Integrand size = 199, antiderivative size = 31 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=2 x+\frac {6}{x+\frac {3 (5-x)}{\left (e^4-\log (2)\right ) \log (x)}} \] Output:
2*x+6/(3*(5-x)/(exp(4)-ln(2))/ln(x)+x)
Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=2 \left (\frac {3}{x}+x+\frac {9 (-5+x)}{x \left (15-3 x+e^4 x \log (x)-x \log (2) \log (x)\right )}\right ) \] Input:
Integrate[(E^4*(90 - 18*x) + 450*x - 180*x^2 + 18*x^3 + (-90 + 18*x)*Log[2 ] + (E^4*(18*x + 60*x^2 - 12*x^3) + (-18*x - 60*x^2 + 12*x^3)*Log[2])*Log[ x] + (E^8*(-6*x + 2*x^3) + E^4*(12*x - 4*x^3)*Log[2] + (-6*x + 2*x^3)*Log[ 2]^2)*Log[x]^2)/(225*x - 90*x^2 + 9*x^3 + (E^4*(30*x^2 - 6*x^3) + (-30*x^2 + 6*x^3)*Log[2])*Log[x] + (E^8*x^3 - 2*E^4*x^3*Log[2] + x^3*Log[2]^2)*Log [x]^2),x]
Output:
2*(3/x + x + (9*(-5 + x))/(x*(15 - 3*x + E^4*x*Log[x] - x*Log[2]*Log[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.
| \(\displaystyle \int \frac {18 x^3+\left (e^8 \left (2 x^3-6 x\right )+\left (2 x^3-6 x\right ) \log ^2(2)+e^4 \left (12 x-4 x^3\right ) \log (2)\right ) \log ^2(x)-180 x^2+\left (e^4 \left (-12 x^3+60 x^2+18 x\right )+\left (12 x^3-60 x^2-18 x\right ) \log (2)\right ) \log (x)+450 x+e^4 (90-18 x)+(18 x-90) \log (2)}{9 x^3+\left (e^8 x^3+x^3 \log ^2(2)-2 e^4 x^3 \log (2)\right ) \log ^2(x)-90 x^2+\left (e^4 \left (30 x^2-6 x^3\right )+\left (6 x^3-30 x^2\right ) \log (2)\right ) \log (x)+225 x} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 \left (x \left (x^2-3\right ) \left (e^4-\log (2)\right )^2 \log ^2(x)-3 x \left (2 x^2-10 x-3\right ) \left (e^4-\log (2)\right ) \log (x)+9 (x-5) \left (x^2-5 x-e^4+\log (2)\right )\right )}{x \left (3 (x-5)-x \left (e^4-\log (2)\right ) \log (x)\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int \frac {-x \left (3-x^2\right ) \left (e^4-\log (2)\right )^2 \log ^2(x)+3 x \left (-2 x^2+10 x+3\right ) \left (e^4-\log (2)\right ) \log (x)+9 (5-x) \left (-x^2+5 x-\log (2)+e^4\right )}{x \left (3 (5-x)+x \left (e^4-\log (2)\right ) \log (x)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (\frac {9 (10-x)}{x^2 \left (e^4 \left (1-\frac {\log (2)}{e^4}\right ) \log (x) x-3 x+15\right )}+\frac {x^2-3}{x^2}+\frac {9 (5-x) \left (x \left (e^4-\log (2)\right )-15\right )}{x^2 \left (e^4 \left (1-\frac {\log (2)}{e^4}\right ) \log (x) x-3 x+15\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (-675 \int \frac {1}{x^2 \left (e^4 \left (1-\frac {\log (2)}{e^4}\right ) \log (x) x-3 x+15\right )^2}dx+90 \int \frac {1}{x^2 \left (e^4 \left (1-\frac {\log (2)}{e^4}\right ) \log (x) x-3 x+15\right )}dx+9 \int \frac {1}{x \left (-e^4 \left (1-\frac {\log (2)}{e^4}\right ) \log (x) x+3 x-15\right )}dx-9 \left (e^4-\log (2)\right ) \int \frac {1}{\left (e^4 \left (1-\frac {\log (2)}{e^4}\right ) \log (x) x-3 x+15\right )^2}dx+45 \left (3+e^4-\log (2)\right ) \int \frac {1}{x \left (e^4 \left (1-\frac {\log (2)}{e^4}\right ) \log (x) x-3 x+15\right )^2}dx+x+\frac {3}{x}\right )\) |
Input:
Int[(E^4*(90 - 18*x) + 450*x - 180*x^2 + 18*x^3 + (-90 + 18*x)*Log[2] + (E ^4*(18*x + 60*x^2 - 12*x^3) + (-18*x - 60*x^2 + 12*x^3)*Log[2])*Log[x] + ( E^8*(-6*x + 2*x^3) + E^4*(12*x - 4*x^3)*Log[2] + (-6*x + 2*x^3)*Log[2]^2)* Log[x]^2)/(225*x - 90*x^2 + 9*x^3 + (E^4*(30*x^2 - 6*x^3) + (-30*x^2 + 6*x ^3)*Log[2])*Log[x] + (E^8*x^3 - 2*E^4*x^3*Log[2] + x^3*Log[2]^2)*Log[x]^2) ,x]
Output:
$Aborted
Time = 2.59 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29
| method | result | size |
| risch | \(\frac {2 x^{2}+6}{x}+\frac {18 x -90}{x \left (x \,{\mathrm e}^{4} \ln \left (x \right )-x \ln \left (2\right ) \ln \left (x \right )-3 x +15\right )}\) | \(40\) |
| default | \(\frac {2 \left (3 \,{\mathrm e}^{4}-3 \ln \left (2\right )\right ) \ln \left (x \right )+2 \left ({\mathrm e}^{4}-\ln \left (2\right )\right ) \ln \left (x \right ) x^{2}-6 x^{2}+10 \left ({\mathrm e}^{4}-\ln \left (2\right )\right ) \ln \left (x \right ) x +150}{x \,{\mathrm e}^{4} \ln \left (x \right )-x \ln \left (2\right ) \ln \left (x \right )-3 x +15}\) | \(67\) |
| norman | \(\frac {\left (6 \,{\mathrm e}^{4}-6 \ln \left (2\right )\right ) \ln \left (x \right )+\left (2 \,{\mathrm e}^{4}-2 \ln \left (2\right )\right ) x^{2} \ln \left (x \right )+\left (10 \,{\mathrm e}^{4}-10 \ln \left (2\right )\right ) x \ln \left (x \right )-6 x^{2}+150}{x \,{\mathrm e}^{4} \ln \left (x \right )-x \ln \left (2\right ) \ln \left (x \right )-3 x +15}\) | \(69\) |
| parallelrisch | \(\frac {6 x^{2} {\mathrm e}^{4} \ln \left (x \right )+450-6 x^{2} \ln \left (2\right ) \ln \left (x \right )+30 x \,{\mathrm e}^{4} \ln \left (x \right )-30 x \ln \left (2\right ) \ln \left (x \right )+18 \,{\mathrm e}^{4} \ln \left (x \right )-18 \ln \left (2\right ) \ln \left (x \right )-18 x^{2}}{3 x \,{\mathrm e}^{4} \ln \left (x \right )-3 x \ln \left (2\right ) \ln \left (x \right )-9 x +45}\) | \(74\) |
Input:
int((((2*x^3-6*x)*ln(2)^2+(-4*x^3+12*x)*exp(4)*ln(2)+(2*x^3-6*x)*exp(4)^2) *ln(x)^2+((12*x^3-60*x^2-18*x)*ln(2)+(-12*x^3+60*x^2+18*x)*exp(4))*ln(x)+( 18*x-90)*ln(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3*ln(2)^2-2*x^3 *exp(4)*ln(2)+x^3*exp(4)^2)*ln(x)^2+((6*x^3-30*x^2)*ln(2)+(-6*x^3+30*x^2)* exp(4))*ln(x)+9*x^3-90*x^2+225*x),x,method=_RETURNVERBOSE)
Output:
2*(x^2+3)/x+18*(-5+x)/x/(x*exp(4)*ln(x)-x*ln(2)*ln(x)-3*x+15)
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (3 \, x^{2} - {\left ({\left (x^{2} + 3\right )} e^{4} - {\left (x^{2} + 3\right )} \log \left (2\right )\right )} \log \left (x\right ) - 15 \, x\right )}}{{\left (x e^{4} - x \log \left (2\right )\right )} \log \left (x\right ) - 3 \, x + 15} \] Input:
integrate((((2*x^3-6*x)*log(2)^2+(-4*x^3+12*x)*exp(4)*log(2)+(2*x^3-6*x)*e xp(4)^2)*log(x)^2+((12*x^3-60*x^2-18*x)*log(2)+(-12*x^3+60*x^2+18*x)*exp(4 ))*log(x)+(18*x-90)*log(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3*l og(2)^2-2*x^3*exp(4)*log(2)+x^3*exp(4)^2)*log(x)^2+((6*x^3-30*x^2)*log(2)+ (-6*x^3+30*x^2)*exp(4))*log(x)+9*x^3-90*x^2+225*x),x, algorithm="fricas")
Output:
-2*(3*x^2 - ((x^2 + 3)*e^4 - (x^2 + 3)*log(2))*log(x) - 15*x)/((x*e^4 - x* log(2))*log(x) - 3*x + 15)
Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=2 x + \frac {18 x - 90}{- 3 x^{2} + 15 x + \left (- x^{2} \log {\left (2 \right )} + x^{2} e^{4}\right ) \log {\left (x \right )}} + \frac {6}{x} \] Input:
integrate((((2*x**3-6*x)*ln(2)**2+(-4*x**3+12*x)*exp(4)*ln(2)+(2*x**3-6*x) *exp(4)**2)*ln(x)**2+((12*x**3-60*x**2-18*x)*ln(2)+(-12*x**3+60*x**2+18*x) *exp(4))*ln(x)+(18*x-90)*ln(2)+(-18*x+90)*exp(4)+18*x**3-180*x**2+450*x)/( (x**3*ln(2)**2-2*x**3*exp(4)*ln(2)+x**3*exp(4)**2)*ln(x)**2+((6*x**3-30*x* *2)*ln(2)+(-6*x**3+30*x**2)*exp(4))*ln(x)+9*x**3-90*x**2+225*x),x)
Output:
2*x + (18*x - 90)/(-3*x**2 + 15*x + (-x**2*log(2) + x**2*exp(4))*log(x)) + 6/x
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (3 \, x^{2} - {\left (x^{2} {\left (e^{4} - \log \left (2\right )\right )} + 3 \, e^{4} - 3 \, \log \left (2\right )\right )} \log \left (x\right ) - 15 \, x\right )}}{x {\left (e^{4} - \log \left (2\right )\right )} \log \left (x\right ) - 3 \, x + 15} \] Input:
integrate((((2*x^3-6*x)*log(2)^2+(-4*x^3+12*x)*exp(4)*log(2)+(2*x^3-6*x)*e xp(4)^2)*log(x)^2+((12*x^3-60*x^2-18*x)*log(2)+(-12*x^3+60*x^2+18*x)*exp(4 ))*log(x)+(18*x-90)*log(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3*l og(2)^2-2*x^3*exp(4)*log(2)+x^3*exp(4)^2)*log(x)^2+((6*x^3-30*x^2)*log(2)+ (-6*x^3+30*x^2)*exp(4))*log(x)+9*x^3-90*x^2+225*x),x, algorithm="maxima")
Output:
-2*(3*x^2 - (x^2*(e^4 - log(2)) + 3*e^4 - 3*log(2))*log(x) - 15*x)/(x*(e^4 - log(2))*log(x) - 3*x + 15)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (28) = 56\).
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.94 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (x^{2} e^{4} \log \left (x\right ) - x^{2} \log \left (2\right ) \log \left (x\right ) - 3 \, x^{2} + 3 \, e^{4} \log \left (x\right ) - 3 \, \log \left (2\right ) \log \left (x\right ) + 15 \, x\right )}}{x e^{4} \log \left (x\right ) - x \log \left (2\right ) \log \left (x\right ) - 3 \, x + 15} \] Input:
integrate((((2*x^3-6*x)*log(2)^2+(-4*x^3+12*x)*exp(4)*log(2)+(2*x^3-6*x)*e xp(4)^2)*log(x)^2+((12*x^3-60*x^2-18*x)*log(2)+(-12*x^3+60*x^2+18*x)*exp(4 ))*log(x)+(18*x-90)*log(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3*l og(2)^2-2*x^3*exp(4)*log(2)+x^3*exp(4)^2)*log(x)^2+((6*x^3-30*x^2)*log(2)+ (-6*x^3+30*x^2)*exp(4))*log(x)+9*x^3-90*x^2+225*x),x, algorithm="giac")
Output:
2*(x^2*e^4*log(x) - x^2*log(2)*log(x) - 3*x^2 + 3*e^4*log(x) - 3*log(2)*lo g(x) + 15*x)/(x*e^4*log(x) - x*log(2)*log(x) - 3*x + 15)
Timed out. \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=\int \frac {450\,x+\ln \left (2\right )\,\left (18\,x-90\right )-{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^8\,\left (6\,x-2\,x^3\right )+{\ln \left (2\right )}^2\,\left (6\,x-2\,x^3\right )-{\mathrm {e}}^4\,\ln \left (2\right )\,\left (12\,x-4\,x^3\right )\right )+\ln \left (x\right )\,\left ({\mathrm {e}}^4\,\left (-12\,x^3+60\,x^2+18\,x\right )-\ln \left (2\right )\,\left (-12\,x^3+60\,x^2+18\,x\right )\right )-180\,x^2+18\,x^3-{\mathrm {e}}^4\,\left (18\,x-90\right )}{225\,x+{\ln \left (x\right )}^2\,\left (x^3\,{\ln \left (2\right )}^2+x^3\,{\mathrm {e}}^8-2\,x^3\,{\mathrm {e}}^4\,\ln \left (2\right )\right )+\ln \left (x\right )\,\left ({\mathrm {e}}^4\,\left (30\,x^2-6\,x^3\right )-\ln \left (2\right )\,\left (30\,x^2-6\,x^3\right )\right )-90\,x^2+9\,x^3} \,d x \] Input:
int((450*x + log(2)*(18*x - 90) - log(x)^2*(exp(8)*(6*x - 2*x^3) + log(2)^ 2*(6*x - 2*x^3) - exp(4)*log(2)*(12*x - 4*x^3)) + log(x)*(exp(4)*(18*x + 6 0*x^2 - 12*x^3) - log(2)*(18*x + 60*x^2 - 12*x^3)) - 180*x^2 + 18*x^3 - ex p(4)*(18*x - 90))/(225*x + log(x)^2*(x^3*log(2)^2 + x^3*exp(8) - 2*x^3*exp (4)*log(2)) + log(x)*(exp(4)*(30*x^2 - 6*x^3) - log(2)*(30*x^2 - 6*x^3)) - 90*x^2 + 9*x^3),x)
Output:
int((450*x + log(2)*(18*x - 90) - log(x)^2*(exp(8)*(6*x - 2*x^3) + log(2)^ 2*(6*x - 2*x^3) - exp(4)*log(2)*(12*x - 4*x^3)) + log(x)*(exp(4)*(18*x + 6 0*x^2 - 12*x^3) - log(2)*(18*x + 60*x^2 - 12*x^3)) - 180*x^2 + 18*x^3 - ex p(4)*(18*x - 90))/(225*x + log(x)^2*(x^3*log(2)^2 + x^3*exp(8) - 2*x^3*exp (4)*log(2)) + log(x)*(exp(4)*(30*x^2 - 6*x^3) - log(2)*(30*x^2 - 6*x^3)) - 90*x^2 + 9*x^3), x)
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.03 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=\frac {2 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right ) x^{2}+6 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )-2 \,\mathrm {log}\left (x \right ) e^{4} x^{2}-6 \,\mathrm {log}\left (x \right ) e^{4}+6 x^{2}-30 x}{x \,\mathrm {log}\left (2\right ) \mathrm {log}\left (x \right )-\mathrm {log}\left (x \right ) e^{4} x +3 x -15} \] Input:
int((((2*x^3-6*x)*log(2)^2+(-4*x^3+12*x)*exp(4)*log(2)+(2*x^3-6*x)*exp(4)^ 2)*log(x)^2+((12*x^3-60*x^2-18*x)*log(2)+(-12*x^3+60*x^2+18*x)*exp(4))*log (x)+(18*x-90)*log(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3*log(2)^ 2-2*x^3*exp(4)*log(2)+x^3*exp(4)^2)*log(x)^2+((6*x^3-30*x^2)*log(2)+(-6*x^ 3+30*x^2)*exp(4))*log(x)+9*x^3-90*x^2+225*x),x)
Output:
(2*(log(x)*log(2)*x**2 + 3*log(x)*log(2) - log(x)*e**4*x**2 - 3*log(x)*e** 4 + 3*x**2 - 15*x))/(log(x)*log(2)*x - log(x)*e**4*x + 3*x - 15)