Integrand size = 37, antiderivative size = 43 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4\right )} \, dx=\text {arctanh}\left (\frac {x}{\sqrt {1+x^2+x^4}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^2+x^4}}\right ) \]
Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4\right )} \, dx=\text {arctanh}\left (\frac {x}{\sqrt {1+x^2+x^4}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^2+x^4}}\right ) \]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.36 (sec) , antiderivative size = 451, normalized size of antiderivative = 10.49, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {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.
\(\displaystyle \int \frac {\left (x^4-1\right ) \sqrt {x^4+x^2+1}}{\left (x^4+1\right ) \left (x^4-x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {\left (2 x^2-1\right ) \sqrt {x^4+x^2+1}}{x^4-x^2+1}-\frac {2 x^2 \sqrt {x^4+x^2+1}}{x^4+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (\sqrt {3}+2 i\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{\left (\sqrt {3}+3 i\right ) \sqrt {x^4+x^2+1}}+\frac {\left (-\sqrt {3}+2 i\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{\left (-\sqrt {3}+3 i\right ) \sqrt {x^4+x^2+1}}-\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{2 \sqrt {x^4+x^2+1}}-\frac {\left (-\sqrt {3}+i\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {3}{4},2 \arctan (x),\frac {1}{4}\right )}{2 \left (\sqrt {3}+3 i\right ) \sqrt {x^4+x^2+1}}-\frac {\left (\sqrt {3}+i\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {3}{4},2 \arctan (x),\frac {1}{4}\right )}{2 \left (-\sqrt {3}+3 i\right ) \sqrt {x^4+x^2+1}}+\text {arctanh}\left (\frac {x}{\sqrt {x^4+x^2+1}}\right )-\frac {\sqrt {\frac {3}{2}} \left (1+i \sqrt {3}\right ) \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^4+x^2+1}}\right )}{\sqrt {3}+3 i}-\frac {\sqrt {\frac {3}{2}} \left (\sqrt {3}+i\right ) \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^4+x^2+1}}\right )}{3+i \sqrt {3}}\) |
ArcTanh[x/Sqrt[1 + x^2 + x^4]] - (Sqrt[3/2]*(I + Sqrt[3])*ArcTanh[(Sqrt[2] *x)/Sqrt[1 + x^2 + x^4]])/(3 + I*Sqrt[3]) - (Sqrt[3/2]*(1 + I*Sqrt[3])*Arc Tanh[(Sqrt[2]*x)/Sqrt[1 + x^2 + x^4]])/(3*I + Sqrt[3]) - (3*(1 + x^2)*Sqrt [(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/4])/(2*Sqrt[1 + x^2 + x^4]) + ((2*I - Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*El lipticF[2*ArcTan[x], 1/4])/((3*I - Sqrt[3])*Sqrt[1 + x^2 + x^4]) + ((2*I + Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x ], 1/4])/((3*I + Sqrt[3])*Sqrt[1 + x^2 + x^4]) - ((I + Sqrt[3])*(1 + x^2)* Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticPi[3/4, 2*ArcTan[x], 1/4])/(2*(3 *I - Sqrt[3])*Sqrt[1 + x^2 + x^4]) - ((I - Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^ 2 + x^4)/(1 + x^2)^2]*EllipticPi[3/4, 2*ArcTan[x], 1/4])/(2*(3*I + Sqrt[3] )*Sqrt[1 + x^2 + x^4])
3.6.46.3.1 Defintions of rubi rules used
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 = 3.62 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.63
method | result | size |
elliptic | \(\frac {\left (\ln \left (\frac {\sqrt {x^{4}+x^{2}+1}\, \sqrt {2}}{2 x}-1\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x^{4}+x^{2}+1}}{x}\right )-\ln \left (1+\frac {\sqrt {x^{4}+x^{2}+1}\, \sqrt {2}}{2 x}\right )\right ) \sqrt {2}}{2}\) | \(70\) |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{4}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}-4 x \sqrt {x^{4}+x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{x^{4}-x^{2}+1}\right )}{2}-\frac {\ln \left (-\frac {-x^{4}+2 x \sqrt {x^{4}+x^{2}+1}-2 x^{2}-1}{x^{4}+1}\right )}{2}\) | \(102\) |
default | \(-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (\sqrt {3}\, x^{2}+\sqrt {3}-x \right ) \sqrt {2}}{2 \sqrt {x^{4}+x^{2}+1}}\right )}{2}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (x +\sqrt {3}\, x^{2}+\sqrt {3}\right ) \sqrt {2}}{2 \sqrt {x^{4}+x^{2}+1}}\right )}{2}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {2}\, x^{2}+\sqrt {2}-x}{\sqrt {x^{4}+x^{2}+1}}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {x +\sqrt {2}\, x^{2}+\sqrt {2}}{\sqrt {x^{4}+x^{2}+1}}\right )}{2}\) | \(124\) |
pseudoelliptic | \(-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (\sqrt {3}\, x^{2}+\sqrt {3}-x \right ) \sqrt {2}}{2 \sqrt {x^{4}+x^{2}+1}}\right )}{2}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (x +\sqrt {3}\, x^{2}+\sqrt {3}\right ) \sqrt {2}}{2 \sqrt {x^{4}+x^{2}+1}}\right )}{2}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {2}\, x^{2}+\sqrt {2}-x}{\sqrt {x^{4}+x^{2}+1}}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {x +\sqrt {2}\, x^{2}+\sqrt {2}}{\sqrt {x^{4}+x^{2}+1}}\right )}{2}\) | \(124\) |
1/2*(ln(1/2*(x^4+x^2+1)^(1/2)*2^(1/2)/x-1)+2^(1/2)*arctanh(1/x*(x^4+x^2+1) ^(1/2))-ln(1+1/2*(x^4+x^2+1)^(1/2)*2^(1/2)/x))*2^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (35) = 70\).
Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.58 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4\right )} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (-\frac {x^{8} + 14 \, x^{6} + 19 \, x^{4} - 4 \, \sqrt {2} {\left (x^{5} + 3 \, x^{3} + x\right )} \sqrt {x^{4} + x^{2} + 1} + 14 \, x^{2} + 1}{x^{8} - 2 \, x^{6} + 3 \, x^{4} - 2 \, x^{2} + 1}\right ) + \frac {1}{2} \, \log \left (-\frac {x^{4} + 2 \, x^{2} + 2 \, \sqrt {x^{4} + x^{2} + 1} x + 1}{x^{4} + 1}\right ) \]
1/4*sqrt(2)*log(-(x^8 + 14*x^6 + 19*x^4 - 4*sqrt(2)*(x^5 + 3*x^3 + x)*sqrt (x^4 + x^2 + 1) + 14*x^2 + 1)/(x^8 - 2*x^6 + 3*x^4 - 2*x^2 + 1)) + 1/2*log (-(x^4 + 2*x^2 + 2*sqrt(x^4 + x^2 + 1)*x + 1)/(x^4 + 1))
\[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4\right )} \, dx=\int \frac {\sqrt {\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}{\left (x^{4} + 1\right ) \left (x^{4} - x^{2} + 1\right )}\, dx \]
Integral(sqrt((x**2 - x + 1)*(x**2 + x + 1))*(x - 1)*(x + 1)*(x**2 + 1)/(( x**4 + 1)*(x**4 - x**2 + 1)), x)
\[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4\right )} \, dx=\int { \frac {\sqrt {x^{4} + x^{2} + 1} {\left (x^{4} - 1\right )}}{{\left (x^{4} - x^{2} + 1\right )} {\left (x^{4} + 1\right )}} \,d x } \]
\[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4\right )} \, dx=\int { \frac {\sqrt {x^{4} + x^{2} + 1} {\left (x^{4} - 1\right )}}{{\left (x^{4} - x^{2} + 1\right )} {\left (x^{4} + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4\right )} \, dx=\int \frac {\left (x^4-1\right )\,\sqrt {x^4+x^2+1}}{\left (x^4+1\right )\,\left (x^4-x^2+1\right )} \,d x \]