Integrand size = 60, antiderivative size = 26 \[ \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx=\frac {25 \log (5) (x+9 \log (5))}{\log ^2\left (\frac {3}{1+\frac {e^x}{4}}\right )} \] Output:
25*ln(5)/ln(3/(1+1/4*exp(x)))^2*(9*ln(5)+x)
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx=\frac {25 \log (5) (x+9 \log (5))}{\log ^2\left (\frac {12}{4+e^x}\right )} \] Input:
Integrate[(E^x*(50*x*Log[5] + 450*Log[5]^2) + (100*Log[5] + 25*E^x*Log[5]) *Log[12/(4 + E^x)])/((4 + E^x)*Log[12/(4 + E^x)]^3),x]
Output:
(25*Log[5]*(x + 9*Log[5]))/Log[12/(4 + E^x)]^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.
\(\displaystyle \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (25 e^x \log (5)+100 \log (5)\right ) \log \left (\frac {12}{e^x+4}\right )}{\left (e^x+4\right ) \log ^3\left (\frac {12}{e^x+4}\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {25 \log (5) \left (2 x+\log \left (\frac {12}{e^x+4}\right )+18 \log (5)\right )}{\log ^3\left (\frac {12}{e^x+4}\right )}-\frac {200 \log (5) (x+9 \log (5))}{\left (e^x+4\right ) \log ^3\left (\frac {12}{e^x+4}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 25 \log (5) \text {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {12}{x+4}\right )}dx,x,e^x\right )+50 \log (5) \int \frac {x}{\log ^3\left (\frac {12}{4+e^x}\right )}dx-200 \log (5) \int \frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )}dx+\frac {225 \log ^2(5)}{\log ^2\left (\frac {12}{e^x+4}\right )}\) |
Input:
Int[(E^x*(50*x*Log[5] + 450*Log[5]^2) + (100*Log[5] + 25*E^x*Log[5])*Log[1 2/(4 + E^x)])/((4 + E^x)*Log[12/(4 + E^x)]^3),x]
Output:
$Aborted
Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {25 x \ln \left (5\right )+225 \ln \left (5\right )^{2}}{\ln \left (\frac {12}{{\mathrm e}^{x}+4}\right )^{2}}\) | \(25\) |
norman | \(\frac {25 x \ln \left (5\right )+225 \ln \left (5\right )^{2}}{\ln \left (\frac {12}{{\mathrm e}^{x}+4}\right )^{2}}\) | \(25\) |
parallelrisch | \(\frac {25 x \ln \left (5\right )+225 \ln \left (5\right )^{2}}{\ln \left (\frac {12}{{\mathrm e}^{x}+4}\right )^{2}}\) | \(25\) |
risch | \(\frac {100 \ln \left (5\right ) \left (9 \ln \left (5\right )+x \right )}{{\left (4 \ln \left (2\right )+2 \ln \left (3\right )-2 \ln \left ({\mathrm e}^{x}+4\right )\right )}^{2}}\) | \(29\) |
Input:
int(((25*exp(x)*ln(5)+100*ln(5))*ln(12/(exp(x)+4))+(450*ln(5)^2+50*x*ln(5) )*exp(x))/(exp(x)+4)/ln(12/(exp(x)+4))^3,x,method=_RETURNVERBOSE)
Output:
(25*x*ln(5)+225*ln(5)^2)/ln(12/(exp(x)+4))^2
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx=\frac {25 \, {\left (x \log \left (5\right ) + 9 \, \log \left (5\right )^{2}\right )}}{\log \left (\frac {12}{e^{x} + 4}\right )^{2}} \] Input:
integrate(((25*exp(x)*log(5)+100*log(5))*log(12/(exp(x)+4))+(450*log(5)^2+ 50*x*log(5))*exp(x))/(exp(x)+4)/log(12/(exp(x)+4))^3,x, algorithm="fricas" )
Output:
25*(x*log(5) + 9*log(5)^2)/log(12/(e^x + 4))^2
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx=\frac {25 x \log {\left (5 \right )} + 225 \log {\left (5 \right )}^{2}}{\log {\left (\frac {12}{e^{x} + 4} \right )}^{2}} \] Input:
integrate(((25*exp(x)*ln(5)+100*ln(5))*ln(12/(exp(x)+4))+(450*ln(5)**2+50* x*ln(5))*exp(x))/(exp(x)+4)/ln(12/(exp(x)+4))**3,x)
Output:
(25*x*log(5) + 225*log(5)**2)/log(12/(exp(x) + 4))**2
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (21) = 42\).
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.54 \[ \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx=\frac {25 \, x \log \left (5\right )}{\log \left (3\right )^{2} + 4 \, \log \left (3\right ) \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 2 \, {\left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )} \log \left (e^{x} + 4\right ) + \log \left (e^{x} + 4\right )^{2}} + \frac {225 \, \log \left (5\right )^{2}}{\log \left (3\right )^{2} + 4 \, \log \left (3\right ) \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 2 \, {\left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )} \log \left (e^{x} + 4\right ) + \log \left (e^{x} + 4\right )^{2}} \] Input:
integrate(((25*exp(x)*log(5)+100*log(5))*log(12/(exp(x)+4))+(450*log(5)^2+ 50*x*log(5))*exp(x))/(exp(x)+4)/log(12/(exp(x)+4))^3,x, algorithm="maxima" )
Output:
25*x*log(5)/(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2 - 2*(log(3) + 2*log(2 ))*log(e^x + 4) + log(e^x + 4)^2) + 225*log(5)^2/(log(3)^2 + 4*log(3)*log( 2) + 4*log(2)^2 - 2*(log(3) + 2*log(2))*log(e^x + 4) + log(e^x + 4)^2)
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx=\frac {25 \, {\left (x \log \left (5\right ) + 9 \, \log \left (5\right )^{2}\right )}}{\log \left (12\right )^{2} - 2 \, \log \left (12\right ) \log \left (e^{x} + 4\right ) + \log \left (e^{x} + 4\right )^{2}} \] Input:
integrate(((25*exp(x)*log(5)+100*log(5))*log(12/(exp(x)+4))+(450*log(5)^2+ 50*x*log(5))*exp(x))/(exp(x)+4)/log(12/(exp(x)+4))^3,x, algorithm="giac")
Output:
25*(x*log(5) + 9*log(5)^2)/(log(12)^2 - 2*log(12)*log(e^x + 4) + log(e^x + 4)^2)
Time = 3.65 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx=\frac {25\,\ln \left (5\right )\,\left (x+9\,\ln \left (5\right )\right )}{{\ln \left (\frac {12}{{\mathrm {e}}^x+4}\right )}^2} \] Input:
int((exp(x)*(50*x*log(5) + 450*log(5)^2) + log(12/(exp(x) + 4))*(100*log(5 ) + 25*exp(x)*log(5)))/(log(12/(exp(x) + 4))^3*(exp(x) + 4)),x)
Output:
(25*log(5)*(x + 9*log(5)))/log(12/(exp(x) + 4))^2
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx=\frac {25 \,\mathrm {log}\left (5\right ) \left (9 \,\mathrm {log}\left (5\right )+x \right )}{\mathrm {log}\left (\frac {12}{e^{x}+4}\right )^{2}} \] Input:
int(((25*exp(x)*log(5)+100*log(5))*log(12/(exp(x)+4))+(450*log(5)^2+50*x*l og(5))*exp(x))/(exp(x)+4)/log(12/(exp(x)+4))^3,x)
Output:
(25*log(5)*(9*log(5) + x))/log(12/(e**x + 4))**2