Integrand size = 85, antiderivative size = 24 \[ \int \frac {10 x+2 x^2-10 x^3-2 x^4+\left (-60+30 x+23 x^2-10 x^3-x^4\right ) \log \left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )}{\left (-20 x^4+10 x^5+x^6\right ) \log ^2\left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )} \, dx=\frac {-\frac {1}{x}+x}{x^2 \log \left (-2+x+\frac {x^2}{10}\right )} \] Output:
(x-1/x)/ln(1/10*x^2+x-2)/x^2
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {10 x+2 x^2-10 x^3-2 x^4+\left (-60+30 x+23 x^2-10 x^3-x^4\right ) \log \left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )}{\left (-20 x^4+10 x^5+x^6\right ) \log ^2\left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )} \, dx=-\frac {1-x^2}{x^3 \log \left (-2+x+\frac {x^2}{10}\right )} \] Input:
Integrate[(10*x + 2*x^2 - 10*x^3 - 2*x^4 + (-60 + 30*x + 23*x^2 - 10*x^3 - x^4)*Log[(-20 + 10*x + x^2)/10])/((-20*x^4 + 10*x^5 + x^6)*Log[(-20 + 10* x + x^2)/10]^2),x]
Output:
-((1 - x^2)/(x^3*Log[-2 + x + x^2/10]))
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.
\(\displaystyle \int \frac {-2 x^4-10 x^3+2 x^2+\left (-x^4-10 x^3+23 x^2+30 x-60\right ) \log \left (\frac {1}{10} \left (x^2+10 x-20\right )\right )+10 x}{\left (x^6+10 x^5-20 x^4\right ) \log ^2\left (\frac {1}{10} \left (x^2+10 x-20\right )\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-2 x^4-10 x^3+2 x^2+\left (-x^4-10 x^3+23 x^2+30 x-60\right ) \log \left (\frac {1}{10} \left (x^2+10 x-20\right )\right )+10 x}{x^4 \left (x^2+10 x-20\right ) \log ^2\left (\frac {1}{10} \left (x^2+10 x-20\right )\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {3-x^2}{x^4 \log \left (\frac {x^2}{10}+x-2\right )}-\frac {2 \left (x^3+5 x^2-x-5\right )}{x^3 \left (x^2+10 x-20\right ) \log ^2\left (\frac {x^2}{10}+x-2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {31 \int \frac {1}{\left (-2 x+6 \sqrt {5}-10\right ) \log ^2\left (\frac {x^2}{10}+x-2\right )}dx}{20 \sqrt {5}}-\frac {7}{20} \int \frac {1}{x^2 \log ^2\left (\frac {x^2}{10}+x-2\right )}dx+\frac {3}{10} \int \frac {1}{x \log ^2\left (\frac {x^2}{10}+x-2\right )}dx-\frac {1}{10} \left (3-\sqrt {5}\right ) \int \frac {1}{\left (2 x-6 \sqrt {5}+10\right ) \log ^2\left (\frac {x^2}{10}+x-2\right )}dx-\frac {1}{10} \left (3+\sqrt {5}\right ) \int \frac {1}{\left (2 x+6 \sqrt {5}+10\right ) \log ^2\left (\frac {x^2}{10}+x-2\right )}dx+\frac {31 \int \frac {1}{\left (2 x+6 \sqrt {5}+10\right ) \log ^2\left (\frac {x^2}{10}+x-2\right )}dx}{20 \sqrt {5}}-\int \frac {1}{x^2 \log \left (\frac {x^2}{10}+x-2\right )}dx+3 \int \frac {1}{x^4 \log \left (\frac {x^2}{10}+x-2\right )}dx-\frac {1}{2} \int \frac {1}{x^3 \log ^2\left (\frac {x^2}{10}+x-2\right )}dx\) |
Input:
Int[(10*x + 2*x^2 - 10*x^3 - 2*x^4 + (-60 + 30*x + 23*x^2 - 10*x^3 - x^4)* Log[(-20 + 10*x + x^2)/10])/((-20*x^4 + 10*x^5 + x^6)*Log[(-20 + 10*x + x^ 2)/10]^2),x]
Output:
$Aborted
Time = 2.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {x^{2}-1}{\ln \left (\frac {1}{10} x^{2}+x -2\right ) x^{3}}\) | \(21\) |
risch | \(\frac {x^{2}-1}{\ln \left (\frac {1}{10} x^{2}+x -2\right ) x^{3}}\) | \(21\) |
parallelrisch | \(\frac {x^{2}-1}{\ln \left (\frac {1}{10} x^{2}+x -2\right ) x^{3}}\) | \(21\) |
Input:
int(((-x^4-10*x^3+23*x^2+30*x-60)*ln(1/10*x^2+x-2)-2*x^4-10*x^3+2*x^2+10*x )/(x^6+10*x^5-20*x^4)/ln(1/10*x^2+x-2)^2,x,method=_RETURNVERBOSE)
Output:
(x^2-1)/ln(1/10*x^2+x-2)/x^3
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {10 x+2 x^2-10 x^3-2 x^4+\left (-60+30 x+23 x^2-10 x^3-x^4\right ) \log \left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )}{\left (-20 x^4+10 x^5+x^6\right ) \log ^2\left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )} \, dx=\frac {x^{2} - 1}{x^{3} \log \left (\frac {1}{10} \, x^{2} + x - 2\right )} \] Input:
integrate(((-x^4-10*x^3+23*x^2+30*x-60)*log(1/10*x^2+x-2)-2*x^4-10*x^3+2*x ^2+10*x)/(x^6+10*x^5-20*x^4)/log(1/10*x^2+x-2)^2,x, algorithm="fricas")
Output:
(x^2 - 1)/(x^3*log(1/10*x^2 + x - 2))
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {10 x+2 x^2-10 x^3-2 x^4+\left (-60+30 x+23 x^2-10 x^3-x^4\right ) \log \left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )}{\left (-20 x^4+10 x^5+x^6\right ) \log ^2\left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )} \, dx=\frac {x^{2} - 1}{x^{3} \log {\left (\frac {x^{2}}{10} + x - 2 \right )}} \] Input:
integrate(((-x**4-10*x**3+23*x**2+30*x-60)*ln(1/10*x**2+x-2)-2*x**4-10*x** 3+2*x**2+10*x)/(x**6+10*x**5-20*x**4)/ln(1/10*x**2+x-2)**2,x)
Output:
(x**2 - 1)/(x**3*log(x**2/10 + x - 2))
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {10 x+2 x^2-10 x^3-2 x^4+\left (-60+30 x+23 x^2-10 x^3-x^4\right ) \log \left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )}{\left (-20 x^4+10 x^5+x^6\right ) \log ^2\left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )} \, dx=-\frac {x^{2} - 1}{x^{3} {\left (\log \left (5\right ) + \log \left (2\right )\right )} - x^{3} \log \left (x^{2} + 10 \, x - 20\right )} \] Input:
integrate(((-x^4-10*x^3+23*x^2+30*x-60)*log(1/10*x^2+x-2)-2*x^4-10*x^3+2*x ^2+10*x)/(x^6+10*x^5-20*x^4)/log(1/10*x^2+x-2)^2,x, algorithm="maxima")
Output:
-(x^2 - 1)/(x^3*(log(5) + log(2)) - x^3*log(x^2 + 10*x - 20))
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {10 x+2 x^2-10 x^3-2 x^4+\left (-60+30 x+23 x^2-10 x^3-x^4\right ) \log \left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )}{\left (-20 x^4+10 x^5+x^6\right ) \log ^2\left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )} \, dx=\frac {x^{2} - 1}{x^{3} \log \left (\frac {1}{10} \, x^{2} + x - 2\right )} \] Input:
integrate(((-x^4-10*x^3+23*x^2+30*x-60)*log(1/10*x^2+x-2)-2*x^4-10*x^3+2*x ^2+10*x)/(x^6+10*x^5-20*x^4)/log(1/10*x^2+x-2)^2,x, algorithm="giac")
Output:
(x^2 - 1)/(x^3*log(1/10*x^2 + x - 2))
Time = 3.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {10 x+2 x^2-10 x^3-2 x^4+\left (-60+30 x+23 x^2-10 x^3-x^4\right ) \log \left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )}{\left (-20 x^4+10 x^5+x^6\right ) \log ^2\left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )} \, dx=\frac {x^2-1}{x^3\,\ln \left (\frac {x^2}{10}+x-2\right )} \] Input:
int(-(log(x + x^2/10 - 2)*(10*x^3 - 23*x^2 - 30*x + x^4 + 60) - 10*x - 2*x ^2 + 10*x^3 + 2*x^4)/(log(x + x^2/10 - 2)^2*(10*x^5 - 20*x^4 + x^6)),x)
Output:
(x^2 - 1)/(x^3*log(x + x^2/10 - 2))
Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {10 x+2 x^2-10 x^3-2 x^4+\left (-60+30 x+23 x^2-10 x^3-x^4\right ) \log \left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )}{\left (-20 x^4+10 x^5+x^6\right ) \log ^2\left (\frac {1}{10} \left (-20+10 x+x^2\right )\right )} \, dx=\frac {x^{2}-1}{\mathrm {log}\left (\frac {1}{10} x^{2}+x -2\right ) x^{3}} \] Input:
int(((-x^4-10*x^3+23*x^2+30*x-60)*log(1/10*x^2+x-2)-2*x^4-10*x^3+2*x^2+10* x)/(x^6+10*x^5-20*x^4)/log(1/10*x^2+x-2)^2,x)
Output:
(x**2 - 1)/(log((x**2 + 10*x - 20)/10)*x**3)