Integrand size = 161, antiderivative size = 30 \[ \int \frac {2 x^2 \log ^2(4)+e^x \left (-6 x^4+\left (6 x^3-18 x^4-6 x^5\right ) \log (4)\right )+\left (-4 x^2 \log (4)+4 x \log ^2(4)+e^x \left (18 x^4+6 x^5+\left (-24 x^3-6 x^4\right ) \log (4)\right )\right ) \log (x)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)}{3 x^2 \log ^2(4)+\left (-6 x^2 \log (4)+6 x \log ^2(4)\right ) \log (x)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)} \, dx=x \left (\frac {2}{3}+\frac {2 e^x x^2}{\log (x)-\frac {\log (4) (x+\log (x))}{x}}\right ) \] Output:
(2/3+2/(ln(x)-2*ln(2)*(x+ln(x))/x)*exp(x)*x^2)*x
Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(30)=60\).
Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.50 \[ \int \frac {2 x^2 \log ^2(4)+e^x \left (-6 x^4+\left (6 x^3-18 x^4-6 x^5\right ) \log (4)\right )+\left (-4 x^2 \log (4)+4 x \log ^2(4)+e^x \left (18 x^4+6 x^5+\left (-24 x^3-6 x^4\right ) \log (4)\right )\right ) \log (x)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)}{3 x^2 \log ^2(4)+\left (-6 x^2 \log (4)+6 x \log ^2(4)\right ) \log (x)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {2}{3} \left (x-\frac {3 e^x x^4 \left (x^2+\log ^2(4)+x \left (4 \log ^2(4)-\log (16)-\log (4) \log (64)\right )\right )}{\left (x^2+x (-2+\log (4)) \log (4)+\log ^2(4)\right ) (x \log (4)+(-x+\log (4)) \log (x))}\right ) \] Input:
Integrate[(2*x^2*Log[4]^2 + E^x*(-6*x^4 + (6*x^3 - 18*x^4 - 6*x^5)*Log[4]) + (-4*x^2*Log[4] + 4*x*Log[4]^2 + E^x*(18*x^4 + 6*x^5 + (-24*x^3 - 6*x^4) *Log[4]))*Log[x] + (2*x^2 - 4*x*Log[4] + 2*Log[4]^2)*Log[x]^2)/(3*x^2*Log[ 4]^2 + (-6*x^2*Log[4] + 6*x*Log[4]^2)*Log[x] + (3*x^2 - 6*x*Log[4] + 3*Log [4]^2)*Log[x]^2),x]
Output:
(2*(x - (3*E^x*x^4*(x^2 + Log[4]^2 + x*(4*Log[4]^2 - Log[16] - Log[4]*Log[ 64])))/((x^2 + x*(-2 + Log[4])*Log[4] + Log[4]^2)*(x*Log[4] + (-x + Log[4] )*Log[x]))))/3
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.
| \(\displaystyle \int \frac {2 x^2 \log ^2(4)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)+e^x \left (\left (-6 x^5-18 x^4+6 x^3\right ) \log (4)-6 x^4\right )+\left (-4 x^2 \log (4)+e^x \left (6 x^5+18 x^4+\left (-6 x^4-24 x^3\right ) \log (4)\right )+4 x \log ^2(4)\right ) \log (x)}{3 x^2 \log ^2(4)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)+\left (6 x \log ^2(4)-6 x^2 \log (4)\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 x^2 \log ^2(4)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)+e^x \left (\left (-6 x^5-18 x^4+6 x^3\right ) \log (4)-6 x^4\right )+\left (-4 x^2 \log (4)+e^x \left (6 x^5+18 x^4+\left (-6 x^4-24 x^3\right ) \log (4)\right )+4 x \log ^2(4)\right ) \log (x)}{3 (x (-\log (x))+x \log (4)+\log (4) \log (x))^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {2 \left (\log ^2(4) x^2+\left (x^2-2 \log (4) x+\log ^2(4)\right ) \log ^2(x)-3 e^x \left (x^4-\left (-x^5-3 x^4+x^3\right ) \log (4)\right )-\left (2 \log (4) x^2-2 \log ^2(4) x-3 e^x \left (x^5+3 x^4-\left (x^4+4 x^3\right ) \log (4)\right )\right ) \log (x)\right )}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{3} \int \frac {\log ^2(4) x^2+\left (x^2-2 \log (4) x+\log ^2(4)\right ) \log ^2(x)-3 e^x \left (x^4-\left (-x^5-3 x^4+x^3\right ) \log (4)\right )-\left (2 \log (4) x^2-2 \log ^2(4) x-3 e^x \left (x^5+3 x^4-\left (x^4+4 x^3\right ) \log (4)\right )\right ) \log (x)}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {2}{3} \int \left (\frac {3 e^x \left (\log (x) x^2-\log (4) x^2+3 \left (1-\frac {2 \log (2)}{3}\right ) \log (x) x-(1+\log (64)) x-\log (256) \log (x)+\log (4)\right ) x^3}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}-\frac {2 \log (4) \log (x) x^2}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}+\frac {\log ^2(4) x^2}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}+\frac {2 \log ^2(4) \log (x) x}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}+\frac {(x-\log (4))^2 \log ^2(x)}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2}{3} \left (-3 \int \frac {e^x x^4}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}dx-3 \int \frac {e^x x^4}{-\log (x) x+\log (4) x+\log (4) \log (x)}dx+3 (1-\log (4)) \log (4) \int \frac {e^x x^3}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}dx-9 \int \frac {e^x x^3}{-\log (x) x+\log (4) x+\log (4) \log (x)}dx-3 \log ^3(4) \int \frac {e^x x^2}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}dx+3 \log (4) \int \frac {e^x x^2}{-\log (x) x+\log (4) x+\log (4) \log (x)}dx-3 \log ^6(4) \int \frac {e^x}{(x-\log (4)) (-\log (x) x+\log (4) x+\log (4) \log (x))^2}dx-3 \log ^5(4) \int \frac {e^x}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}dx-3 \log ^4(4) \int \frac {e^x x}{(-\log (x) x+\log (4) x+\log (4) \log (x))^2}dx+3 \log ^4(4) \int \frac {e^x}{(x-\log (4)) (-\log (x) x+\log (4) x+\log (4) \log (x))}dx+3 \log ^3(4) \int \frac {e^x}{-\log (x) x+\log (4) x+\log (4) \log (x)}dx+3 \log ^2(4) \int \frac {e^x x}{-\log (x) x+\log (4) x+\log (4) \log (x)}dx+x\right )\) |
Input:
Int[(2*x^2*Log[4]^2 + E^x*(-6*x^4 + (6*x^3 - 18*x^4 - 6*x^5)*Log[4]) + (-4 *x^2*Log[4] + 4*x*Log[4]^2 + E^x*(18*x^4 + 6*x^5 + (-24*x^3 - 6*x^4)*Log[4 ]))*Log[x] + (2*x^2 - 4*x*Log[4] + 2*Log[4]^2)*Log[x]^2)/(3*x^2*Log[4]^2 + (-6*x^2*Log[4] + 6*x*Log[4]^2)*Log[x] + (3*x^2 - 6*x*Log[4] + 3*Log[4]^2) *Log[x]^2),x]
Output:
$Aborted
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03
\[\frac {2 x}{3}-\frac {2 x^{4} {\mathrm e}^{x}}{2 \ln \left (2\right ) \ln \left (x \right )+2 x \ln \left (2\right )-x \ln \left (x \right )}\]
Input:
int(((8*ln(2)^2-8*x*ln(2)+2*x^2)*ln(x)^2+((2*(-6*x^4-24*x^3)*ln(2)+6*x^5+1 8*x^4)*exp(x)+16*x*ln(2)^2-8*x^2*ln(2))*ln(x)+(2*(-6*x^5-18*x^4+6*x^3)*ln( 2)-6*x^4)*exp(x)+8*x^2*ln(2)^2)/((12*ln(2)^2-12*x*ln(2)+3*x^2)*ln(x)^2+(24 *x*ln(2)^2-12*x^2*ln(2))*ln(x)+12*x^2*ln(2)^2),x)
Output:
2/3*x-2*x^4*exp(x)/(2*ln(2)*ln(x)+2*x*ln(2)-x*ln(x))
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {2 x^2 \log ^2(4)+e^x \left (-6 x^4+\left (6 x^3-18 x^4-6 x^5\right ) \log (4)\right )+\left (-4 x^2 \log (4)+4 x \log ^2(4)+e^x \left (18 x^4+6 x^5+\left (-24 x^3-6 x^4\right ) \log (4)\right )\right ) \log (x)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)}{3 x^2 \log ^2(4)+\left (-6 x^2 \log (4)+6 x \log ^2(4)\right ) \log (x)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (3 \, x^{4} e^{x} - 2 \, x^{2} \log \left (2\right ) + {\left (x^{2} - 2 \, x \log \left (2\right )\right )} \log \left (x\right )\right )}}{3 \, {\left (2 \, x \log \left (2\right ) - {\left (x - 2 \, \log \left (2\right )\right )} \log \left (x\right )\right )}} \] Input:
integrate(((8*log(2)^2-8*x*log(2)+2*x^2)*log(x)^2+((2*(-6*x^4-24*x^3)*log( 2)+6*x^5+18*x^4)*exp(x)+16*x*log(2)^2-8*x^2*log(2))*log(x)+(2*(-6*x^5-18*x ^4+6*x^3)*log(2)-6*x^4)*exp(x)+8*x^2*log(2)^2)/((12*log(2)^2-12*x*log(2)+3 *x^2)*log(x)^2+(24*x*log(2)^2-12*x^2*log(2))*log(x)+12*x^2*log(2)^2),x, al gorithm="fricas")
Output:
-2/3*(3*x^4*e^x - 2*x^2*log(2) + (x^2 - 2*x*log(2))*log(x))/(2*x*log(2) - (x - 2*log(2))*log(x))
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {2 x^2 \log ^2(4)+e^x \left (-6 x^4+\left (6 x^3-18 x^4-6 x^5\right ) \log (4)\right )+\left (-4 x^2 \log (4)+4 x \log ^2(4)+e^x \left (18 x^4+6 x^5+\left (-24 x^3-6 x^4\right ) \log (4)\right )\right ) \log (x)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)}{3 x^2 \log ^2(4)+\left (-6 x^2 \log (4)+6 x \log ^2(4)\right ) \log (x)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {2 x^{4} e^{x}}{x \log {\left (x \right )} - 2 x \log {\left (2 \right )} - 2 \log {\left (2 \right )} \log {\left (x \right )}} + \frac {2 x}{3} \] Input:
integrate(((8*ln(2)**2-8*x*ln(2)+2*x**2)*ln(x)**2+((2*(-6*x**4-24*x**3)*ln (2)+6*x**5+18*x**4)*exp(x)+16*x*ln(2)**2-8*x**2*ln(2))*ln(x)+(2*(-6*x**5-1 8*x**4+6*x**3)*ln(2)-6*x**4)*exp(x)+8*x**2*ln(2)**2)/((12*ln(2)**2-12*x*ln (2)+3*x**2)*ln(x)**2+(24*x*ln(2)**2-12*x**2*ln(2))*ln(x)+12*x**2*ln(2)**2) ,x)
Output:
2*x**4*exp(x)/(x*log(x) - 2*x*log(2) - 2*log(2)*log(x)) + 2*x/3
Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {2 x^2 \log ^2(4)+e^x \left (-6 x^4+\left (6 x^3-18 x^4-6 x^5\right ) \log (4)\right )+\left (-4 x^2 \log (4)+4 x \log ^2(4)+e^x \left (18 x^4+6 x^5+\left (-24 x^3-6 x^4\right ) \log (4)\right )\right ) \log (x)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)}{3 x^2 \log ^2(4)+\left (-6 x^2 \log (4)+6 x \log ^2(4)\right ) \log (x)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (3 \, x^{4} e^{x} - 2 \, x^{2} \log \left (2\right ) + {\left (x^{2} - 2 \, x \log \left (2\right )\right )} \log \left (x\right )\right )}}{3 \, {\left (2 \, x \log \left (2\right ) - {\left (x - 2 \, \log \left (2\right )\right )} \log \left (x\right )\right )}} \] Input:
integrate(((8*log(2)^2-8*x*log(2)+2*x^2)*log(x)^2+((2*(-6*x^4-24*x^3)*log( 2)+6*x^5+18*x^4)*exp(x)+16*x*log(2)^2-8*x^2*log(2))*log(x)+(2*(-6*x^5-18*x ^4+6*x^3)*log(2)-6*x^4)*exp(x)+8*x^2*log(2)^2)/((12*log(2)^2-12*x*log(2)+3 *x^2)*log(x)^2+(24*x*log(2)^2-12*x^2*log(2))*log(x)+12*x^2*log(2)^2),x, al gorithm="maxima")
Output:
-2/3*(3*x^4*e^x - 2*x^2*log(2) + (x^2 - 2*x*log(2))*log(x))/(2*x*log(2) - (x - 2*log(2))*log(x))
Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {2 x^2 \log ^2(4)+e^x \left (-6 x^4+\left (6 x^3-18 x^4-6 x^5\right ) \log (4)\right )+\left (-4 x^2 \log (4)+4 x \log ^2(4)+e^x \left (18 x^4+6 x^5+\left (-24 x^3-6 x^4\right ) \log (4)\right )\right ) \log (x)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)}{3 x^2 \log ^2(4)+\left (-6 x^2 \log (4)+6 x \log ^2(4)\right ) \log (x)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (3 \, x^{4} e^{x} - 2 \, x^{2} \log \left (2\right ) + x^{2} \log \left (x\right ) - 2 \, x \log \left (2\right ) \log \left (x\right )\right )}}{3 \, {\left (2 \, x \log \left (2\right ) - x \log \left (x\right ) + 2 \, \log \left (2\right ) \log \left (x\right )\right )}} \] Input:
integrate(((8*log(2)^2-8*x*log(2)+2*x^2)*log(x)^2+((2*(-6*x^4-24*x^3)*log( 2)+6*x^5+18*x^4)*exp(x)+16*x*log(2)^2-8*x^2*log(2))*log(x)+(2*(-6*x^5-18*x ^4+6*x^3)*log(2)-6*x^4)*exp(x)+8*x^2*log(2)^2)/((12*log(2)^2-12*x*log(2)+3 *x^2)*log(x)^2+(24*x*log(2)^2-12*x^2*log(2))*log(x)+12*x^2*log(2)^2),x, al gorithm="giac")
Output:
-2/3*(3*x^4*e^x - 2*x^2*log(2) + x^2*log(x) - 2*x*log(2)*log(x))/(2*x*log( 2) - x*log(x) + 2*log(2)*log(x))
Timed out. \[ \int \frac {2 x^2 \log ^2(4)+e^x \left (-6 x^4+\left (6 x^3-18 x^4-6 x^5\right ) \log (4)\right )+\left (-4 x^2 \log (4)+4 x \log ^2(4)+e^x \left (18 x^4+6 x^5+\left (-24 x^3-6 x^4\right ) \log (4)\right )\right ) \log (x)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)}{3 x^2 \log ^2(4)+\left (-6 x^2 \log (4)+6 x \log ^2(4)\right ) \log (x)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)} \, dx=\int \frac {8\,x^2\,{\ln \left (2\right )}^2+\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (18\,x^4-2\,\ln \left (2\right )\,\left (6\,x^4+24\,x^3\right )+6\,x^5\right )+16\,x\,{\ln \left (2\right )}^2-8\,x^2\,\ln \left (2\right )\right )-{\mathrm {e}}^x\,\left (2\,\ln \left (2\right )\,\left (6\,x^5+18\,x^4-6\,x^3\right )+6\,x^4\right )+{\ln \left (x\right )}^2\,\left (2\,x^2-8\,\ln \left (2\right )\,x+8\,{\ln \left (2\right )}^2\right )}{12\,x^2\,{\ln \left (2\right )}^2+{\ln \left (x\right )}^2\,\left (3\,x^2-12\,\ln \left (2\right )\,x+12\,{\ln \left (2\right )}^2\right )+\ln \left (x\right )\,\left (24\,x\,{\ln \left (2\right )}^2-12\,x^2\,\ln \left (2\right )\right )} \,d x \] Input:
int((8*x^2*log(2)^2 + log(x)*(exp(x)*(18*x^4 - 2*log(2)*(24*x^3 + 6*x^4) + 6*x^5) + 16*x*log(2)^2 - 8*x^2*log(2)) - exp(x)*(2*log(2)*(18*x^4 - 6*x^3 + 6*x^5) + 6*x^4) + log(x)^2*(8*log(2)^2 - 8*x*log(2) + 2*x^2))/(12*x^2*l og(2)^2 + log(x)^2*(12*log(2)^2 - 12*x*log(2) + 3*x^2) + log(x)*(24*x*log( 2)^2 - 12*x^2*log(2))),x)
Output:
int((8*x^2*log(2)^2 + log(x)*(exp(x)*(18*x^4 - 2*log(2)*(24*x^3 + 6*x^4) + 6*x^5) + 16*x*log(2)^2 - 8*x^2*log(2)) - exp(x)*(2*log(2)*(18*x^4 - 6*x^3 + 6*x^5) + 6*x^4) + log(x)^2*(8*log(2)^2 - 8*x*log(2) + 2*x^2))/(12*x^2*l og(2)^2 + log(x)^2*(12*log(2)^2 - 12*x*log(2) + 3*x^2) + log(x)*(24*x*log( 2)^2 - 12*x^2*log(2))), x)
Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {2 x^2 \log ^2(4)+e^x \left (-6 x^4+\left (6 x^3-18 x^4-6 x^5\right ) \log (4)\right )+\left (-4 x^2 \log (4)+4 x \log ^2(4)+e^x \left (18 x^4+6 x^5+\left (-24 x^3-6 x^4\right ) \log (4)\right )\right ) \log (x)+\left (2 x^2-4 x \log (4)+2 \log ^2(4)\right ) \log ^2(x)}{3 x^2 \log ^2(4)+\left (-6 x^2 \log (4)+6 x \log ^2(4)\right ) \log (x)+\left (3 x^2-6 x \log (4)+3 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {2 x \left (-3 e^{x} x^{3}+2 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )-\mathrm {log}\left (x \right ) x +2 \,\mathrm {log}\left (2\right ) x \right )}{6 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )-3 \,\mathrm {log}\left (x \right ) x +6 \,\mathrm {log}\left (2\right ) x} \] Input:
int(((8*log(2)^2-8*x*log(2)+2*x^2)*log(x)^2+((2*(-6*x^4-24*x^3)*log(2)+6*x ^5+18*x^4)*exp(x)+16*x*log(2)^2-8*x^2*log(2))*log(x)+(2*(-6*x^5-18*x^4+6*x ^3)*log(2)-6*x^4)*exp(x)+8*x^2*log(2)^2)/((12*log(2)^2-12*x*log(2)+3*x^2)* log(x)^2+(24*x*log(2)^2-12*x^2*log(2))*log(x)+12*x^2*log(2)^2),x)
Output:
(2*x*( - 3*e**x*x**3 + 2*log(x)*log(2) - log(x)*x + 2*log(2)*x))/(3*(2*log (x)*log(2) - log(x)*x + 2*log(2)*x))