Integrand size = 163, antiderivative size = 27 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {9}{5 \left (\frac {x^2}{4}+\log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )} \] Output:
9/(5/4*x^2+5*ln(x*ln((exp(x)+x)*exp(x))))
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )} \] Input:
Integrate[(-144*x - 288*E^x*x - 144*x^2 + (-144*x - 72*x^3 + E^x*(-144 - 7 2*x^2))*Log[E^(2*x) + E^x*x])/((5*E^x*x^5 + 5*x^6)*Log[E^(2*x) + E^x*x] + (40*E^x*x^3 + 40*x^4)*Log[E^(2*x) + E^x*x]*Log[x*Log[E^(2*x) + E^x*x]] + ( 80*E^x*x + 80*x^2)*Log[E^(2*x) + E^x*x]*Log[x*Log[E^(2*x) + E^x*x]]^2),x]
Output:
36/(5*(x^2 + 4*Log[x*Log[E^x*(E^x + 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.
\(\displaystyle \int \frac {-144 x^2+\left (-72 x^3+e^x \left (-72 x^2-144\right )-144 x\right ) \log \left (e^x x+e^{2 x}\right )-288 e^x x-144 x}{\left (80 x^2+80 e^x x\right ) \log \left (e^x x+e^{2 x}\right ) \log ^2\left (x \log \left (e^x x+e^{2 x}\right )\right )+\left (5 x^6+5 e^x x^5\right ) \log \left (e^x x+e^{2 x}\right )+\left (40 x^4+40 e^x x^3\right ) \log \left (e^x x+e^{2 x}\right ) \log \left (x \log \left (e^x x+e^{2 x}\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {72 \left (-\left (x+e^x\right ) \left (x^2+2\right ) \log \left (e^x \left (x+e^x\right )\right )-2 x \left (x+2 e^x+1\right )\right )}{5 x \left (x+e^x\right ) \log \left (e^x \left (x+e^x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {72}{5} \int -\frac {2 x \left (x+2 e^x+1\right )+\left (x+e^x\right ) \left (x^2+2\right ) \log \left (e^x \left (x+e^x\right )\right )}{x \left (x+e^x\right ) \log \left (e^x \left (x+e^x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {72}{5} \int \frac {2 x \left (x+2 e^x+1\right )+\left (x+e^x\right ) \left (x^2+2\right ) \log \left (e^x \left (x+e^x\right )\right )}{x \left (x+e^x\right ) \log \left (e^x \left (x+e^x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {72}{5} \int \left (\frac {\log \left (e^x \left (x+e^x\right )\right ) x^2+4 x+2 \log \left (e^x \left (x+e^x\right )\right )}{x \log \left (e^x \left (x+e^x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}-\frac {2 (x-1)}{\left (x+e^x\right ) \log \left (e^x \left (x+e^x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {72}{5} \left (2 \int \frac {1}{x \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}dx+\int \frac {x}{\left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}dx+4 \int \frac {1}{\log \left (e^x \left (x+e^x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}dx+2 \int \frac {1}{\left (x+e^x\right ) \log \left (e^x \left (x+e^x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}dx-2 \int \frac {x}{\left (x+e^x\right ) \log \left (e^x \left (x+e^x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (x+e^x\right )\right )\right )\right )^2}dx\right )\) |
Input:
Int[(-144*x - 288*E^x*x - 144*x^2 + (-144*x - 72*x^3 + E^x*(-144 - 72*x^2) )*Log[E^(2*x) + E^x*x])/((5*E^x*x^5 + 5*x^6)*Log[E^(2*x) + E^x*x] + (40*E^ x*x^3 + 40*x^4)*Log[E^(2*x) + E^x*x]*Log[x*Log[E^(2*x) + E^x*x]] + (80*E^x *x + 80*x^2)*Log[E^(2*x) + E^x*x]*Log[x*Log[E^(2*x) + E^x*x]]^2),x]
Output:
$Aborted
Time = 25.83 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
parallelrisch | \(\frac {36}{5 \left (x^{2}+4 \ln \left (x \ln \left (\left ({\mathrm e}^{x}+x \right ) {\mathrm e}^{x}\right )\right )\right )}\) | \(22\) |
Input:
int((((-72*x^2-144)*exp(x)-72*x^3-144*x)*ln(exp(x)^2+exp(x)*x)-288*exp(x)* x-144*x^2-144*x)/((80*exp(x)*x+80*x^2)*ln(exp(x)^2+exp(x)*x)*ln(x*ln(exp(x )^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*ln(exp(x)^2+exp(x)*x)*ln(x*ln(exp( x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*ln(exp(x)^2+exp(x)*x)),x,method=_RETU RNVERBOSE)
Output:
36/5/(x^2+4*ln(x*ln((exp(x)+x)*exp(x))))
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x \log \left (x e^{x} + e^{\left (2 \, x\right )}\right )\right )\right )}} \] Input:
integrate((((-72*x^2-144)*exp(x)-72*x^3-144*x)*log(exp(x)^2+exp(x)*x)-288* exp(x)*x-144*x^2-144*x)/((80*exp(x)*x+80*x^2)*log(exp(x)^2+exp(x)*x)*log(x *log(exp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*log(exp(x)^2+exp(x)*x)*l og(x*log(exp(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*log(exp(x)^2+exp(x)*x)), x, algorithm="fricas")
Output:
36/5/(x^2 + 4*log(x*log(x*e^x + e^(2*x))))
Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 x^{2} + 20 \log {\left (x \log {\left (x e^{x} + e^{2 x} \right )} \right )}} \] Input:
integrate((((-72*x**2-144)*exp(x)-72*x**3-144*x)*ln(exp(x)**2+exp(x)*x)-28 8*exp(x)*x-144*x**2-144*x)/((80*exp(x)*x+80*x**2)*ln(exp(x)**2+exp(x)*x)*l n(x*ln(exp(x)**2+exp(x)*x))**2+(40*exp(x)*x**3+40*x**4)*ln(exp(x)**2+exp(x )*x)*ln(x*ln(exp(x)**2+exp(x)*x))+(5*x**5*exp(x)+5*x**6)*ln(exp(x)**2+exp( x)*x)),x)
Output:
36/(5*x**2 + 20*log(x*log(x*exp(x) + exp(2*x))))
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x + \log \left (x + e^{x}\right )\right ) + 4 \, \log \left (x\right )\right )}} \] Input:
integrate((((-72*x^2-144)*exp(x)-72*x^3-144*x)*log(exp(x)^2+exp(x)*x)-288* exp(x)*x-144*x^2-144*x)/((80*exp(x)*x+80*x^2)*log(exp(x)^2+exp(x)*x)*log(x *log(exp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*log(exp(x)^2+exp(x)*x)*l og(x*log(exp(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*log(exp(x)^2+exp(x)*x)), x, algorithm="maxima")
Output:
36/5/(x^2 + 4*log(x + log(x + e^x)) + 4*log(x))
Time = 1.82 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x^{2} + x \log \left (x + e^{x}\right )\right )\right )}} \] Input:
integrate((((-72*x^2-144)*exp(x)-72*x^3-144*x)*log(exp(x)^2+exp(x)*x)-288* exp(x)*x-144*x^2-144*x)/((80*exp(x)*x+80*x^2)*log(exp(x)^2+exp(x)*x)*log(x *log(exp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*log(exp(x)^2+exp(x)*x)*l og(x*log(exp(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*log(exp(x)^2+exp(x)*x)), x, algorithm="giac")
Output:
36/5/(x^2 + 4*log(x^2 + x*log(x + e^x)))
Time = 4.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5\,\left (4\,\ln \left (x\,\ln \left ({\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^x\right )\right )+x^2\right )} \] Input:
int(-(144*x + log(exp(2*x) + x*exp(x))*(144*x + exp(x)*(72*x^2 + 144) + 72 *x^3) + 288*x*exp(x) + 144*x^2)/(log(exp(2*x) + x*exp(x))*(5*x^5*exp(x) + 5*x^6) + log(x*log(exp(2*x) + x*exp(x)))*log(exp(2*x) + x*exp(x))*(40*x^3* exp(x) + 40*x^4) + log(x*log(exp(2*x) + x*exp(x)))^2*log(exp(2*x) + x*exp( x))*(80*x*exp(x) + 80*x^2)),x)
Output:
36/(5*(4*log(x*log(exp(2*x) + x*exp(x))) + x^2))
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{20 \,\mathrm {log}\left (\mathrm {log}\left (e^{2 x}+e^{x} x \right ) x \right )+5 x^{2}} \] Input:
int((((-72*x^2-144)*exp(x)-72*x^3-144*x)*log(exp(x)^2+exp(x)*x)-288*exp(x) *x-144*x^2-144*x)/((80*exp(x)*x+80*x^2)*log(exp(x)^2+exp(x)*x)*log(x*log(e xp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*log(exp(x)^2+exp(x)*x)*log(x*l og(exp(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*log(exp(x)^2+exp(x)*x)),x)
Output:
36/(5*(4*log(log(e**(2*x) + e**x*x)*x) + x**2))