Integrand size = 139, antiderivative size = 28 \[ \int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx=\log \left (2+\frac {5 e^x}{3}+x+e^{5+x} \log \left (x+\frac {e^x}{\log (2)}\right )\right ) \] Output:
ln(2+5/3*exp(x)+x+exp(5+x)*ln(exp(x)/ln(2)+x))
\[ \int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx=\int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx \] Input:
Integrate[(5*E^(2*x) + 3*x*Log[2] + E^(5 + x)*(3*E^x + 3*Log[2]) + E^x*(3 + 5*x*Log[2]) + E^(5 + x)*(3*E^x + 3*x*Log[2])*Log[(E^x + x*Log[2])/Log[2] ])/(5*E^(2*x) + (6*x + 3*x^2)*Log[2] + E^x*(6 + 3*x + 5*x*Log[2]) + E^(5 + x)*(3*E^x + 3*x*Log[2])*Log[(E^x + x*Log[2])/Log[2]]),x]
Output:
Integrate[(5*E^(2*x) + 3*x*Log[2] + E^(5 + x)*(3*E^x + 3*Log[2]) + E^x*(3 + 5*x*Log[2]) + E^(5 + x)*(3*E^x + 3*x*Log[2])*Log[(E^x + x*Log[2])/Log[2] ])/(5*E^(2*x) + (6*x + 3*x^2)*Log[2] + E^x*(6 + 3*x + 5*x*Log[2]) + E^(5 + x)*(3*E^x + 3*x*Log[2])*Log[(E^x + x*Log[2])/Log[2]]), 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.
| \(\displaystyle \int \frac {5 e^{2 x}+3 x \log (2)+e^{x+5} \left (3 e^x+3 \log (2)\right )+e^x (5 x \log (2)+3)+e^{x+5} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{\left (3 x^2+6 x\right ) \log (2)+5 e^{2 x}+e^x (3 x+5 x \log (2)+6)+e^{x+5} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {5 e^{2 x}+3 x \log (2)+e^{x+5} \left (3 e^x+3 \log (2)\right )+e^x (5 x \log (2)+3)+e^{x+5} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{\left (e^x+x \log (2)\right ) \left (3 x+5 e^x+3 e^{x+5} \log \left (x+\frac {e^x}{\log (2)}\right )+6\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {3 \left (9 e^{10} x^2 \log (2) \log ^2\left (x+\frac {e^x}{\log (2)}\right )-9 e^5 x^2 \left (1-\frac {\log (1024)}{3}\right ) \log \left (x+\frac {e^x}{\log (2)}\right )-15 x^2 \left (1+\frac {3 e^5}{5}-\frac {5 \log (2)}{3}\right )+9 e^{10} x \log (2) \log ^2\left (x+\frac {e^x}{\log (2)}\right )-27 e^5 x \left (1+\frac {1}{9} \left (-e^5 \log (8)-\log (1024)\right )\right ) \log \left (x+\frac {e^x}{\log (2)}\right )-45 x \left (1+\frac {4}{5} e^5 \left (1-\frac {5}{12} \log (2) \left (1+\frac {\log (32)}{e^5 \log (8)}\right )\right )\right )-18 e^5 \left (1-e^5 \log (2)\right ) \log \left (x+\frac {e^x}{\log (2)}\right )-30 \left (1-\frac {1}{5} e^5 (\log (32)-6)\right )\right )}{\left (3 e^5 \log \left (x+\frac {e^x}{\log (2)}\right )+5\right ) \left (3 x+5 e^x+3 e^{x+5} \log \left (x+\frac {e^x}{\log (2)}\right )+6\right ) \left (-3 e^5 x \log (2) \log \left (x+\frac {e^x}{\log (2)}\right )+3 x \left (1-\frac {\log (32)}{3}\right )+6\right )}+\frac {3 e^5 (x-1) x \log ^2(2)}{\left (e^x+x \log (2)\right ) \left (-3 e^5 x \log (2) \log \left (x+\frac {e^x}{\log (2)}\right )+3 x \left (1-\frac {\log (32)}{3}\right )+6\right )}+\frac {3 e^5 \log \left (x+\frac {e^x}{\log (2)}\right )+5 \left (1+\frac {3 e^5}{5}\right )}{3 e^5 \log \left (x+\frac {e^x}{\log (2)}\right )+5}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {3 \left (9 e^{10} x^2 \log (2) \log ^2\left (x+\frac {e^x}{\log (2)}\right )-9 e^5 x^2 \left (1-\frac {\log (1024)}{3}\right ) \log \left (x+\frac {e^x}{\log (2)}\right )-15 x^2 \left (1+\frac {3 e^5}{5}-\frac {5 \log (2)}{3}\right )+9 e^{10} x \log (2) \log ^2\left (x+\frac {e^x}{\log (2)}\right )-27 e^5 x \left (1+\frac {1}{9} \left (-e^5 \log (8)-\log (1024)\right )\right ) \log \left (x+\frac {e^x}{\log (2)}\right )-45 x \left (1+\frac {4}{5} e^5 \left (1-\frac {5}{12} \log (2) \left (1+\frac {\log (32)}{e^5 \log (8)}\right )\right )\right )-18 e^5 \left (1-e^5 \log (2)\right ) \log \left (x+\frac {e^x}{\log (2)}\right )-30 \left (1-\frac {1}{5} e^5 (\log (32)-6)\right )\right )}{\left (3 e^5 \log \left (x+\frac {e^x}{\log (2)}\right )+5\right ) \left (3 x+5 e^x+3 e^{x+5} \log \left (x+\frac {e^x}{\log (2)}\right )+6\right ) \left (-3 e^5 x \log (2) \log \left (x+\frac {e^x}{\log (2)}\right )+3 x \left (1-\frac {\log (32)}{3}\right )+6\right )}+\frac {3 e^5 (x-1) x \log ^2(2)}{\left (e^x+x \log (2)\right ) \left (-3 e^5 x \log (2) \log \left (x+\frac {e^x}{\log (2)}\right )+3 x \left (1-\frac {\log (32)}{3}\right )+6\right )}+\frac {3 e^5 \log \left (x+\frac {e^x}{\log (2)}\right )+5 \left (1+\frac {3 e^5}{5}\right )}{3 e^5 \log \left (x+\frac {e^x}{\log (2)}\right )+5}\right )dx\) |
Input:
Int[(5*E^(2*x) + 3*x*Log[2] + E^(5 + x)*(3*E^x + 3*Log[2]) + E^x*(3 + 5*x* Log[2]) + E^(5 + x)*(3*E^x + 3*x*Log[2])*Log[(E^x + x*Log[2])/Log[2]])/(5* E^(2*x) + (6*x + 3*x^2)*Log[2] + E^x*(6 + 3*x + 5*x*Log[2]) + E^(5 + x)*(3 *E^x + 3*x*Log[2])*Log[(E^x + x*Log[2])/Log[2]]),x]
Output:
$Aborted
Time = 2.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
| method | result | size |
| parallelrisch | \(\ln \left ({\mathrm e}^{5+x} \ln \left (\frac {{\mathrm e}^{x}+x \ln \left (2\right )}{\ln \left (2\right )}\right )+x +\frac {5 \,{\mathrm e}^{x}}{3}+2\right )\) | \(27\) |
| norman | \(\ln \left (3 \,{\mathrm e}^{5} {\mathrm e}^{x} \ln \left (\frac {{\mathrm e}^{x}+x \ln \left (2\right )}{\ln \left (2\right )}\right )+3 x +5 \,{\mathrm e}^{x}+6\right )\) | \(30\) |
| risch | \(x +\ln \left (\ln \left (\frac {{\mathrm e}^{x}+x \ln \left (2\right )}{\ln \left (2\right )}\right )+\frac {\left (3 x +5 \,{\mathrm e}^{x}+6\right ) {\mathrm e}^{-x -5}}{3}\right )\) | \(35\) |
Input:
int(((3*exp(x)+3*x*ln(2))*exp(5+x)*ln((exp(x)+x*ln(2))/ln(2))+(3*exp(x)+3* ln(2))*exp(5+x)+5*exp(x)^2+(5*x*ln(2)+3)*exp(x)+3*x*ln(2))/((3*exp(x)+3*x* ln(2))*exp(5+x)*ln((exp(x)+x*ln(2))/ln(2))+5*exp(x)^2+(5*x*ln(2)+6+3*x)*ex p(x)+(3*x^2+6*x)*ln(2)),x,method=_RETURNVERBOSE)
Output:
ln(exp(5+x)*ln((exp(x)+x*ln(2))/ln(2))+x+5/3*exp(x)+2)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (23) = 46\).
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx=x + \log \left ({\left (3 \, {\left (x + 2\right )} e^{5} + 3 \, e^{\left (x + 10\right )} \log \left (\frac {{\left (x e^{5} \log \left (2\right ) + e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}}{\log \left (2\right )}\right ) + 5 \, e^{\left (x + 5\right )}\right )} e^{\left (-x - 5\right )}\right ) \] Input:
integrate(((3*exp(x)+3*x*log(2))*exp(5+x)*log((exp(x)+x*log(2))/log(2))+(3 *exp(x)+3*log(2))*exp(5+x)+5*exp(x)^2+(5*x*log(2)+3)*exp(x)+3*x*log(2))/(( 3*exp(x)+3*x*log(2))*exp(5+x)*log((exp(x)+x*log(2))/log(2))+5*exp(x)^2+(5* x*log(2)+6+3*x)*exp(x)+(3*x^2+6*x)*log(2)),x, algorithm="fricas")
Output:
x + log((3*(x + 2)*e^5 + 3*e^(x + 10)*log((x*e^5*log(2) + e^(x + 5))*e^(-5 )/log(2)) + 5*e^(x + 5))*e^(-x - 5))
Time = 0.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx=x + \log {\left (\frac {\left (3 x + 5 e^{x} + 6\right ) e^{- x}}{3 e^{5}} + \log {\left (\frac {x \log {\left (2 \right )} + e^{x}}{\log {\left (2 \right )}} \right )} \right )} \] Input:
integrate(((3*exp(x)+3*x*ln(2))*exp(5+x)*ln((exp(x)+x*ln(2))/ln(2))+(3*exp (x)+3*ln(2))*exp(5+x)+5*exp(x)**2+(5*x*ln(2)+3)*exp(x)+3*x*ln(2))/((3*exp( x)+3*x*ln(2))*exp(5+x)*ln((exp(x)+x*ln(2))/ln(2))+5*exp(x)**2+(5*x*ln(2)+6 +3*x)*exp(x)+(3*x**2+6*x)*ln(2)),x)
Output:
x + log((3*x + 5*exp(x) + 6)*exp(-5)*exp(-x)/3 + log((x*log(2) + exp(x))/l og(2)))
Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx=x + \log \left (-\frac {1}{3} \, {\left ({\left (3 \, e^{5} \log \left (\log \left (2\right )\right ) - 5\right )} e^{x} - 3 \, e^{\left (x + 5\right )} \log \left (x \log \left (2\right ) + e^{x}\right ) - 3 \, x - 6\right )} e^{\left (-x - 5\right )}\right ) \] Input:
integrate(((3*exp(x)+3*x*log(2))*exp(5+x)*log((exp(x)+x*log(2))/log(2))+(3 *exp(x)+3*log(2))*exp(5+x)+5*exp(x)^2+(5*x*log(2)+3)*exp(x)+3*x*log(2))/(( 3*exp(x)+3*x*log(2))*exp(5+x)*log((exp(x)+x*log(2))/log(2))+5*exp(x)^2+(5* x*log(2)+6+3*x)*exp(x)+(3*x^2+6*x)*log(2)),x, algorithm="maxima")
Output:
x + log(-1/3*((3*e^5*log(log(2)) - 5)*e^x - 3*e^(x + 5)*log(x*log(2) + e^x ) - 3*x - 6)*e^(-x - 5))
Time = 0.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx=\log \left (3 \, e^{\left (x + 5\right )} \log \left (x \log \left (2\right ) + e^{x}\right ) - 3 \, e^{\left (x + 5\right )} \log \left (\log \left (2\right )\right ) + 3 \, x + 5 \, e^{x} + 6\right ) \] Input:
integrate(((3*exp(x)+3*x*log(2))*exp(5+x)*log((exp(x)+x*log(2))/log(2))+(3 *exp(x)+3*log(2))*exp(5+x)+5*exp(x)^2+(5*x*log(2)+3)*exp(x)+3*x*log(2))/(( 3*exp(x)+3*x*log(2))*exp(5+x)*log((exp(x)+x*log(2))/log(2))+5*exp(x)^2+(5* x*log(2)+6+3*x)*exp(x)+(3*x^2+6*x)*log(2)),x, algorithm="giac")
Output:
log(3*e^(x + 5)*log(x*log(2) + e^x) - 3*e^(x + 5)*log(log(2)) + 3*x + 5*e^ x + 6)
Timed out. \[ \int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx=\int \frac {5\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{x+5}\,\left (3\,\ln \left (2\right )+3\,{\mathrm {e}}^x\right )+3\,x\,\ln \left (2\right )+{\mathrm {e}}^x\,\left (5\,x\,\ln \left (2\right )+3\right )+\ln \left (\frac {{\mathrm {e}}^x+x\,\ln \left (2\right )}{\ln \left (2\right )}\right )\,{\mathrm {e}}^{x+5}\,\left (3\,{\mathrm {e}}^x+3\,x\,\ln \left (2\right )\right )}{5\,{\mathrm {e}}^{2\,x}+\ln \left (2\right )\,\left (3\,x^2+6\,x\right )+{\mathrm {e}}^x\,\left (3\,x+5\,x\,\ln \left (2\right )+6\right )+\ln \left (\frac {{\mathrm {e}}^x+x\,\ln \left (2\right )}{\ln \left (2\right )}\right )\,{\mathrm {e}}^{x+5}\,\left (3\,{\mathrm {e}}^x+3\,x\,\ln \left (2\right )\right )} \,d x \] Input:
int((5*exp(2*x) + exp(x + 5)*(3*log(2) + 3*exp(x)) + 3*x*log(2) + exp(x)*( 5*x*log(2) + 3) + log((exp(x) + x*log(2))/log(2))*exp(x + 5)*(3*exp(x) + 3 *x*log(2)))/(5*exp(2*x) + log(2)*(6*x + 3*x^2) + exp(x)*(3*x + 5*x*log(2) + 6) + log((exp(x) + x*log(2))/log(2))*exp(x + 5)*(3*exp(x) + 3*x*log(2))) ,x)
Output:
int((5*exp(2*x) + exp(x + 5)*(3*log(2) + 3*exp(x)) + 3*x*log(2) + exp(x)*( 5*x*log(2) + 3) + log((exp(x) + x*log(2))/log(2))*exp(x + 5)*(3*exp(x) + 3 *x*log(2)))/(5*exp(2*x) + log(2)*(6*x + 3*x^2) + exp(x)*(3*x + 5*x*log(2) + 6) + log((exp(x) + x*log(2))/log(2))*exp(x + 5)*(3*exp(x) + 3*x*log(2))) , x)
Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {5 e^{2 x}+3 x \log (2)+e^{5+x} \left (3 e^x+3 \log (2)\right )+e^x (3+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )}{5 e^{2 x}+\left (6 x+3 x^2\right ) \log (2)+e^x (6+3 x+5 x \log (2))+e^{5+x} \left (3 e^x+3 x \log (2)\right ) \log \left (\frac {e^x+x \log (2)}{\log (2)}\right )} \, dx=\mathrm {log}\left (3 e^{x} \mathrm {log}\left (\frac {e^{x}+\mathrm {log}\left (2\right ) x}{\mathrm {log}\left (2\right )}\right ) e^{5}+5 e^{x}+3 x +6\right ) \] Input:
int(((3*exp(x)+3*x*log(2))*exp(5+x)*log((exp(x)+x*log(2))/log(2))+(3*exp(x )+3*log(2))*exp(5+x)+5*exp(x)^2+(5*x*log(2)+3)*exp(x)+3*x*log(2))/((3*exp( x)+3*x*log(2))*exp(5+x)*log((exp(x)+x*log(2))/log(2))+5*exp(x)^2+(5*x*log( 2)+6+3*x)*exp(x)+(3*x^2+6*x)*log(2)),x)
Output:
log(3*e**x*log((e**x + log(2)*x)/log(2))*e**5 + 5*e**x + 3*x + 6)