Integrand size = 85, antiderivative size = 26 \[ \int \frac {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx=\frac {2-x}{-2-e^x+30 \left (2 x+x^2\right ) \log (5)} \] Output:
(2-x)/(30*(x^2+2*x)*ln(5)-exp(x)-2)
Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx=\frac {-2+x}{2+e^x-60 x \log (5)-30 x^2 \log (5)} \] Input:
Integrate[(2 + E^x*(3 - x) + (-120 - 120*x + 30*x^2)*Log[5])/(4 + E^(2*x) + (-240*x - 120*x^2)*Log[5] + (3600*x^2 + 3600*x^3 + 900*x^4)*Log[5]^2 + E ^x*(4 + (-120*x - 60*x^2)*Log[5])),x]
Output:
(-2 + x)/(2 + E^x - 60*x*Log[5] - 30*x^2*Log[5])
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.
\(\displaystyle \int \frac {\left (30 x^2-120 x-120\right ) \log (5)+e^x (3-x)+2}{\left (-120 x^2-240 x\right ) \log (5)+e^x \left (\left (-60 x^2-120 x\right ) \log (5)+4\right )+\left (900 x^4+3600 x^3+3600 x^2\right ) \log ^2(5)+e^{2 x}+4} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (15 x^2 \log (5)-60 x \log (5)+1-60 \log (5)\right )-e^x (x-3)}{\left (-30 x^2 \log (5)+e^x-60 x \log (5)+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x-3}{30 x^2 \log (5)-e^x+60 x \log (5)-2}-\frac {2 (x-2) \left (15 x^2 \log (5)-1-30 \log (5)\right )}{\left (30 x^2 \log (5)-e^x+60 x \log (5)-2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \int \frac {1}{-30 \log (5) x^2-60 \log (5) x+e^x+2}dx-4 (1+30 \log (5)) \int \frac {1}{\left (30 \log (5) x^2+60 \log (5) x-e^x-2\right )^2}dx+2 (1+30 \log (5)) \int \frac {x}{\left (30 \log (5) x^2+60 \log (5) x-e^x-2\right )^2}dx+60 \log (5) \int \frac {x^2}{\left (30 \log (5) x^2+60 \log (5) x-e^x-2\right )^2}dx+\int \frac {x}{30 \log (5) x^2+60 \log (5) x-e^x-2}dx-30 \log (5) \int \frac {x^3}{\left (30 \log (5) x^2+60 \log (5) x-e^x-2\right )^2}dx\) |
Input:
Int[(2 + E^x*(3 - x) + (-120 - 120*x + 30*x^2)*Log[5])/(4 + E^(2*x) + (-24 0*x - 120*x^2)*Log[5] + (3600*x^2 + 3600*x^3 + 900*x^4)*Log[5]^2 + E^x*(4 + (-120*x - 60*x^2)*Log[5])),x]
Output:
$Aborted
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {-2+x}{30 x^{2} \ln \left (5\right )+60 x \ln \left (5\right )-{\mathrm e}^{x}-2}\) | \(26\) |
parallelrisch | \(-\frac {-2+x}{30 x^{2} \ln \left (5\right )+60 x \ln \left (5\right )-{\mathrm e}^{x}-2}\) | \(26\) |
norman | \(\frac {2-x}{30 x^{2} \ln \left (5\right )+60 x \ln \left (5\right )-{\mathrm e}^{x}-2}\) | \(27\) |
Input:
int(((-x+3)*exp(x)+(30*x^2-120*x-120)*ln(5)+2)/(exp(x)^2+((-60*x^2-120*x)* ln(5)+4)*exp(x)+(900*x^4+3600*x^3+3600*x^2)*ln(5)^2+(-120*x^2-240*x)*ln(5) +4),x,method=_RETURNVERBOSE)
Output:
-(-2+x)/(30*x^2*ln(5)+60*x*ln(5)-exp(x)-2)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx=-\frac {x - 2}{30 \, {\left (x^{2} + 2 \, x\right )} \log \left (5\right ) - e^{x} - 2} \] Input:
integrate(((3-x)*exp(x)+(30*x^2-120*x-120)*log(5)+2)/(exp(x)^2+((-60*x^2-1 20*x)*log(5)+4)*exp(x)+(900*x^4+3600*x^3+3600*x^2)*log(5)^2+(-120*x^2-240* x)*log(5)+4),x, algorithm="fricas")
Output:
-(x - 2)/(30*(x^2 + 2*x)*log(5) - e^x - 2)
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx=\frac {x - 2}{- 30 x^{2} \log {\left (5 \right )} - 60 x \log {\left (5 \right )} + e^{x} + 2} \] Input:
integrate(((3-x)*exp(x)+(30*x**2-120*x-120)*ln(5)+2)/(exp(x)**2+((-60*x**2 -120*x)*ln(5)+4)*exp(x)+(900*x**4+3600*x**3+3600*x**2)*ln(5)**2+(-120*x**2 -240*x)*ln(5)+4),x)
Output:
(x - 2)/(-30*x**2*log(5) - 60*x*log(5) + exp(x) + 2)
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx=-\frac {x - 2}{30 \, x^{2} \log \left (5\right ) + 60 \, x \log \left (5\right ) - e^{x} - 2} \] Input:
integrate(((3-x)*exp(x)+(30*x^2-120*x-120)*log(5)+2)/(exp(x)^2+((-60*x^2-1 20*x)*log(5)+4)*exp(x)+(900*x^4+3600*x^3+3600*x^2)*log(5)^2+(-120*x^2-240* x)*log(5)+4),x, algorithm="maxima")
Output:
-(x - 2)/(30*x^2*log(5) + 60*x*log(5) - e^x - 2)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx=-\frac {x - 2}{30 \, x^{2} \log \left (5\right ) + 60 \, x \log \left (5\right ) - e^{x} - 2} \] Input:
integrate(((3-x)*exp(x)+(30*x^2-120*x-120)*log(5)+2)/(exp(x)^2+((-60*x^2-1 20*x)*log(5)+4)*exp(x)+(900*x^4+3600*x^3+3600*x^2)*log(5)^2+(-120*x^2-240* x)*log(5)+4),x, algorithm="giac")
Output:
-(x - 2)/(30*x^2*log(5) + 60*x*log(5) - e^x - 2)
Time = 0.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx=\frac {x-2}{{\mathrm {e}}^x-60\,x\,\ln \left (5\right )-30\,x^2\,\ln \left (5\right )+2} \] Input:
int(-(exp(x)*(x - 3) + log(5)*(120*x - 30*x^2 + 120) - 2)/(exp(2*x) + log( 5)^2*(3600*x^2 + 3600*x^3 + 900*x^4) - log(5)*(240*x + 120*x^2) - exp(x)*( log(5)*(120*x + 60*x^2) - 4) + 4),x)
Output:
(x - 2)/(exp(x) - 60*x*log(5) - 30*x^2*log(5) + 2)
Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx=\frac {x -2}{e^{x}-30 \,\mathrm {log}\left (5\right ) x^{2}-60 \,\mathrm {log}\left (5\right ) x +2} \] Input:
int(((3-x)*exp(x)+(30*x^2-120*x-120)*log(5)+2)/(exp(x)^2+((-60*x^2-120*x)* log(5)+4)*exp(x)+(900*x^4+3600*x^3+3600*x^2)*log(5)^2+(-120*x^2-240*x)*log (5)+4),x)
Output:
(x - 2)/(e**x - 30*log(5)*x**2 - 60*log(5)*x + 2)