Integrand size = 147, antiderivative size = 29 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=\frac {8 x}{5 (2+x) \left (1-x+\frac {e^x}{\log (4)}-\log (4)\right )} \] Output:
8/5*x/(1/2*exp(x)/ln(2)-x+1-2*ln(2))/(2+x)
Time = 1.55 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\frac {8 x \left (x^2 \log (4)+2 \log ^2(4)+x \log ^2(4)-\log (256)\right )}{5 (2+x)^2 (-2+x+\log (4)) \left (-e^x+\log (4) (-1+x+\log (4))\right )} \] Input:
Integrate[(E^x*(16 - 16*x - 8*x^2)*Log[4] + (16 + 8*x^2)*Log[4]^2 - 16*Log [4]^3)/(E^(2*x)*(20 + 20*x + 5*x^2) + (20 - 20*x - 15*x^2 + 10*x^3 + 5*x^4 )*Log[4]^2 + (-40 + 30*x^2 + 10*x^3)*Log[4]^3 + (20 + 20*x + 5*x^2)*Log[4] ^4 + E^x*((40 - 30*x^2 - 10*x^3)*Log[4] + (-40 - 40*x - 10*x^2)*Log[4]^2)) ,x]
Output:
(-8*x*(x^2*Log[4] + 2*Log[4]^2 + x*Log[4]^2 - Log[256]))/(5*(2 + x)^2*(-2 + x + Log[4])*(-E^x + Log[4]*(-1 + x + Log[4])))
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.
| \(\displaystyle \int \frac {\left (8 x^2+16\right ) \log ^2(4)+e^x \left (-8 x^2-16 x+16\right ) \log (4)-16 \log ^3(4)}{e^{2 x} \left (5 x^2+20 x+20\right )+\left (5 x^2+20 x+20\right ) \log ^4(4)+\left (10 x^3+30 x^2-40\right ) \log ^3(4)+e^x \left (\left (-10 x^2-40 x-40\right ) \log ^2(4)+\left (-10 x^3-30 x^2+40\right ) \log (4)\right )+\left (5 x^4+10 x^3-15 x^2-20 x+20\right ) \log ^2(4)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {8 \log (4) \left (-e^x \left (x^2+2 x-2\right )+x^2 \log (4)-2 \log ^2(4) \left (1-\frac {\log (16)}{2 \log ^2(4)}\right )\right )}{5 (x+2)^2 \left (e^x-\log (4) (x-1+\log (4))\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {8}{5} \log (4) \int \frac {\log (4) x^2+e^x \left (-x^2-2 x+2\right )+\log (16)-2 \log ^2(4)}{(x+2)^2 \left (\log (4) (-x-\log (4)+1)+e^x\right )^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \frac {8}{5} \log (4) \int \frac {\log (4) x^2+e^x \left (-x^2-2 x+2\right )-2 \log ^2(4) \left (1-\frac {\log (16)}{2 \log ^2(4)}\right )}{(x+2)^2 \left (\log (4) (-x-\log (4)+1)+e^x\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {8}{5} \log (4) \int \left (\frac {-x^2-2 x+2}{(x+2)^2 \left (-\log (4) x+e^x+(1-\log (4)) \log (4)\right )}+\frac {x \left (-\log (4) x^2-\log ^2(4) x+\log (256)-2 \log ^2(4)\right )}{(x+2)^2 \left (-\log (4) x+e^x+(1-\log (4)) \log (4)\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {8}{5} \log (4) \left (-\frac {1}{2} \left (2 \log ^2(4)-\log (65536)\right ) \int \frac {1}{\left (-\log (4) x+e^x+(1-\log (4)) \log (4)\right )^2}dx+2 \left (\log ^2(4)-\log (256)\right ) \int \frac {1}{(x+2) \left (-\log (4) x+e^x+(1-\log (4)) \log (4)\right )^2}dx+\int \frac {1}{\log (4) x-e^x-(1-\log (4)) \log (4)}dx-\log (4) \int \frac {x}{\left (-\log (4) x+e^x+(1-\log (4)) \log (4)\right )^2}dx+2 \int \frac {1}{(x+2)^2 \left (-\log (4) x+e^x+(1-\log (4)) \log (4)\right )}dx+2 \int \frac {1}{(x+2) \left (-\log (4) x+e^x+(1-\log (4)) \log (4)\right )}dx\right )\) |
Input:
Int[(E^x*(16 - 16*x - 8*x^2)*Log[4] + (16 + 8*x^2)*Log[4]^2 - 16*Log[4]^3) /(E^(2*x)*(20 + 20*x + 5*x^2) + (20 - 20*x - 15*x^2 + 10*x^3 + 5*x^4)*Log[ 4]^2 + (-40 + 30*x^2 + 10*x^3)*Log[4]^3 + (20 + 20*x + 5*x^2)*Log[4]^4 + E ^x*((40 - 30*x^2 - 10*x^3)*Log[4] + (-40 - 40*x - 10*x^2)*Log[4]^2)),x]
Output:
$Aborted
Time = 0.46 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
| method | result | size |
| norman | \(-\frac {16 x \ln \left (2\right )}{5 \left (2+x \right ) \left (4 \ln \left (2\right )^{2}+2 x \ln \left (2\right )-2 \ln \left (2\right )-{\mathrm e}^{x}\right )}\) | \(33\) |
| risch | \(-\frac {16 x \ln \left (2\right )}{5 \left (2+x \right ) \left (4 \ln \left (2\right )^{2}+2 x \ln \left (2\right )-2 \ln \left (2\right )-{\mathrm e}^{x}\right )}\) | \(33\) |
| parallelrisch | \(-\frac {16 x \ln \left (2\right )}{5 \left (4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+8 \ln \left (2\right )^{2}+2 x \ln \left (2\right )-{\mathrm e}^{x} x -4 \ln \left (2\right )-2 \,{\mathrm e}^{x}\right )}\) | \(47\) |
Input:
int((2*(-8*x^2-16*x+16)*ln(2)*exp(x)-128*ln(2)^3+4*(8*x^2+16)*ln(2)^2)/((5 *x^2+20*x+20)*exp(x)^2+(4*(-10*x^2-40*x-40)*ln(2)^2+2*(-10*x^3-30*x^2+40)* ln(2))*exp(x)+16*(5*x^2+20*x+20)*ln(2)^4+8*(10*x^3+30*x^2-40)*ln(2)^3+4*(5 *x^4+10*x^3-15*x^2-20*x+20)*ln(2)^2),x,method=_RETURNVERBOSE)
Output:
-16/5*x*ln(2)/(2+x)/(4*ln(2)^2+2*x*ln(2)-2*ln(2)-exp(x))
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\frac {16 \, x \log \left (2\right )}{5 \, {\left (4 \, {\left (x + 2\right )} \log \left (2\right )^{2} - {\left (x + 2\right )} e^{x} + 2 \, {\left (x^{2} + x - 2\right )} \log \left (2\right )\right )}} \] Input:
integrate((2*(-8*x^2-16*x+16)*log(2)*exp(x)-128*log(2)^3+4*(8*x^2+16)*log( 2)^2)/((5*x^2+20*x+20)*exp(x)^2+(4*(-10*x^2-40*x-40)*log(2)^2+2*(-10*x^3-3 0*x^2+40)*log(2))*exp(x)+16*(5*x^2+20*x+20)*log(2)^4+8*(10*x^3+30*x^2-40)* log(2)^3+4*(5*x^4+10*x^3-15*x^2-20*x+20)*log(2)^2),x, algorithm="fricas")
Output:
-16/5*x*log(2)/(4*(x + 2)*log(2)^2 - (x + 2)*e^x + 2*(x^2 + x - 2)*log(2))
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).
Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=\frac {16 x \log {\left (2 \right )}}{- 10 x^{2} \log {\left (2 \right )} - 20 x \log {\left (2 \right )}^{2} - 10 x \log {\left (2 \right )} + \left (5 x + 10\right ) e^{x} - 40 \log {\left (2 \right )}^{2} + 20 \log {\left (2 \right )}} \] Input:
integrate((2*(-8*x**2-16*x+16)*ln(2)*exp(x)-128*ln(2)**3+4*(8*x**2+16)*ln( 2)**2)/((5*x**2+20*x+20)*exp(x)**2+(4*(-10*x**2-40*x-40)*ln(2)**2+2*(-10*x **3-30*x**2+40)*ln(2))*exp(x)+16*(5*x**2+20*x+20)*ln(2)**4+8*(10*x**3+30*x **2-40)*ln(2)**3+4*(5*x**4+10*x**3-15*x**2-20*x+20)*ln(2)**2),x)
Output:
16*x*log(2)/(-10*x**2*log(2) - 20*x*log(2)**2 - 10*x*log(2) + (5*x + 10)*e xp(x) - 40*log(2)**2 + 20*log(2))
Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\frac {16 \, x \log \left (2\right )}{5 \, {\left (2 \, x^{2} \log \left (2\right ) + 2 \, {\left (2 \, \log \left (2\right )^{2} + \log \left (2\right )\right )} x - {\left (x + 2\right )} e^{x} + 8 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right )\right )}} \] Input:
integrate((2*(-8*x^2-16*x+16)*log(2)*exp(x)-128*log(2)^3+4*(8*x^2+16)*log( 2)^2)/((5*x^2+20*x+20)*exp(x)^2+(4*(-10*x^2-40*x-40)*log(2)^2+2*(-10*x^3-3 0*x^2+40)*log(2))*exp(x)+16*(5*x^2+20*x+20)*log(2)^4+8*(10*x^3+30*x^2-40)* log(2)^3+4*(5*x^4+10*x^3-15*x^2-20*x+20)*log(2)^2),x, algorithm="maxima")
Output:
-16/5*x*log(2)/(2*x^2*log(2) + 2*(2*log(2)^2 + log(2))*x - (x + 2)*e^x + 8 *log(2)^2 - 4*log(2))
Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\frac {16 \, x \log \left (2\right )}{5 \, {\left (2 \, x^{2} \log \left (2\right ) + 4 \, x \log \left (2\right )^{2} - x e^{x} + 2 \, x \log \left (2\right ) + 8 \, \log \left (2\right )^{2} - 2 \, e^{x} - 4 \, \log \left (2\right )\right )}} \] Input:
integrate((2*(-8*x^2-16*x+16)*log(2)*exp(x)-128*log(2)^3+4*(8*x^2+16)*log( 2)^2)/((5*x^2+20*x+20)*exp(x)^2+(4*(-10*x^2-40*x-40)*log(2)^2+2*(-10*x^3-3 0*x^2+40)*log(2))*exp(x)+16*(5*x^2+20*x+20)*log(2)^4+8*(10*x^3+30*x^2-40)* log(2)^3+4*(5*x^4+10*x^3-15*x^2-20*x+20)*log(2)^2),x, algorithm="giac")
Output:
-16/5*x*log(2)/(2*x^2*log(2) + 4*x*log(2)^2 - x*e^x + 2*x*log(2) + 8*log(2 )^2 - 2*e^x - 4*log(2))
Timed out. \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\int \frac {128\,{\ln \left (2\right )}^3-4\,{\ln \left (2\right )}^2\,\left (8\,x^2+16\right )+2\,{\mathrm {e}}^x\,\ln \left (2\right )\,\left (8\,x^2+16\,x-16\right )}{4\,{\ln \left (2\right )}^2\,\left (5\,x^4+10\,x^3-15\,x^2-20\,x+20\right )+{\mathrm {e}}^{2\,x}\,\left (5\,x^2+20\,x+20\right )+16\,{\ln \left (2\right )}^4\,\left (5\,x^2+20\,x+20\right )-{\mathrm {e}}^x\,\left (2\,\ln \left (2\right )\,\left (10\,x^3+30\,x^2-40\right )+4\,{\ln \left (2\right )}^2\,\left (10\,x^2+40\,x+40\right )\right )+8\,{\ln \left (2\right )}^3\,\left (10\,x^3+30\,x^2-40\right )} \,d x \] Input:
int(-(128*log(2)^3 - 4*log(2)^2*(8*x^2 + 16) + 2*exp(x)*log(2)*(16*x + 8*x ^2 - 16))/(4*log(2)^2*(10*x^3 - 15*x^2 - 20*x + 5*x^4 + 20) + exp(2*x)*(20 *x + 5*x^2 + 20) + 16*log(2)^4*(20*x + 5*x^2 + 20) - exp(x)*(2*log(2)*(30* x^2 + 10*x^3 - 40) + 4*log(2)^2*(40*x + 10*x^2 + 40)) + 8*log(2)^3*(30*x^2 + 10*x^3 - 40)),x)
Output:
-int((128*log(2)^3 - 4*log(2)^2*(8*x^2 + 16) + 2*exp(x)*log(2)*(16*x + 8*x ^2 - 16))/(4*log(2)^2*(10*x^3 - 15*x^2 - 20*x + 5*x^4 + 20) + exp(2*x)*(20 *x + 5*x^2 + 20) + 16*log(2)^4*(20*x + 5*x^2 + 20) - exp(x)*(2*log(2)*(30* x^2 + 10*x^3 - 40) + 4*log(2)^2*(40*x + 10*x^2 + 40)) + 8*log(2)^3*(30*x^2 + 10*x^3 - 40)), x)
Time = 0.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=\frac {16 \,\mathrm {log}\left (2\right ) x}{5 e^{x} x +10 e^{x}-20 \mathrm {log}\left (2\right )^{2} x -40 \mathrm {log}\left (2\right )^{2}-10 \,\mathrm {log}\left (2\right ) x^{2}-10 \,\mathrm {log}\left (2\right ) x +20 \,\mathrm {log}\left (2\right )} \] Input:
int((2*(-8*x^2-16*x+16)*log(2)*exp(x)-128*log(2)^3+4*(8*x^2+16)*log(2)^2)/ ((5*x^2+20*x+20)*exp(x)^2+(4*(-10*x^2-40*x-40)*log(2)^2+2*(-10*x^3-30*x^2+ 40)*log(2))*exp(x)+16*(5*x^2+20*x+20)*log(2)^4+8*(10*x^3+30*x^2-40)*log(2) ^3+4*(5*x^4+10*x^3-15*x^2-20*x+20)*log(2)^2),x)
Output:
(16*log(2)*x)/(5*(e**x*x + 2*e**x - 4*log(2)**2*x - 8*log(2)**2 - 2*log(2) *x**2 - 2*log(2)*x + 4*log(2)))