Integrand size = 27, antiderivative size = 201 \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\frac {1}{2} \sqrt {\frac {1}{10} \left (11+5 \sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-11+5 \sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{10} \left (11+5 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-11+5 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right ) \]
1/20*(110+50*5^(1/2))^(1/2)*arctan(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4 ))-1/20*(-110+50*5^(1/2))^(1/2)*arctan(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+1)^( 1/4))+1/20*(110+50*5^(1/2))^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+ 1)^(1/4))-1/20*(-110+50*5^(1/2))^(1/2)*arctanh(1/2*(2+2*5^(1/2))^(1/2)*x/( x^4+1)^(1/4))
Time = 0.80 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.88 \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\frac {\sqrt {11+5 \sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )-\sqrt {-11+5 \sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\sqrt {11+5 \sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )-\sqrt {-11+5 \sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {10}} \]
(Sqrt[11 + 5*Sqrt[5]]*ArcTan[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] - Sqrt[-11 + 5*Sqrt[5]]*ArcTan[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] + Sqrt[11 + 5*Sqrt[5]]*ArcTanh[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] - Sqrt[-11 + 5*Sqrt[5]]*ArcTanh[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] )/(2*Sqrt[10])
Time = 0.57 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, 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.
\(\displaystyle \int \frac {2 x^4-1}{\sqrt [4]{x^4+1} \left (x^8+x^4-1\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {2-\frac {4}{\sqrt {5}}}{\left (2 x^4-\sqrt {5}+1\right ) \sqrt [4]{x^4+1}}+\frac {2+\frac {4}{\sqrt {5}}}{\left (2 x^4+\sqrt {5}+1\right ) \sqrt [4]{x^4+1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}+\frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}\) |
(((123 + 55*Sqrt[5])/2)^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4) ^(1/4)])/(2*Sqrt[5]) - (((123 - 55*Sqrt[5])/2)^(1/4)*ArcTan[(((3 + Sqrt[5] )/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5]) + (((123 + 55*Sqrt[5])/2)^(1/4 )*ArcTanh[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5]) - (((1 23 - 55*Sqrt[5])/2)^(1/4)*ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1 /4)])/(2*Sqrt[5])
3.25.71.3.1 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 = 11.98 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(\frac {\sqrt {5}\, \left (\left (\sqrt {5}-3\right ) \left (\operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )-\arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )\right ) \sqrt {-2+2 \sqrt {5}}+\sqrt {2+2 \sqrt {5}}\, \left (3+\sqrt {5}\right ) \left (\operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )-\arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )\right )\right )}{40}\) | \(131\) |
trager | \(\text {Expression too large to display}\) | \(1571\) |
1/40*5^(1/2)*((5^(1/2)-3)*(arctanh(2/(2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))- arctan(2/(2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4)))*(-2+2*5^(1/2))^(1/2)+(2+2*5 ^(1/2))^(1/2)*(3+5^(1/2))*(arctanh(2/(-2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4)) -arctan(2/(-2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))))
Leaf count of result is larger than twice the leaf count of optimal. 1309 vs. \(2 (125) = 250\).
Time = 17.46 (sec) , antiderivative size = 1309, normalized size of antiderivative = 6.51 \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\text {Too large to display} \]
1/80*sqrt(10)*sqrt(-5*sqrt(5) + 11)*log((sqrt(10)*(10*x^6 + 15*x^2 + sqrt( 5)*(4*x^6 + 7*x^2))*sqrt(x^4 + 1)*sqrt(-5*sqrt(5) + 11) - sqrt(10)*(10*x^8 + 20*x^4 + sqrt(5)*(5*x^8 + 9*x^4 + 2) + 5)*sqrt(-5*sqrt(5) + 11) + 10*(2 *x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/4) - 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x ^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) - 1/80*sqrt(10)*sqrt(-5*sqrt(5) + 1 1)*log(-(sqrt(10)*(10*x^6 + 15*x^2 + sqrt(5)*(4*x^6 + 7*x^2))*sqrt(x^4 + 1 )*sqrt(-5*sqrt(5) + 11) - sqrt(10)*(10*x^8 + 20*x^4 + sqrt(5)*(5*x^8 + 9*x ^4 + 2) + 5)*sqrt(-5*sqrt(5) + 11) - 10*(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^ (3/4) + 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) - 1/80*sqrt(10)*sqrt(-5*sqrt(5) - 11)*log((sqrt(10)*(10*x^6 + 15*x^ 2 - sqrt(5)*(4*x^6 + 7*x^2))*sqrt(x^4 + 1)*sqrt(-5*sqrt(5) - 11) + sqrt(10 )*(10*x^8 + 20*x^4 - sqrt(5)*(5*x^8 + 9*x^4 + 2) + 5)*sqrt(-5*sqrt(5) - 11 ) + 10*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3/4) + 10*(x^7 + 3*x^3 - sqrt(5) *(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) + 1/80*sqrt(10)*sqrt(-5*sq rt(5) - 11)*log(-(sqrt(10)*(10*x^6 + 15*x^2 - sqrt(5)*(4*x^6 + 7*x^2))*sqr t(x^4 + 1)*sqrt(-5*sqrt(5) - 11) + sqrt(10)*(10*x^8 + 20*x^4 - sqrt(5)*(5* x^8 + 9*x^4 + 2) + 5)*sqrt(-5*sqrt(5) - 11) - 10*(2*x^5 - sqrt(5)*x + x)*( x^4 + 1)^(3/4) - 10*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/( x^8 + x^4 - 1)) - 1/80*sqrt(10)*sqrt(5*sqrt(5) - 11)*log((10*(2*x^5 + sqrt (5)*x + x)*(x^4 + 1)^(3/4) + (sqrt(10)*(10*x^6 + 15*x^2 + sqrt(5)*(4*x^...
Timed out. \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{4} - 1}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{4} - 1}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int \frac {2\,x^4-1}{{\left (x^4+1\right )}^{1/4}\,\left (x^8+x^4-1\right )} \,d x \]