Integrand size = 64, antiderivative size = 24 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {18 \left (2+\frac {1}{16} \left (-3+x+\log ^2\left (2+e^{15+x}\right )\right )\right )}{x} \] Output:
18*(29/16+1/16*ln(exp(x+15)+2)^2+1/16*x)/x
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9}{8} \left (\frac {29}{x}+\frac {\log ^2\left (2+e^{15+x}\right )}{x}\right ) \] Input:
Integrate[(-522 - 261*E^(15 + x) + 18*E^(15 + x)*x*Log[2 + E^(15 + x)] + ( -18 - 9*E^(15 + x))*Log[2 + E^(15 + x)]^2)/(16*x^2 + 8*E^(15 + x)*x^2),x]
Output:
(9*(29/x + Log[2 + E^(15 + x)]^2/x))/8
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.
\(\displaystyle \int \frac {-261 e^{x+15}+\left (-9 e^{x+15}-18\right ) \log ^2\left (e^{x+15}+2\right )+18 e^{x+15} x \log \left (e^{x+15}+2\right )-522}{8 e^{x+15} x^2+16 x^2} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-261 e^{x+15}+\left (-9 e^{x+15}-18\right ) \log ^2\left (e^{x+15}+2\right )+18 e^{x+15} x \log \left (e^{x+15}+2\right )-522}{8 \left (e^{x+15}+2\right ) x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int -\frac {9 \left (\left (2+e^{x+15}\right ) \log ^2\left (2+e^{x+15}\right )-2 e^{x+15} x \log \left (2+e^{x+15}\right )+29 e^{x+15}+58\right )}{\left (2+e^{x+15}\right ) x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {9}{8} \int \frac {\left (2+e^{x+15}\right ) \log ^2\left (2+e^{x+15}\right )-2 e^{x+15} x \log \left (2+e^{x+15}\right )+29 e^{x+15}+58}{\left (2+e^{x+15}\right ) x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {9}{8} \int \left (\frac {4 \log \left (2+e^{x+15}\right )}{\left (2+e^{x+15}\right ) x}+\frac {\log ^2\left (2+e^{x+15}\right )-2 x \log \left (2+e^{x+15}\right )+29}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {9}{8} \left (\int \frac {\log ^2\left (2+e^{x+15}\right )}{x^2}dx-4 \int \frac {e^{x+15} \int \frac {1}{e^{x+15} x+2 x}dx}{2+e^{x+15}}dx+4 \log \left (e^{x+15}+2\right ) \int \frac {1}{\left (2+e^{x+15}\right ) x}dx-2 \int \frac {\log \left (2+e^{x+15}\right )}{x}dx-\frac {29}{x}\right )\) |
Input:
Int[(-522 - 261*E^(15 + x) + 18*E^(15 + x)*x*Log[2 + E^(15 + x)] + (-18 - 9*E^(15 + x))*Log[2 + E^(15 + x)]^2)/(16*x^2 + 8*E^(15 + x)*x^2),x]
Output:
$Aborted
Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75
method | result | size |
norman | \(\frac {\frac {261}{8}+\frac {9 \ln \left ({\mathrm e}^{x +15}+2\right )^{2}}{8}}{x}\) | \(18\) |
parallelrisch | \(\frac {261+9 \ln \left ({\mathrm e}^{x +15}+2\right )^{2}}{8 x}\) | \(19\) |
risch | \(\frac {9 \ln \left ({\mathrm e}^{x +15}+2\right )^{2}}{8 x}+\frac {261}{8 x}\) | \(21\) |
Input:
int(((-9*exp(x+15)-18)*ln(exp(x+15)+2)^2+18*x*exp(x+15)*ln(exp(x+15)+2)-26 1*exp(x+15)-522)/(8*x^2*exp(x+15)+16*x^2),x,method=_RETURNVERBOSE)
Output:
(261/8+9/8*ln(exp(x+15)+2)^2)/x
Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \] Input:
integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+1 5)+2)-261*exp(x+15)-522)/(8*x^2*exp(x+15)+16*x^2),x, algorithm="fricas")
Output:
9/8*(log(e^(x + 15) + 2)^2 + 29)/x
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9 \log {\left (e^{x + 15} + 2 \right )}^{2}}{8 x} + \frac {261}{8 x} \] Input:
integrate(((-9*exp(x+15)-18)*ln(exp(x+15)+2)**2+18*x*exp(x+15)*ln(exp(x+15 )+2)-261*exp(x+15)-522)/(8*x**2*exp(x+15)+16*x**2),x)
Output:
9*log(exp(x + 15) + 2)**2/(8*x) + 261/(8*x)
Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \] Input:
integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+1 5)+2)-261*exp(x+15)-522)/(8*x^2*exp(x+15)+16*x^2),x, algorithm="maxima")
Output:
9/8*(log(e^(x + 15) + 2)^2 + 29)/x
Time = 0.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \] Input:
integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+1 5)+2)-261*exp(x+15)-522)/(8*x^2*exp(x+15)+16*x^2),x, algorithm="giac")
Output:
9/8*(log(e^(x + 15) + 2)^2 + 29)/x
Time = 0.64 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9\,\left ({\ln \left ({\mathrm {e}}^{15}\,{\mathrm {e}}^x+2\right )}^2+29\right )}{8\,x} \] Input:
int(-(261*exp(x + 15) + log(exp(x + 15) + 2)^2*(9*exp(x + 15) + 18) - 18*x *log(exp(x + 15) + 2)*exp(x + 15) + 522)/(8*x^2*exp(x + 15) + 16*x^2),x)
Output:
(9*(log(exp(15)*exp(x) + 2)^2 + 29))/(8*x)
Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {\frac {9 \mathrm {log}\left (e^{x} e^{15}+2\right )^{2}}{8}+\frac {261}{8}}{x} \] Input:
int(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+15)+2)- 261*exp(x+15)-522)/(8*x^2*exp(x+15)+16*x^2),x)
Output:
(9*(log(e**x*e**15 + 2)**2 + 29))/(8*x)