Integrand size = 28, antiderivative size = 87 \[ \int \frac {x^2 \left (4+x^3\right )}{\left (1+x^3\right )^{3/4} \left (1+x^3+x^4\right )} \, dx=-\sqrt {2} \arctan \left (\frac {-\frac {x^2}{\sqrt {2}}+\frac {\sqrt {1+x^3}}{\sqrt {2}}}{x \sqrt [4]{1+x^3}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{1+x^3}}{x^2+\sqrt {1+x^3}}\right ) \]
-2^(1/2)*arctan((-1/2*2^(1/2)*x^2+1/2*(x^3+1)^(1/2)*2^(1/2))/x/(x^3+1)^(1/ 4))-2^(1/2)*arctanh(2^(1/2)*x*(x^3+1)^(1/4)/(x^2+(x^3+1)^(1/2)))
Time = 2.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int \frac {x^2 \left (4+x^3\right )}{\left (1+x^3\right )^{3/4} \left (1+x^3+x^4\right )} \, dx=-\sqrt {2} \left (\arctan \left (\frac {-x^2+\sqrt {1+x^3}}{\sqrt {2} x \sqrt [4]{1+x^3}}\right )+\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{1+x^3}}{x^2+\sqrt {1+x^3}}\right )\right ) \]
-(Sqrt[2]*(ArcTan[(-x^2 + Sqrt[1 + x^3])/(Sqrt[2]*x*(1 + x^3)^(1/4))] + Ar cTanh[(Sqrt[2]*x*(1 + x^3)^(1/4))/(x^2 + Sqrt[1 + x^3])]))
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.
\(\displaystyle \int \frac {x^2 \left (x^3+4\right )}{\left (x^3+1\right )^{3/4} \left (x^4+x^3+1\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x}{\left (x^3+1\right )^{3/4}}-\frac {1}{\left (x^3+1\right )^{3/4}}+\frac {x^3+4 x^2-x+1}{\left (x^3+1\right )^{3/4} \left (x^4+x^3+1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{\left (x^3+1\right )^{3/4} \left (x^4+x^3+1\right )}dx-\int \frac {x}{\left (x^3+1\right )^{3/4} \left (x^4+x^3+1\right )}dx+\int \frac {x^3}{\left (x^3+1\right )^{3/4} \left (x^4+x^3+1\right )}dx+4 \int \frac {x^2}{\left (x^3+1\right )^{3/4} \left (x^4+x^3+1\right )}dx-x \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-x^3\right )+\frac {1}{2} x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {3}{4},\frac {5}{3},-x^3\right )\) |
3.12.83.3.1 Defintions of rubi rules used
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.84 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.39
method | result | size |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{4}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{3}-2 \left (x^{3}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{3}+2 \sqrt {x^{3}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{2}-2 \left (x^{3}+1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}}{x^{4}+x^{3}+1}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {2 \sqrt {x^{3}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}+2 \left (x^{3}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{4}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{3}-2 \left (x^{3}+1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{4}+x^{3}+1}\right )\) | \(208\) |
RootOf(_Z^4+1)*ln(-(RootOf(_Z^4+1)^3*x^4-RootOf(_Z^4+1)^3*x^3-2*(x^3+1)^(1 /4)*RootOf(_Z^4+1)^2*x^3+2*(x^3+1)^(1/2)*RootOf(_Z^4+1)*x^2-2*(x^3+1)^(3/4 )*x-RootOf(_Z^4+1)^3)/(x^4+x^3+1))+RootOf(_Z^4+1)^3*ln(-(2*(x^3+1)^(1/2)*R ootOf(_Z^4+1)^3*x^2+2*(x^3+1)^(1/4)*RootOf(_Z^4+1)^2*x^3+RootOf(_Z^4+1)*x^ 4-RootOf(_Z^4+1)*x^3-2*(x^3+1)^(3/4)*x-RootOf(_Z^4+1))/(x^4+x^3+1))
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.21 \[ \int \frac {x^2 \left (4+x^3\right )}{\left (1+x^3\right )^{3/4} \left (1+x^3+x^4\right )} \, dx=-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \log \left (\frac {\left (i + 1\right ) \, \sqrt {2} x + 2 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \log \left (\frac {-\left (i - 1\right ) \, \sqrt {2} x + 2 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \log \left (\frac {\left (i - 1\right ) \, \sqrt {2} x + 2 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \log \left (\frac {-\left (i + 1\right ) \, \sqrt {2} x + 2 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{x}\right ) \]
-(1/2*I + 1/2)*sqrt(2)*log(((I + 1)*sqrt(2)*x + 2*(x^3 + 1)^(1/4))/x) + (1 /2*I - 1/2)*sqrt(2)*log((-(I - 1)*sqrt(2)*x + 2*(x^3 + 1)^(1/4))/x) - (1/2 *I - 1/2)*sqrt(2)*log(((I - 1)*sqrt(2)*x + 2*(x^3 + 1)^(1/4))/x) + (1/2*I + 1/2)*sqrt(2)*log((-(I + 1)*sqrt(2)*x + 2*(x^3 + 1)^(1/4))/x)
Timed out. \[ \int \frac {x^2 \left (4+x^3\right )}{\left (1+x^3\right )^{3/4} \left (1+x^3+x^4\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^2 \left (4+x^3\right )}{\left (1+x^3\right )^{3/4} \left (1+x^3+x^4\right )} \, dx=\int { \frac {{\left (x^{3} + 4\right )} x^{2}}{{\left (x^{4} + x^{3} + 1\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {x^2 \left (4+x^3\right )}{\left (1+x^3\right )^{3/4} \left (1+x^3+x^4\right )} \, dx=\int { \frac {{\left (x^{3} + 4\right )} x^{2}}{{\left (x^{4} + x^{3} + 1\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (4+x^3\right )}{\left (1+x^3\right )^{3/4} \left (1+x^3+x^4\right )} \, dx=\int \frac {x^2\,\left (x^3+4\right )}{{\left (x^3+1\right )}^{3/4}\,\left (x^4+x^3+1\right )} \,d x \]