Integrand size = 68, antiderivative size = 24 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 e^4}{\log \left (\frac {2}{1+\frac {x}{5+e^x+x}}\right )} \] Output:
5*exp(4)/ln(2/(1+x/(exp(x)+5+x)))
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 e^4}{\log \left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \] Input:
Integrate[(25*E^4 + E^(4 + x)*(5 - 5*x))/((25 + E^(2*x) + 15*x + 2*x^2 + E ^x*(10 + 3*x))*Log[(10 + 2*E^x + 2*x)/(5 + E^x + 2*x)]^2),x]
Output:
(5*E^4)/Log[(2*(5 + E^x + x))/(5 + E^x + 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.
\(\displaystyle \int \frac {e^{x+4} (5-5 x)+25 e^4}{\left (2 x^2+15 x+e^{2 x}+e^x (3 x+10)+25\right ) \log ^2\left (\frac {2 x+2 e^x+10}{2 x+e^x+5}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 e^4 \left (-e^x x+e^x+5\right )}{\left (2 x^2+15 x+e^{2 x}+e^x (3 x+10)+25\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 e^4 \int \frac {-e^x x+e^x+5}{\left (2 x^2+15 x+e^{2 x}+e^x (3 x+10)+25\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 5 e^4 \int \left (\frac {x+4}{\left (x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}-\frac {2 x+3}{\left (2 x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 e^4 \left (4 \int \frac {1}{\left (x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx+\int \frac {x}{\left (x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx-3 \int \frac {1}{\left (2 x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx-2 \int \frac {x}{\left (2 x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx\right )\) |
Input:
Int[(25*E^4 + E^(4 + x)*(5 - 5*x))/((25 + E^(2*x) + 15*x + 2*x^2 + E^x*(10 + 3*x))*Log[(10 + 2*E^x + 2*x)/(5 + E^x + 2*x)]^2),x]
Output:
$Aborted
Time = 0.54 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {5 \,{\mathrm e}^{4}}{\ln \left (\frac {2 \,{\mathrm e}^{x}+2 x +10}{{\mathrm e}^{x}+5+2 x}\right )}\) | \(24\) |
risch | \(\frac {10 i {\mathrm e}^{4}}{\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5+x \right )\right ) \operatorname {csgn}\left (\frac {i}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )}^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )}^{3}+2 i \ln \left ({\mathrm e}^{x}+5+x \right )-2 i \ln \left (\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x \right )}\) | \(161\) |
Input:
int(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x^2+15* x+25)/ln((2*exp(x)+2*x+10)/(exp(x)+5+2*x))^2,x,method=_RETURNVERBOSE)
Output:
5*exp(4)/ln(2*(exp(x)+5+x)/(exp(x)+5+2*x))
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 \, e^{4}}{\log \left (\frac {2 \, {\left ({\left (x + 5\right )} e^{4} + e^{\left (x + 4\right )}\right )}}{{\left (2 \, x + 5\right )} e^{4} + e^{\left (x + 4\right )}}\right )} \] Input:
integrate(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x ^2+15*x+25)/log((2*exp(x)+2*x+10)/(exp(x)+5+2*x))^2,x, algorithm="fricas")
Output:
5*e^4/log(2*((x + 5)*e^4 + e^(x + 4))/((2*x + 5)*e^4 + e^(x + 4)))
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 e^{4}}{\log {\left (\frac {2 x + 2 e^{x} + 10}{2 x + e^{x} + 5} \right )}} \] Input:
integrate(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)**2+(3*x+10)*exp(x)+2* x**2+15*x+25)/ln((2*exp(x)+2*x+10)/(exp(x)+5+2*x))**2,x)
Output:
5*exp(4)/log((2*x + 2*exp(x) + 10)/(2*x + exp(x) + 5))
Time = 0.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 \, e^{4}}{\log \left (2\right ) - \log \left (2 \, x + e^{x} + 5\right ) + \log \left (x + e^{x} + 5\right )} \] Input:
integrate(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x ^2+15*x+25)/log((2*exp(x)+2*x+10)/(exp(x)+5+2*x))^2,x, algorithm="maxima")
Output:
5*e^4/(log(2) - log(2*x + e^x + 5) + log(x + e^x + 5))
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 \, e^{4}}{\log \left (2\right ) - \log \left (2 \, x + e^{x} + 5\right ) + \log \left (x + e^{x} + 5\right )} \] Input:
integrate(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x ^2+15*x+25)/log((2*exp(x)+2*x+10)/(exp(x)+5+2*x))^2,x, algorithm="giac")
Output:
5*e^4/(log(2) - log(2*x + e^x + 5) + log(x + e^x + 5))
Time = 4.82 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5\,{\mathrm {e}}^4}{\ln \left (\frac {2\,x+2\,{\mathrm {e}}^x+10}{2\,x+{\mathrm {e}}^x+5}\right )} \] Input:
int((25*exp(4) - exp(4)*exp(x)*(5*x - 5))/(log((2*x + 2*exp(x) + 10)/(2*x + exp(x) + 5))^2*(15*x + exp(2*x) + exp(x)*(3*x + 10) + 2*x^2 + 25)),x)
Output:
(5*exp(4))/log((2*x + 2*exp(x) + 10)/(2*x + exp(x) + 5))
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 e^{4}}{\mathrm {log}\left (\frac {2 e^{x}+2 x +10}{e^{x}+2 x +5}\right )} \] Input:
int(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x^2+15* x+25)/log((2*exp(x)+2*x+10)/(exp(x)+5+2*x))^2,x)
Output:
(5*e**4)/log((2*e**x + 2*x + 10)/(e**x + 2*x + 5))