Integrand size = 75, antiderivative size = 24 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {x}{-1+x-\frac {x}{3 \log \left (5 e^{x+x^2}\right )}} \] Output:
x/(x-1-1/3*x/ln(5*exp(x^2+x)))
Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3 \left (x+3 \log \left (5 e^{x+x^2}\right )\right )}{-3 x+9 (-1+x) \log \left (5 e^{x+x^2}\right )} \] Input:
Integrate[(-3*x^2 - 6*x^3 - 9*Log[5*E^(x + x^2)]^2)/(x^2 + (6*x - 6*x^2)*L og[5*E^(x + x^2)] + (9 - 18*x + 9*x^2)*Log[5*E^(x + x^2)]^2),x]
Output:
(3*(x + 3*Log[5*E^(x + x^2)]))/(-3*x + 9*(-1 + x)*Log[5*E^(x + 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.
\(\displaystyle \int \frac {-6 x^3-3 x^2-9 \log ^2\left (5 e^{x^2+x}\right )}{x^2+\left (9 x^2-18 x+9\right ) \log ^2\left (5 e^{x^2+x}\right )+\left (6 x-6 x^2\right ) \log \left (5 e^{x^2+x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 \left (-2 x^3-x^2-3 \log ^2\left (5 e^{x^2+x}\right )\right )}{\left (-3 x \log \left (5 e^{x^2+x}\right )+3 \log \left (5 e^{x^2+x}\right )+x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int -\frac {2 x^3+x^2+3 \log ^2\left (5 e^{x^2+x}\right )}{\left (-3 \log \left (5 e^{x^2+x}\right ) x+x+3 \log \left (5 e^{x^2+x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -3 \int \frac {2 x^3+x^2+3 \log ^2\left (5 e^{x^2+x}\right )}{\left (-3 \log \left (5 e^{x^2+x}\right ) x+x+3 \log \left (5 e^{x^2+x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -3 \int \left (\frac {\left (6 x^3-9 x^2+4\right ) x^2}{3 (x-1)^2 \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}+\frac {2 x}{3 (x-1)^2 \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )}+\frac {1}{3 (x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \left (\frac {1}{3} \int \frac {1}{\left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\frac {1}{3} \int \frac {1}{(x-1)^2 \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\frac {2}{3} \int \frac {1}{(x-1) \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\int \frac {x^2}{\left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\frac {2}{3} \int \frac {1}{(x-1)^2 \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )}dx+\frac {2}{3} \int \frac {1}{(x-1) \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )}dx+2 \int \frac {x^3}{\left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\frac {1}{3 (1-x)}\right )\) |
Input:
Int[(-3*x^2 - 6*x^3 - 9*Log[5*E^(x + x^2)]^2)/(x^2 + (6*x - 6*x^2)*Log[5*E ^(x + x^2)] + (9 - 18*x + 9*x^2)*Log[5*E^(x + x^2)]^2),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(46\) vs. \(2(21)=42\).
Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96
method | result | size |
parallelrisch | \(-\frac {-9 x -27 \ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )}{9 \left (3 \ln \left (5 \,{\mathrm e}^{x^{2}+x}\right ) x -3 \ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )-x \right )}\) | \(47\) |
risch | \(\frac {1}{-1+x}+\frac {2 i x^{2}}{\left (-1+x \right ) \left (6 i x \ln \left (5\right )+6 i x \ln \left ({\mathrm e}^{\left (1+x \right ) x}\right )-6 i \ln \left (5\right )-2 i x -6 i \ln \left ({\mathrm e}^{\left (1+x \right ) x}\right )\right )}\) | \(57\) |
default | \(-\frac {3 \left (-\frac {x}{9}-\frac {\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )}{3}\right )}{x^{3}+x \left (\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )-x^{2}-x \right )-\frac {x}{3}-\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )+x^{2}}\) | \(61\) |
Input:
int((-9*ln(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*ln(5*exp(x^2+x))^2 +(-6*x^2+6*x)*ln(5*exp(x^2+x))+x^2),x,method=_RETURNVERBOSE)
Output:
-1/9*(-9*x-27*ln(5*exp(x^2+x)))/(3*ln(5*exp(x^2+x))*x-3*ln(5*exp(x^2+x))-x )
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3 \, x^{2} + 4 \, x + 3 \, \log \left (5\right )}{3 \, x^{3} + 3 \, {\left (x - 1\right )} \log \left (5\right ) - 4 \, x} \] Input:
integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x ^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+x))+x^2),x, algorithm="fricas")
Output:
(3*x^2 + 4*x + 3*log(5))/(3*x^3 + 3*(x - 1)*log(5) - 4*x)
Time = 0.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=- \frac {- 3 x^{2} - 4 x - 3 \log {\left (5 \right )}}{3 x^{3} + x \left (-4 + 3 \log {\left (5 \right )}\right ) - 3 \log {\left (5 \right )}} \] Input:
integrate((-9*ln(5*exp(x**2+x))**2-6*x**3-3*x**2)/((9*x**2-18*x+9)*ln(5*ex p(x**2+x))**2+(-6*x**2+6*x)*ln(5*exp(x**2+x))+x**2),x)
Output:
-(-3*x**2 - 4*x - 3*log(5))/(3*x**3 + x*(-4 + 3*log(5)) - 3*log(5))
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3 \, x^{2} + 4 \, x + 3 \, \log \left (5\right )}{3 \, x^{3} + x {\left (3 \, \log \left (5\right ) - 4\right )} - 3 \, \log \left (5\right )} \] Input:
integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x ^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+x))+x^2),x, algorithm="maxima")
Output:
(3*x^2 + 4*x + 3*log(5))/(3*x^3 + x*(3*log(5) - 4) - 3*log(5))
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3 \, x^{2} + 4 \, x + 3 \, \log \left (5\right )}{3 \, x^{3} + 3 \, x \log \left (5\right ) - 4 \, x - 3 \, \log \left (5\right )} \] Input:
integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x ^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+x))+x^2),x, algorithm="giac")
Output:
(3*x^2 + 4*x + 3*log(5))/(3*x^3 + 3*x*log(5) - 4*x - 3*log(5))
Time = 0.62 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3\,x^2+4\,x+\ln \left (125\right )}{3\,x^3+\left (\ln \left (125\right )-4\right )\,x-\ln \left (125\right )} \] Input:
int(-(9*log(5*exp(x + x^2))^2 + 3*x^2 + 6*x^3)/(log(5*exp(x + x^2))^2*(9*x ^2 - 18*x + 9) + log(5*exp(x + x^2))*(6*x - 6*x^2) + x^2),x)
Output:
(4*x + log(125) + 3*x^2)/(x*(log(125) - 4) - log(125) + 3*x^3)
\[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=-9 \left (\int \frac {\mathrm {log}\left (5 e^{x^{2}+x}\right )^{2}}{9 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2} x^{2}-18 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2} x +9 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2}-6 \,\mathrm {log}\left (5 e^{x^{2}+x}\right ) x^{2}+6 \,\mathrm {log}\left (5 e^{x^{2}+x}\right ) x +x^{2}}d x \right )-6 \left (\int \frac {x^{3}}{9 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2} x^{2}-18 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2} x +9 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2}-6 \,\mathrm {log}\left (5 e^{x^{2}+x}\right ) x^{2}+6 \,\mathrm {log}\left (5 e^{x^{2}+x}\right ) x +x^{2}}d x \right )-3 \left (\int \frac {x^{2}}{9 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2} x^{2}-18 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2} x +9 \mathrm {log}\left (5 e^{x^{2}+x}\right )^{2}-6 \,\mathrm {log}\left (5 e^{x^{2}+x}\right ) x^{2}+6 \,\mathrm {log}\left (5 e^{x^{2}+x}\right ) x +x^{2}}d x \right ) \] Input:
int((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x^2+x)) ^2+(-6*x^2+6*x)*log(5*exp(x^2+x))+x^2),x)
Output:
3*( - 3*int(log(5*e**(x**2 + x))**2/(9*log(5*e**(x**2 + x))**2*x**2 - 18*l og(5*e**(x**2 + x))**2*x + 9*log(5*e**(x**2 + x))**2 - 6*log(5*e**(x**2 + x))*x**2 + 6*log(5*e**(x**2 + x))*x + x**2),x) - 2*int(x**3/(9*log(5*e**(x **2 + x))**2*x**2 - 18*log(5*e**(x**2 + x))**2*x + 9*log(5*e**(x**2 + x))* *2 - 6*log(5*e**(x**2 + x))*x**2 + 6*log(5*e**(x**2 + x))*x + x**2),x) - i nt(x**2/(9*log(5*e**(x**2 + x))**2*x**2 - 18*log(5*e**(x**2 + x))**2*x + 9 *log(5*e**(x**2 + x))**2 - 6*log(5*e**(x**2 + x))*x**2 + 6*log(5*e**(x**2 + x))*x + x**2),x))