Integrand size = 111, antiderivative size = 23 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \left (2+e^x\right ) x \log (5 x-\log (4 (-3+x))) \] Output:
1/5*(2+exp(x))*x*ln(-ln(4*x-12)+5*x)
Time = 1.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \left (2+e^x\right ) x \log (5 x-\log (4 (-3+x))) \] Input:
Integrate[(32*x - 10*x^2 + E^x*(16*x - 5*x^2) + (30*x - 10*x^2 + E^x*(15*x + 10*x^2 - 5*x^3) + (-6 + 2*x + E^x*(-3 - 2*x + x^2))*Log[-12 + 4*x])*Log [5*x - Log[-12 + 4*x]])/(75*x - 25*x^2 + (-15 + 5*x)*Log[-12 + 4*x]),x]
Output:
((2 + E^x)*x*Log[5*x - Log[4*(-3 + x)]])/5
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.
\(\displaystyle \int \frac {-10 x^2+e^x \left (16 x-5 x^2\right )+\left (-10 x^2+\left (e^x \left (x^2-2 x-3\right )+2 x-6\right ) \log (4 x-12)+e^x \left (-5 x^3+10 x^2+15 x\right )+30 x\right ) \log (5 x-\log (4 x-12))+32 x}{-25 x^2+75 x+(5 x-15) \log (4 x-12)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-10 x^2+e^x \left (16 x-5 x^2\right )+\left (-10 x^2+\left (e^x \left (x^2-2 x-3\right )+2 x-6\right ) \log (4 x-12)+e^x \left (-5 x^3+10 x^2+15 x\right )+30 x\right ) \log (5 x-\log (4 x-12))+32 x}{5 (3-x) (5 x-\log (4 (x-3)))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {-10 x^2+32 x+e^x \left (16 x-5 x^2\right )+\left (-10 x^2+30 x+5 e^x \left (-x^3+2 x^2+3 x\right )-\left (-2 x+e^x \left (-x^2+2 x+3\right )+6\right ) \log (4 x-12)\right ) \log (5 x-\log (4 x-12))}{(3-x) (5 x-\log (-4 (3-x)))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {10 \log (5 x-\log (4 (x-3))) x^2}{(x-3) (5 x-\log (4 (x-3)))}+\frac {10 x^2}{(x-3) (5 x-\log (4 (x-3)))}+\frac {2 \log (4 x-12) \log (5 x-\log (4 (x-3))) x}{(3-x) (5 x-\log (4 (x-3)))}-\frac {30 \log (5 x-\log (4 (x-3))) x}{(x-3) (5 x-\log (4 (x-3)))}-\frac {32 x}{(x-3) (5 x-\log (4 (x-3)))}+\frac {6 \log (4 x-12) \log (5 x-\log (4 (x-3)))}{(x-3) (5 x-\log (4 (x-3)))}+\frac {e^x \left (5 \log (5 x-\log (4 (x-3))) x^3-\log (4 (x-3)) \log (5 x-\log (4 (x-3))) x^2-10 \log (5 x-\log (4 (x-3))) x^2+5 x^2+2 \log (4 (x-3)) \log (5 x-\log (4 (x-3))) x-15 \log (5 x-\log (4 (x-3))) x-16 x+3 \log (4 (x-3)) \log (5 x-\log (4 (x-3)))\right )}{(x-3) (5 x-\log (4 (x-3)))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-2 \int \frac {1}{5 x-\log (4 (x-3))}dx-6 \int \frac {1}{(x-3) (5 x-\log (4 (x-3)))}dx-3 \int \frac {e^x}{(x-3) (5 x-\log (4 (x-3)))}dx+10 \int \frac {x}{5 x-\log (4 (x-3))}dx+5 \int \frac {e^x x}{5 x-\log (4 (x-3))}dx+\int \frac {e^x}{\log (4 (x-3))-5 x}dx+\int e^x \log (5 x-\log (4 (x-3)))dx+\int e^x x \log (5 x-\log (4 (x-3)))dx+10 \int \frac {x \log (5 x-\log (4 (x-3)))}{5 x-\log (4 (x-3))}dx+6 \int \frac {\log (4 x-12) \log (5 x-\log (4 (x-3)))}{(3-x) (5 x-\log (4 (x-3)))}dx+6 \int \frac {\log (4 x-12) \log (5 x-\log (4 (x-3)))}{(x-3) (5 x-\log (4 (x-3)))}dx+2 \int \frac {\log (4 x-12) \log (5 x-\log (4 (x-3)))}{\log (4 (x-3))-5 x}dx\right )\) |
Input:
Int[(32*x - 10*x^2 + E^x*(16*x - 5*x^2) + (30*x - 10*x^2 + E^x*(15*x + 10* x^2 - 5*x^3) + (-6 + 2*x + E^x*(-3 - 2*x + x^2))*Log[-12 + 4*x])*Log[5*x - Log[-12 + 4*x]])/(75*x - 25*x^2 + (-15 + 5*x)*Log[-12 + 4*x]),x]
Output:
$Aborted
Time = 7.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\left (\frac {2 x}{5}+\frac {{\mathrm e}^{x} x}{5}\right ) \ln \left (-\ln \left (4 x -12\right )+5 x \right )\) | \(24\) |
parallelrisch | \(\frac {\ln \left (-\ln \left (4 x -12\right )+5 x \right ) {\mathrm e}^{x} x}{5}+\frac {2 \ln \left (-\ln \left (4 x -12\right )+5 x \right ) x}{5}-\frac {3 \ln \left (-\frac {\ln \left (4 x -12\right )}{5}+x \right )}{5}+\frac {3 \ln \left (-\ln \left (4 x -12\right )+5 x \right )}{5}\) | \(64\) |
Input:
int(((((x^2-2*x-3)*exp(x)+2*x-6)*ln(4*x-12)+(-5*x^3+10*x^2+15*x)*exp(x)-10 *x^2+30*x)*ln(-ln(4*x-12)+5*x)+(-5*x^2+16*x)*exp(x)-10*x^2+32*x)/((5*x-15) *ln(4*x-12)-25*x^2+75*x),x,method=_RETURNVERBOSE)
Output:
(2/5*x+1/5*exp(x)*x)*ln(-ln(4*x-12)+5*x)
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \, {\left (x e^{x} + 2 \, x\right )} \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) \] Input:
integrate(((((x^2-2*x-3)*exp(x)+2*x-6)*log(4*x-12)+(-5*x^3+10*x^2+15*x)*ex p(x)-10*x^2+30*x)*log(-log(4*x-12)+5*x)+(-5*x^2+16*x)*exp(x)-10*x^2+32*x)/ ((5*x-15)*log(4*x-12)-25*x^2+75*x),x, algorithm="fricas")
Output:
1/5*(x*e^x + 2*x)*log(5*x - log(4*x - 12))
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).
Time = 10.70 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {x e^{x} \log {\left (5 x - \log {\left (4 x - 12 \right )} \right )}}{5} + \left (\frac {2 x}{5} - \frac {3}{5}\right ) \log {\left (5 x - \log {\left (4 x - 12 \right )} \right )} + \frac {3 \log {\left (- 5 x + \log {\left (4 x - 12 \right )} \right )}}{5} \] Input:
integrate(((((x**2-2*x-3)*exp(x)+2*x-6)*ln(4*x-12)+(-5*x**3+10*x**2+15*x)* exp(x)-10*x**2+30*x)*ln(-ln(4*x-12)+5*x)+(-5*x**2+16*x)*exp(x)-10*x**2+32* x)/((5*x-15)*ln(4*x-12)-25*x**2+75*x),x)
Output:
x*exp(x)*log(5*x - log(4*x - 12))/5 + (2*x/5 - 3/5)*log(5*x - log(4*x - 12 )) + 3*log(-5*x + log(4*x - 12))/5
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \, {\left (x e^{x} + 2 \, x\right )} \log \left (5 \, x - 2 \, \log \left (2\right ) - \log \left (x - 3\right )\right ) \] Input:
integrate(((((x^2-2*x-3)*exp(x)+2*x-6)*log(4*x-12)+(-5*x^3+10*x^2+15*x)*ex p(x)-10*x^2+30*x)*log(-log(4*x-12)+5*x)+(-5*x^2+16*x)*exp(x)-10*x^2+32*x)/ ((5*x-15)*log(4*x-12)-25*x^2+75*x),x, algorithm="maxima")
Output:
1/5*(x*e^x + 2*x)*log(5*x - 2*log(2) - log(x - 3))
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \, x e^{x} \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) + \frac {2}{5} \, x \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) \] Input:
integrate(((((x^2-2*x-3)*exp(x)+2*x-6)*log(4*x-12)+(-5*x^3+10*x^2+15*x)*ex p(x)-10*x^2+30*x)*log(-log(4*x-12)+5*x)+(-5*x^2+16*x)*exp(x)-10*x^2+32*x)/ ((5*x-15)*log(4*x-12)-25*x^2+75*x),x, algorithm="giac")
Output:
1/5*x*e^x*log(5*x - log(4*x - 12)) + 2/5*x*log(5*x - log(4*x - 12))
Time = 0.62 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {x\,\ln \left (5\,x-\ln \left (4\,x-12\right )\right )\,\left ({\mathrm {e}}^x+2\right )}{5} \] Input:
int((32*x + log(5*x - log(4*x - 12))*(30*x - 10*x^2 + exp(x)*(15*x + 10*x^ 2 - 5*x^3) - log(4*x - 12)*(exp(x)*(2*x - x^2 + 3) - 2*x + 6)) + exp(x)*(1 6*x - 5*x^2) - 10*x^2)/(75*x + log(4*x - 12)*(5*x - 15) - 25*x^2),x)
Output:
(x*log(5*x - log(4*x - 12))*(exp(x) + 2))/5
Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.78 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {e^{x} \mathrm {log}\left (-\mathrm {log}\left (4 x -12\right )+5 x \right ) x}{5}-\frac {32 \,\mathrm {log}\left (\mathrm {log}\left (4 x -12\right )-5 x \right )}{25}+\frac {2 \,\mathrm {log}\left (-\mathrm {log}\left (4 x -12\right )+5 x \right ) x}{5}+\frac {32 \,\mathrm {log}\left (-\mathrm {log}\left (4 x -12\right )+5 x \right )}{25} \] Input:
int(((((x^2-2*x-3)*exp(x)+2*x-6)*log(4*x-12)+(-5*x^3+10*x^2+15*x)*exp(x)-1 0*x^2+30*x)*log(-log(4*x-12)+5*x)+(-5*x^2+16*x)*exp(x)-10*x^2+32*x)/((5*x- 15)*log(4*x-12)-25*x^2+75*x),x)
Output:
(5*e**x*log( - log(4*x - 12) + 5*x)*x - 32*log(log(4*x - 12) - 5*x) + 10*l og( - log(4*x - 12) + 5*x)*x + 32*log( - log(4*x - 12) + 5*x))/25