Integrand size = 29, antiderivative size = 193 \[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=-\frac {1}{2} \sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-1+x^4}}\right ) \]
-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 = 0.55 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.87 \[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\frac {-\sqrt {-1+\sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )+\sqrt {1+\sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )-\sqrt {-1+\sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )+\sqrt {1+\sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2}} \]
(-(Sqrt[-1 + Sqrt[5]]*ArcTan[(Sqrt[(-1 + Sqrt[5])/2]*x)/(-1 + x^4)^(1/4)]) + Sqrt[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 = 0.53 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, 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 |
\(\displaystyle \int \left (\frac {2}{\left (2 x^4-\sqrt {5}-1\right ) \sqrt [4]{x^4-1}}+\frac {2}{\left (2 x^4+\sqrt {5}-1\right ) \sqrt [4]{x^4-1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4-1}}\right )\) |
-1/2*(((3 - Sqrt[5])/2)^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x)/(-1 + x^4 )^(1/4)]) + (((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 + Sqr t[5])/2)^(1/4)*x)/(-1 + x^4)^(1/4)])/2
3.25.2.3.1 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 = 131, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {\sqrt {5}\, \left (\left (5+\sqrt {5}\right ) \left (\operatorname {arctanh}\left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )-\arctan \left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )\right ) \sqrt {-2+2 \sqrt {5}}+\sqrt {2+2 \sqrt {5}}\, \left (-5+\sqrt {5}\right ) \left (\operatorname {arctanh}\left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )-\arctan \left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )\right )\right )}{40}\) | \(131\) |
trager | \(\text {Expression too large to display}\) | \(1623\) |
1/40*5^(1/2)*((5+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)))*(-2+2*5^(1/2))^(1/2)+(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) )-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. 1285 vs. \(2 (125) = 250\).
Time = 15.55 (sec) , antiderivative size = 1285, normalized size of antiderivative = 6.66 \[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\text {Too large to display} \]
1/16*sqrt(2)*sqrt(-sqrt(5) + 1)*log((sqrt(x^4 - 1)*(sqrt(5)*sqrt(2)*x^2 + sqrt(2)*(2*x^6 - x^2))*sqrt(-sqrt(5) + 1) + 2*(2*x^5 + sqrt(5)*x - x)*(x^4 - 1)^(3/4) - (sqrt(5)*sqrt(2)*(x^8 - x^4) + sqrt(2)*(2*x^4 - 1))*sqrt(-sq rt(5) + 1) + 2*(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(-(sqrt(x^4 - 1)*(sqrt(5)* sqrt(2)*x^2 + sqrt(2)*(2*x^6 - x^2))*sqrt(-sqrt(5) + 1) - 2*(2*x^5 + sqrt( 5)*x - x)*(x^4 - 1)^(3/4) - (sqrt(5)*sqrt(2)*(x^8 - x^4) + sqrt(2)*(2*x^4 - 1))*sqrt(-sqrt(5) + 1) - 2*(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((sqrt(x^4 - 1)*(sqrt(5)*sqrt(2)*x^2 - sqrt(2)*(2*x^6 - x^2))*sqrt(-sqrt(5) - 1) + 2*( 2*x^5 - sqrt(5)*x - x)*(x^4 - 1)^(3/4) + (sqrt(5)*sqrt(2)*(x^8 - x^4) - sq rt(2)*(2*x^4 - 1))*sqrt(-sqrt(5) - 1) - 2*(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)*lo g(-(sqrt(x^4 - 1)*(sqrt(5)*sqrt(2)*x^2 - sqrt(2)*(2*x^6 - x^2))*sqrt(-sqrt (5) - 1) - 2*(2*x^5 - sqrt(5)*x - x)*(x^4 - 1)^(3/4) + (sqrt(5)*sqrt(2)*(x ^8 - x^4) - sqrt(2)*(2*x^4 - 1))*sqrt(-sqrt(5) - 1) + 2*(x^7 - 3*x^3 + sqr t(5)*(x^7 - x^3))*(x^4 - 1)^(1/4))/(x^8 - x^4 - 1)) - 1/16*sqrt(2)*sqrt(sq rt(5) - 1)*log((2*(2*x^5 + sqrt(5)*x - x)*(x^4 - 1)^(3/4) + (sqrt(5)*sqrt( 2)*(x^8 - x^4) + sqrt(2)*(2*x^4 - 1) + sqrt(x^4 - 1)*(sqrt(5)*sqrt(2)*x^2 + sqrt(2)*(2*x^6 - x^2)))*sqrt(sqrt(5) - 1) - 2*(x^7 - 3*x^3 - sqrt(5)*...
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. \[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=-\int \frac {2\,x^4-1}{{\left (x^4-1\right )}^{1/4}\,\left (-x^8+x^4+1\right )} \,d x \]