Integrand size = 42, antiderivative size = 53 \[ \int \frac {\left (-2+x^3\right ) \sqrt {1+x^3} \left (2-x^2+2 x^3\right )}{x^4 \left (1-3 x^2+x^3\right )} \, dx=\frac {2 \sqrt {1+x^3} \left (2+15 x^2+2 x^3\right )}{3 x^3}-10 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} x}{\sqrt {1+x^3}}\right ) \]
Time = 0.84 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2+x^3\right ) \sqrt {1+x^3} \left (2-x^2+2 x^3\right )}{x^4 \left (1-3 x^2+x^3\right )} \, dx=\frac {2 \sqrt {1+x^3} \left (2+15 x^2+2 x^3\right )}{3 x^3}-10 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} x}{\sqrt {1+x^3}}\right ) \]
(2*Sqrt[1 + x^3]*(2 + 15*x^2 + 2*x^3))/(3*x^3) - 10*Sqrt[3]*ArcTanh[(Sqrt[ 3]*x)/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.
\(\displaystyle \int \frac {\left (x^3-2\right ) \sqrt {x^3+1} \left (2 x^3-x^2+2\right )}{x^4 \left (x^3-3 x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \sqrt {x^3+1}}{x}-\frac {4 \sqrt {x^3+1}}{x^4}+\frac {15 \sqrt {x^3+1} (x-2)}{x^3-3 x^2+1}-\frac {10 \sqrt {x^3+1}}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -30 \int \frac {\sqrt {x^3+1}}{x^3-3 x^2+1}dx+15 \int \frac {x \sqrt {x^3+1}}{x^3-3 x^2+1}dx-\frac {10 \sqrt {2} 3^{3/4} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {15 \sqrt [4]{3} \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {10 \sqrt {x^3+1}}{x}-\frac {30 \sqrt {x^3+1}}{x+\sqrt {3}+1}+\frac {4 \sqrt {x^3+1}}{3 x^3}+\frac {4 \sqrt {x^3+1}}{3}\) |
3.7.65.3.1 Defintions of rubi rules used
Time = 2.97 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {10 x^{5}+10 x^{2}+\frac {8}{3} x^{3}+\frac {4}{3}+\frac {4}{3} x^{6}}{\sqrt {x^{3}+1}\, x^{3}}-10 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {x^{3}+1}\, \sqrt {3}}{3 x}\right )\) | \(57\) |
default | \(\frac {-30 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {x^{3}+1}\, \sqrt {3}}{3 x}\right ) x^{3}+4 \sqrt {x^{3}+1}\, x^{3}+30 x^{2} \sqrt {x^{3}+1}+4 \sqrt {x^{3}+1}}{3 x^{3}}\) | \(64\) |
pseudoelliptic | \(\frac {-30 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {x^{3}+1}\, \sqrt {3}}{3 x}\right ) x^{3}+4 \sqrt {x^{3}+1}\, x^{3}+30 x^{2} \sqrt {x^{3}+1}+4 \sqrt {x^{3}+1}}{3 x^{3}}\) | \(64\) |
trager | \(\frac {2 \sqrt {x^{3}+1}\, \left (2 x^{3}+15 x^{2}+2\right )}{3 x^{3}}-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{2}+6 x \sqrt {x^{3}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )}{x^{3}-3 x^{2}+1}\right )\) | \(87\) |
elliptic | \(\frac {4 \sqrt {x^{3}+1}}{3}+\frac {4 \sqrt {x^{3}+1}}{3 x^{3}}+\frac {10 \sqrt {x^{3}+1}}{x}+\frac {30 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+5 \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{3}-3 \textit {\_Z}^{2}+1\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha \left (-\underline {\hspace {1.25 ex}}\alpha ^{2}+4 \underline {\hspace {1.25 ex}}\alpha -4\right ) \left (3-i \sqrt {3}\right ) \sqrt {\frac {1+x}{3-i \sqrt {3}}}\, \sqrt {\frac {-1+2 x -i \sqrt {3}}{-3-i \sqrt {3}}}\, \sqrt {\frac {-1+2 x +i \sqrt {3}}{-3+i \sqrt {3}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\underline {\hspace {1.25 ex}}\alpha ^{2}}{2}-2 \underline {\hspace {1.25 ex}}\alpha +2-\frac {i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}}{6}+\frac {2 i \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha }{3}-\frac {2 i \sqrt {3}}{3}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}\right )\) | \(330\) |
2/3*(2*x^6+15*x^5+4*x^3+15*x^2+2)/x^3/(x^3+1)^(1/2)-10*3^(1/2)*arctanh(1/3 *(x^3+1)^(1/2)/x*3^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (43) = 86\).
Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.11 \[ \int \frac {\left (-2+x^3\right ) \sqrt {1+x^3} \left (2-x^2+2 x^3\right )}{x^4 \left (1-3 x^2+x^3\right )} \, dx=\frac {15 \, \sqrt {3} x^{3} \log \left (-\frac {x^{6} + 18 \, x^{5} + 9 \, x^{4} + 2 \, x^{3} - 4 \, \sqrt {3} {\left (x^{4} + 3 \, x^{3} + x\right )} \sqrt {x^{3} + 1} + 18 \, x^{2} + 1}{x^{6} - 6 \, x^{5} + 9 \, x^{4} + 2 \, x^{3} - 6 \, x^{2} + 1}\right ) + 4 \, {\left (2 \, x^{3} + 15 \, x^{2} + 2\right )} \sqrt {x^{3} + 1}}{6 \, x^{3}} \]
1/6*(15*sqrt(3)*x^3*log(-(x^6 + 18*x^5 + 9*x^4 + 2*x^3 - 4*sqrt(3)*(x^4 + 3*x^3 + x)*sqrt(x^3 + 1) + 18*x^2 + 1)/(x^6 - 6*x^5 + 9*x^4 + 2*x^3 - 6*x^ 2 + 1)) + 4*(2*x^3 + 15*x^2 + 2)*sqrt(x^3 + 1))/x^3
\[ \int \frac {\left (-2+x^3\right ) \sqrt {1+x^3} \left (2-x^2+2 x^3\right )}{x^4 \left (1-3 x^2+x^3\right )} \, dx=\int \frac {\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x^{3} - 2\right ) \left (2 x^{3} - x^{2} + 2\right )}{x^{4} \left (x^{3} - 3 x^{2} + 1\right )}\, dx \]
Integral(sqrt((x + 1)*(x**2 - x + 1))*(x**3 - 2)*(2*x**3 - x**2 + 2)/(x**4 *(x**3 - 3*x**2 + 1)), x)
\[ \int \frac {\left (-2+x^3\right ) \sqrt {1+x^3} \left (2-x^2+2 x^3\right )}{x^4 \left (1-3 x^2+x^3\right )} \, dx=\int { \frac {{\left (2 \, x^{3} - x^{2} + 2\right )} \sqrt {x^{3} + 1} {\left (x^{3} - 2\right )}}{{\left (x^{3} - 3 \, x^{2} + 1\right )} x^{4}} \,d x } \]
\[ \int \frac {\left (-2+x^3\right ) \sqrt {1+x^3} \left (2-x^2+2 x^3\right )}{x^4 \left (1-3 x^2+x^3\right )} \, dx=\int { \frac {{\left (2 \, x^{3} - x^{2} + 2\right )} \sqrt {x^{3} + 1} {\left (x^{3} - 2\right )}}{{\left (x^{3} - 3 \, x^{2} + 1\right )} x^{4}} \,d x } \]
Time = 6.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.43 \[ \int \frac {\left (-2+x^3\right ) \sqrt {1+x^3} \left (2-x^2+2 x^3\right )}{x^4 \left (1-3 x^2+x^3\right )} \, dx=5\,\sqrt {3}\,\ln \left (\frac {3\,x^2+x^3-2\,\sqrt {3}\,x\,\sqrt {x^3+1}+1}{x^3-3\,x^2+1}\right )+\frac {4\,\sqrt {x^3+1}}{3}+\frac {10\,\sqrt {x^3+1}}{x}+\frac {4\,\sqrt {x^3+1}}{3\,x^3} \]