Integrand size = 20, antiderivative size = 129 \[ \int \frac {1-x^4}{1-3 x^4+x^8} \, dx=\frac {\arctan \left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}+\frac {\arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {\text {arctanh}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}+\frac {\text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}} \]
arctan(x*2^(1/2)/(5^(1/2)-1)^(1/2))/(-10+10*5^(1/2))^(1/2)+arctanh(x*2^(1/ 2)/(5^(1/2)-1)^(1/2))/(-10+10*5^(1/2))^(1/2)+arctan(x*2^(1/2)/(5^(1/2)+1)^ (1/2))/(10+10*5^(1/2))^(1/2)+arctanh(x*2^(1/2)/(5^(1/2)+1)^(1/2))/(10+10*5 ^(1/2))^(1/2)
Time = 0.06 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00 \[ \int \frac {1-x^4}{1-3 x^4+x^8} \, dx=\frac {\arctan \left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}+\frac {\arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {\text {arctanh}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}+\frac {\text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}} \]
ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x]/Sqrt[10*(-1 + Sqrt[5])] + ArcTan[Sqrt[2/( 1 + Sqrt[5])]*x]/Sqrt[10*(1 + Sqrt[5])] + ArcTanh[Sqrt[2/(-1 + Sqrt[5])]*x ]/Sqrt[10*(-1 + Sqrt[5])] + ArcTanh[Sqrt[2/(1 + Sqrt[5])]*x]/Sqrt[10*(1 + Sqrt[5])]
Time = 0.25 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1749, 1406, 216, 220}
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.
| \(\displaystyle \int \frac {1-x^4}{x^8-3 x^4+1} \, dx\) |
| \(\Big \downarrow \) 1749 |
| \(\displaystyle -\frac {1}{2} \int \frac {1}{x^4-x^2-1}dx-\frac {1}{2} \int \frac {1}{x^4+x^2-1}dx\) |
| \(\Big \downarrow \) 1406 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{x^2+\frac {1}{2} \left (-1+\sqrt {5}\right )}dx}{\sqrt {5}}-\frac {\int \frac {1}{x^2+\frac {1}{2} \left (-1-\sqrt {5}\right )}dx}{\sqrt {5}}\right )+\frac {1}{2} \left (\frac {\int \frac {1}{x^2+\frac {1}{2} \left (1+\sqrt {5}\right )}dx}{\sqrt {5}}-\frac {\int \frac {1}{x^2+\frac {1}{2} \left (1-\sqrt {5}\right )}dx}{\sqrt {5}}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{2} \left (\sqrt {\frac {2}{5 \left (\sqrt {5}-1\right )}} \arctan \left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )-\frac {\int \frac {1}{x^2+\frac {1}{2} \left (-1-\sqrt {5}\right )}dx}{\sqrt {5}}\right )+\frac {1}{2} \left (\sqrt {\frac {2}{5 \left (1+\sqrt {5}\right )}} \arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )-\frac {\int \frac {1}{x^2+\frac {1}{2} \left (1-\sqrt {5}\right )}dx}{\sqrt {5}}\right )\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {1}{2} \left (\sqrt {\frac {2}{5 \left (1+\sqrt {5}\right )}} \arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )+\sqrt {\frac {2}{5 \left (\sqrt {5}-1\right )}} \text {arctanh}\left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )\right )+\frac {1}{2} \left (\sqrt {\frac {2}{5 \left (\sqrt {5}-1\right )}} \arctan \left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )+\sqrt {\frac {2}{5 \left (1+\sqrt {5}\right )}} \text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )\right )\) |
(Sqrt[2/(5*(1 + Sqrt[5]))]*ArcTan[Sqrt[2/(1 + Sqrt[5])]*x] + Sqrt[2/(5*(-1 + Sqrt[5]))]*ArcTanh[Sqrt[2/(-1 + Sqrt[5])]*x])/2 + (Sqrt[2/(5*(-1 + Sqrt [5]))]*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x] + Sqrt[2/(5*(1 + Sqrt[5]))]*ArcTan h[Sqrt[2/(1 + Sqrt[5])]*x])/2
3.1.27.3.1 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[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^ 2 - 4*a*c, 2]}, Simp[c/q Int[1/(b/2 - q/2 + c*x^2), x], x] - Simp[c/q I nt[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c , 0] && PosQ[b^2 - 4*a*c]
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x _Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x^(n/2) + x^n, x], x], x] + Simp[e/(2*c) Int[1/Simp[d/e - q*x^(n/2) + x^n, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && IGtQ[n/2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d, e*Rt[a/c, 2]]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.50
| method | result | size |
| risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}-5 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (-5 \textit {\_R}^{3}+3 \textit {\_R} +x \right )\right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+5 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (5 \textit {\_R}^{3}+3 \textit {\_R} +x \right )\right )}{4}\) | \(64\) |
| default | \(\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2 \sqrt {5}+2}}\right )}{5 \sqrt {2 \sqrt {5}+2}}+\frac {\sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2 \sqrt {5}-2}}\right )}{5 \sqrt {2 \sqrt {5}-2}}+\frac {\sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2 \sqrt {5}+2}}\right )}{5 \sqrt {2 \sqrt {5}+2}}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2 \sqrt {5}-2}}\right )}{5 \sqrt {2 \sqrt {5}-2}}\) | \(110\) |
1/4*sum(_R*ln(-5*_R^3+3*_R+x),_R=RootOf(25*_Z^4-5*_Z^2-1))+1/4*sum(_R*ln(5 *_R^3+3*_R+x),_R=RootOf(25*_Z^4+5*_Z^2-1))
Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (93) = 186\).
Time = 0.29 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.21 \[ \int \frac {1-x^4}{1-3 x^4+x^8} \, dx=\frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} - 1} \log \left (\sqrt {10} {\left (\sqrt {5} + 5\right )} \sqrt {\sqrt {5} - 1} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} - 1} \log \left (-\sqrt {10} {\left (\sqrt {5} + 5\right )} \sqrt {\sqrt {5} - 1} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} + 1} \log \left (\sqrt {10} \sqrt {\sqrt {5} + 1} {\left (\sqrt {5} - 5\right )} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} + 1} \log \left (-\sqrt {10} \sqrt {\sqrt {5} + 1} {\left (\sqrt {5} - 5\right )} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} + 1} \log \left (\sqrt {10} {\left (\sqrt {5} + 5\right )} \sqrt {-\sqrt {5} + 1} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} + 1} \log \left (-\sqrt {10} {\left (\sqrt {5} + 5\right )} \sqrt {-\sqrt {5} + 1} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} - 1} \log \left (\sqrt {10} {\left (\sqrt {5} - 5\right )} \sqrt {-\sqrt {5} - 1} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} - 1} \log \left (-\sqrt {10} {\left (\sqrt {5} - 5\right )} \sqrt {-\sqrt {5} - 1} + 20 \, x\right ) \]
1/40*sqrt(10)*sqrt(sqrt(5) - 1)*log(sqrt(10)*(sqrt(5) + 5)*sqrt(sqrt(5) - 1) + 20*x) - 1/40*sqrt(10)*sqrt(sqrt(5) - 1)*log(-sqrt(10)*(sqrt(5) + 5)*s qrt(sqrt(5) - 1) + 20*x) - 1/40*sqrt(10)*sqrt(sqrt(5) + 1)*log(sqrt(10)*sq rt(sqrt(5) + 1)*(sqrt(5) - 5) + 20*x) + 1/40*sqrt(10)*sqrt(sqrt(5) + 1)*lo g(-sqrt(10)*sqrt(sqrt(5) + 1)*(sqrt(5) - 5) + 20*x) + 1/40*sqrt(10)*sqrt(- sqrt(5) + 1)*log(sqrt(10)*(sqrt(5) + 5)*sqrt(-sqrt(5) + 1) + 20*x) - 1/40* sqrt(10)*sqrt(-sqrt(5) + 1)*log(-sqrt(10)*(sqrt(5) + 5)*sqrt(-sqrt(5) + 1) + 20*x) - 1/40*sqrt(10)*sqrt(-sqrt(5) - 1)*log(sqrt(10)*(sqrt(5) - 5)*sqr t(-sqrt(5) - 1) + 20*x) + 1/40*sqrt(10)*sqrt(-sqrt(5) - 1)*log(-sqrt(10)*( sqrt(5) - 5)*sqrt(-sqrt(5) - 1) + 20*x)
Time = 0.68 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.40 \[ \int \frac {1-x^4}{1-3 x^4+x^8} \, dx=- \operatorname {RootSum} {\left (6400 t^{4} - 80 t^{2} - 1, \left ( t \mapsto t \log {\left (25600 t^{5} - 16 t + x \right )} \right )\right )} - \operatorname {RootSum} {\left (6400 t^{4} + 80 t^{2} - 1, \left ( t \mapsto t \log {\left (25600 t^{5} - 16 t + x \right )} \right )\right )} \]
-RootSum(6400*_t**4 - 80*_t**2 - 1, Lambda(_t, _t*log(25600*_t**5 - 16*_t + x))) - RootSum(6400*_t**4 + 80*_t**2 - 1, Lambda(_t, _t*log(25600*_t**5 - 16*_t + x)))
\[ \int \frac {1-x^4}{1-3 x^4+x^8} \, dx=\int { -\frac {x^{4} - 1}{x^{8} - 3 \, x^{4} + 1} \,d x } \]
Time = 0.48 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.14 \[ \int \frac {1-x^4}{1-3 x^4+x^8} \, dx=\frac {1}{20} \, \sqrt {10 \, \sqrt {5} - 10} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) + \frac {1}{20} \, \sqrt {10 \, \sqrt {5} + 10} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) + \frac {1}{40} \, \sqrt {10 \, \sqrt {5} - 10} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) - \frac {1}{40} \, \sqrt {10 \, \sqrt {5} - 10} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {10 \, \sqrt {5} + 10} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{40} \, \sqrt {10 \, \sqrt {5} + 10} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) \]
1/20*sqrt(10*sqrt(5) - 10)*arctan(x/sqrt(1/2*sqrt(5) + 1/2)) + 1/20*sqrt(1 0*sqrt(5) + 10)*arctan(x/sqrt(1/2*sqrt(5) - 1/2)) + 1/40*sqrt(10*sqrt(5) - 10)*log(abs(x + sqrt(1/2*sqrt(5) + 1/2))) - 1/40*sqrt(10*sqrt(5) - 10)*lo g(abs(x - sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(10*sqrt(5) + 10)*log(abs(x + sqrt(1/2*sqrt(5) - 1/2))) - 1/40*sqrt(10*sqrt(5) + 10)*log(abs(x - sqrt (1/2*sqrt(5) - 1/2)))
Time = 0.10 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.09 \[ \int \frac {1-x^4}{1-3 x^4+x^8} \, dx=-\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {\sqrt {5}-1}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}-7\right )}-\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {\sqrt {5}-1}\,7{}\mathrm {i}}{10\,\left (3\,\sqrt {5}-7\right )}\right )\,\sqrt {\sqrt {5}-1}\,1{}\mathrm {i}}{20}-\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}+7\right )}+\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,7{}\mathrm {i}}{10\,\left (3\,\sqrt {5}+7\right )}\right )\,\sqrt {\sqrt {5}+1}\,1{}\mathrm {i}}{20}+\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}-7\right )}-\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,7{}\mathrm {i}}{10\,\left (3\,\sqrt {5}-7\right )}\right )\,\sqrt {1-\sqrt {5}}\,1{}\mathrm {i}}{20}+\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {-\sqrt {5}-1}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}+7\right )}+\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {-\sqrt {5}-1}\,7{}\mathrm {i}}{10\,\left (3\,\sqrt {5}+7\right )}\right )\,\sqrt {-\sqrt {5}-1}\,1{}\mathrm {i}}{20} \]
(10^(1/2)*atan((10^(1/2)*x*(1 - 5^(1/2))^(1/2)*3i)/(2*(3*5^(1/2) - 7)) - ( 5^(1/2)*10^(1/2)*x*(1 - 5^(1/2))^(1/2)*7i)/(10*(3*5^(1/2) - 7)))*(1 - 5^(1 /2))^(1/2)*1i)/20 - (10^(1/2)*atan((10^(1/2)*x*(5^(1/2) + 1)^(1/2)*3i)/(2* (3*5^(1/2) + 7)) + (5^(1/2)*10^(1/2)*x*(5^(1/2) + 1)^(1/2)*7i)/(10*(3*5^(1 /2) + 7)))*(5^(1/2) + 1)^(1/2)*1i)/20 - (10^(1/2)*atan((10^(1/2)*x*(5^(1/2 ) - 1)^(1/2)*3i)/(2*(3*5^(1/2) - 7)) - (5^(1/2)*10^(1/2)*x*(5^(1/2) - 1)^( 1/2)*7i)/(10*(3*5^(1/2) - 7)))*(5^(1/2) - 1)^(1/2)*1i)/20 + (10^(1/2)*atan ((10^(1/2)*x*(- 5^(1/2) - 1)^(1/2)*3i)/(2*(3*5^(1/2) + 7)) + (5^(1/2)*10^( 1/2)*x*(- 5^(1/2) - 1)^(1/2)*7i)/(10*(3*5^(1/2) + 7)))*(- 5^(1/2) - 1)^(1/ 2)*1i)/20