Integrand size = 124, antiderivative size = 32 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=-e^{5 x}+4 \log \left (-3 e^{-x}+x\right ) \log \left (-x+\frac {x^2}{3}\right ) \] Output:
4*ln(1/3*x^2-x)*ln(x-3/exp(x))-exp(5*x)
Leaf count is larger than twice the leaf count of optimal. \(95\) vs. \(2(32)=64\).
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.97 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=-e^{5 x}-4 x \log \left (\frac {1}{3} (-3+x) x\right )+4 \log (3-x) \left (x+\log \left (-3 e^{-x}+x\right )-\log \left (-3+e^x x\right )\right )+4 \log (x) \left (x+\log \left (-3 e^{-x}+x\right )-\log \left (-3+e^x x\right )\right )+4 \log \left (\frac {1}{3} (-3+x) x\right ) \log \left (-3+e^x x\right ) \] Input:
Integrate[(E^(5*x)*(-45*x + 15*x^2 + E^x*(15*x^2 - 5*x^3)) + (36 - 24*x + E^x*(-12*x + 8*x^2))*Log[(-3 + E^x*x)/E^x] + (-36*x + 12*x^2 + E^x*(-12*x + 4*x^2))*Log[(-3*x + x^2)/3])/(9*x - 3*x^2 + E^x*(-3*x^2 + x^3)),x]
Output:
-E^(5*x) - 4*x*Log[((-3 + x)*x)/3] + 4*Log[3 - x]*(x + Log[-3/E^x + x] - L og[-3 + E^x*x]) + 4*Log[x]*(x + Log[-3/E^x + x] - Log[-3 + E^x*x]) + 4*Log [((-3 + x)*x)/3]*Log[-3 + E^x*x]
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.
| \(\displaystyle \int \frac {\left (e^x \left (8 x^2-12 x\right )-24 x+36\right ) \log \left (e^{-x} \left (e^x x-3\right )\right )+\left (12 x^2+e^x \left (4 x^2-12 x\right )-36 x\right ) \log \left (\frac {1}{3} \left (x^2-3 x\right )\right )+e^{5 x} \left (15 x^2+e^x \left (15 x^2-5 x^3\right )-45 x\right )}{-3 x^2+e^x \left (x^3-3 x^2\right )+9 x} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (-5 e^{5 x}+\frac {4 \left (e^x+3\right ) \log \left (\frac {1}{3} (x-3) x\right )}{e^x x-3}+\frac {4 (2 x-3) \log \left (x-3 e^{-x}\right )}{(x-3) x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -12 \int \frac {\int \frac {1}{e^x x-3}dx}{x-3}dx-12 \int \frac {\int \frac {1}{e^x x-3}dx}{x}dx-12 \int \frac {\int \frac {1}{x \left (e^x x-3\right )}dx}{x-3}dx-12 \int \frac {\int \frac {1}{x \left (e^x x-3\right )}dx}{x}dx+12 \log \left (-\frac {1}{3} (3-x) x\right ) \int \frac {1}{e^x x-3}dx+12 \log \left (-\frac {1}{3} (3-x) x\right ) \int \frac {1}{x \left (e^x x-3\right )}dx+4 \int \frac {\log \left (x-3 e^{-x}\right )}{x-3}dx+4 \int \frac {\log \left (x-3 e^{-x}\right )}{x}dx+4 \operatorname {PolyLog}\left (2,1-\frac {x}{3}\right )-e^{5 x}-2 \log ^2(x)+4 \log \left (-\frac {1}{3} (3-x) x\right ) \log (x)-4 \log (3) \log (x-3)\) |
Input:
Int[(E^(5*x)*(-45*x + 15*x^2 + E^x*(15*x^2 - 5*x^3)) + (36 - 24*x + E^x*(- 12*x + 8*x^2))*Log[(-3 + E^x*x)/E^x] + (-36*x + 12*x^2 + E^x*(-12*x + 4*x^ 2))*Log[(-3*x + x^2)/3])/(9*x - 3*x^2 + E^x*(-3*x^2 + x^3)),x]
Output:
$Aborted
Time = 67.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00
| method | result | size |
| parallelrisch | \(4 \ln \left (\left ({\mathrm e}^{x} x -3\right ) {\mathrm e}^{-x}\right ) \ln \left (\frac {1}{3} x^{2}-x \right )-{\mathrm e}^{5 x}\) | \(32\) |
Input:
int((((8*x^2-12*x)*exp(x)-24*x+36)*ln((exp(x)*x-3)/exp(x))+((4*x^2-12*x)*e xp(x)+12*x^2-36*x)*ln(1/3*x^2-x)+((-5*x^3+15*x^2)*exp(x)+15*x^2-45*x)*exp( 5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x,method=_RETURNVERBOSE)
Output:
4*ln((exp(x)*x-3)/exp(x))*ln(1/3*x^2-x)-exp(5*x)
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=4 \, \log \left (\frac {1}{3} \, x^{2} - x\right ) \log \left ({\left (x e^{x} - 3\right )} e^{\left (-x\right )}\right ) - e^{\left (5 \, x\right )} \] Input:
integrate((((8*x^2-12*x)*exp(x)-24*x+36)*log((exp(x)*x-3)/exp(x))+((4*x^2- 12*x)*exp(x)+12*x^2-36*x)*log(1/3*x^2-x)+((-5*x^3+15*x^2)*exp(x)+15*x^2-45 *x)*exp(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x, algorithm="fricas")
Output:
4*log(1/3*x^2 - x)*log((x*e^x - 3)*e^(-x)) - e^(5*x)
Time = 0.41 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=- e^{5 x} + 4 \log {\left (\left (x e^{x} - 3\right ) e^{- x} \right )} \log {\left (\frac {x^{2}}{3} - x \right )} \] Input:
integrate((((8*x**2-12*x)*exp(x)-24*x+36)*ln((exp(x)*x-3)/exp(x))+((4*x**2 -12*x)*exp(x)+12*x**2-36*x)*ln(1/3*x**2-x)+((-5*x**3+15*x**2)*exp(x)+15*x* *2-45*x)*exp(5*x))/((x**3-3*x**2)*exp(x)-3*x**2+9*x),x)
Output:
-exp(5*x) + 4*log((x*exp(x) - 3)*exp(-x))*log(x**2/3 - x)
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (28) = 56\).
Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=4 \, x \log \left (3\right ) + 4 \, {\left (\log \left (x - 3\right ) + \log \left (x\right )\right )} \log \left (x e^{x} - 3\right ) - 4 \, x \log \left (x - 3\right ) - 4 \, {\left (x + \log \left (3\right )\right )} \log \left (x\right ) - 4 \, \log \left (3\right ) \log \left (\frac {x e^{x} - 3}{x}\right ) - e^{\left (5 \, x\right )} \] Input:
integrate((((8*x^2-12*x)*exp(x)-24*x+36)*log((exp(x)*x-3)/exp(x))+((4*x^2- 12*x)*exp(x)+12*x^2-36*x)*log(1/3*x^2-x)+((-5*x^3+15*x^2)*exp(x)+15*x^2-45 *x)*exp(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x, algorithm="maxima")
Output:
4*x*log(3) + 4*(log(x - 3) + log(x))*log(x*e^x - 3) - 4*x*log(x - 3) - 4*( x + log(3))*log(x) - 4*log(3)*log((x*e^x - 3)/x) - e^(5*x)
\[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=\int { -\frac {5 \, {\left (3 \, x^{2} - {\left (x^{3} - 3 \, x^{2}\right )} e^{x} - 9 \, x\right )} e^{\left (5 \, x\right )} + 4 \, {\left (3 \, x^{2} + {\left (x^{2} - 3 \, x\right )} e^{x} - 9 \, x\right )} \log \left (\frac {1}{3} \, x^{2} - x\right ) + 4 \, {\left ({\left (2 \, x^{2} - 3 \, x\right )} e^{x} - 6 \, x + 9\right )} \log \left ({\left (x e^{x} - 3\right )} e^{\left (-x\right )}\right )}{3 \, x^{2} - {\left (x^{3} - 3 \, x^{2}\right )} e^{x} - 9 \, x} \,d x } \] Input:
integrate((((8*x^2-12*x)*exp(x)-24*x+36)*log((exp(x)*x-3)/exp(x))+((4*x^2- 12*x)*exp(x)+12*x^2-36*x)*log(1/3*x^2-x)+((-5*x^3+15*x^2)*exp(x)+15*x^2-45 *x)*exp(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x, algorithm="giac")
Output:
integrate(-(5*(3*x^2 - (x^3 - 3*x^2)*e^x - 9*x)*e^(5*x) + 4*(3*x^2 + (x^2 - 3*x)*e^x - 9*x)*log(1/3*x^2 - x) + 4*((2*x^2 - 3*x)*e^x - 6*x + 9)*log(( x*e^x - 3)*e^(-x)))/(3*x^2 - (x^3 - 3*x^2)*e^x - 9*x), x)
Time = 7.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=4\,\ln \left (\frac {x^2}{3}-x\right )\,\ln \left (x-3\,{\mathrm {e}}^{-x}\right )-{\mathrm {e}}^{5\,x} \] Input:
int((log(x^2/3 - x)*(36*x + exp(x)*(12*x - 4*x^2) - 12*x^2) - exp(5*x)*(ex p(x)*(15*x^2 - 5*x^3) - 45*x + 15*x^2) + log(exp(-x)*(x*exp(x) - 3))*(24*x + exp(x)*(12*x - 8*x^2) - 36))/(exp(x)*(3*x^2 - x^3) - 9*x + 3*x^2),x)
Output:
4*log(x^2/3 - x)*log(x - 3*exp(-x)) - exp(5*x)
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=-e^{5 x}+4 \,\mathrm {log}\left (\frac {1}{3} x^{2}-x \right ) \mathrm {log}\left (\frac {e^{x} x -3}{e^{x}}\right ) \] Input:
int((((8*x^2-12*x)*exp(x)-24*x+36)*log((exp(x)*x-3)/exp(x))+((4*x^2-12*x)* exp(x)+12*x^2-36*x)*log(1/3*x^2-x)+((-5*x^3+15*x^2)*exp(x)+15*x^2-45*x)*ex p(5*x))/((x^3-3*x^2)*exp(x)-3*x^2+9*x),x)
Output:
- e**(5*x) + 4*log((x**2 - 3*x)/3)*log((e**x*x - 3)/e**x)