Integrand size = 90, antiderivative size = 25 \[ \int \frac {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx=x+\frac {1}{2} \left (x-\frac {20 \left (e^x+x\right )^2}{\log (-3+12 x)}\right ) \] Output:
3/2*x-10*(exp(x)+x)^2/ln(12*x-3)
Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx=\frac {3 x}{2}-\frac {10 \left (e^x+x\right )^2}{\log (-3+12 x)} \] Input:
Integrate[(80*E^(2*x) + 160*E^x*x + 80*x^2 + (E^(2*x)*(40 - 160*x) + 40*x - 160*x^2 + E^x*(40 - 120*x - 160*x^2))*Log[-3 + 12*x] + (-3 + 12*x)*Log[- 3 + 12*x]^2)/((-2 + 8*x)*Log[-3 + 12*x]^2),x]
Output:
(3*x)/2 - (10*(E^x + x)^2)/Log[-3 + 12*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.
\(\displaystyle \int \frac {80 x^2+\left (-160 x^2+e^x \left (-160 x^2-120 x+40\right )+40 x+e^{2 x} (40-160 x)\right ) \log (12 x-3)+160 e^x x+80 e^{2 x}+(12 x-3) \log ^2(12 x-3)}{(8 x-2) \log ^2(12 x-3)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {40 \left (x+e^x\right )^2}{(4 x-1) \log ^2(12 x-3)}-\frac {20 \left (e^x+1\right ) \left (x+e^x\right )}{\log (12 x-3)}+\frac {3}{2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 20 \int \frac {e^x}{\log ^2(12 x-3)}dx+20 \int \frac {e^x}{(4 x-1) \log ^2(12 x-3)}dx+40 \int \frac {e^{2 x}}{(4 x-1) \log ^2(12 x-3)}dx-25 \int \frac {e^x}{\log (12 x-3)}dx-20 \int \frac {e^{2 x}}{\log (12 x-3)}dx-\frac {5}{3} \int \frac {e^x (12 x-3)}{\log (12 x-3)}dx+\frac {3 x}{2}-\frac {5 (1-4 x)^2}{8 \log (12 x-3)}+\frac {5 (1-4 x)}{4 \log (12 x-3)}-\frac {5}{8 \log (12 x-3)}\) |
Input:
Int[(80*E^(2*x) + 160*E^x*x + 80*x^2 + (E^(2*x)*(40 - 160*x) + 40*x - 160* x^2 + E^x*(40 - 120*x - 160*x^2))*Log[-3 + 12*x] + (-3 + 12*x)*Log[-3 + 12 *x]^2)/((-2 + 8*x)*Log[-3 + 12*x]^2),x]
Output:
$Aborted
Time = 17.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {3 x}{2}-\frac {10 \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}\right )}{\ln \left (12 x -3\right )}\) | \(28\) |
parallelrisch | \(-\frac {320 x^{2}+640 \,{\mathrm e}^{x} x -48 \ln \left (12 x -3\right ) x +320 \,{\mathrm e}^{2 x}-6 \ln \left (12 x -3\right )}{32 \ln \left (12 x -3\right )}\) | \(45\) |
default | \(\frac {3 x}{2}-\frac {10 \,{\mathrm e}^{2 x}}{\ln \left (12 x -3\right )}-\frac {20 \,{\mathrm e}^{x} x}{\ln \left (12 x -3\right )}-\frac {5 \left (12 x -3\right )^{2}}{72 \ln \left (12 x -3\right )}-\frac {5 \left (12 x -3\right )}{12 \ln \left (12 x -3\right )}-\frac {5}{8 \ln \left (12 x -3\right )}\) | \(74\) |
parts | \(\frac {3 x}{2}-\frac {10 \,{\mathrm e}^{2 x}}{\ln \left (12 x -3\right )}-\frac {20 \,{\mathrm e}^{x} x}{\ln \left (12 x -3\right )}-\frac {5 \left (12 x -3\right )^{2}}{72 \ln \left (12 x -3\right )}-\frac {5 \left (12 x -3\right )}{12 \ln \left (12 x -3\right )}-\frac {5}{8 \ln \left (12 x -3\right )}\) | \(74\) |
Input:
int(((12*x-3)*ln(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40)*exp(x )-160*x^2+40*x)*ln(12*x-3)+80*exp(x)^2+160*exp(x)*x+80*x^2)/(8*x-2)/ln(12* x-3)^2,x,method=_RETURNVERBOSE)
Output:
3/2*x-10*(exp(2*x)+2*exp(x)*x+x^2)/ln(12*x-3)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx=-\frac {20 \, x^{2} + 40 \, x e^{x} - 3 \, x \log \left (12 \, x - 3\right ) + 20 \, e^{\left (2 \, x\right )}}{2 \, \log \left (12 \, x - 3\right )} \] Input:
integrate(((12*x-3)*log(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40 )*exp(x)-160*x^2+40*x)*log(12*x-3)+80*exp(x)^2+160*exp(x)*x+80*x^2)/(8*x-2 )/log(12*x-3)^2,x, algorithm="fricas")
Output:
-1/2*(20*x^2 + 40*x*e^x - 3*x*log(12*x - 3) + 20*e^(2*x))/log(12*x - 3)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (19) = 38\).
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx=- \frac {10 x^{2}}{\log {\left (12 x - 3 \right )}} + \frac {3 x}{2} + \frac {- 20 x e^{x} \log {\left (12 x - 3 \right )} - 10 e^{2 x} \log {\left (12 x - 3 \right )}}{\log {\left (12 x - 3 \right )}^{2}} \] Input:
integrate(((12*x-3)*ln(12*x-3)**2+((-160*x+40)*exp(x)**2+(-160*x**2-120*x+ 40)*exp(x)-160*x**2+40*x)*ln(12*x-3)+80*exp(x)**2+160*exp(x)*x+80*x**2)/(8 *x-2)/ln(12*x-3)**2,x)
Output:
-10*x**2/log(12*x - 3) + 3*x/2 + (-20*x*exp(x)*log(12*x - 3) - 10*exp(2*x) *log(12*x - 3))/log(12*x - 3)**2
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).
Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx=-\frac {20 \, x^{2} + 40 \, x e^{x} - 3 \, x \log \left (3\right ) - 3 \, x \log \left (4 \, x - 1\right ) + 20 \, e^{\left (2 \, x\right )}}{2 \, {\left (\log \left (3\right ) + \log \left (4 \, x - 1\right )\right )}} \] Input:
integrate(((12*x-3)*log(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40 )*exp(x)-160*x^2+40*x)*log(12*x-3)+80*exp(x)^2+160*exp(x)*x+80*x^2)/(8*x-2 )/log(12*x-3)^2,x, algorithm="maxima")
Output:
-1/2*(20*x^2 + 40*x*e^x - 3*x*log(3) - 3*x*log(4*x - 1) + 20*e^(2*x))/(log (3) + log(4*x - 1))
Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx=-\frac {20 \, x^{2} + 40 \, x e^{x} - 3 \, x \log \left (12 \, x - 3\right ) + 20 \, e^{\left (2 \, x\right )}}{2 \, \log \left (12 \, x - 3\right )} \] Input:
integrate(((12*x-3)*log(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40 )*exp(x)-160*x^2+40*x)*log(12*x-3)+80*exp(x)^2+160*exp(x)*x+80*x^2)/(8*x-2 )/log(12*x-3)^2,x, algorithm="giac")
Output:
-1/2*(20*x^2 + 40*x*e^x - 3*x*log(12*x - 3) + 20*e^(2*x))/log(12*x - 3)
Time = 2.88 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx=\frac {3\,x}{2}-\frac {10\,{\mathrm {e}}^{2\,x}+20\,x\,{\mathrm {e}}^x+10\,x^2}{\ln \left (12\,x-3\right )} \] Input:
int((80*exp(2*x) + log(12*x - 3)^2*(12*x - 3) + 160*x*exp(x) + 80*x^2 - lo g(12*x - 3)*(exp(x)*(120*x + 160*x^2 - 40) - 40*x + exp(2*x)*(160*x - 40) + 160*x^2))/(log(12*x - 3)^2*(8*x - 2)),x)
Output:
(3*x)/2 - (10*exp(2*x) + 20*x*exp(x) + 10*x^2)/log(12*x - 3)
Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx=\frac {-20 e^{2 x}-40 e^{x} x +3 \,\mathrm {log}\left (12 x -3\right ) x -20 x^{2}}{2 \,\mathrm {log}\left (12 x -3\right )} \] Input:
int(((12*x-3)*log(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40)*exp( x)-160*x^2+40*x)*log(12*x-3)+80*exp(x)^2+160*exp(x)*x+80*x^2)/(8*x-2)/log( 12*x-3)^2,x)
Output:
( - 20*e**(2*x) - 40*e**x*x + 3*log(12*x - 3)*x - 20*x**2)/(2*log(12*x - 3 ))