Integrand size = 22, antiderivative size = 67 \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \]
-1/2*x/(x^4+1)^(1/2)-1/8*arctan(2^(1/2)*x/(x^4+1)^(1/2))*2^(1/2)-1/8*arcta nh(2^(1/2)*x/(x^4+1)^(1/2))*2^(1/2)
Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \]
-1/2*x/Sqrt[1 + x^4] - ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(4*Sqrt[2]) - Arc Tanh[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(4*Sqrt[2])
Time = 0.54 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1388, 7276, 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.
\(\displaystyle \int \frac {x^8+1}{\sqrt {x^4+1} \left (x^8-1\right )} \, dx\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \int \frac {x^8+1}{\left (x^4-1\right ) \left (x^4+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {x^4}{\left (x^4+1\right )^{3/2}}+\frac {2}{\left (x^4-1\right ) \left (x^4+1\right )^{3/2}}+\frac {1}{\left (x^4+1\right )^{3/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}}-\frac {x}{2 \sqrt {x^4+1}}\) |
-1/2*x/Sqrt[1 + x^4] - ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(4*Sqrt[2]) - Arc Tanh[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(4*Sqrt[2])
3.9.90.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 5.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {x}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right )}{16}\) | \(73\) |
elliptic | \(\frac {\left (\frac {\ln \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-1\right )}{8}-\frac {\sqrt {2}\, x}{2 \sqrt {x^{4}+1}}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{8}\right ) \sqrt {2}}{2}\) | \(78\) |
trager | \(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (x -1\right ) \left (1+x \right )}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{8}\) | \(83\) |
default | \(\frac {\sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) x^{4}-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right ) x^{4}+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right ) x^{4}+2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+4 \sqrt {2}\, x \sqrt {x^{4}+1}-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right )}{16 \left (x^{2}+x \sqrt {2}+1\right ) \left (x \sqrt {2}-x^{2}-1\right )}\) | \(166\) |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) x^{4}-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right ) x^{4}+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right ) x^{4}+2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+4 \sqrt {2}\, x \sqrt {x^{4}+1}-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right )}{16 \left (x^{2}+x \sqrt {2}+1\right ) \left (x \sqrt {2}-x^{2}-1\right )}\) | \(166\) |
-1/2*x/(x^4+1)^(1/2)-1/16*2^(1/2)*(2*arctan(2^(1/2)*x/(x^4+1)^(1/2))+arcta nh((x^2-x+1)*2^(1/2)/(x^4+1)^(1/2))-arctanh((x^2+x+1)*2^(1/2)/(x^4+1)^(1/2 )))
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.34 \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {2 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) + 8 \, \sqrt {x^{4} + 1} x}{16 \, {\left (x^{4} + 1\right )}} \]
-1/16*(2*sqrt(2)*(x^4 + 1)*arctan(sqrt(2)*x/sqrt(x^4 + 1)) - sqrt(2)*(x^4 + 1)*log((x^4 - 2*sqrt(2)*sqrt(x^4 + 1)*x + 2*x^2 + 1)/(x^4 - 2*x^2 + 1)) + 8*sqrt(x^4 + 1)*x)/(x^4 + 1)
\[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=\int \frac {x^{8} + 1}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=\int { \frac {x^{8} + 1}{{\left (x^{8} - 1\right )} \sqrt {x^{4} + 1}} \,d x } \]
\[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=\int { \frac {x^{8} + 1}{{\left (x^{8} - 1\right )} \sqrt {x^{4} + 1}} \,d x } \]
Timed out. \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=\int \frac {x^8+1}{\sqrt {x^4+1}\,\left (x^8-1\right )} \,d x \]