Integrand size = 62, antiderivative size = 26 \[ \int \frac {e^{-1-3 x} \log ^2\left (-x+x^2\right ) \left ((-9+18 x) \log (x)+\left (-3+3 x+\left (9-9 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )\right )}{-x^4+x^5} \, dx=\frac {3 e^{-1+3 \left (-x+\log \left (\log \left (-x+x^2\right )\right )\right )} \log (x)}{x^3} \] Output:
3*ln(x)/x^3/exp(1)*exp(3*ln(ln(x^2-x))-3*x)
Time = 0.45 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-1-3 x} \log ^2\left (-x+x^2\right ) \left ((-9+18 x) \log (x)+\left (-3+3 x+\left (9-9 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )\right )}{-x^4+x^5} \, dx=\frac {3 e^{-1-3 x} \log (x) \log ^3((-1+x) x)}{x^3} \] Input:
Integrate[(E^(-1 - 3*x)*Log[-x + x^2]^2*((-9 + 18*x)*Log[x] + (-3 + 3*x + (9 - 9*x^2)*Log[x])*Log[-x + x^2]))/(-x^4 + x^5),x]
Output:
(3*E^(-1 - 3*x)*Log[x]*Log[(-1 + x)*x]^3)/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.
\(\displaystyle \int \frac {e^{-3 x-1} \log ^2\left (x^2-x\right ) \left (\left (\left (9-9 x^2\right ) \log (x)+3 x-3\right ) \log \left (x^2-x\right )+(18 x-9) \log (x)\right )}{x^5-x^4} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-3 x-1} \log ^2\left (x^2-x\right ) \left (\left (\left (9-9 x^2\right ) \log (x)+3 x-3\right ) \log \left (x^2-x\right )+(18 x-9) \log (x)\right )}{(x-1) x^4}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-3 x-1} \log ^2((x-1) x) \left (-\left (\left (9-9 x^2\right ) \log (x)+3 x-3\right ) \log \left (x^2-x\right )-((18 x-9) \log (x))\right )}{(1-x) x^4}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {9 e^{-3 x-1} (2 x-1) \log (x) \log ^2((x-1) x)}{(x-1) x^4}-\frac {3 e^{-3 x-1} (3 x \log (x)+3 \log (x)-1) \log ^3((x-1) x)}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \int \frac {e^{-3 x-1} \log ^3((x-1) x)}{x^4}dx-9 \int \frac {e^{-3 x-1} \log (x) \log ^3((x-1) x)}{x^4}dx+9 \int \frac {e^{-3 x-1} \log (x) \log ^2((x-1) x)}{x^4}dx-9 \int \frac {e^{-3 x-1} \log (x) \log ^3((x-1) x)}{x^3}dx-9 \int \frac {e^{-3 x-1} \log (x) \log ^2((x-1) x)}{x^3}dx-9 \int \frac {e^{-3 x-1} \log (x) \log ^2((x-1) x)}{x^2}dx+9 \int \frac {e^{-3 x-1} \log (x) \log ^2((x-1) x)}{x-1}dx-9 \int \frac {e^{-3 x-1} \log (x) \log ^2((x-1) x)}{x}dx\) |
Input:
Int[(E^(-1 - 3*x)*Log[-x + x^2]^2*((-9 + 18*x)*Log[x] + (-3 + 3*x + (9 - 9 *x^2)*Log[x])*Log[-x + x^2]))/(-x^4 + x^5),x]
Output:
$Aborted
Time = 12.54 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {3 \ln \left (x \right ) {\mathrm e}^{-1} {\mathrm e}^{3 \ln \left (\ln \left (x^{2}-x \right )\right )-3 x}}{x^{3}}\) | \(28\) |
risch | \(\text {Expression too large to display}\) | \(1414\) |
Input:
int((((-9*x^2+9)*ln(x)+3*x-3)*ln(x^2-x)+(18*x-9)*ln(x))*exp(3*ln(ln(x^2-x) )-3*x)/(x^5-x^4)/exp(1)/ln(x^2-x),x,method=_RETURNVERBOSE)
Output:
3*ln(x)/x^3/exp(1)*exp(3*ln(ln(x^2-x))-3*x)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-1-3 x} \log ^2\left (-x+x^2\right ) \left ((-9+18 x) \log (x)+\left (-3+3 x+\left (9-9 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )\right )}{-x^4+x^5} \, dx=\frac {3 \, e^{\left (-3 \, x + 3 \, \log \left (\log \left (x^{2} - x\right )\right ) - 1\right )} \log \left (x\right )}{x^{3}} \] Input:
integrate((((-9*x^2+9)*log(x)+3*x-3)*log(x^2-x)+(18*x-9)*log(x))*exp(3*log (log(x^2-x))-3*x)/(x^5-x^4)/exp(1)/log(x^2-x),x, algorithm="fricas")
Output:
3*e^(-3*x + 3*log(log(x^2 - x)) - 1)*log(x)/x^3
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-1-3 x} \log ^2\left (-x+x^2\right ) \left ((-9+18 x) \log (x)+\left (-3+3 x+\left (9-9 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )\right )}{-x^4+x^5} \, dx=\frac {3 e^{- 3 x} \log {\left (x \right )} \log {\left (x^{2} - x \right )}^{3}}{e x^{3}} \] Input:
integrate((((-9*x**2+9)*ln(x)+3*x-3)*ln(x**2-x)+(18*x-9)*ln(x))*exp(3*ln(l n(x**2-x))-3*x)/(x**5-x**4)/exp(1)/ln(x**2-x),x)
Output:
3*exp(-1)*exp(-3*x)*log(x)*log(x**2 - x)**3/x**3
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (24) = 48\).
Time = 0.15 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {e^{-1-3 x} \log ^2\left (-x+x^2\right ) \left ((-9+18 x) \log (x)+\left (-3+3 x+\left (9-9 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )\right )}{-x^4+x^5} \, dx=\frac {3 \, {\left (e^{\left (-3 \, x\right )} \log \left (x - 1\right )^{3} \log \left (x\right ) + 3 \, e^{\left (-3 \, x\right )} \log \left (x - 1\right )^{2} \log \left (x\right )^{2} + 3 \, e^{\left (-3 \, x\right )} \log \left (x - 1\right ) \log \left (x\right )^{3} + e^{\left (-3 \, x\right )} \log \left (x\right )^{4}\right )} e^{\left (-1\right )}}{x^{3}} \] Input:
integrate((((-9*x^2+9)*log(x)+3*x-3)*log(x^2-x)+(18*x-9)*log(x))*exp(3*log (log(x^2-x))-3*x)/(x^5-x^4)/exp(1)/log(x^2-x),x, algorithm="maxima")
Output:
3*(e^(-3*x)*log(x - 1)^3*log(x) + 3*e^(-3*x)*log(x - 1)^2*log(x)^2 + 3*e^( -3*x)*log(x - 1)*log(x)^3 + e^(-3*x)*log(x)^4)*e^(-1)/x^3
\[ \int \frac {e^{-1-3 x} \log ^2\left (-x+x^2\right ) \left ((-9+18 x) \log (x)+\left (-3+3 x+\left (9-9 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )\right )}{-x^4+x^5} \, dx=\int { -\frac {3 \, {\left ({\left (3 \, {\left (x^{2} - 1\right )} \log \left (x\right ) - x + 1\right )} \log \left (x^{2} - x\right ) - 3 \, {\left (2 \, x - 1\right )} \log \left (x\right )\right )} e^{\left (-3 \, x + 3 \, \log \left (\log \left (x^{2} - x\right )\right ) - 1\right )}}{{\left (x^{5} - x^{4}\right )} \log \left (x^{2} - x\right )} \,d x } \] Input:
integrate((((-9*x^2+9)*log(x)+3*x-3)*log(x^2-x)+(18*x-9)*log(x))*exp(3*log (log(x^2-x))-3*x)/(x^5-x^4)/exp(1)/log(x^2-x),x, algorithm="giac")
Output:
integrate(-3*((3*(x^2 - 1)*log(x) - x + 1)*log(x^2 - x) - 3*(2*x - 1)*log( x))*e^(-3*x + 3*log(log(x^2 - x)) - 1)/((x^5 - x^4)*log(x^2 - x)), x)
Timed out. \[ \int \frac {e^{-1-3 x} \log ^2\left (-x+x^2\right ) \left ((-9+18 x) \log (x)+\left (-3+3 x+\left (9-9 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )\right )}{-x^4+x^5} \, dx=\int -\frac {{\mathrm {e}}^{-1}\,{\mathrm {e}}^{3\,\ln \left (\ln \left (x^2-x\right )\right )-3\,x}\,\left (\ln \left (x\right )\,\left (18\,x-9\right )-\ln \left (x^2-x\right )\,\left (\ln \left (x\right )\,\left (9\,x^2-9\right )-3\,x+3\right )\right )}{\ln \left (x^2-x\right )\,\left (x^4-x^5\right )} \,d x \] Input:
int(-(exp(-1)*exp(3*log(log(x^2 - x)) - 3*x)*(log(x)*(18*x - 9) - log(x^2 - x)*(log(x)*(9*x^2 - 9) - 3*x + 3)))/(log(x^2 - x)*(x^4 - x^5)),x)
Output:
int(-(exp(-1)*exp(3*log(log(x^2 - x)) - 3*x)*(log(x)*(18*x - 9) - log(x^2 - x)*(log(x)*(9*x^2 - 9) - 3*x + 3)))/(log(x^2 - x)*(x^4 - x^5)), x)
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-1-3 x} \log ^2\left (-x+x^2\right ) \left ((-9+18 x) \log (x)+\left (-3+3 x+\left (9-9 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )\right )}{-x^4+x^5} \, dx=\frac {3 \mathrm {log}\left (x^{2}-x \right )^{3} \mathrm {log}\left (x \right )}{e^{3 x} e \,x^{3}} \] Input:
int((((-9*x^2+9)*log(x)+3*x-3)*log(x^2-x)+(18*x-9)*log(x))*exp(3*log(log(x ^2-x))-3*x)/(x^5-x^4)/exp(1)/log(x^2-x),x)
Output:
(3*log(x**2 - x)**3*log(x))/(e**(3*x)*e*x**3)