Integrand size = 29, antiderivative size = 82 \[ \int \frac {-1+3 x^4}{\left (1-x+x^4\right ) \sqrt [3]{x^2+x^6}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x^2+x^6}}\right )+\log \left (-x+\sqrt [3]{x^2+x^6}\right )-\frac {1}{2} \log \left (x^2+x \sqrt [3]{x^2+x^6}+\left (x^2+x^6\right )^{2/3}\right ) \]
-3^(1/2)*arctan(3^(1/2)*x/(x+2*(x^6+x^2)^(1/3)))+ln(-x+(x^6+x^2)^(1/3))-1/ 2*ln(x^2+x*(x^6+x^2)^(1/3)+(x^6+x^2)^(2/3))
\[ \int \frac {-1+3 x^4}{\left (1-x+x^4\right ) \sqrt [3]{x^2+x^6}} \, dx=\int \frac {-1+3 x^4}{\left (1-x+x^4\right ) \sqrt [3]{x^2+x^6}} \, dx \]
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.
| \(\displaystyle \int \frac {3 x^4-1}{\left (x^4-x+1\right ) \sqrt [3]{x^6+x^2}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \sqrt [3]{x^4+1} \int -\frac {1-3 x^4}{x^{2/3} \sqrt [3]{x^4+1} \left (x^4-x+1\right )}dx}{\sqrt [3]{x^6+x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \sqrt [3]{x^4+1} \int \frac {1-3 x^4}{x^{2/3} \sqrt [3]{x^4+1} \left (x^4-x+1\right )}dx}{\sqrt [3]{x^6+x^2}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^4+1} \int \frac {1-3 x^4}{\sqrt [3]{x^4+1} \left (x^4-x+1\right )}d\sqrt [3]{x}}{\sqrt [3]{x^6+x^2}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^4+1} \int \left (\frac {4-3 x}{\sqrt [3]{x^4+1} \left (x^4-x+1\right )}-\frac {3}{\sqrt [3]{x^4+1}}\right )d\sqrt [3]{x}}{\sqrt [3]{x^6+x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^4+1} \left (4 \int \frac {1}{\sqrt [3]{x^4+1} \left (x^4-x+1\right )}d\sqrt [3]{x}-3 \int \frac {x}{\sqrt [3]{x^4+1} \left (x^4-x+1\right )}d\sqrt [3]{x}-3 \sqrt [3]{x} \operatorname {Hypergeometric2F1}\left (\frac {1}{12},\frac {1}{3},\frac {13}{12},-x^4\right )\right )}{\sqrt [3]{x^6+x^2}}\) |
3.12.4.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 6.90 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04
| method | result | size |
| pseudoelliptic | \(\ln \left (\frac {\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{3}}-x}{x}\right )-\frac {\ln \left (\frac {\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {2}{3}}+\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )\) | \(85\) |
| trager | \(\ln \left (-\frac {4747 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{5}-6570 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{5}-5848 x^{5}-9494 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+4747 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +12039 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {2}{3}}+9873 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{3}} x -20089 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-6570 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x +21912 \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-12039 x \left (x^{6}+x^{2}\right )^{\frac {1}{3}}-8772 x^{2}-5848 x}{x \left (x^{4}-x +1\right )}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {2924 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{5}+13519 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{5}+14241 x^{5}-5848 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+2924 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +12039 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-21912 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{3}} x -6570 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+13519 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x -9873 \left (x^{6}+x^{2}\right )^{\frac {2}{3}}-12039 x \left (x^{6}+x^{2}\right )^{\frac {1}{3}}+4747 x^{2}+14241 x}{x \left (x^{4}-x +1\right )}\right )\) | \(337\) |
ln(((x^2*(x^4+1))^(1/3)-x)/x)-1/2*ln(((x^2*(x^4+1))^(2/3)+(x^2*(x^4+1))^(1 /3)*x+x^2)/x^2)+3^(1/2)*arctan(1/3*(2*(x^2*(x^4+1))^(1/3)+x)*3^(1/2)/x)
Time = 1.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.37 \[ \int \frac {-1+3 x^4}{\left (1-x+x^4\right ) \sqrt [3]{x^2+x^6}} \, dx=-\sqrt {3} \arctan \left (-\frac {2 \, \sqrt {3} {\left (x^{6} + x^{2}\right )}^{\frac {1}{3}} x + \sqrt {3} {\left (x^{5} - x^{2} + x\right )} - 2 \, \sqrt {3} {\left (x^{6} + x^{2}\right )}^{\frac {2}{3}}}{3 \, {\left (x^{5} + x^{2} + x\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{5} - x^{2} + 3 \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{3}} x + x - 3 \, {\left (x^{6} + x^{2}\right )}^{\frac {2}{3}}}{x^{5} - x^{2} + x}\right ) \]
-sqrt(3)*arctan(-1/3*(2*sqrt(3)*(x^6 + x^2)^(1/3)*x + sqrt(3)*(x^5 - x^2 + x) - 2*sqrt(3)*(x^6 + x^2)^(2/3))/(x^5 + x^2 + x)) + 1/2*log((x^5 - x^2 + 3*(x^6 + x^2)^(1/3)*x + x - 3*(x^6 + x^2)^(2/3))/(x^5 - x^2 + x))
Timed out. \[ \int \frac {-1+3 x^4}{\left (1-x+x^4\right ) \sqrt [3]{x^2+x^6}} \, dx=\text {Timed out} \]
\[ \int \frac {-1+3 x^4}{\left (1-x+x^4\right ) \sqrt [3]{x^2+x^6}} \, dx=\int { \frac {3 \, x^{4} - 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{4} - x + 1\right )}} \,d x } \]
\[ \int \frac {-1+3 x^4}{\left (1-x+x^4\right ) \sqrt [3]{x^2+x^6}} \, dx=\int { \frac {3 \, x^{4} - 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{4} - x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {-1+3 x^4}{\left (1-x+x^4\right ) \sqrt [3]{x^2+x^6}} \, dx=\int \frac {3\,x^4-1}{{\left (x^6+x^2\right )}^{1/3}\,\left (x^4-x+1\right )} \,d x \]