Integrand size = 119, antiderivative size = 30 \[ \int \frac {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx=-\frac {3 x}{\left (-x+\frac {x^2}{e^{10}}\right ) \log \left (\frac {5}{2}\right ) \left (3+\log ^2(x)\right )} \] Output:
3*x/(ln(x)^2+3)/ln(2/5)/(x^2/exp(5)^2-x)
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx=\frac {3 e^{10}}{\left (e^{10}-x\right ) \log \left (\frac {5}{2}\right ) \left (3+\log ^2(x)\right )} \] Input:
Integrate[(-9*E^10*x + (6*E^20 - 6*E^10*x)*Log[x] - 3*E^10*x*Log[x]^2)/(-( (9*E^20*x - 18*E^10*x^2 + 9*x^3)*Log[5/2]) - (6*E^20*x - 12*E^10*x^2 + 6*x ^3)*Log[5/2]*Log[x]^2 - (E^20*x - 2*E^10*x^2 + x^3)*Log[5/2]*Log[x]^4),x]
Output:
(3*E^10)/((E^10 - x)*Log[5/2]*(3 + Log[x]^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.
| \(\displaystyle \int \frac {-9 e^{10} x-3 e^{10} x \log ^2(x)+\left (6 e^{20}-6 e^{10} x\right ) \log (x)}{-\left (\left (x^3-2 e^{10} x^2+e^{20} x\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)\right )-\left (6 x^3-12 e^{10} x^2+6 e^{20} x\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (9 x^3-18 e^{10} x^2+9 e^{20} x\right ) \log \left (\frac {5}{2}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {3 e^{10} \left (3 x+x \log ^2(x)-2 \left (e^{10}-x\right ) \log (x)\right )}{\left (e^{10}-x\right )^2 x \log \left (\frac {5}{2}\right ) \left (\log ^2(x)+3\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 e^{10} \int \frac {x \log ^2(x)-2 \left (e^{10}-x\right ) \log (x)+3 x}{\left (e^{10}-x\right )^2 x \left (\log ^2(x)+3\right )^2}dx}{\log \left (\frac {5}{2}\right )}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {3 e^{10} \int \left (\frac {1}{\left (e^{10}-x\right )^2 \left (\log ^2(x)+3\right )}-\frac {2 \log (x)}{\left (e^{10}-x\right ) x \left (\log ^2(x)+3\right )^2}\right )dx}{\log \left (\frac {5}{2}\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 e^{10} \left (-\frac {2 \int \frac {\log (x)}{\left (e^{10}-x\right ) \left (\log ^2(x)+3\right )^2}dx}{e^{10}}+\int \frac {1}{\left (e^{10}-x\right )^2 \left (\log ^2(x)+3\right )}dx+\frac {1}{e^{10} \left (\log ^2(x)+3\right )}\right )}{\log \left (\frac {5}{2}\right )}\) |
Input:
Int[(-9*E^10*x + (6*E^20 - 6*E^10*x)*Log[x] - 3*E^10*x*Log[x]^2)/(-((9*E^2 0*x - 18*E^10*x^2 + 9*x^3)*Log[5/2]) - (6*E^20*x - 12*E^10*x^2 + 6*x^3)*Lo g[5/2]*Log[x]^2 - (E^20*x - 2*E^10*x^2 + x^3)*Log[5/2]*Log[x]^4),x]
Output:
$Aborted
Time = 0.96 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
| method | result | size |
| parallelrisch | \(-\frac {3 \,{\mathrm e}^{10}}{\ln \left (\frac {2}{5}\right ) \left (\ln \left (x \right )^{2}+3\right ) \left ({\mathrm e}^{10}-x \right )}\) | \(29\) |
| risch | \(-\frac {3 \,{\mathrm e}^{10}}{\left (\ln \left (2\right )-\ln \left (5\right )\right ) \left (\ln \left (x \right )^{2}+3\right ) \left ({\mathrm e}^{10}-x \right )}\) | \(30\) |
| default | \(-\frac {3 \,{\mathrm e}^{10}}{\left (\ln \left (2\right )-\ln \left (5\right )\right ) \left (\ln \left (x \right )^{2}+3\right ) \left ({\mathrm e}^{10}-x \right )}\) | \(34\) |
| norman | \(-\frac {3 \,{\mathrm e}^{10}}{\left (\ln \left (2\right )-\ln \left (5\right )\right ) \left (\ln \left (x \right )^{2}+3\right ) \left ({\mathrm e}^{10}-x \right )}\) | \(34\) |
Input:
int((-3*x*exp(5)^2*ln(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*ln(x)-9*x*exp(5)^2)/( (x*exp(5)^4-2*x^2*exp(5)^2+x^3)*ln(2/5)*ln(x)^4+(6*x*exp(5)^4-12*x^2*exp(5 )^2+6*x^3)*ln(2/5)*ln(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x^3)*ln(2/5)),x ,method=_RETURNVERBOSE)
Output:
-3*exp(5)^2/ln(2/5)/(ln(x)^2+3)/(exp(5)^2-x)
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx=\frac {3 \, e^{10}}{{\left (x - e^{10}\right )} \log \left (\frac {2}{5}\right ) \log \left (x\right )^{2} + 3 \, {\left (x - e^{10}\right )} \log \left (\frac {2}{5}\right )} \] Input:
integrate((-3*x*exp(5)^2*log(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*log(x)-9*x*exp (5)^2)/((x*exp(5)^4-2*x^2*exp(5)^2+x^3)*log(2/5)*log(x)^4+(6*x*exp(5)^4-12 *x^2*exp(5)^2+6*x^3)*log(2/5)*log(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x^3 )*log(2/5)),x, algorithm="fricas")
Output:
3*e^10/((x - e^10)*log(2/5)*log(x)^2 + 3*(x - e^10)*log(2/5))
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (22) = 44\).
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10 \[ \int \frac {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx=\frac {3 e^{10}}{- 3 x \log {\left (5 \right )} + 3 x \log {\left (2 \right )} + \left (- x \log {\left (5 \right )} + x \log {\left (2 \right )} - e^{10} \log {\left (2 \right )} + e^{10} \log {\left (5 \right )}\right ) \log {\left (x \right )}^{2} - 3 e^{10} \log {\left (2 \right )} + 3 e^{10} \log {\left (5 \right )}} \] Input:
integrate((-3*x*exp(5)**2*ln(x)**2+(6*exp(5)**4-6*x*exp(5)**2)*ln(x)-9*x*e xp(5)**2)/((x*exp(5)**4-2*x**2*exp(5)**2+x**3)*ln(2/5)*ln(x)**4+(6*x*exp(5 )**4-12*x**2*exp(5)**2+6*x**3)*ln(2/5)*ln(x)**2+(9*x*exp(5)**4-18*x**2*exp (5)**2+9*x**3)*ln(2/5)),x)
Output:
3*exp(10)/(-3*x*log(5) + 3*x*log(2) + (-x*log(5) + x*log(2) - exp(10)*log( 2) + exp(10)*log(5))*log(x)**2 - 3*exp(10)*log(2) + 3*exp(10)*log(5))
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx=-\frac {3 \, e^{10}}{{\left (x {\left (\log \left (5\right ) - \log \left (2\right )\right )} - {\left (\log \left (5\right ) - \log \left (2\right )\right )} e^{10}\right )} \log \left (x\right )^{2} + 3 \, x {\left (\log \left (5\right ) - \log \left (2\right )\right )} - 3 \, {\left (\log \left (5\right ) - \log \left (2\right )\right )} e^{10}} \] Input:
integrate((-3*x*exp(5)^2*log(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*log(x)-9*x*exp (5)^2)/((x*exp(5)^4-2*x^2*exp(5)^2+x^3)*log(2/5)*log(x)^4+(6*x*exp(5)^4-12 *x^2*exp(5)^2+6*x^3)*log(2/5)*log(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x^3 )*log(2/5)),x, algorithm="maxima")
Output:
-3*e^10/((x*(log(5) - log(2)) - (log(5) - log(2))*e^10)*log(x)^2 + 3*x*(lo g(5) - log(2)) - 3*(log(5) - log(2))*e^10)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (27) = 54\).
Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.17 \[ \int \frac {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx=-\frac {6 \, e^{10}}{x \log \left (5\right ) \log \left (x\right )^{2} - e^{10} \log \left (5\right ) \log \left (x\right )^{2} - x \log \left (2\right ) \log \left (x\right )^{2} + e^{10} \log \left (2\right ) \log \left (x\right )^{2} + 3 \, x \log \left (5\right ) - 3 \, e^{10} \log \left (5\right ) - 3 \, x \log \left (2\right ) + 3 \, e^{10} \log \left (2\right )} \] Input:
integrate((-3*x*exp(5)^2*log(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*log(x)-9*x*exp (5)^2)/((x*exp(5)^4-2*x^2*exp(5)^2+x^3)*log(2/5)*log(x)^4+(6*x*exp(5)^4-12 *x^2*exp(5)^2+6*x^3)*log(2/5)*log(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x^3 )*log(2/5)),x, algorithm="giac")
Output:
-6*e^10/(x*log(5)*log(x)^2 - e^10*log(5)*log(x)^2 - x*log(2)*log(x)^2 + e^ 10*log(2)*log(x)^2 + 3*x*log(5) - 3*e^10*log(5) - 3*x*log(2) + 3*e^10*log( 2))
Time = 3.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx=\frac {3\,{\mathrm {e}}^{10}}{\ln \left (\frac {2}{5}\right )\,\left (x-{\mathrm {e}}^{10}\right )\,\left ({\ln \left (x\right )}^2+3\right )} \] Input:
int(-(9*x*exp(10) - log(x)*(6*exp(20) - 6*x*exp(10)) + 3*x*exp(10)*log(x)^ 2)/(log(2/5)*(9*x*exp(20) - 18*x^2*exp(10) + 9*x^3) + log(2/5)*log(x)^4*(x *exp(20) - 2*x^2*exp(10) + x^3) + log(2/5)*log(x)^2*(6*x*exp(20) - 12*x^2* exp(10) + 6*x^3)),x)
Output:
(3*exp(10))/(log(2/5)*(x - exp(10))*(log(x)^2 + 3))
Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx=\frac {\mathrm {log}\left (x \right )^{2} e^{10}-\mathrm {log}\left (x \right )^{2} x -3 x}{\mathrm {log}\left (\frac {2}{5}\right ) \left (\mathrm {log}\left (x \right )^{2} e^{10}-\mathrm {log}\left (x \right )^{2} x +3 e^{10}-3 x \right )} \] Input:
int((-3*x*exp(5)^2*log(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*log(x)-9*x*exp(5)^2) /((x*exp(5)^4-2*x^2*exp(5)^2+x^3)*log(2/5)*log(x)^4+(6*x*exp(5)^4-12*x^2*e xp(5)^2+6*x^3)*log(2/5)*log(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x^3)*log( 2/5)),x)
Output:
(log(x)**2*e**10 - log(x)**2*x - 3*x)/(log(2/5)*(log(x)**2*e**10 - log(x)* *2*x + 3*e**10 - 3*x))