Integrand size = 68, antiderivative size = 24 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 e^4}{\log \left (\frac {2}{1+\frac {x}{5+e^x+x}}\right )} \]
Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 e^4}{\log \left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \]
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]
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 |
\(\displaystyle 5 e^4 \left (4 \int \frac {1}{\left (x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx+\int \frac {x}{\left (x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx-3 \int \frac {1}{\left (2 x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx-2 \int \frac {x}{\left (2 x+e^x+5\right ) \log ^2\left (\frac {2 \left (x+e^x+5\right )}{2 x+e^x+5}\right )}dx\right )\) |
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]
3.15.70.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.47 (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 | \(\frac {10 i {\mathrm e}^{4}}{\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5+x \right )\right ) \operatorname {csgn}\left (\frac {i}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )}^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )}^{3}+2 i \ln \left ({\mathrm e}^{x}+5+x \right )-2 i \ln \left (\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x \right )}\) | \(161\) |
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)
Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 \, e^{4}}{\log \left (\frac {2 \, {\left ({\left (x + 5\right )} e^{4} + e^{\left (x + 4\right )}\right )}}{{\left (2 \, x + 5\right )} e^{4} + e^{\left (x + 4\right )}}\right )} \]
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=\
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 e^{4}}{\log {\left (\frac {2 x + 2 e^{x} + 10}{2 x + e^{x} + 5} \right )}} \]
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)
Time = 0.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 \, e^{4}}{\log \left (2\right ) - \log \left (2 \, x + e^{x} + 5\right ) + \log \left (x + e^{x} + 5\right )} \]
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=\
Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5 \, e^{4}}{\log \left (2\right ) - \log \left (2 \, x + e^{x} + 5\right ) + \log \left (x + e^{x} + 5\right )} \]
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=\
Time = 12.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx=\frac {5\,{\mathrm {e}}^4}{\ln \left (\frac {2\,x+2\,{\mathrm {e}}^x+10}{2\,x+{\mathrm {e}}^x+5}\right )} \]