Integrand size = 28, antiderivative size = 105 \[ \int \frac {x \left (3+5 x^2\right )}{\sqrt [3]{1+x^2} \left (-1+x^3+x^5\right )} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} x \sqrt [3]{1+x^2}}{2+x \sqrt [3]{1+x^2}}\right )+2 \text {arctanh}\left (1-2 x \sqrt [3]{1+x^2}\right )+\frac {1}{2} \log \left (x^2 \left (1+x^2\right )^{2/3}\right )-\frac {1}{2} \log \left (1+x \sqrt [3]{1+x^2}+x^2 \left (1+x^2\right )^{2/3}\right ) \]
3^(1/2)*arctan(3^(1/2)*x*(x^2+1)^(1/3)/(2+x*(x^2+1)^(1/3)))-2*arctanh(-1+2 *x*(x^2+1)^(1/3))+1/2*ln(x^2*(x^2+1)^(2/3))-1/2*ln(1+x*(x^2+1)^(1/3)+x^2*( x^2+1)^(2/3))
Time = 1.87 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {x \left (3+5 x^2\right )}{\sqrt [3]{1+x^2} \left (-1+x^3+x^5\right )} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} x \sqrt [3]{1+x^2}}{2+x \sqrt [3]{1+x^2}}\right )+2 \text {arctanh}\left (1-2 x \sqrt [3]{1+x^2}\right )+\frac {1}{2} \log \left (x^2 \left (1+x^2\right )^{2/3}\right )-\frac {1}{2} \log \left (1+x \sqrt [3]{1+x^2}+x^2 \left (1+x^2\right )^{2/3}\right ) \]
Sqrt[3]*ArcTan[(Sqrt[3]*x*(1 + x^2)^(1/3))/(2 + x*(1 + x^2)^(1/3))] + 2*Ar cTanh[1 - 2*x*(1 + x^2)^(1/3)] + Log[x^2*(1 + x^2)^(2/3)]/2 - Log[1 + x*(1 + x^2)^(1/3) + x^2*(1 + x^2)^(2/3)]/2
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.
\(\displaystyle \int \frac {x \left (5 x^2+3\right )}{\sqrt [3]{x^2+1} \left (x^5+x^3-1\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 x^3}{\sqrt [3]{x^2+1} \left (x^5+x^3-1\right )}+\frac {3 x}{\sqrt [3]{x^2+1} \left (x^5+x^3-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \int \frac {x}{\sqrt [3]{x^2+1} \left (x^5+x^3-1\right )}dx+5 \int \frac {x^3}{\sqrt [3]{x^2+1} \left (x^5+x^3-1\right )}dx\) |
3.16.10.3.1 Defintions of rubi rules used
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.08 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.58
method | result | size |
trager | \(\ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{5}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{5}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}} x -3 x \left (x^{2}+1\right )^{\frac {1}{3}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+2}{x^{5}+x^{3}-1}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{5}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{5}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-x^{5}+2 x^{2} \left (x^{2}+1\right )^{\frac {2}{3}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}} x -x^{3}-x \left (x^{2}+1\right )^{\frac {1}{3}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-1}{x^{5}+x^{3}-1}\right )\) | \(271\) |
ln(-(-RootOf(_Z^2+_Z+1)^2*x^5-2*RootOf(_Z^2+_Z+1)*x^5+3*RootOf(_Z^2+_Z+1)* (x^2+1)^(2/3)*x^2-RootOf(_Z^2+_Z+1)^2*x^3-2*RootOf(_Z^2+_Z+1)*x^3-3*RootOf (_Z^2+_Z+1)*(x^2+1)^(1/3)*x-3*x*(x^2+1)^(1/3)+RootOf(_Z^2+_Z+1)+2)/(x^5+x^ 3-1))+RootOf(_Z^2+_Z+1)*ln((-RootOf(_Z^2+_Z+1)^2*x^5-2*RootOf(_Z^2+_Z+1)*x ^5+RootOf(_Z^2+_Z+1)*(x^2+1)^(2/3)*x^2-RootOf(_Z^2+_Z+1)^2*x^3-x^5+2*x^2*( x^2+1)^(2/3)-2*RootOf(_Z^2+_Z+1)*x^3+RootOf(_Z^2+_Z+1)*(x^2+1)^(1/3)*x-x^3 -x*(x^2+1)^(1/3)-RootOf(_Z^2+_Z+1)-1)/(x^5+x^3-1))
Time = 1.33 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.98 \[ \int \frac {x \left (3+5 x^2\right )}{\sqrt [3]{1+x^2} \left (-1+x^3+x^5\right )} \, dx=-\sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} {\left (x^{2} + 1\right )}^{\frac {2}{3}} x^{2} - 4 \, \sqrt {3} {\left (x^{2} + 1\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (x^{5} + x^{3}\right )}}{x^{5} + x^{3} + 8}\right ) + \frac {1}{2} \, \log \left (\frac {x^{5} + x^{3} - 3 \, {\left (x^{2} + 1\right )}^{\frac {2}{3}} x^{2} + 3 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} x - 1}{x^{5} + x^{3} - 1}\right ) \]
-sqrt(3)*arctan((2*sqrt(3)*(x^2 + 1)^(2/3)*x^2 - 4*sqrt(3)*(x^2 + 1)^(1/3) *x - sqrt(3)*(x^5 + x^3))/(x^5 + x^3 + 8)) + 1/2*log((x^5 + x^3 - 3*(x^2 + 1)^(2/3)*x^2 + 3*(x^2 + 1)^(1/3)*x - 1)/(x^5 + x^3 - 1))
\[ \int \frac {x \left (3+5 x^2\right )}{\sqrt [3]{1+x^2} \left (-1+x^3+x^5\right )} \, dx=\int \frac {x \left (5 x^{2} + 3\right )}{\sqrt [3]{x^{2} + 1} \left (x^{5} + x^{3} - 1\right )}\, dx \]
\[ \int \frac {x \left (3+5 x^2\right )}{\sqrt [3]{1+x^2} \left (-1+x^3+x^5\right )} \, dx=\int { \frac {{\left (5 \, x^{2} + 3\right )} x}{{\left (x^{5} + x^{3} - 1\right )} {\left (x^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {x \left (3+5 x^2\right )}{\sqrt [3]{1+x^2} \left (-1+x^3+x^5\right )} \, dx=\int { \frac {{\left (5 \, x^{2} + 3\right )} x}{{\left (x^{5} + x^{3} - 1\right )} {\left (x^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {x \left (3+5 x^2\right )}{\sqrt [3]{1+x^2} \left (-1+x^3+x^5\right )} \, dx=\int \frac {x\,\left (5\,x^2+3\right )}{{\left (x^2+1\right )}^{1/3}\,\left (x^5+x^3-1\right )} \,d x \]