, algorithm="maxima")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) + 1/15*(5*x^2 - 3)/x^5 + 1/4*log(x^2 + x + 1) - 1/4*log(x^2 - x + 1)
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^6 \left (1+x^2+x^4\right )} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {5 \, x^{2} - 3}{15 \, x^{5}} + \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(1/x^6/(x^4+x^2+1),x, algorithm="giac")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) + 1/15*(5*x^2 - 3)/x^5 + 1/4*log(x^2 + x + 1) - 1/4*log(x^2 - x + 1)
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^6 \left (1+x^2+x^4\right )} \, dx=\mathrm {atanh}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\mathrm {atanh}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {\frac {x^2}{3}-\frac {1}{5}}{x^5} \] Input:
int(1/(x^6*(x^2 + x^4 + 1)),x)
Output:
atanh((2*x)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/6 - 1/2) + atanh((2*x)/(3^(1/2 )*1i + 1))*((3^(1/2)*1i)/6 + 1/2) + (x^2/3 - 1/5)/x^5
Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^6 \left (1+x^2+x^4\right )} \, dx=\frac {10 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) x^{5}+10 \sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) x^{5}-15 \,\mathrm {log}\left (x^{2}-x +1\right ) x^{5}+15 \,\mathrm {log}\left (x^{2}+x +1\right ) x^{5}+20 x^{2}-12}{60 x^{5}} \] Input:
int(1/x^6/(x^4+x^2+1),x)
Output:
(10*sqrt(3)*atan((2*x - 1)/sqrt(3))*x**5 + 10*sqrt(3)*atan((2*x + 1)/sqrt( 3))*x**5 - 15*log(x**2 - x + 1)*x**5 + 15*log(x**2 + x + 1)*x**5 + 20*x**2 - 12)/(60*x**5)