Integrand size = 89, antiderivative size = 24 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {1}{5} e^{\frac {x^2}{x+\frac {2}{x^2+\log (5)}}} \] Output:
1/5*exp(x^2/(2/(x^2+ln(5))+x))
\[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx \] Input:
Integrate[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x^4)*Log[5] + x^2*Log[5]^2))/(20 + 20*x^3 + 5*x^6 + (20*x + 10*x^4)*L og[5] + 5*x^2*Log[5]^2),x]
Output:
Integrate[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x^4)*Log[5] + x^2*Log[5]^2))/(20 + 20*x^3 + 5*x^6 + (20*x + 10*x^4)*L og[5] + 5*x^2*Log[5]^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.
\(\displaystyle \int \frac {e^{\frac {x^4+x^2 \log (5)}{x^3+x \log (5)+2}} \left (x^6+\left (2 x^4+4 x\right ) \log (5)+8 x^3+x^2 \log ^2(5)\right )}{5 x^6+\left (10 x^4+20 x\right ) \log (5)+20 x^3+5 x^2 \log ^2(5)+20} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \frac {e^{\frac {x^4+x^2 \log (5)}{x^3+x \log (5)+2}} \left (x^6+\left (2 x^4+4 x\right ) \log (5)+8 x^3+x^2 \log ^2(5)\right )}{5 \left (x^3+x \log (5)+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}} \left (x^6+8 x^3+\log ^2(5) x^2+2 \left (x^4+2 x\right ) \log (5)\right )}{\left (x^3+\log (5) x+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (-\frac {4\ 5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}} (\log (5) x+3)}{\left (x^3+\log (5) x+2\right )^2}+5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}+\frac {4\ 5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}}{x^3+\log (5) x+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\int 5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}dx-12 \int \frac {5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}}{\left (x^3+\log (5) x+2\right )^2}dx-4 \log (5) \int \frac {5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}} x}{\left (x^3+\log (5) x+2\right )^2}dx+4 \int \frac {5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}}{x^3+\log (5) x+2}dx\right )\) |
Input:
Int[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x ^4)*Log[5] + x^2*Log[5]^2))/(20 + 20*x^3 + 5*x^6 + (20*x + 10*x^4)*Log[5] + 5*x^2*Log[5]^2),x]
Output:
$Aborted
Time = 0.74 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {{\mathrm e}^{\frac {x^{2} \left (x^{2}+\ln \left (5\right )\right )}{x \ln \left (5\right )+x^{3}+2}}}{5}\) | \(25\) |
risch | \(\frac {{\mathrm e}^{\frac {x^{2} \left (x^{2}+\ln \left (5\right )\right )}{x \ln \left (5\right )+x^{3}+2}}}{5}\) | \(25\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {x^{2} \left (x^{2}+\ln \left (5\right )\right )}{x \ln \left (5\right )+x^{3}+2}}}{5}\) | \(25\) |
norman | \(\frac {\frac {x^{3} {\mathrm e}^{\frac {x^{2} \ln \left (5\right )+x^{4}}{x \ln \left (5\right )+x^{3}+2}}}{5}+\frac {x \ln \left (5\right ) {\mathrm e}^{\frac {x^{2} \ln \left (5\right )+x^{4}}{x \ln \left (5\right )+x^{3}+2}}}{5}+\frac {2 \,{\mathrm e}^{\frac {x^{2} \ln \left (5\right )+x^{4}}{x \ln \left (5\right )+x^{3}+2}}}{5}}{x \ln \left (5\right )+x^{3}+2}\) | \(95\) |
Input:
int((x^2*ln(5)^2+(2*x^4+4*x)*ln(5)+x^6+8*x^3)*exp((x^2*ln(5)+x^4)/(x*ln(5) +x^3+2))/(5*x^2*ln(5)^2+(10*x^4+20*x)*ln(5)+5*x^6+20*x^3+20),x,method=_RET URNVERBOSE)
Output:
1/5*exp(x^2*(x^2+ln(5))/(x*ln(5)+x^3+2))
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {1}{5} \, e^{\left (\frac {x^{4} + x^{2} \log \left (5\right )}{x^{3} + x \log \left (5\right ) + 2}\right )} \] Input:
integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4) /(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x , algorithm="fricas")
Output:
1/5*e^((x^4 + x^2*log(5))/(x^3 + x*log(5) + 2))
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {e^{\frac {x^{4} + x^{2} \log {\left (5 \right )}}{x^{3} + x \log {\left (5 \right )} + 2}}}{5} \] Input:
integrate((x**2*ln(5)**2+(2*x**4+4*x)*ln(5)+x**6+8*x**3)*exp((x**2*ln(5)+x **4)/(x*ln(5)+x**3+2))/(5*x**2*ln(5)**2+(10*x**4+20*x)*ln(5)+5*x**6+20*x** 3+20),x)
Output:
exp((x**4 + x**2*log(5))/(x**3 + x*log(5) + 2))/5
Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {1}{5} \, e^{\left (x - \frac {2 \, x}{x^{3} + x \log \left (5\right ) + 2}\right )} \] Input:
integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4) /(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x , algorithm="maxima")
Output:
1/5*e^(x - 2*x/(x^3 + x*log(5) + 2))
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {1}{5} \, e^{\left (\frac {x^{4}}{x^{3} + x \log \left (5\right ) + 2} + \frac {x^{2} \log \left (5\right )}{x^{3} + x \log \left (5\right ) + 2}\right )} \] Input:
integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4) /(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x , algorithm="giac")
Output:
1/5*e^(x^4/(x^3 + x*log(5) + 2) + x^2*log(5)/(x^3 + x*log(5) + 2))
Time = 4.67 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {{\mathrm {e}}^{\frac {x^4}{x^3+\ln \left (5\right )\,x+2}}\,{\mathrm {e}}^{\frac {x^2\,\ln \left (5\right )}{x^3+\ln \left (5\right )\,x+2}}}{5} \] Input:
int((exp((x^2*log(5) + x^4)/(x*log(5) + x^3 + 2))*(x^2*log(5)^2 + log(5)*( 4*x + 2*x^4) + 8*x^3 + x^6))/(5*x^2*log(5)^2 + log(5)*(20*x + 10*x^4) + 20 *x^3 + 5*x^6 + 20),x)
Output:
(exp(x^4/(x*log(5) + x^3 + 2))*exp((x^2*log(5))/(x*log(5) + x^3 + 2)))/5
Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {e^{x}}{5 e^{\frac {2 x}{\mathrm {log}\left (5\right ) x +x^{3}+2}}} \] Input:
int((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4)/(x*lo g(5)+x^3+2))/(5*x^2*log(5)^2+(10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x)
Output:
e**x/(5*e**((2*x)/(log(5)*x + x**3 + 2)))