Integrand size = 39, antiderivative size = 45 \[ \int \frac {\sqrt {1+x^2+x^5} \left (-2+3 x^5\right )}{\left (1+x^5\right ) \left (1-x^2+x^5\right )} \, dx=2 \text {arctanh}\left (\frac {x}{\sqrt {1+x^2+x^5}}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^2+x^5}}\right ) \] Output:
2*arctanh(x/(x^5+x^2+1)^(1/2))-2*2^(1/2)*arctanh(2^(1/2)*x/(x^5+x^2+1)^(1/ 2))
Time = 1.37 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1+x^2+x^5} \left (-2+3 x^5\right )}{\left (1+x^5\right ) \left (1-x^2+x^5\right )} \, dx=2 \text {arctanh}\left (\frac {x}{\sqrt {1+x^2+x^5}}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^2+x^5}}\right ) \] Input:
Integrate[(Sqrt[1 + x^2 + x^5]*(-2 + 3*x^5))/((1 + x^5)*(1 - x^2 + x^5)),x ]
Output:
2*ArcTanh[x/Sqrt[1 + x^2 + x^5]] - 2*Sqrt[2]*ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^2 + x^5]]
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.
\(\displaystyle \int \frac {\sqrt {x^5+x^2+1} \left (3 x^5-2\right )}{\left (x^5+1\right ) \left (x^5-x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {\sqrt {x^5+x^2+1}}{x+1}+\frac {\left (5 x^3-2\right ) \sqrt {x^5+x^2+1}}{x^5-x^2+1}+\frac {\sqrt {x^5+x^2+1} \left (-x^3-3 x^2+2 x-1\right )}{x^4-x^3+x^2-x+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {\sqrt {x^5+x^2+1}}{x+1}dx-2 \int \frac {\sqrt {x^5+x^2+1}}{x^5-x^2+1}dx+5 \int \frac {x^3 \sqrt {x^5+x^2+1}}{x^5-x^2+1}dx+\int \frac {\sqrt {x^5+x^2+1}}{-x^4+x^3-x^2+x-1}dx+2 \int \frac {x \sqrt {x^5+x^2+1}}{x^4-x^3+x^2-x+1}dx-3 \int \frac {x^2 \sqrt {x^5+x^2+1}}{x^4-x^3+x^2-x+1}dx-\int \frac {x^3 \sqrt {x^5+x^2+1}}{x^4-x^3+x^2-x+1}dx\) |
Input:
Int[(Sqrt[1 + x^2 + x^5]*(-2 + 3*x^5))/((1 + x^5)*(1 - x^2 + x^5)),x]
Output:
$Aborted
Time = 0.45 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.42
method | result | size |
pseudoelliptic | \(\ln \left (\frac {x +\sqrt {x^{5}+x^{2}+1}}{x}\right )-2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x^{5}+x^{2}+1}\, \sqrt {2}}{2 x}\right )-\ln \left (\frac {\sqrt {x^{5}+x^{2}+1}-x}{x}\right )\) | \(64\) |
trager | \(-\ln \left (-\frac {-x^{5}+2 \sqrt {x^{5}+x^{2}+1}\, x -2 x^{2}-1}{\left (1+x \right ) \left (x^{4}-x^{3}+x^{2}-x +1\right )}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{5}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}-4 \sqrt {x^{5}+x^{2}+1}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{x^{5}-x^{2}+1}\right )\) | \(117\) |
Input:
int((x^5+x^2+1)^(1/2)*(3*x^5-2)/(x^5+1)/(x^5-x^2+1),x,method=_RETURNVERBOS E)
Output:
ln((x+(x^5+x^2+1)^(1/2))/x)-2*2^(1/2)*arctanh(1/2*(x^5+x^2+1)^(1/2)/x*2^(1 /2))-ln(((x^5+x^2+1)^(1/2)-x)/x)
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (37) = 74\).
Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.56 \[ \int \frac {\sqrt {1+x^2+x^5} \left (-2+3 x^5\right )}{\left (1+x^5\right ) \left (1-x^2+x^5\right )} \, dx=\frac {1}{2} \, \sqrt {2} \log \left (\frac {x^{10} + 14 \, x^{7} + 2 \, x^{5} + 17 \, x^{4} - 4 \, \sqrt {2} {\left (x^{6} + 3 \, x^{3} + x\right )} \sqrt {x^{5} + x^{2} + 1} + 14 \, x^{2} + 1}{x^{10} - 2 \, x^{7} + 2 \, x^{5} + x^{4} - 2 \, x^{2} + 1}\right ) + \log \left (\frac {x^{5} + 2 \, x^{2} + 2 \, \sqrt {x^{5} + x^{2} + 1} x + 1}{x^{5} + 1}\right ) \] Input:
integrate((x^5+x^2+1)^(1/2)*(3*x^5-2)/(x^5+1)/(x^5-x^2+1),x, algorithm="fr icas")
Output:
1/2*sqrt(2)*log((x^10 + 14*x^7 + 2*x^5 + 17*x^4 - 4*sqrt(2)*(x^6 + 3*x^3 + x)*sqrt(x^5 + x^2 + 1) + 14*x^2 + 1)/(x^10 - 2*x^7 + 2*x^5 + x^4 - 2*x^2 + 1)) + log((x^5 + 2*x^2 + 2*sqrt(x^5 + x^2 + 1)*x + 1)/(x^5 + 1))
Timed out. \[ \int \frac {\sqrt {1+x^2+x^5} \left (-2+3 x^5\right )}{\left (1+x^5\right ) \left (1-x^2+x^5\right )} \, dx=\text {Timed out} \] Input:
integrate((x**5+x**2+1)**(1/2)*(3*x**5-2)/(x**5+1)/(x**5-x**2+1),x)
Output:
Timed out
\[ \int \frac {\sqrt {1+x^2+x^5} \left (-2+3 x^5\right )}{\left (1+x^5\right ) \left (1-x^2+x^5\right )} \, dx=\int { \frac {{\left (3 \, x^{5} - 2\right )} \sqrt {x^{5} + x^{2} + 1}}{{\left (x^{5} - x^{2} + 1\right )} {\left (x^{5} + 1\right )}} \,d x } \] Input:
integrate((x^5+x^2+1)^(1/2)*(3*x^5-2)/(x^5+1)/(x^5-x^2+1),x, algorithm="ma xima")
Output:
integrate((3*x^5 - 2)*sqrt(x^5 + x^2 + 1)/((x^5 - x^2 + 1)*(x^5 + 1)), x)
\[ \int \frac {\sqrt {1+x^2+x^5} \left (-2+3 x^5\right )}{\left (1+x^5\right ) \left (1-x^2+x^5\right )} \, dx=\int { \frac {{\left (3 \, x^{5} - 2\right )} \sqrt {x^{5} + x^{2} + 1}}{{\left (x^{5} - x^{2} + 1\right )} {\left (x^{5} + 1\right )}} \,d x } \] Input:
integrate((x^5+x^2+1)^(1/2)*(3*x^5-2)/(x^5+1)/(x^5-x^2+1),x, algorithm="gi ac")
Output:
integrate((3*x^5 - 2)*sqrt(x^5 + x^2 + 1)/((x^5 - x^2 + 1)*(x^5 + 1)), x)
Time = 11.00 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.71 \[ \int \frac {\sqrt {1+x^2+x^5} \left (-2+3 x^5\right )}{\left (1+x^5\right ) \left (1-x^2+x^5\right )} \, dx=\ln \left (\frac {2\,x\,\sqrt {x^5+x^2+1}+2\,x^2+x^5+1}{x^5+1}\right )+\sqrt {2}\,\ln \left (\frac {3\,x^2+x^5-2\,\sqrt {2}\,x\,\sqrt {x^5+x^2+1}+1}{x^5-x^2+1}\right ) \] Input:
int(((3*x^5 - 2)*(x^2 + x^5 + 1)^(1/2))/((x^5 + 1)*(x^5 - x^2 + 1)),x)
Output:
log((2*x*(x^2 + x^5 + 1)^(1/2) + 2*x^2 + x^5 + 1)/(x^5 + 1)) + 2^(1/2)*log ((3*x^2 + x^5 - 2*2^(1/2)*x*(x^2 + x^5 + 1)^(1/2) + 1)/(x^5 - x^2 + 1))
\[ \int \frac {\sqrt {1+x^2+x^5} \left (-2+3 x^5\right )}{\left (1+x^5\right ) \left (1-x^2+x^5\right )} \, dx=-2 \left (\int \frac {\sqrt {x^{5}+x^{2}+1}}{x^{10}-x^{7}+2 x^{5}-x^{2}+1}d x \right )+3 \left (\int \frac {\sqrt {x^{5}+x^{2}+1}\, x^{5}}{x^{10}-x^{7}+2 x^{5}-x^{2}+1}d x \right ) \] Input:
int((x^5+x^2+1)^(1/2)*(3*x^5-2)/(x^5+1)/(x^5-x^2+1),x)
Output:
- 2*int(sqrt(x**5 + x**2 + 1)/(x**10 - x**7 + 2*x**5 - x**2 + 1),x) + 3*i nt((sqrt(x**5 + x**2 + 1)*x**5)/(x**10 - x**7 + 2*x**5 - x**2 + 1),x)