Integrand size = 29, antiderivative size = 53 \[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\frac {1}{3} \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2+x^4}}\right )+\frac {1}{3} \text {arctanh}\left (\frac {x}{\sqrt {1-x^2+x^4}}\right ) \]
Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\frac {1}{3} \left (\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt {1-x^2+x^4}}\right )\right ) \]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.31 (sec) , antiderivative size = 371, normalized size of antiderivative = 7.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2019, 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.
\(\displaystyle \int \frac {x^6+1}{\sqrt {x^4-x^2+1} \left (1-x^6\right )} \, dx\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle \int \frac {\left (x^2+1\right ) \sqrt {x^4-x^2+1}}{1-x^6}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {\sqrt {x^4-x^2+1}}{3 (1-x)}+\frac {\sqrt {x^4-x^2+1}}{3 (x+1)}+\frac {\left (1-\sqrt [3]{-1}\right ) \sqrt {x^4-x^2+1}}{6 \left (1-\sqrt [3]{-1} x\right )}+\frac {\left (1-\sqrt [3]{-1}\right ) \sqrt {x^4-x^2+1}}{6 \left (\sqrt [3]{-1} x+1\right )}+\frac {\left (1+(-1)^{2/3}\right ) \sqrt {x^4-x^2+1}}{6 \left (1-(-1)^{2/3} x\right )}+\frac {\left (1+(-1)^{2/3}\right ) \sqrt {x^4-x^2+1}}{6 \left ((-1)^{2/3} x+1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4-x^2+1}}\right )+\frac {\left (1+3 i \sqrt {3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {3}{4}\right )}{12 \sqrt {x^4-x^2+1}}+\frac {\left (1-3 i \sqrt {3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {3}{4}\right )}{12 \sqrt {x^4-x^2+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {3}{4}\right )}{6 \sqrt {x^4-x^2+1}}-\frac {\left (1-(-1)^{2/3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{4},2 \arctan (x),\frac {3}{4}\right )}{6 \left (1+(-1)^{2/3}\right ) \sqrt {x^4-x^2+1}}-\frac {i \left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{4},2 \arctan (x),\frac {3}{4}\right )}{2 \sqrt {3} \sqrt {x^4-x^2+1}}+\frac {1}{3} \text {arctanh}\left (\frac {x}{\sqrt {x^4-x^2+1}}\right )\) |
(Sqrt[2]*ArcTan[(Sqrt[2]*x)/Sqrt[1 - x^2 + x^4]])/3 + ArcTanh[x/Sqrt[1 - x ^2 + x^4]]/3 - ((1 + x^2)*Sqrt[(1 - x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*Ar cTan[x], 3/4])/(6*Sqrt[1 - x^2 + x^4]) + ((1 - (3*I)*Sqrt[3])*(1 + x^2)*Sq rt[(1 - x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 3/4])/(12*Sqrt[1 - x^2 + x^4]) + ((1 + (3*I)*Sqrt[3])*(1 + x^2)*Sqrt[(1 - x^2 + x^4)/(1 + x^2 )^2]*EllipticF[2*ArcTan[x], 3/4])/(12*Sqrt[1 - x^2 + x^4]) - ((I/2)*(1 + x ^2)*Sqrt[(1 - x^2 + x^4)/(1 + x^2)^2]*EllipticPi[1/4, 2*ArcTan[x], 3/4])/( Sqrt[3]*Sqrt[1 - x^2 + x^4]) - ((1 - (-1)^(2/3))*(1 + x^2)*Sqrt[(1 - x^2 + x^4)/(1 + x^2)^2]*EllipticPi[1/4, 2*ArcTan[x], 3/4])/(6*(1 + (-1)^(2/3))* Sqrt[1 - x^2 + x^4])
3.7.78.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 = 6.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98
method | result | size |
elliptic | \(\frac {\left (-\frac {2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}-x^{2}+1}}{2 x}\right )}{3}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x^{4}-x^{2}+1}}{x}\right )}{3}\right ) \sqrt {2}}{2}\) | \(52\) |
trager | \(-\frac {\ln \left (-\frac {\sqrt {x^{4}-x^{2}+1}-x}{\left (x -1\right ) \left (1+x \right )}\right )}{3}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{4}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}+4 \sqrt {x^{4}-x^{2}+1}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{\left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )}\right )}{6}\) | \(105\) |
default | \(-\frac {\operatorname {arctanh}\left (\frac {2 x^{2}+3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )}{6}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (x^{2}-3 x +1\right ) \sqrt {2}}{2 \sqrt {x^{4}-x^{2}+1}}\right )}{6}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (x^{2}+3 x +1\right ) \sqrt {2}}{2 \sqrt {x^{4}-x^{2}+1}}\right )}{6}+\frac {\operatorname {arctanh}\left (\frac {2 x^{2}-3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )}{6}\) | \(116\) |
pseudoelliptic | \(-\frac {\operatorname {arctanh}\left (\frac {2 x^{2}+3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )}{6}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (x^{2}-3 x +1\right ) \sqrt {2}}{2 \sqrt {x^{4}-x^{2}+1}}\right )}{6}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (x^{2}+3 x +1\right ) \sqrt {2}}{2 \sqrt {x^{4}-x^{2}+1}}\right )}{6}+\frac {\operatorname {arctanh}\left (\frac {2 x^{2}-3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )}{6}\) | \(116\) |
1/2*(-2/3*arctan(1/2*2^(1/2)/x*(x^4-x^2+1)^(1/2))+1/3*2^(1/2)*arctanh((x^4 -x^2+1)^(1/2)/x))*2^(1/2)
Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.17 \[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\frac {1}{6} \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {x^{4} - x^{2} + 1} x}{x^{4} - 3 \, x^{2} + 1}\right ) + \frac {1}{3} \, \log \left (\frac {x + \sqrt {x^{4} - x^{2} + 1}}{x^{2} - 1}\right ) \]
1/6*sqrt(2)*arctan(2*sqrt(2)*sqrt(x^4 - x^2 + 1)*x/(x^4 - 3*x^2 + 1)) + 1/ 3*log((x + sqrt(x^4 - x^2 + 1))/(x^2 - 1))
\[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=- \int \frac {\sqrt {x^{4} - x^{2} + 1}}{x^{6} - 1}\, dx - \int \frac {x^{2} \sqrt {x^{4} - x^{2} + 1}}{x^{6} - 1}\, dx \]
-Integral(sqrt(x**4 - x**2 + 1)/(x**6 - 1), x) - Integral(x**2*sqrt(x**4 - x**2 + 1)/(x**6 - 1), x)
\[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\int { -\frac {x^{6} + 1}{{\left (x^{6} - 1\right )} \sqrt {x^{4} - x^{2} + 1}} \,d x } \]
\[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\int { -\frac {x^{6} + 1}{{\left (x^{6} - 1\right )} \sqrt {x^{4} - x^{2} + 1}} \,d x } \]
Timed out. \[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\int -\frac {x^6+1}{\left (x^6-1\right )\,\sqrt {x^4-x^2+1}} \,d x \]