Integrand size = 109, antiderivative size = 26 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=e^{e^x} \log \left (5 \left (x+\frac {1}{45} x^2 (x+\log (3)) \log (2 x)\right )\right ) \] Output:
exp(exp(x))*ln(5*x+1/9*ln(2*x)*x^2*(ln(3)+x))
\[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=\int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx \] Input:
Integrate[(E^E^x*(45 + x^2 + x*Log[3] + (3*x^2 + 2*x*Log[3])*Log[2*x]) + E ^E^x*(45*E^x*x + E^x*(x^3 + x^2*Log[3])*Log[2*x])*Log[(45*x + (x^3 + x^2*L og[3])*Log[2*x])/9])/(45*x + (x^3 + x^2*Log[3])*Log[2*x]),x]
Output:
Integrate[(E^E^x*(45 + x^2 + x*Log[3] + (3*x^2 + 2*x*Log[3])*Log[2*x]) + E ^E^x*(45*E^x*x + E^x*(x^3 + x^2*Log[3])*Log[2*x])*Log[(45*x + (x^3 + x^2*L og[3])*Log[2*x])/9])/(45*x + (x^3 + x^2*Log[3])*Log[2*x]), x]
Time = 7.49 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {e^{e^x} \left (x^2+\left (3 x^2+2 x \log (3)\right ) \log (2 x)+x \log (3)+45\right )+e^{e^x} \left (e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)+45 e^x x\right ) \log \left (\frac {1}{9} \left (\left (x^3+x^2 \log (3)\right ) \log (2 x)+45 x\right )\right )}{\left (x^3+x^2 \log (3)\right ) \log (2 x)+45 x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{e^x} \left (x^2+3 x^2 \log (2 x)+x \log (9) \log (2 x)+x \log (3)+45\right )}{x \left (x^2 \log (2 x)+x \log (3) \log (2 x)+45\right )}+e^{x+e^x} \log \left (\frac {1}{9} x (x (x+\log (3)) \log (2 x)+45)\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{e^x} \log \left (\frac {1}{9} x (x (x+\log (3)) \log (2 x)+45)\right )\) |
Input:
Int[(E^E^x*(45 + x^2 + x*Log[3] + (3*x^2 + 2*x*Log[3])*Log[2*x]) + E^E^x*( 45*E^x*x + E^x*(x^3 + x^2*Log[3])*Log[2*x])*Log[(45*x + (x^3 + x^2*Log[3]) *Log[2*x])/9])/(45*x + (x^3 + x^2*Log[3])*Log[2*x]),x]
Output:
E^E^x*Log[(x*(45 + x*(x + Log[3])*Log[2*x]))/9]
\[\int \frac {\left (\left (x^{2} \ln \left (3\right )+x^{3}\right ) {\mathrm e}^{x} \ln \left (2 x \right )+45 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{{\mathrm e}^{x}} \ln \left (\frac {\left (x^{2} \ln \left (3\right )+x^{3}\right ) \ln \left (2 x \right )}{9}+5 x \right )+\left (\left (2 x \ln \left (3\right )+3 x^{2}\right ) \ln \left (2 x \right )+x \ln \left (3\right )+x^{2}+45\right ) {\mathrm e}^{{\mathrm e}^{x}}}{\left (x^{2} \ln \left (3\right )+x^{3}\right ) \ln \left (2 x \right )+45 x}d x\]
Input:
int((((x^2*ln(3)+x^3)*exp(x)*ln(2*x)+45*exp(x)*x)*exp(exp(x))*ln(1/9*(x^2* ln(3)+x^3)*ln(2*x)+5*x)+((2*x*ln(3)+3*x^2)*ln(2*x)+x*ln(3)+x^2+45)*exp(exp (x)))/((x^2*ln(3)+x^3)*ln(2*x)+45*x),x)
Output:
int((((x^2*ln(3)+x^3)*exp(x)*ln(2*x)+45*exp(x)*x)*exp(exp(x))*ln(1/9*(x^2* ln(3)+x^3)*ln(2*x)+5*x)+((2*x*ln(3)+3*x^2)*ln(2*x)+x*ln(3)+x^2+45)*exp(exp (x)))/((x^2*ln(3)+x^3)*ln(2*x)+45*x),x)
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=e^{\left (e^{x}\right )} \log \left (\frac {1}{9} \, {\left (x^{3} + x^{2} \log \left (3\right )\right )} \log \left (2 \, x\right ) + 5 \, x\right ) \] Input:
integrate((((x^2*log(3)+x^3)*exp(x)*log(2*x)+45*exp(x)*x)*exp(exp(x))*log( 1/9*(x^2*log(3)+x^3)*log(2*x)+5*x)+((2*x*log(3)+3*x^2)*log(2*x)+x*log(3)+x ^2+45)*exp(exp(x)))/((x^2*log(3)+x^3)*log(2*x)+45*x),x, algorithm="fricas" )
Output:
e^(e^x)*log(1/9*(x^3 + x^2*log(3))*log(2*x) + 5*x)
Timed out. \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=\text {Timed out} \] Input:
integrate((((x**2*ln(3)+x**3)*exp(x)*ln(2*x)+45*exp(x)*x)*exp(exp(x))*ln(1 /9*(x**2*ln(3)+x**3)*ln(2*x)+5*x)+((2*x*ln(3)+3*x**2)*ln(2*x)+x*ln(3)+x**2 +45)*exp(exp(x)))/((x**2*ln(3)+x**3)*ln(2*x)+45*x),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=-{\left (2 \, \log \left (3\right ) - \log \left (x\right )\right )} e^{\left (e^{x}\right )} + e^{\left (e^{x}\right )} \log \left (x^{2} \log \left (2\right ) + x \log \left (3\right ) \log \left (2\right ) + {\left (x^{2} + x \log \left (3\right )\right )} \log \left (x\right ) + 45\right ) \] Input:
integrate((((x^2*log(3)+x^3)*exp(x)*log(2*x)+45*exp(x)*x)*exp(exp(x))*log( 1/9*(x^2*log(3)+x^3)*log(2*x)+5*x)+((2*x*log(3)+3*x^2)*log(2*x)+x*log(3)+x ^2+45)*exp(exp(x)))/((x^2*log(3)+x^3)*log(2*x)+45*x),x, algorithm="maxima" )
Output:
-(2*log(3) - log(x))*e^(e^x) + e^(e^x)*log(x^2*log(2) + x*log(3)*log(2) + (x^2 + x*log(3))*log(x) + 45)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=-{\left (2 \, e^{\left (x + e^{x}\right )} \log \left (3\right ) - e^{\left (x + e^{x}\right )} \log \left (x^{2} \log \left (2 \, x\right ) + x \log \left (3\right ) \log \left (2 \, x\right ) + 45\right ) - e^{\left (x + e^{x}\right )} \log \left (x\right )\right )} e^{\left (-x\right )} \] Input:
integrate((((x^2*log(3)+x^3)*exp(x)*log(2*x)+45*exp(x)*x)*exp(exp(x))*log( 1/9*(x^2*log(3)+x^3)*log(2*x)+5*x)+((2*x*log(3)+3*x^2)*log(2*x)+x*log(3)+x ^2+45)*exp(exp(x)))/((x^2*log(3)+x^3)*log(2*x)+45*x),x, algorithm="giac")
Output:
-(2*e^(x + e^x)*log(3) - e^(x + e^x)*log(x^2*log(2*x) + x*log(3)*log(2*x) + 45) - e^(x + e^x)*log(x))*e^(-x)
Time = 3.48 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=\ln \left (5\,x+\frac {\ln \left (2\,x\right )\,\left (x^3+\ln \left (3\right )\,x^2\right )}{9}\right )\,{\mathrm {e}}^{{\mathrm {e}}^x} \] Input:
int((exp(exp(x))*(x*log(3) + log(2*x)*(2*x*log(3) + 3*x^2) + x^2 + 45) + l og(5*x + (log(2*x)*(x^2*log(3) + x^3))/9)*exp(exp(x))*(45*x*exp(x) + log(2 *x)*exp(x)*(x^2*log(3) + x^3)))/(45*x + log(2*x)*(x^2*log(3) + x^3)),x)
Output:
log(5*x + (log(2*x)*(x^2*log(3) + x^3))/9)*exp(exp(x))
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=e^{e^{x}} \mathrm {log}\left (\frac {\mathrm {log}\left (2 x \right ) \mathrm {log}\left (3\right ) x^{2}}{9}+\frac {\mathrm {log}\left (2 x \right ) x^{3}}{9}+5 x \right ) \] Input:
int((((x^2*log(3)+x^3)*exp(x)*log(2*x)+45*exp(x)*x)*exp(exp(x))*log(1/9*(x ^2*log(3)+x^3)*log(2*x)+5*x)+((2*x*log(3)+3*x^2)*log(2*x)+x*log(3)+x^2+45) *exp(exp(x)))/((x^2*log(3)+x^3)*log(2*x)+45*x),x)
Output:
e**(e**x)*log((log(2*x)*log(3)*x**2 + log(2*x)*x**3 + 45*x)/9)