Integrand size = 16, antiderivative size = 67 \[ \int \frac {1}{x^6 \left (1-x^2+x^4\right )} \, dx=-\frac {1}{5 x^5}-\frac {1}{3 x^3}+\frac {1}{2} \arctan \left (\sqrt {3}-2 x\right )-\frac {1}{2} \arctan \left (\sqrt {3}+2 x\right )-\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{2 \sqrt {3}} \] Output:
-1/5/x^5-1/3/x^3-1/2*arctan(-3^(1/2)+2*x)-1/2*arctan(3^(1/2)+2*x)-1/6*arct anh(3^(1/2)*x/(x^2+1))*3^(1/2)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x^6 \left (1-x^2+x^4\right )} \, dx=-\frac {6+10 x^2+5 i \sqrt {-6-6 i \sqrt {3}} x^5 \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )-5 i \sqrt {-6+6 i \sqrt {3}} x^5 \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )}{30 x^5} \] Input:
Integrate[1/(x^6*(1 - x^2 + x^4)),x]
Output:
-1/30*(6 + 10*x^2 + (5*I)*Sqrt[-6 - (6*I)*Sqrt[3]]*x^5*ArcTan[((1 - I*Sqrt [3])*x)/2] - (5*I)*Sqrt[-6 + (6*I)*Sqrt[3]]*x^5*ArcTan[((1 + I*Sqrt[3])*x) /2])/x^5
Time = 0.48 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.54, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, 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 |
\(\displaystyle \frac {1}{5} \int \frac {5 \left (1-x^2\right )}{x^4 \left (x^4-x^2+1\right )}dx-\frac {1}{5 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1-x^2}{x^4 \left (x^4-x^2+1\right )}dx-\frac {1}{5 x^5}\) |
\(\Big \downarrow \) 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 |
\(\displaystyle -\frac {\int \frac {\sqrt {3}-x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int \frac {x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {1}{5 x^5}-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle -\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2-\sqrt {3} x+1}dx-\frac {1}{2} \int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2+\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {1}{5 x^5}-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2-\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2+\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {1}{5 x^5}-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -\frac {\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx-\sqrt {3} \int \frac {1}{-\left (2 x-\sqrt {3}\right )^2-1}d\left (2 x-\sqrt {3}\right )}{2 \sqrt {3}}-\frac {\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx-\sqrt {3} \int \frac {1}{-\left (2 x+\sqrt {3}\right )^2-1}d\left (2 x+\sqrt {3}\right )}{2 \sqrt {3}}-\frac {1}{5 x^5}-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx-\sqrt {3} \arctan \left (\sqrt {3}-2 x\right )}{2 \sqrt {3}}-\frac {\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx+\sqrt {3} \arctan \left (2 x+\sqrt {3}\right )}{2 \sqrt {3}}-\frac {1}{5 x^5}-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle -\frac {-\sqrt {3} \arctan \left (\sqrt {3}-2 x\right )-\frac {1}{2} \log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\sqrt {3} \arctan \left (2 x+\sqrt {3}\right )+\frac {1}{2} \log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {1}{5 x^5}-\frac {1}{3 x^3}\) |
Input:
Int[1/(x^6*(1 - x^2 + x^4)),x]
Output:
-1/5*1/x^5 - 1/(3*x^3) - (-(Sqrt[3]*ArcTan[Sqrt[3] - 2*x]) - Log[1 - Sqrt[ 3]*x + x^2]/2)/(2*Sqrt[3]) - (Sqrt[3]*ArcTan[Sqrt[3] + 2*x] + Log[1 + Sqrt [3]*x + x^2]/2)/(2*Sqrt[3])
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])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66
method | result | size |
risch | \(\frac {-\frac {x^{2}}{3}-\frac {1}{5}}{x^{5}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (3 \textit {\_R}^{3}-\textit {\_R} +x \right )\right )}{2}\) | \(44\) |
default | \(-\frac {1}{5 x^{5}}-\frac {1}{3 x^{3}}+\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{12}-\frac {\arctan \left (2 x -\sqrt {3}\right )}{2}-\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{12}-\frac {\arctan \left (2 x +\sqrt {3}\right )}{2}\) | \(67\) |
Input:
int(1/x^6/(x^4-x^2+1),x,method=_RETURNVERBOSE)
Output:
(-1/3*x^2-1/5)/x^5+1/2*sum(_R*ln(3*_R^3-_R+x),_R=RootOf(9*_Z^4+3*_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^6 \left (1-x^2+x^4\right )} \, dx=-\frac {5 \, \sqrt {3} x^{5} \log \left (x^{2} + \sqrt {3} x + 1\right ) - 5 \, \sqrt {3} x^{5} \log \left (x^{2} - \sqrt {3} x + 1\right ) + 30 \, x^{5} \arctan \left (2 \, x + \sqrt {3}\right ) - 30 \, x^{5} \arctan \left (-2 \, x + \sqrt {3}\right ) + 20 \, x^{2} + 12}{60 \, x^{5}} \] Input:
integrate(1/x^6/(x^4-x^2+1),x, algorithm="fricas")
Output:
-1/60*(5*sqrt(3)*x^5*log(x^2 + sqrt(3)*x + 1) - 5*sqrt(3)*x^5*log(x^2 - sq rt(3)*x + 1) + 30*x^5*arctan(2*x + sqrt(3)) - 30*x^5*arctan(-2*x + sqrt(3) ) + 20*x^2 + 12)/x^5
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^6 \left (1-x^2+x^4\right )} \, dx=\frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} - \frac {\operatorname {atan}{\left (2 x - \sqrt {3} \right )}}{2} - \frac {\operatorname {atan}{\left (2 x + \sqrt {3} \right )}}{2} + \frac {- 5 x^{2} - 3}{15 x^{5}} \] Input:
integrate(1/x**6/(x**4-x**2+1),x)
Output:
sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 - sqrt(3)*log(x**2 + sqrt(3)*x + 1)/1 2 - atan(2*x - sqrt(3))/2 - atan(2*x + sqrt(3))/2 + (-5*x**2 - 3)/(15*x**5 )
\[ \int \frac {1}{x^6 \left (1-x^2+x^4\right )} \, dx=\int { \frac {1}{{\left (x^{4} - x^{2} + 1\right )} x^{6}} \,d x } \] Input:
integrate(1/x^6/(x^4-x^2+1),x, algorithm="maxima")
Output:
-1/15*(5*x^2 + 3)/x^5 - integrate(1/(x^4 - x^2 + 1), x)
Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^6 \left (1-x^2+x^4\right )} \, dx=-\frac {1}{6} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{6} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) - \frac {5 \, x^{2} + 3}{15 \, x^{5}} \] Input:
integrate(1/x^6/(x^4-x^2+1),x, algorithm="giac")
Output:
-1/6*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/6*sqrt(3)*log(x^2 - sqrt(3)*x + 1) - 1/15*(5*x^2 + 3)/x^5
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^6 \left (1-x^2+x^4\right )} \, dx=-\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\frac {\frac {x^2}{3}+\frac {1}{5}}{x^5} \] Input:
int(1/(x^6*(x^4 - x^2 + 1)),x)
Output:
- atan((2*x)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/6 - 1/2) - atan((2*x)/(3^(1/2 )*1i + 1))*((3^(1/2)*1i)/6 + 1/2) - (x^2/3 + 1/5)/x^5
Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^6 \left (1-x^2+x^4\right )} \, dx=\frac {30 \mathit {atan} \left (\sqrt {3}-2 x \right ) x^{5}-30 \mathit {atan} \left (\sqrt {3}+2 x \right ) x^{5}+5 \sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right ) x^{5}-5 \sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right ) x^{5}-20 x^{2}-12}{60 x^{5}} \] Input:
int(1/x^6/(x^4-x^2+1),x)
Output:
(30*atan(sqrt(3) - 2*x)*x**5 - 30*atan(sqrt(3) + 2*x)*x**5 + 5*sqrt(3)*log ( - sqrt(3)*x + x**2 + 1)*x**5 - 5*sqrt(3)*log(sqrt(3)*x + x**2 + 1)*x**5 - 20*x**2 - 12)/(60*x**5)