Integrand size = 159, antiderivative size = 23 \[ \int \frac {\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )+\left (-16 x-4 x^2+(-8-2 x) \log (8)+\left (8 x+2 x^2+(8+2 x) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx=(4+x) \left (e^5+\log ^2\left (\frac {x}{\log (-12 x (x+\log (8)))}\right )\right ) \] Output:
(ln(x/ln(-12*x*(3*ln(2)+x)))^2+exp(5))*(4+x)
Time = 0.46 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )+\left (-16 x-4 x^2+(-8-2 x) \log (8)+\left (8 x+2 x^2+(8+2 x) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx=e^5 x+(4+x) \log ^2\left (\frac {x}{\log (-12 x (x+\log (8)))}\right ) \] Input:
Integrate[((E^5*x^2 + E^5*x*Log[8])*Log[-12*x^2 - 12*x*Log[8]] + (-16*x - 4*x^2 + (-8 - 2*x)*Log[8] + (8*x + 2*x^2 + (8 + 2*x)*Log[8])*Log[-12*x^2 - 12*x*Log[8]])*Log[x/Log[-12*x^2 - 12*x*Log[8]]] + (x^2 + x*Log[8])*Log[-1 2*x^2 - 12*x*Log[8]]*Log[x/Log[-12*x^2 - 12*x*Log[8]]]^2)/((x^2 + x*Log[8] )*Log[-12*x^2 - 12*x*Log[8]]),x]
Output:
E^5*x + (4 + x)*Log[x/Log[-12*x*(x + Log[8])]]^2
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.
| \(\displaystyle \int \frac {\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (-4 x^2+\left (2 x^2+8 x+(2 x+8) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )-16 x+(-2 x-8) \log (8)\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (-4 x^2+\left (2 x^2+8 x+(2 x+8) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )-16 x+(-2 x-8) \log (8)\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )}{x (x+\log (8)) \log \left (-12 x^2-12 x \log (8)\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^5 x^2+x (x+\log (8)) \log ^2\left (\frac {x}{\log (-12 x (x+\log (8)))}\right )+e^5 x \log (8)+\frac {2 (x+4) (-2 x+(x+\log (8)) \log (-12 x (x+\log (8)))-\log (8)) \log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{\log (-12 x (x+\log (8)))}}{x (x+\log (8))}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\log ^2\left (\frac {x}{\log (-12 x (x+\log (8)))}\right )+\frac {2 (x+4) (-2 x+x \log (-12 x (x+\log (8)))+\log (8) \log (-12 x (x+\log (8)))-\log (8)) \log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{x (x+\log (8)) \log (-12 x (x+\log (8)))}+e^5\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \log ^2\left (\frac {x}{\log (-12 x (x+\log (8)))}\right )dx+\frac {2 (4-\log (8)) \int \frac {-2 x-\log (8)}{(x+\log (8)) \log (-12 x (x+\log (8)))}dx}{\log (8)}-\frac {8 \int \frac {-2 x-\log (8)}{(x+\log (8)) \log (-12 x (x+\log (8)))}dx}{\log (8)}+8 \int \frac {\log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{x}dx+\frac {4 (4-\log (8)) \int \frac {\log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{\log (-12 x (x+\log (8)))}dx}{\log (8)}-\frac {16 \int \frac {\log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{\log (-12 x (x+\log (8)))}dx}{\log (8)}-8 \int \frac {\log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{x \log (-12 x (x+\log (8)))}dx-2 (4-\log (8)) \int \frac {\log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{(x+\log (8)) \log (-12 x (x+\log (8)))}dx+e^5 x-\frac {2 x (4-\log (8)) \log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{\log (8)}+\frac {8 x \log \left (\frac {x}{\log (-12 x (x+\log (8)))}\right )}{\log (8)}+\frac {2 x (4-\log (8))}{\log (8)}-\frac {8 x}{\log (8)}\) |
Input:
Int[((E^5*x^2 + E^5*x*Log[8])*Log[-12*x^2 - 12*x*Log[8]] + (-16*x - 4*x^2 + (-8 - 2*x)*Log[8] + (8*x + 2*x^2 + (8 + 2*x)*Log[8])*Log[-12*x^2 - 12*x* Log[8]])*Log[x/Log[-12*x^2 - 12*x*Log[8]]] + (x^2 + x*Log[8])*Log[-12*x^2 - 12*x*Log[8]]*Log[x/Log[-12*x^2 - 12*x*Log[8]]]^2)/((x^2 + x*Log[8])*Log[ -12*x^2 - 12*x*Log[8]]),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(49\) vs. \(2(24)=48\).
Time = 3.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.17
| method | result | size |
| parallelrisch | \(\ln \left (\frac {x}{\ln \left (-12 x \left (3 \ln \left (2\right )+x \right )\right )}\right )^{2} x -6 \,{\mathrm e}^{5} \ln \left (2\right )+x \,{\mathrm e}^{5}+4 \ln \left (\frac {x}{\ln \left (-12 x \left (3 \ln \left (2\right )+x \right )\right )}\right )^{2}\) | \(50\) |
Input:
int(((3*x*ln(2)+x^2)*ln(-36*x*ln(2)-12*x^2)*ln(x/ln(-36*x*ln(2)-12*x^2))^2 +((3*(2*x+8)*ln(2)+2*x^2+8*x)*ln(-36*x*ln(2)-12*x^2)+3*(-2*x-8)*ln(2)-4*x^ 2-16*x)*ln(x/ln(-36*x*ln(2)-12*x^2))+(3*x*exp(5)*ln(2)+x^2*exp(5))*ln(-36* x*ln(2)-12*x^2))/(3*x*ln(2)+x^2)/ln(-36*x*ln(2)-12*x^2),x,method=_RETURNVE RBOSE)
Output:
ln(x/ln(-12*x*(3*ln(2)+x)))^2*x-6*exp(5)*ln(2)+x*exp(5)+4*ln(x/ln(-12*x*(3 *ln(2)+x)))^2
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )+\left (-16 x-4 x^2+(-8-2 x) \log (8)+\left (8 x+2 x^2+(8+2 x) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx={\left (x + 4\right )} \log \left (\frac {x}{\log \left (-12 \, x^{2} - 36 \, x \log \left (2\right )\right )}\right )^{2} + x e^{5} \] Input:
integrate(((3*x*log(2)+x^2)*log(-36*x*log(2)-12*x^2)*log(x/log(-36*x*log(2 )-12*x^2))^2+((3*(2*x+8)*log(2)+2*x^2+8*x)*log(-36*x*log(2)-12*x^2)+3*(-2* x-8)*log(2)-4*x^2-16*x)*log(x/log(-36*x*log(2)-12*x^2))+(3*x*exp(5)*log(2) +x^2*exp(5))*log(-36*x*log(2)-12*x^2))/(3*x*log(2)+x^2)/log(-36*x*log(2)-1 2*x^2),x, algorithm="fricas")
Output:
(x + 4)*log(x/log(-12*x^2 - 36*x*log(2)))^2 + x*e^5
Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )+\left (-16 x-4 x^2+(-8-2 x) \log (8)+\left (8 x+2 x^2+(8+2 x) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx=x e^{5} + \left (x + 4\right ) \log {\left (\frac {x}{\log {\left (- 12 x^{2} - 36 x \log {\left (2 \right )} \right )}} \right )}^{2} \] Input:
integrate(((3*x*ln(2)+x**2)*ln(-36*x*ln(2)-12*x**2)*ln(x/ln(-36*x*ln(2)-12 *x**2))**2+((3*(2*x+8)*ln(2)+2*x**2+8*x)*ln(-36*x*ln(2)-12*x**2)+3*(-2*x-8 )*ln(2)-4*x**2-16*x)*ln(x/ln(-36*x*ln(2)-12*x**2))+(3*x*exp(5)*ln(2)+x**2* exp(5))*ln(-36*x*ln(2)-12*x**2))/(3*x*ln(2)+x**2)/ln(-36*x*ln(2)-12*x**2), x)
Output:
x*exp(5) + (x + 4)*log(x/log(-12*x**2 - 36*x*log(2)))**2
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 4.09 \[ \int \frac {\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )+\left (-16 x-4 x^2+(-8-2 x) \log (8)+\left (8 x+2 x^2+(8+2 x) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx={\left (x + 4\right )} \log \left (i \, \pi + \log \left (3\right ) + 2 \, \log \left (2\right ) + \log \left (x + 3 \, \log \left (2\right )\right ) + \log \left (x\right )\right )^{2} + 3 \, e^{5} \log \left (2\right ) \log \left (x + 3 \, \log \left (2\right )\right ) - 2 \, {\left (x + 4\right )} \log \left (i \, \pi + \log \left (3\right ) + 2 \, \log \left (2\right ) + \log \left (x + 3 \, \log \left (2\right )\right ) + \log \left (x\right )\right ) \log \left (x\right ) + {\left (x + 4\right )} \log \left (x\right )^{2} - {\left (3 \, \log \left (2\right ) \log \left (x + 3 \, \log \left (2\right )\right ) - x\right )} e^{5} \] Input:
integrate(((3*x*log(2)+x^2)*log(-36*x*log(2)-12*x^2)*log(x/log(-36*x*log(2 )-12*x^2))^2+((3*(2*x+8)*log(2)+2*x^2+8*x)*log(-36*x*log(2)-12*x^2)+3*(-2* x-8)*log(2)-4*x^2-16*x)*log(x/log(-36*x*log(2)-12*x^2))+(3*x*exp(5)*log(2) +x^2*exp(5))*log(-36*x*log(2)-12*x^2))/(3*x*log(2)+x^2)/log(-36*x*log(2)-1 2*x^2),x, algorithm="maxima")
Output:
(x + 4)*log(I*pi + log(3) + 2*log(2) + log(x + 3*log(2)) + log(x))^2 + 3*e ^5*log(2)*log(x + 3*log(2)) - 2*(x + 4)*log(I*pi + log(3) + 2*log(2) + log (x + 3*log(2)) + log(x))*log(x) + (x + 4)*log(x)^2 - (3*log(2)*log(x + 3*l og(2)) - x)*e^5
\[ \int \frac {\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )+\left (-16 x-4 x^2+(-8-2 x) \log (8)+\left (8 x+2 x^2+(8+2 x) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx=\int { \frac {{\left (x^{2} + 3 \, x \log \left (2\right )\right )} \log \left (-12 \, x^{2} - 36 \, x \log \left (2\right )\right ) \log \left (\frac {x}{\log \left (-12 \, x^{2} - 36 \, x \log \left (2\right )\right )}\right )^{2} + {\left (x^{2} e^{5} + 3 \, x e^{5} \log \left (2\right )\right )} \log \left (-12 \, x^{2} - 36 \, x \log \left (2\right )\right ) - 2 \, {\left (2 \, x^{2} + 3 \, {\left (x + 4\right )} \log \left (2\right ) - {\left (x^{2} + 3 \, {\left (x + 4\right )} \log \left (2\right ) + 4 \, x\right )} \log \left (-12 \, x^{2} - 36 \, x \log \left (2\right )\right ) + 8 \, x\right )} \log \left (\frac {x}{\log \left (-12 \, x^{2} - 36 \, x \log \left (2\right )\right )}\right )}{{\left (x^{2} + 3 \, x \log \left (2\right )\right )} \log \left (-12 \, x^{2} - 36 \, x \log \left (2\right )\right )} \,d x } \] Input:
integrate(((3*x*log(2)+x^2)*log(-36*x*log(2)-12*x^2)*log(x/log(-36*x*log(2 )-12*x^2))^2+((3*(2*x+8)*log(2)+2*x^2+8*x)*log(-36*x*log(2)-12*x^2)+3*(-2* x-8)*log(2)-4*x^2-16*x)*log(x/log(-36*x*log(2)-12*x^2))+(3*x*exp(5)*log(2) +x^2*exp(5))*log(-36*x*log(2)-12*x^2))/(3*x*log(2)+x^2)/log(-36*x*log(2)-1 2*x^2),x, algorithm="giac")
Output:
integrate(((x^2 + 3*x*log(2))*log(-12*x^2 - 36*x*log(2))*log(x/log(-12*x^2 - 36*x*log(2)))^2 + (x^2*e^5 + 3*x*e^5*log(2))*log(-12*x^2 - 36*x*log(2)) - 2*(2*x^2 + 3*(x + 4)*log(2) - (x^2 + 3*(x + 4)*log(2) + 4*x)*log(-12*x^ 2 - 36*x*log(2)) + 8*x)*log(x/log(-12*x^2 - 36*x*log(2))))/((x^2 + 3*x*log (2))*log(-12*x^2 - 36*x*log(2))), x)
Timed out. \[ \int \frac {\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )+\left (-16 x-4 x^2+(-8-2 x) \log (8)+\left (8 x+2 x^2+(8+2 x) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx=\int \frac {\ln \left (-12\,x^2-36\,\ln \left (2\right )\,x\right )\,\left (x^2+3\,\ln \left (2\right )\,x\right )\,{\ln \left (\frac {x}{\ln \left (-12\,x^2-36\,\ln \left (2\right )\,x\right )}\right )}^2+\left (\ln \left (-12\,x^2-36\,\ln \left (2\right )\,x\right )\,\left (8\,x+3\,\ln \left (2\right )\,\left (2\,x+8\right )+2\,x^2\right )-3\,\ln \left (2\right )\,\left (2\,x+8\right )-16\,x-4\,x^2\right )\,\ln \left (\frac {x}{\ln \left (-12\,x^2-36\,\ln \left (2\right )\,x\right )}\right )+\ln \left (-12\,x^2-36\,\ln \left (2\right )\,x\right )\,\left ({\mathrm {e}}^5\,x^2+3\,{\mathrm {e}}^5\,\ln \left (2\right )\,x\right )}{\ln \left (-12\,x^2-36\,\ln \left (2\right )\,x\right )\,\left (x^2+3\,\ln \left (2\right )\,x\right )} \,d x \] Input:
int((log(- 36*x*log(2) - 12*x^2)*(x^2*exp(5) + 3*x*exp(5)*log(2)) - log(x/ log(- 36*x*log(2) - 12*x^2))*(16*x + 3*log(2)*(2*x + 8) - log(- 36*x*log(2 ) - 12*x^2)*(8*x + 3*log(2)*(2*x + 8) + 2*x^2) + 4*x^2) + log(- 36*x*log(2 ) - 12*x^2)*log(x/log(- 36*x*log(2) - 12*x^2))^2*(3*x*log(2) + x^2))/(log( - 36*x*log(2) - 12*x^2)*(3*x*log(2) + x^2)),x)
Output:
int((log(- 36*x*log(2) - 12*x^2)*(x^2*exp(5) + 3*x*exp(5)*log(2)) - log(x/ log(- 36*x*log(2) - 12*x^2))*(16*x + 3*log(2)*(2*x + 8) - log(- 36*x*log(2 ) - 12*x^2)*(8*x + 3*log(2)*(2*x + 8) + 2*x^2) + 4*x^2) + log(- 36*x*log(2 ) - 12*x^2)*log(x/log(- 36*x*log(2) - 12*x^2))^2*(3*x*log(2) + x^2))/(log( - 36*x*log(2) - 12*x^2)*(3*x*log(2) + x^2)), x)
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09 \[ \int \frac {\left (e^5 x^2+e^5 x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )+\left (-16 x-4 x^2+(-8-2 x) \log (8)+\left (8 x+2 x^2+(8+2 x) \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )\right ) \log \left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )+\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right ) \log ^2\left (\frac {x}{\log \left (-12 x^2-12 x \log (8)\right )}\right )}{\left (x^2+x \log (8)\right ) \log \left (-12 x^2-12 x \log (8)\right )} \, dx={\mathrm {log}\left (\frac {x}{\mathrm {log}\left (-36 \,\mathrm {log}\left (2\right ) x -12 x^{2}\right )}\right )}^{2} x +4 {\mathrm {log}\left (\frac {x}{\mathrm {log}\left (-36 \,\mathrm {log}\left (2\right ) x -12 x^{2}\right )}\right )}^{2}+e^{5} x \] Input:
int(((3*x*log(2)+x^2)*log(-36*x*log(2)-12*x^2)*log(x/log(-36*x*log(2)-12*x ^2))^2+((3*(2*x+8)*log(2)+2*x^2+8*x)*log(-36*x*log(2)-12*x^2)+3*(-2*x-8)*l og(2)-4*x^2-16*x)*log(x/log(-36*x*log(2)-12*x^2))+(3*x*exp(5)*log(2)+x^2*e xp(5))*log(-36*x*log(2)-12*x^2))/(3*x*log(2)+x^2)/log(-36*x*log(2)-12*x^2) ,x)
Output:
log(x/log( - 36*log(2)*x - 12*x**2))**2*x + 4*log(x/log( - 36*log(2)*x - 1 2*x**2))**2 + e**5*x