Integrand size = 14, antiderivative size = 67 \[ \int \frac {x^6}{1+x^2+x^4} \, dx=-x+\frac {x^3}{3}-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{2} \text {arctanh}\left (\frac {x}{1+x^2}\right ) \] Output:
-x+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 = 81, normalized size of antiderivative = 1.21 \[ \int \frac {x^6}{1+x^2+x^4} \, dx=\frac {1}{6} \left (2 x \left (-3+x^2\right )+i \sqrt {6-6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x\right )-i \sqrt {6+6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x\right )\right ) \] Input:
Integrate[x^6/(1 + x^2 + x^4),x]
Output:
(2*x*(-3 + x^2) + I*Sqrt[6 - (6*I)*Sqrt[3]]*ArcTan[((-I + Sqrt[3])*x)/2] - I*Sqrt[6 + (6*I)*Sqrt[3]]*ArcTan[((I + Sqrt[3])*x)/2])/6
Time = 0.45 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.21, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {1442, 27, 1602, 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.
\(\displaystyle \int \frac {x^6}{x^4+x^2+1} \, dx\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle \frac {x^3}{3}-\frac {1}{3} \int \frac {3 x^2 \left (x^2+1\right )}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^3}{3}-\int \frac {x^2 \left (x^2+1\right )}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle \int \frac {1}{x^4+x^2+1}dx+\frac {x^3}{3}-x\) |
\(\Big \downarrow \) 1407 |
\(\displaystyle \frac {1}{2} \int \frac {1-x}{x^2-x+1}dx+\frac {1}{2} \int \frac {x+1}{x^2+x+1}dx+\frac {x^3}{3}-x\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2+x+1}dx+\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {x^3}{3}-x\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2+x+1}dx+\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {x^3}{3}-x\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx-\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )+\frac {x^3}{3}-x\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx+\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {x^3}{3}-x\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (x^2+x+1\right )\right )+\frac {x^3}{3}-x\) |
Input:
Int[x^6/(1 + x^2 + x^4),x]
Output:
-x + x^3/3 + (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^3*(d*x)^(m - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[d^4/(c*(m + 4*p + 1)) Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x ] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 3] && NeQ[m + 4*p + 1, 0] && 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[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {x^{3}}{3}-x +\frac {\ln \left (x^{2}+x +1\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (x^{2}-x +1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}\) | \(62\) |
risch | \(\frac {x^{3}}{3}-x +\frac {\ln \left (4 x^{2}+4 x +4\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (4 x^{2}-4 x +4\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}\) | \(68\) |
Input:
int(x^6/(x^4+x^2+1),x,method=_RETURNVERBOSE)
Output:
1/3*x^3-x+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))
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \frac {x^6}{1+x^2+x^4} \, dx=\frac {1}{3} \, x^{3} + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - x + \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^6/(x^4+x^2+1),x, algorithm="fricas")
Output:
1/3*x^3 + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1 /3*sqrt(3)*(2*x - 1)) - x + 1/4*log(x^2 + x + 1) - 1/4*log(x^2 - x + 1)
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.13 \[ \int \frac {x^6}{1+x^2+x^4} \, dx=\frac {x^{3}}{3} - x - \frac {\log {\left (x^{2} - x + 1 \right )}}{4} + \frac {\log {\left (x^{2} + x + 1 \right )}}{4} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \] Input:
integrate(x**6/(x**4+x**2+1),x)
Output:
x**3/3 - x - log(x**2 - x + 1)/4 + log(x**2 + x + 1)/4 + sqrt(3)*atan(2*sq rt(3)*x/3 - sqrt(3)/3)/6 + sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/6
Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \frac {x^6}{1+x^2+x^4} \, dx=\frac {1}{3} \, x^{3} + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - x + \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^6/(x^4+x^2+1),x, algorithm="maxima")
Output:
1/3*x^3 + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1 /3*sqrt(3)*(2*x - 1)) - x + 1/4*log(x^2 + x + 1) - 1/4*log(x^2 - x + 1)
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \frac {x^6}{1+x^2+x^4} \, dx=\frac {1}{3} \, x^{3} + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - x + \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^6/(x^4+x^2+1),x, algorithm="giac")
Output:
1/3*x^3 + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1 /3*sqrt(3)*(2*x - 1)) - x + 1/4*log(x^2 + x + 1) - 1/4*log(x^2 - x + 1)
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.85 \[ \int \frac {x^6}{1+x^2+x^4} \, dx=\frac {x^3}{3}+\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{6}+\frac {1}{2}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{6}-\frac {1}{2}{}\mathrm {i}\right )-x \] Input:
int(x^6/(x^2 + x^4 + 1),x)
Output:
atan((x*2i)/(3^(1/2)*1i - 1))*(3^(1/2)/6 + 1i/2) - x + atan((x*2i)/(3^(1/2 )*1i + 1))*(3^(1/2)/6 - 1i/2) + x^3/3
Time = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {x^6}{1+x^2+x^4} \, dx=\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{6}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )}{6}-\frac {\mathrm {log}\left (x^{2}-x +1\right )}{4}+\frac {\mathrm {log}\left (x^{2}+x +1\right )}{4}+\frac {x^{3}}{3}-x \] Input:
int(x^6/(x^4+x^2+1),x)
Output:
(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) + 4*x**3 - 12*x)/12