Integrand size = 16, antiderivative size = 170 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=x-\frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2 \sqrt {5}}+\frac {\sqrt [4]{984-440 \sqrt {5}} \arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{4 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \text {arctanh}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2 \sqrt {5}}+\frac {\sqrt [4]{984-440 \sqrt {5}} \text {arctanh}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{4 \sqrt {5}} \]
x+1/20*arctan(1/2*x*(3+5^(1/2))^(1/4)*2^(3/4))*(984-440*5^(1/2))^(1/4)*5^( 1/2)+1/20*arctanh(1/2*x*(3+5^(1/2))^(1/4)*2^(3/4))*(984-440*5^(1/2))^(1/4) *5^(1/2)-1/20*arctan(2^(1/4)*x*(1/(3+5^(1/2)))^(1/4))*(123+55*5^(1/2))^(1/ 4)*2^(3/4)*5^(1/2)-1/20*arctanh(2^(1/4)*x*(1/(3+5^(1/2)))^(1/4))*(123+55*5 ^(1/2))^(1/4)*2^(3/4)*5^(1/2)
Time = 0.34 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.94 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=x+\frac {\left (-2+\sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\left (2+\sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {\left (-2+\sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\left (2+\sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}} \]
x + ((-2 + Sqrt[5])*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x])/Sqrt[10*(-1 + Sqrt[5 ])] - ((2 + Sqrt[5])*ArcTan[Sqrt[2/(1 + Sqrt[5])]*x])/Sqrt[10*(1 + Sqrt[5] )] + ((-2 + Sqrt[5])*ArcTanh[Sqrt[2/(-1 + Sqrt[5])]*x])/Sqrt[10*(-1 + Sqrt [5])] - ((2 + Sqrt[5])*ArcTanh[Sqrt[2/(1 + Sqrt[5])]*x])/Sqrt[10*(1 + Sqrt [5])]
Time = 0.30 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1703, 1752, 756, 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.
| \(\displaystyle \int \frac {x^8}{x^8-3 x^4+1} \, dx\) |
| \(\Big \downarrow \) 1703 |
| \(\displaystyle x-\int \frac {1-3 x^4}{x^8-3 x^4+1}dx\) |
| \(\Big \downarrow \) 1752 |
| \(\displaystyle \frac {1}{10} \left (15+7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (-3-\sqrt {5}\right )}dx+\frac {1}{10} \left (15-7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (-3+\sqrt {5}\right )}dx+x\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {1}{10} \left (15-7 \sqrt {5}\right ) \left (-\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2}dx}{\sqrt {3-\sqrt {5}}}-\frac {\int \frac {1}{\sqrt {2} x^2+\sqrt {3-\sqrt {5}}}dx}{\sqrt {3-\sqrt {5}}}\right )+\frac {1}{10} \left (15+7 \sqrt {5}\right ) \left (-\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2}dx}{\sqrt {3+\sqrt {5}}}-\frac {\int \frac {1}{\sqrt {2} x^2+\sqrt {3+\sqrt {5}}}dx}{\sqrt {3+\sqrt {5}}}\right )+x\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{10} \left (15-7 \sqrt {5}\right ) \left (-\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2}dx}{\sqrt {3-\sqrt {5}}}-\frac {\arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{\sqrt [4]{2} \left (3-\sqrt {5}\right )^{3/4}}\right )+\frac {1}{10} \left (15+7 \sqrt {5}\right ) \left (-\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2}dx}{\sqrt {3+\sqrt {5}}}-\frac {\arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{\sqrt [4]{2} \left (3+\sqrt {5}\right )^{3/4}}\right )+x\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{10} \left (15+7 \sqrt {5}\right ) \left (-\frac {\arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{\sqrt [4]{2} \left (3+\sqrt {5}\right )^{3/4}}-\frac {\text {arctanh}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{\sqrt [4]{2} \left (3+\sqrt {5}\right )^{3/4}}\right )+\frac {1}{10} \left (15-7 \sqrt {5}\right ) \left (-\frac {\arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{\sqrt [4]{2} \left (3-\sqrt {5}\right )^{3/4}}-\frac {\text {arctanh}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{\sqrt [4]{2} \left (3-\sqrt {5}\right )^{3/4}}\right )+x\) |
x + ((15 + 7*Sqrt[5])*(-(ArcTan[(2/(3 + Sqrt[5]))^(1/4)*x]/(2^(1/4)*(3 + S qrt[5])^(3/4))) - ArcTanh[(2/(3 + Sqrt[5]))^(1/4)*x]/(2^(1/4)*(3 + Sqrt[5] )^(3/4))))/10 + ((15 - 7*Sqrt[5])*(-(ArcTan[((3 + Sqrt[5])/2)^(1/4)*x]/(2^ (1/4)*(3 - Sqrt[5])^(3/4))) - ArcTanh[((3 + Sqrt[5])/2)^(1/4)*x]/(2^(1/4)* (3 - Sqrt[5])^(3/4))))/10
3.4.96.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) 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^(2*n - 1)*(d*x)^(m - 2*n + 1)*((a + b*x^n + c*x^(2*n))^( p + 1)/(c*(m + 2*n*p + 1))), x] - Simp[d^(2*n)/(c*(m + 2*n*p + 1)) Int[(d *x)^(m - 2*n)*Simp[a*(m - 2*n + 1) + b*(m + n*(p - 1) + 1)*x^n, x]*(a + b*x ^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[n2, 2*n] && N eQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1] && NeQ[m + 2*n*p + 1, 0 ] && IntegerQ[p]
Int[((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[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) I nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 , 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] || !IGtQ[n/2, 0])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.41
| method | result | size |
| risch | \(x +\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+55 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (15 \textit {\_R}^{3}+29 \textit {\_R} +5 x \right )\right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}-55 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (-15 \textit {\_R}^{3}+29 \textit {\_R} +5 x \right )\right )}{4}\) | \(69\) |
| default | \(x -\frac {\left (2+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2 \sqrt {5}+2}}\right )}{5 \sqrt {2 \sqrt {5}+2}}+\frac {\left (\sqrt {5}-2\right ) \sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2 \sqrt {5}-2}}\right )}{5 \sqrt {2 \sqrt {5}-2}}-\frac {\left (2+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2 \sqrt {5}+2}}\right )}{5 \sqrt {2 \sqrt {5}+2}}+\frac {\left (\sqrt {5}-2\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2 \sqrt {5}-2}}\right )}{5 \sqrt {2 \sqrt {5}-2}}\) | \(131\) |
x+1/4*sum(_R*ln(15*_R^3+29*_R+5*x),_R=RootOf(25*_Z^4+55*_Z^2-1))+1/4*sum(_ R*ln(-15*_R^3+29*_R+5*x),_R=RootOf(25*_Z^4-55*_Z^2-1))
Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (118) = 236\).
Time = 0.25 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.87 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=\frac {1}{40} \, \sqrt {10} \sqrt {5 \, \sqrt {5} - 11} \log \left (\sqrt {10} \sqrt {5 \, \sqrt {5} - 11} {\left (3 \, \sqrt {5} + 5\right )} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {5 \, \sqrt {5} - 11} \log \left (-\sqrt {10} \sqrt {5 \, \sqrt {5} - 11} {\left (3 \, \sqrt {5} + 5\right )} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {5 \, \sqrt {5} + 11} \log \left (\sqrt {10} \sqrt {5 \, \sqrt {5} + 11} {\left (3 \, \sqrt {5} - 5\right )} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {5 \, \sqrt {5} + 11} \log \left (-\sqrt {10} \sqrt {5 \, \sqrt {5} + 11} {\left (3 \, \sqrt {5} - 5\right )} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-5 \, \sqrt {5} + 11} \log \left (\sqrt {10} {\left (3 \, \sqrt {5} + 5\right )} \sqrt {-5 \, \sqrt {5} + 11} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-5 \, \sqrt {5} + 11} \log \left (-\sqrt {10} {\left (3 \, \sqrt {5} + 5\right )} \sqrt {-5 \, \sqrt {5} + 11} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-5 \, \sqrt {5} - 11} \log \left (\sqrt {10} {\left (3 \, \sqrt {5} - 5\right )} \sqrt {-5 \, \sqrt {5} - 11} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-5 \, \sqrt {5} - 11} \log \left (-\sqrt {10} {\left (3 \, \sqrt {5} - 5\right )} \sqrt {-5 \, \sqrt {5} - 11} + 20 \, x\right ) + x \]
1/40*sqrt(10)*sqrt(5*sqrt(5) - 11)*log(sqrt(10)*sqrt(5*sqrt(5) - 11)*(3*sq rt(5) + 5) + 20*x) - 1/40*sqrt(10)*sqrt(5*sqrt(5) - 11)*log(-sqrt(10)*sqrt (5*sqrt(5) - 11)*(3*sqrt(5) + 5) + 20*x) - 1/40*sqrt(10)*sqrt(5*sqrt(5) + 11)*log(sqrt(10)*sqrt(5*sqrt(5) + 11)*(3*sqrt(5) - 5) + 20*x) + 1/40*sqrt( 10)*sqrt(5*sqrt(5) + 11)*log(-sqrt(10)*sqrt(5*sqrt(5) + 11)*(3*sqrt(5) - 5 ) + 20*x) + 1/40*sqrt(10)*sqrt(-5*sqrt(5) + 11)*log(sqrt(10)*(3*sqrt(5) + 5)*sqrt(-5*sqrt(5) + 11) + 20*x) - 1/40*sqrt(10)*sqrt(-5*sqrt(5) + 11)*log (-sqrt(10)*(3*sqrt(5) + 5)*sqrt(-5*sqrt(5) + 11) + 20*x) - 1/40*sqrt(10)*s qrt(-5*sqrt(5) - 11)*log(sqrt(10)*(3*sqrt(5) - 5)*sqrt(-5*sqrt(5) - 11) + 20*x) + 1/40*sqrt(10)*sqrt(-5*sqrt(5) - 11)*log(-sqrt(10)*(3*sqrt(5) - 5)* sqrt(-5*sqrt(5) - 11) + 20*x) + x
Time = 0.73 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.34 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=x + \operatorname {RootSum} {\left (6400 t^{4} - 880 t^{2} - 1, \left ( t \mapsto t \log {\left (- \frac {15360 t^{5}}{11} + \frac {1288 t}{55} + x \right )} \right )\right )} + \operatorname {RootSum} {\left (6400 t^{4} + 880 t^{2} - 1, \left ( t \mapsto t \log {\left (- \frac {15360 t^{5}}{11} + \frac {1288 t}{55} + x \right )} \right )\right )} \]
x + RootSum(6400*_t**4 - 880*_t**2 - 1, Lambda(_t, _t*log(-15360*_t**5/11 + 1288*_t/55 + x))) + RootSum(6400*_t**4 + 880*_t**2 - 1, Lambda(_t, _t*lo g(-15360*_t**5/11 + 1288*_t/55 + x)))
\[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=\int { \frac {x^{8}}{x^{8} - 3 \, x^{4} + 1} \,d x } \]
x + 1/2*integrate((2*x^2 + 1)/(x^4 - x^2 - 1), x) - 1/2*integrate((2*x^2 - 1)/(x^4 + x^2 - 1), x)
Time = 0.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.87 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=-\frac {1}{20} \, \sqrt {50 \, \sqrt {5} + 110} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) + \frac {1}{20} \, \sqrt {50 \, \sqrt {5} - 110} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{40} \, \sqrt {50 \, \sqrt {5} + 110} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {50 \, \sqrt {5} + 110} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {50 \, \sqrt {5} - 110} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{40} \, \sqrt {50 \, \sqrt {5} - 110} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) + x \]
-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) + 11 0)*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))) + x
Time = 8.57 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.45 \[ \int \frac {x^8}{1-3 x^4+x^8} \, dx=x-\frac {\mathrm {atan}\left (\frac {x\,\sqrt {-50\,\sqrt {5}-110}\,55{}\mathrm {i}}{2\,\left (275\,\sqrt {5}+605\right )}+\frac {\sqrt {5}\,x\,\sqrt {-50\,\sqrt {5}-110}\,33{}\mathrm {i}}{2\,\left (275\,\sqrt {5}+605\right )}\right )\,\sqrt {-50\,\sqrt {5}-110}\,1{}\mathrm {i}}{20}-\frac {\mathrm {atan}\left (\frac {x\,\sqrt {110-50\,\sqrt {5}}\,55{}\mathrm {i}}{2\,\left (275\,\sqrt {5}-605\right )}-\frac {\sqrt {5}\,x\,\sqrt {110-50\,\sqrt {5}}\,33{}\mathrm {i}}{2\,\left (275\,\sqrt {5}-605\right )}\right )\,\sqrt {110-50\,\sqrt {5}}\,1{}\mathrm {i}}{20}+\frac {\mathrm {atan}\left (\frac {x\,\sqrt {50\,\sqrt {5}-110}\,55{}\mathrm {i}}{2\,\left (275\,\sqrt {5}-605\right )}-\frac {\sqrt {5}\,x\,\sqrt {50\,\sqrt {5}-110}\,33{}\mathrm {i}}{2\,\left (275\,\sqrt {5}-605\right )}\right )\,\sqrt {50\,\sqrt {5}-110}\,1{}\mathrm {i}}{20}+\frac {\mathrm {atan}\left (\frac {x\,\sqrt {50\,\sqrt {5}+110}\,55{}\mathrm {i}}{2\,\left (275\,\sqrt {5}+605\right )}+\frac {\sqrt {5}\,x\,\sqrt {50\,\sqrt {5}+110}\,33{}\mathrm {i}}{2\,\left (275\,\sqrt {5}+605\right )}\right )\,\sqrt {50\,\sqrt {5}+110}\,1{}\mathrm {i}}{20} \]
x - (atan((x*(- 50*5^(1/2) - 110)^(1/2)*55i)/(2*(275*5^(1/2) + 605)) + (5^ (1/2)*x*(- 50*5^(1/2) - 110)^(1/2)*33i)/(2*(275*5^(1/2) + 605)))*(- 50*5^( 1/2) - 110)^(1/2)*1i)/20 - (atan((x*(110 - 50*5^(1/2))^(1/2)*55i)/(2*(275* 5^(1/2) - 605)) - (5^(1/2)*x*(110 - 50*5^(1/2))^(1/2)*33i)/(2*(275*5^(1/2) - 605)))*(110 - 50*5^(1/2))^(1/2)*1i)/20 + (atan((x*(50*5^(1/2) - 110)^(1 /2)*55i)/(2*(275*5^(1/2) - 605)) - (5^(1/2)*x*(50*5^(1/2) - 110)^(1/2)*33i )/(2*(275*5^(1/2) - 605)))*(50*5^(1/2) - 110)^(1/2)*1i)/20 + (atan((x*(50* 5^(1/2) + 110)^(1/2)*55i)/(2*(275*5^(1/2) + 605)) + (5^(1/2)*x*(50*5^(1/2) + 110)^(1/2)*33i)/(2*(275*5^(1/2) + 605)))*(50*5^(1/2) + 110)^(1/2)*1i)/2 0