Integrand size = 32, antiderivative size = 217 \[ \int \frac {-1+2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right ) \]
arctan(x/(x^4+1)^(1/4))-1/20*(-10+10*5^(1/2))^(1/2)*arctan(1/2*(-2+2*5^(1/ 2))^(1/2)*x/(x^4+1)^(1/4))-1/20*(10+10*5^(1/2))^(1/2)*arctan(1/2*(2+2*5^(1 /2))^(1/2)*x/(x^4+1)^(1/4))+arctanh(x/(x^4+1)^(1/4))-1/20*(-10+10*5^(1/2)) ^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))-1/20*(10+10*5^(1/ 2))^(1/2)*arctanh(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))
Time = 0.53 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.93 \[ \int \frac {-1+2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\frac {1}{20} \left (20 \arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\sqrt {10 \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )-\sqrt {10 \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+20 \text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\sqrt {10 \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )-\sqrt {10 \left (1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )\right ) \]
(20*ArcTan[x/(1 + x^4)^(1/4)] - Sqrt[10*(-1 + Sqrt[5])]*ArcTan[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] - Sqrt[10*(1 + Sqrt[5])]*ArcTan[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] + 20*ArcTanh[x/(1 + x^4)^(1/4)] - Sqrt[1 0*(-1 + Sqrt[5])]*ArcTanh[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] - Sq rt[10*(1 + Sqrt[5])]*ArcTanh[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)])/2 0
Time = 0.51 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {7279, 2009}
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.
| \(\displaystyle \int \frac {2 x^8+2 x^4-1}{\sqrt [4]{x^4+1} \left (x^8+x^4-1\right )} \, dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {2}{\sqrt [4]{x^4+1}}+\frac {1}{\left (x^8+x^4-1\right ) \sqrt [4]{x^4+1}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \arctan \left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {\sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {\sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}\) |
ArcTan[x/(1 + x^4)^(1/4)] - (((3 - Sqrt[5])/2)^(1/4)*ArcTan[((2/(3 + Sqrt[ 5]))^(1/4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5]) - (((3 + Sqrt[5])/2)^(1/4)*Arc Tan[(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5]) + ArcTanh[x/ (1 + x^4)^(1/4)] - (((3 - Sqrt[5])/2)^(1/4)*ArcTanh[((2/(3 + Sqrt[5]))^(1/ 4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5]) - (((3 + Sqrt[5])/2)^(1/4)*ArcTanh[((( 3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5])
3.26.65.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 2.72 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.93
| method | result | size |
| pseudoelliptic | \(\frac {\sqrt {2+2 \sqrt {5}}\, \left (-5+\sqrt {5}\right ) \operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )}{40}-\frac {\sqrt {-2+2 \sqrt {5}}\, \left (5+\sqrt {5}\right ) \operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )}{40}-\frac {\sqrt {2+2 \sqrt {5}}\, \left (-5+\sqrt {5}\right ) \arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )}{40}+\frac {\sqrt {-2+2 \sqrt {5}}\, \left (5+\sqrt {5}\right ) \arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )}{40}+\frac {\ln \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}+x}{x}\right )}{2}-\arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )-\frac {\ln \left (\frac {-x +\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )}{2}\) | \(202\) |
1/40*(2+2*5^(1/2))^(1/2)*(-5+5^(1/2))*arctanh(2/(-2+2*5^(1/2))^(1/2)/x*(x^ 4+1)^(1/4))-1/40*(-2+2*5^(1/2))^(1/2)*(5+5^(1/2))*arctanh(2/(2+2*5^(1/2))^ (1/2)/x*(x^4+1)^(1/4))-1/40*(2+2*5^(1/2))^(1/2)*(-5+5^(1/2))*arctan(2/(-2+ 2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))+1/40*(-2+2*5^(1/2))^(1/2)*(5+5^(1/2))*ar ctan(2/(2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))+1/2*ln(((x^4+1)^(1/4)+x)/x)-ar ctan((x^4+1)^(1/4)/x)-1/2*ln((-x+(x^4+1)^(1/4))/x)
Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (145) = 290\).
Time = 0.26 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.87 \[ \int \frac {-1+2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=-\frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} + 1} \log \left (\frac {\sqrt {10} \sqrt {5} x \sqrt {\sqrt {5} + 1} + 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} + 1} \log \left (-\frac {\sqrt {10} \sqrt {5} x \sqrt {\sqrt {5} + 1} - 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} - 1} \log \left (\frac {\sqrt {10} \sqrt {5} x \sqrt {\sqrt {5} - 1} + 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} - 1} \log \left (-\frac {\sqrt {10} \sqrt {5} x \sqrt {\sqrt {5} - 1} - 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} + 1} \log \left (\frac {\sqrt {10} \sqrt {5} x \sqrt {-\sqrt {5} + 1} + 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} + 1} \log \left (-\frac {\sqrt {10} \sqrt {5} x \sqrt {-\sqrt {5} + 1} - 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} - 1} \log \left (\frac {\sqrt {10} \sqrt {5} x \sqrt {-\sqrt {5} - 1} + 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} - 1} \log \left (-\frac {\sqrt {10} \sqrt {5} x \sqrt {-\sqrt {5} - 1} - 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) \]
-1/40*sqrt(10)*sqrt(sqrt(5) + 1)*log((sqrt(10)*sqrt(5)*x*sqrt(sqrt(5) + 1) + 10*(x^4 + 1)^(1/4))/x) + 1/40*sqrt(10)*sqrt(sqrt(5) + 1)*log(-(sqrt(10) *sqrt(5)*x*sqrt(sqrt(5) + 1) - 10*(x^4 + 1)^(1/4))/x) - 1/40*sqrt(10)*sqrt (sqrt(5) - 1)*log((sqrt(10)*sqrt(5)*x*sqrt(sqrt(5) - 1) + 10*(x^4 + 1)^(1/ 4))/x) + 1/40*sqrt(10)*sqrt(sqrt(5) - 1)*log(-(sqrt(10)*sqrt(5)*x*sqrt(sqr t(5) - 1) - 10*(x^4 + 1)^(1/4))/x) + 1/40*sqrt(10)*sqrt(-sqrt(5) + 1)*log( (sqrt(10)*sqrt(5)*x*sqrt(-sqrt(5) + 1) + 10*(x^4 + 1)^(1/4))/x) - 1/40*sqr t(10)*sqrt(-sqrt(5) + 1)*log(-(sqrt(10)*sqrt(5)*x*sqrt(-sqrt(5) + 1) - 10* (x^4 + 1)^(1/4))/x) + 1/40*sqrt(10)*sqrt(-sqrt(5) - 1)*log((sqrt(10)*sqrt( 5)*x*sqrt(-sqrt(5) - 1) + 10*(x^4 + 1)^(1/4))/x) - 1/40*sqrt(10)*sqrt(-sqr t(5) - 1)*log(-(sqrt(10)*sqrt(5)*x*sqrt(-sqrt(5) - 1) - 10*(x^4 + 1)^(1/4) )/x) - arctan((x^4 + 1)^(1/4)/x) + 1/2*log((x + (x^4 + 1)^(1/4))/x) - 1/2* log(-(x - (x^4 + 1)^(1/4))/x)
Timed out. \[ \int \frac {-1+2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-1+2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{8} + 2 \, x^{4} - 1}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {-1+2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{8} + 2 \, x^{4} - 1}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {-1+2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int \frac {2\,x^8+2\,x^4-1}{{\left (x^4+1\right )}^{1/4}\,\left (x^8+x^4-1\right )} \,d x \]