Integrand size = 16, antiderivative size = 180 \[ \int \frac {1}{x^2 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{x}+\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2 \sqrt {5}}-\frac {\left (3+\sqrt {5}\right )^{5/4} \arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{4 \sqrt [4]{2} \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \text {arctanh}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2 \sqrt {5}}+\frac {\left (3+\sqrt {5}\right )^{5/4} \text {arctanh}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{4 \sqrt [4]{2} \sqrt {5}} \] Output:
-1/x+1/10*(123/2-55/2*5^(1/2))^(1/4)*arctan(2^(1/4)*(1/(3+5^(1/2)))^(1/4)* x)*5^(1/2)-1/40*(3+5^(1/2))^(5/4)*arctan((3/2+1/2*5^(1/2))^(1/4)*x)*2^(3/4 )*5^(1/2)-1/10*(123/2-55/2*5^(1/2))^(1/4)*arctanh(2^(1/4)*(1/(3+5^(1/2)))^ (1/4)*x)*5^(1/2)+1/40*(3+5^(1/2))^(5/4)*arctanh((3/2+1/2*5^(1/2))^(1/4)*x) *2^(3/4)*5^(1/2)
Time = 0.19 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^2 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{x}-\frac {\left (3+\sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{2 \sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\left (-3+\sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {\left (3+\sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{2 \sqrt {10 \left (-1+\sqrt {5}\right )}}+\frac {\left (-3+\sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}} \] Input:
Integrate[1/(x^2*(1 - 3*x^4 + x^8)),x]
Output:
-x^(-1) - ((3 + Sqrt[5])*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x])/(2*Sqrt[10*(-1 + Sqrt[5])]) - ((-3 + Sqrt[5])*ArcTan[Sqrt[2/(1 + Sqrt[5])]*x])/(2*Sqrt[10 *(1 + Sqrt[5])]) + ((3 + Sqrt[5])*ArcTanh[Sqrt[2/(-1 + Sqrt[5])]*x])/(2*Sq rt[10*(-1 + Sqrt[5])]) + ((-3 + Sqrt[5])*ArcTanh[Sqrt[2/(1 + Sqrt[5])]*x]) /(2*Sqrt[10*(1 + Sqrt[5])])
Time = 0.31 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1704, 1834, 27, 827, 216, 219}
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.
| \(\displaystyle \int \frac {1}{x^2 \left (x^8-3 x^4+1\right )} \, dx\) |
| \(\Big \downarrow \) 1704 |
| \(\displaystyle \int \frac {x^2 \left (3-x^4\right )}{x^8-3 x^4+1}dx-\frac {1}{x}\) |
| \(\Big \downarrow \) 1834 |
| \(\displaystyle -\frac {1}{10} \left (5+3 \sqrt {5}\right ) \int -\frac {2 x^2}{-2 x^4-\sqrt {5}+3}dx-\frac {1}{10} \left (5-3 \sqrt {5}\right ) \int -\frac {2 x^2}{-2 x^4+\sqrt {5}+3}dx-\frac {1}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (5+3 \sqrt {5}\right ) \int \frac {x^2}{-2 x^4-\sqrt {5}+3}dx+\frac {1}{5} \left (5-3 \sqrt {5}\right ) \int \frac {x^2}{-2 x^4+\sqrt {5}+3}dx-\frac {1}{x}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {1}{5} \left (5+3 \sqrt {5}\right ) \left (\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2}dx}{2 \sqrt {2}}-\frac {\int \frac {1}{\sqrt {2} x^2+\sqrt {3-\sqrt {5}}}dx}{2 \sqrt {2}}\right )+\frac {1}{5} \left (5-3 \sqrt {5}\right ) \left (\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2}dx}{2 \sqrt {2}}-\frac {\int \frac {1}{\sqrt {2} x^2+\sqrt {3+\sqrt {5}}}dx}{2 \sqrt {2}}\right )-\frac {1}{x}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{5} \left (5+3 \sqrt {5}\right ) \left (\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2}dx}{2 \sqrt {2}}-\frac {\arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2\ 2^{3/4} \sqrt [4]{3-\sqrt {5}}}\right )+\frac {1}{5} \left (5-3 \sqrt {5}\right ) \left (\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2}dx}{2 \sqrt {2}}-\frac {\arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2\ 2^{3/4} \sqrt [4]{3+\sqrt {5}}}\right )-\frac {1}{x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{5} \left (5-3 \sqrt {5}\right ) \left (\frac {\text {arctanh}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2\ 2^{3/4} \sqrt [4]{3+\sqrt {5}}}-\frac {\arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2\ 2^{3/4} \sqrt [4]{3+\sqrt {5}}}\right )+\frac {1}{5} \left (5+3 \sqrt {5}\right ) \left (\frac {\text {arctanh}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2\ 2^{3/4} \sqrt [4]{3-\sqrt {5}}}-\frac {\arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2\ 2^{3/4} \sqrt [4]{3-\sqrt {5}}}\right )-\frac {1}{x}\) |
Input:
Int[1/(x^2*(1 - 3*x^4 + x^8)),x]
Output:
-x^(-1) + ((5 - 3*Sqrt[5])*(-1/2*ArcTan[(2/(3 + Sqrt[5]))^(1/4)*x]/(2^(3/4 )*(3 + Sqrt[5])^(1/4)) + ArcTanh[(2/(3 + Sqrt[5]))^(1/4)*x]/(2*2^(3/4)*(3 + Sqrt[5])^(1/4))))/5 + ((5 + 3*Sqrt[5])*(-1/2*ArcTan[((3 + Sqrt[5])/2)^(1 /4)*x]/(2^(3/4)*(3 - Sqrt[5])^(1/4)) + ArcTanh[((3 + Sqrt[5])/2)^(1/4)*x]/ (2*2^(3/4)*(3 - Sqrt[5])^(1/4))))/5
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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[((d_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_ Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*d*(m + 1) )), x] - Simp[1/(a*d^n*(m + 1)) Int[(d*x)^(m + n)*(b*(m + n*(p + 1) + 1) + c*(m + 2*n*(p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{ a, b, c, d, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntegerQ[p]
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ [{a, b, c, d, e, f, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n , 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.41
| method | result | size |
| risch | \(-\frac {1}{x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+55 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (-20 \textit {\_R}^{3}-47 \textit {\_R} +5 x \right )\right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}-55 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (-20 \textit {\_R}^{3}+47 \textit {\_R} +5 x \right )\right )}{4}\) | \(73\) |
| default | \(-\frac {1}{x}+\frac {\left (\sqrt {5}-3\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2+2 \sqrt {5}}}\right )}{10 \sqrt {2+2 \sqrt {5}}}-\frac {\left (3+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {-2+2 \sqrt {5}}}\right )}{10 \sqrt {-2+2 \sqrt {5}}}+\frac {\left (3+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {-2+2 \sqrt {5}}}\right )}{10 \sqrt {-2+2 \sqrt {5}}}-\frac {\left (\sqrt {5}-3\right ) \sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2+2 \sqrt {5}}}\right )}{10 \sqrt {2+2 \sqrt {5}}}\) | \(135\) |
Input:
int(1/x^2/(x^8-3*x^4+1),x,method=_RETURNVERBOSE)
Output:
-1/x+1/4*sum(_R*ln(-20*_R^3-47*_R+5*x),_R=RootOf(25*_Z^4+55*_Z^2-1))+1/4*s um(_R*ln(-20*_R^3+47*_R+5*x),_R=RootOf(25*_Z^4-55*_Z^2-1))
Time = 0.08 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^2 \left (1-3 x^4+x^8\right )} \, dx=-\frac {2 \, x \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} \arctan \left (\frac {1}{2} \, {\left (3 \, \sqrt {5} x - 5 \, x\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}}\right ) - 2 \, x \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} \arctan \left (\frac {1}{2} \, {\left (3 \, \sqrt {5} x + 5 \, x\right )} \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}}\right ) + x \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} \log \left ({\left (2 \, \sqrt {5} - 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} + x\right ) - x \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} \log \left (-{\left (2 \, \sqrt {5} - 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {11}{10}} + x\right ) + x \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} \log \left ({\left (2 \, \sqrt {5} + 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} + x\right ) - x \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} \log \left (-{\left (2 \, \sqrt {5} + 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {11}{10}} + x\right ) + 4}{4 \, x} \] Input:
integrate(1/x^2/(x^8-3*x^4+1),x, algorithm="fricas")
Output:
-1/4*(2*x*sqrt(1/2*sqrt(5) + 11/10)*arctan(1/2*(3*sqrt(5)*x - 5*x)*sqrt(1/ 2*sqrt(5) + 11/10)) - 2*x*sqrt(1/2*sqrt(5) - 11/10)*arctan(1/2*(3*sqrt(5)* x + 5*x)*sqrt(1/2*sqrt(5) - 11/10)) + x*sqrt(1/2*sqrt(5) + 11/10)*log((2*s qrt(5) - 5)*sqrt(1/2*sqrt(5) + 11/10) + x) - x*sqrt(1/2*sqrt(5) + 11/10)*l og(-(2*sqrt(5) - 5)*sqrt(1/2*sqrt(5) + 11/10) + x) + x*sqrt(1/2*sqrt(5) - 11/10)*log((2*sqrt(5) + 5)*sqrt(1/2*sqrt(5) - 11/10) + x) - x*sqrt(1/2*sqr t(5) - 11/10)*log(-(2*sqrt(5) + 5)*sqrt(1/2*sqrt(5) - 11/10) + x) + 4)/x
Time = 0.85 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^2 \left (1-3 x^4+x^8\right )} \, dx=\operatorname {RootSum} {\left (6400 t^{4} - 880 t^{2} - 1, \left ( t \mapsto t \log {\left (\frac {19251200 t^{7}}{11} - \frac {369792 t^{3}}{11} + x \right )} \right )\right )} + \operatorname {RootSum} {\left (6400 t^{4} + 880 t^{2} - 1, \left ( t \mapsto t \log {\left (\frac {19251200 t^{7}}{11} - \frac {369792 t^{3}}{11} + x \right )} \right )\right )} - \frac {1}{x} \] Input:
integrate(1/x**2/(x**8-3*x**4+1),x)
Output:
RootSum(6400*_t**4 - 880*_t**2 - 1, Lambda(_t, _t*log(19251200*_t**7/11 - 369792*_t**3/11 + x))) + RootSum(6400*_t**4 + 880*_t**2 - 1, Lambda(_t, _t *log(19251200*_t**7/11 - 369792*_t**3/11 + x))) - 1/x
\[ \int \frac {1}{x^2 \left (1-3 x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} - 3 \, x^{4} + 1\right )} x^{2}} \,d x } \] Input:
integrate(1/x^2/(x^8-3*x^4+1),x, algorithm="maxima")
Output:
-1/x - 1/2*integrate((x^2 + 2)/(x^4 + x^2 - 1), x) - 1/2*integrate((x^2 - 2)/(x^4 - x^2 - 1), x)
Time = 0.19 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^2 \left (1-3 x^4+x^8\right )} \, dx=\frac {1}{20} \, \sqrt {50 \, \sqrt {5} - 110} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) - \frac {1}{20} \, \sqrt {50 \, \sqrt {5} + 110} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{40} \, \sqrt {50 \, \sqrt {5} - 110} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {50 \, \sqrt {5} - 110} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {50 \, \sqrt {5} + 110} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{40} \, \sqrt {50 \, \sqrt {5} + 110} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{x} \] Input:
integrate(1/x^2/(x^8-3*x^4+1),x, algorithm="giac")
Output:
1/20*sqrt(50*sqrt(5) - 110)*arctan(x/sqrt(1/2*sqrt(5) + 1/2)) - 1/20*sqrt( 50*sqrt(5) + 110)*arctan(x/sqrt(1/2*sqrt(5) - 1/2)) - 1/40*sqrt(50*sqrt(5) - 110)*log(abs(x + sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(50*sqrt(5) - 110 )*log(abs(x - sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(50*sqrt(5) + 110)*log( abs(x + sqrt(1/2*sqrt(5) - 1/2))) - 1/40*sqrt(50*sqrt(5) + 110)*log(abs(x - sqrt(1/2*sqrt(5) - 1/2))) - 1/x
Time = 19.27 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x^2 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{x}-\frac {\mathrm {atan}\left (\frac {x\,\sqrt {-50\,\sqrt {5}-110}\,1155{}\mathrm {i}}{2\,\left (3025\,\sqrt {5}+6765\right )}+\frac {\sqrt {5}\,x\,\sqrt {-50\,\sqrt {5}-110}\,517{}\mathrm {i}}{2\,\left (3025\,\sqrt {5}+6765\right )}\right )\,\sqrt {-50\,\sqrt {5}-110}\,1{}\mathrm {i}}{20}+\frac {\mathrm {atan}\left (\frac {x\,\sqrt {110-50\,\sqrt {5}}\,1155{}\mathrm {i}}{2\,\left (3025\,\sqrt {5}-6765\right )}-\frac {\sqrt {5}\,x\,\sqrt {110-50\,\sqrt {5}}\,517{}\mathrm {i}}{2\,\left (3025\,\sqrt {5}-6765\right )}\right )\,\sqrt {110-50\,\sqrt {5}}\,1{}\mathrm {i}}{20}+\frac {\mathrm {atan}\left (\frac {x\,\sqrt {50\,\sqrt {5}-110}\,1155{}\mathrm {i}}{2\,\left (3025\,\sqrt {5}-6765\right )}-\frac {\sqrt {5}\,x\,\sqrt {50\,\sqrt {5}-110}\,517{}\mathrm {i}}{2\,\left (3025\,\sqrt {5}-6765\right )}\right )\,\sqrt {50\,\sqrt {5}-110}\,1{}\mathrm {i}}{20}-\frac {\mathrm {atan}\left (\frac {x\,\sqrt {50\,\sqrt {5}+110}\,1155{}\mathrm {i}}{2\,\left (3025\,\sqrt {5}+6765\right )}+\frac {\sqrt {5}\,x\,\sqrt {50\,\sqrt {5}+110}\,517{}\mathrm {i}}{2\,\left (3025\,\sqrt {5}+6765\right )}\right )\,\sqrt {50\,\sqrt {5}+110}\,1{}\mathrm {i}}{20} \] Input:
int(1/(x^2*(x^8 - 3*x^4 + 1)),x)
Output:
(atan((x*(110 - 50*5^(1/2))^(1/2)*1155i)/(2*(3025*5^(1/2) - 6765)) - (5^(1 /2)*x*(110 - 50*5^(1/2))^(1/2)*517i)/(2*(3025*5^(1/2) - 6765)))*(110 - 50* 5^(1/2))^(1/2)*1i)/20 - (atan((x*(- 50*5^(1/2) - 110)^(1/2)*1155i)/(2*(302 5*5^(1/2) + 6765)) + (5^(1/2)*x*(- 50*5^(1/2) - 110)^(1/2)*517i)/(2*(3025* 5^(1/2) + 6765)))*(- 50*5^(1/2) - 110)^(1/2)*1i)/20 + (atan((x*(50*5^(1/2) - 110)^(1/2)*1155i)/(2*(3025*5^(1/2) - 6765)) - (5^(1/2)*x*(50*5^(1/2) - 110)^(1/2)*517i)/(2*(3025*5^(1/2) - 6765)))*(50*5^(1/2) - 110)^(1/2)*1i)/2 0 - (atan((x*(50*5^(1/2) + 110)^(1/2)*1155i)/(2*(3025*5^(1/2) + 6765)) + ( 5^(1/2)*x*(50*5^(1/2) + 110)^(1/2)*517i)/(2*(3025*5^(1/2) + 6765)))*(50*5^ (1/2) + 110)^(1/2)*1i)/20 - 1/x
\[ \int \frac {1}{x^2 \left (1-3 x^4+x^8\right )} \, dx=\int \frac {1}{x^{10}-3 x^{6}+x^{2}}d x \] Input:
int(1/x^2/(x^8-3*x^4+1),x)
Output:
int(1/(x**10 - 3*x**6 + x**2),x)