Integrand size = 25, antiderivative size = 193 \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right ) \]
1/4*(2+2*5^(1/2))^(1/2)*arctan(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))+1 /4*(-2+2*5^(1/2))^(1/2)*arctan(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))+1/ 4*(2+2*5^(1/2))^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))+1/ 4*(-2+2*5^(1/2))^(1/2)*arctanh(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))
Time = 0.52 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.86 \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\frac {\sqrt {1+\sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\sqrt {-1+\sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\sqrt {1+\sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\sqrt {-1+\sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}} \]
(Sqrt[1 + Sqrt[5]]*ArcTan[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] + Sq rt[-1 + Sqrt[5]]*ArcTan[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] + Sqrt[ 1 + Sqrt[5]]*ArcTanh[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] + Sqrt[-1 + Sqrt[5]]*ArcTanh[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)])/(2*Sqrt[2] )
Time = 0.54 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.
\(\displaystyle \int \frac {x^4-2}{\sqrt [4]{x^4+1} \left (x^8+x^4-1\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {1-\sqrt {5}}{\left (2 x^4-\sqrt {5}+1\right ) \sqrt [4]{x^4+1}}+\frac {1+\sqrt {5}}{\left (2 x^4+\sqrt {5}+1\right ) \sqrt [4]{x^4+1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )\) |
(((3 + Sqrt[5])/2)^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4 )])/2 + (((3 - Sqrt[5])/2)^(1/4)*ArcTan[(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x ^4)^(1/4)])/2 + (((3 + Sqrt[5])/2)^(1/4)*ArcTanh[((2/(3 + Sqrt[5]))^(1/4)* x)/(1 + x^4)^(1/4)])/2 + (((3 - Sqrt[5])/2)^(1/4)*ArcTanh[(((3 + Sqrt[5])/ 2)^(1/4)*x)/(1 + x^4)^(1/4)])/2
3.25.4.3.1 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 = 10.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right ) \sqrt {-2+2 \sqrt {5}}}{4}-\frac {\arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right ) \sqrt {2+2 \sqrt {5}}}{4}-\frac {\arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right ) \sqrt {-2+2 \sqrt {5}}}{4}+\frac {\operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right ) \sqrt {2+2 \sqrt {5}}}{4}\) | \(134\) |
trager | \(\text {Expression too large to display}\) | \(1644\) |
1/4*arctanh(2/(2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))*(-2+2*5^(1/2))^(1/2)-1/ 4*arctan(2/(-2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))*(2+2*5^(1/2))^(1/2)-1/4*a rctan(2/(2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))*(-2+2*5^(1/2))^(1/2)+1/4*arct anh(2/(-2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))*(2+2*5^(1/2))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 1307 vs. \(2 (125) = 250\).
Time = 14.76 (sec) , antiderivative size = 1307, normalized size of antiderivative = 6.77 \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\text {Too large to display} \]
-1/16*sqrt(2)*sqrt(-sqrt(5) + 1)*log((2*sqrt(x^4 + 1)*(sqrt(5)*sqrt(2)*(x^ 6 + x^2) + sqrt(2)*(x^6 + 3*x^2))*sqrt(-sqrt(5) + 1) + 4*(2*x^5 + sqrt(5)* x + x)*(x^4 + 1)^(3/4) - (sqrt(5)*sqrt(2)*(x^8 + 3*x^4 + 1) + sqrt(2)*(5*x ^8 + 7*x^4 + 1))*sqrt(-sqrt(5) + 1) - 4*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3) )*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) + 1/16*sqrt(2)*sqrt(-sqrt(5) + 1)*log( -(2*sqrt(x^4 + 1)*(sqrt(5)*sqrt(2)*(x^6 + x^2) + sqrt(2)*(x^6 + 3*x^2))*sq rt(-sqrt(5) + 1) - 4*(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/4) - (sqrt(5)*sq rt(2)*(x^8 + 3*x^4 + 1) + sqrt(2)*(5*x^8 + 7*x^4 + 1))*sqrt(-sqrt(5) + 1) + 4*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) - 1/16*sqrt(2)*sqrt(-sqrt(5) - 1)*log((2*sqrt(x^4 + 1)*(sqrt(5)*sqrt(2)*(x ^6 + x^2) - sqrt(2)*(x^6 + 3*x^2))*sqrt(-sqrt(5) - 1) + 4*(2*x^5 - sqrt(5) *x + x)*(x^4 + 1)^(3/4) + (sqrt(5)*sqrt(2)*(x^8 + 3*x^4 + 1) - sqrt(2)*(5* x^8 + 7*x^4 + 1))*sqrt(-sqrt(5) - 1) + 4*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3 ))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) + 1/16*sqrt(2)*sqrt(-sqrt(5) - 1)*log (-(2*sqrt(x^4 + 1)*(sqrt(5)*sqrt(2)*(x^6 + x^2) - sqrt(2)*(x^6 + 3*x^2))*s qrt(-sqrt(5) - 1) - 4*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3/4) + (sqrt(5)*s qrt(2)*(x^8 + 3*x^4 + 1) - sqrt(2)*(5*x^8 + 7*x^4 + 1))*sqrt(-sqrt(5) - 1) - 4*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) + 1/16*sqrt(2)*sqrt(sqrt(5) - 1)*log((4*(2*x^5 + sqrt(5)*x + x)*(x^4 + 1) ^(3/4) + (sqrt(5)*sqrt(2)*(x^8 + 3*x^4 + 1) + sqrt(2)*(5*x^8 + 7*x^4 + ...
Timed out. \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {x^{4} - 2}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {x^{4} - 2}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int \frac {x^4-2}{{\left (x^4+1\right )}^{1/4}\,\left (x^8+x^4-1\right )} \,d x \]