Integrand size = 59, antiderivative size = 29 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-1+e^{e^{e^5}}-e^{2+\frac {x}{4+\frac {23+x}{\log (2)}}} \] Output:
exp(exp(exp(5)))-1-exp(x/(4+(x+23)/ln(2))+2)
Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-\frac {2^{\frac {x}{23+x+\log (16)}} e^2 \log (2) (23+\log (16))}{\log (2) \log (16)+\log (8388608)} \] Input:
Integrate[(E^((46 + 2*x + (8 + x)*Log[2])/(23 + x + 4*Log[2]))*(-23*Log[2] - 4*Log[2]^2))/(529 + 46*x + x^2 + (184 + 8*x)*Log[2] + 16*Log[2]^2),x]
Output:
-((2^(x/(23 + x + Log[16]))*E^2*Log[2]*(23 + Log[16]))/(Log[2]*Log[16] + L og[8388608]))
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.
\(\displaystyle \int \frac {\left (-4 \log ^2(2)-23 \log (2)\right ) e^{\frac {2 x+(x+8) \log (2)+46}{x+23+4 \log (2)}}}{x^2+46 x+(8 x+184) \log (2)+529+16 \log ^2(2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\log (2) (23+\log (16)) \int \frac {2^{\frac {x+8}{x+4 \log (2)+23}} e^{\frac {2 (x+23)}{x+\log (16)+23}}}{x^2+46 x+16 \log ^2(2)+8 (x+23) \log (2)+529}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle -\log (2) (23+\log (16)) \int \frac {2^{\frac {x+8}{x+4 \log (2)+23}} e^{\frac {2 (x+23)}{x+\log (16)+23}}}{(x+\log (16)+23)^2}dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle -\log (2) (23+\log (16)) \int \frac {2^{\frac {x+8}{x+4 \log (2)+23}} e^{\frac {2 (x+23)}{x+\log (16)+23}}}{(x+\log (16)+23)^2}dx\) |
Input:
Int[(E^((46 + 2*x + (8 + x)*Log[2])/(23 + x + 4*Log[2]))*(-23*Log[2] - 4*L og[2]^2))/(529 + 46*x + x^2 + (184 + 8*x)*Log[2] + 16*Log[2]^2),x]
Output:
$Aborted
Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
gosper | \(-{\mathrm e}^{\frac {x \ln \left (2\right )+8 \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}\) | \(27\) |
parallelrisch | \(\frac {\left (-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )\right ) {\mathrm e}^{\frac {\left (x +8\right ) \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}}{\ln \left (2\right ) \left (4 \ln \left (2\right )+23\right )}\) | \(47\) |
default | \(-\frac {\left (-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )\right )^{2} {\mathrm e}^{\frac {-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )}{4 \ln \left (2\right )+x +23}+\ln \left (2\right )+2}}{\ln \left (2\right )^{2} \left (16 \ln \left (2\right )^{2}+184 \ln \left (2\right )+529\right )}\) | \(60\) |
norman | \(\frac {\left (-4 \ln \left (2\right )-23\right ) {\mathrm e}^{\frac {\left (x +8\right ) \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}-x \,{\mathrm e}^{\frac {\left (x +8\right ) \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}}{4 \ln \left (2\right )+x +23}\) | \(66\) |
derivativedivides | \(\frac {\left (4 \ln \left (2\right )^{2}+23 \ln \left (2\right )\right ) \left (-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )\right ) {\mathrm e}^{\frac {-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )}{4 \ln \left (2\right )+x +23}+\ln \left (2\right )+2}}{\ln \left (2\right )^{2} \left (16 \ln \left (2\right )^{2}+184 \ln \left (2\right )+529\right )}\) | \(68\) |
risch | \(-\frac {4 \ln \left (2\right ) {\mathrm e}^{\frac {x \ln \left (2\right )+8 \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}}{4 \ln \left (2\right )+23}-\frac {23 \,{\mathrm e}^{\frac {x \ln \left (2\right )+8 \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}}{4 \ln \left (2\right )+23}\) | \(72\) |
Input:
int((-4*ln(2)^2-23*ln(2))*exp(((x+8)*ln(2)+2*x+46)/(4*ln(2)+x+23))/(16*ln( 2)^2+(8*x+184)*ln(2)+x^2+46*x+529),x,method=_RETURNVERBOSE)
Output:
-exp((x*ln(2)+8*ln(2)+2*x+46)/(4*ln(2)+x+23))
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-e^{\left (\frac {{\left (x + 8\right )} \log \left (2\right ) + 2 \, x + 46}{x + 4 \, \log \left (2\right ) + 23}\right )} \] Input:
integrate((-4*log(2)^2-23*log(2))*exp(((x+8)*log(2)+2*x+46)/(4*log(2)+x+23 ))/(16*log(2)^2+(8*x+184)*log(2)+x^2+46*x+529),x, algorithm="fricas")
Output:
-e^(((x + 8)*log(2) + 2*x + 46)/(x + 4*log(2) + 23))
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=- e^{\frac {2 x + \left (x + 8\right ) \log {\left (2 \right )} + 46}{x + 4 \log {\left (2 \right )} + 23}} \] Input:
integrate((-4*ln(2)**2-23*ln(2))*exp(((x+8)*ln(2)+2*x+46)/(4*ln(2)+x+23))/ (16*ln(2)**2+(8*x+184)*ln(2)+x**2+46*x+529),x)
Output:
-exp((2*x + (x + 8)*log(2) + 46)/(x + 4*log(2) + 23))
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-2 \, e^{\left (-\frac {4 \, \log \left (2\right )^{2}}{x + 4 \, \log \left (2\right ) + 23} - \frac {23 \, \log \left (2\right )}{x + 4 \, \log \left (2\right ) + 23} + 2\right )} \] Input:
integrate((-4*log(2)^2-23*log(2))*exp(((x+8)*log(2)+2*x+46)/(4*log(2)+x+23 ))/(16*log(2)^2+(8*x+184)*log(2)+x^2+46*x+529),x, algorithm="maxima")
Output:
-2*e^(-4*log(2)^2/(x + 4*log(2) + 23) - 23*log(2)/(x + 4*log(2) + 23) + 2)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-e^{\left (\frac {x \log \left (2\right )}{x + 4 \, \log \left (2\right ) + 23} + \frac {2 \, x}{x + 4 \, \log \left (2\right ) + 23} + \frac {8 \, \log \left (2\right )}{x + 4 \, \log \left (2\right ) + 23} + \frac {46}{x + 4 \, \log \left (2\right ) + 23}\right )} \] Input:
integrate((-4*log(2)^2-23*log(2))*exp(((x+8)*log(2)+2*x+46)/(4*log(2)+x+23 ))/(16*log(2)^2+(8*x+184)*log(2)+x^2+46*x+529),x, algorithm="giac")
Output:
-e^(x*log(2)/(x + 4*log(2) + 23) + 2*x/(x + 4*log(2) + 23) + 8*log(2)/(x + 4*log(2) + 23) + 46/(x + 4*log(2) + 23))
Time = 3.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-2^{\frac {x+8}{x+\ln \left (16\right )+23}}\,{\mathrm {e}}^{\frac {2\,x}{x+\ln \left (16\right )+23}+\frac {46}{x+\ln \left (16\right )+23}} \] Input:
int(-(exp((2*x + log(2)*(x + 8) + 46)/(x + 4*log(2) + 23))*(23*log(2) + 4* log(2)^2))/(46*x + log(2)*(8*x + 184) + 16*log(2)^2 + x^2 + 529),x)
Output:
-2^((x + 8)/(x + log(16) + 23))*exp((2*x)/(x + log(16) + 23) + 46/(x + log (16) + 23))
Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-e^{\frac {\mathrm {log}\left (2\right ) x}{4 \,\mathrm {log}\left (2\right )+x +23}} e^{2} \] Input:
int((-4*log(2)^2-23*log(2))*exp(((x+8)*log(2)+2*x+46)/(4*log(2)+x+23))/(16 *log(2)^2+(8*x+184)*log(2)+x^2+46*x+529),x)
Output:
- e**((log(2)*x)/(4*log(2) + x + 23))*e**2