Integrand size = 249, antiderivative size = 27 \[ \int \frac {e^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx=x \left (-5-x+\log \left (2 x \left (2+\log \left (5 \left (2+\frac {4}{e^{25}+x}\right )\right )\right )\right )\right ) \] Output:
(ln(x*(4+2*ln(10+20/(exp(25)+x))))-5-x)*x
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx=-5 x-x^2+x \log \left (2 x \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )\right ) \] Input:
Integrate[(E^50*(-8 - 4*x) - 18*x - 16*x^2 - 4*x^3 + E^25*(-16 - 24*x - 8* x^2) + (E^50*(-4 - 2*x) - 8*x - 8*x^2 - 2*x^3 + E^25*(-8 - 12*x - 4*x^2))* Log[(20 + 10*E^25 + 10*x)/(E^25 + x)] + (2*E^50 + 4*x + 2*x^2 + E^25*(4 + 4*x) + (E^50 + 2*x + x^2 + E^25*(2 + 2*x))*Log[(20 + 10*E^25 + 10*x)/(E^25 + x)])*Log[4*x + 2*x*Log[(20 + 10*E^25 + 10*x)/(E^25 + x)]])/(2*E^50 + 4* x + 2*x^2 + E^25*(4 + 4*x) + (E^50 + 2*x + x^2 + E^25*(2 + 2*x))*Log[(20 + 10*E^25 + 10*x)/(E^25 + x)]),x]
Output:
-5*x - x^2 + x*Log[2*x*(2 + Log[(10*(2 + E^25 + x))/(E^25 + 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.
| \(\displaystyle \int \frac {-4 x^3-16 x^2+e^{25} \left (-8 x^2-24 x-16\right )+\left (2 x^2+\left (x^2+2 x+e^{25} (2 x+2)+e^{50}\right ) \log \left (\frac {10 x+10 e^{25}+20}{x+e^{25}}\right )+4 x+e^{25} (4 x+4)+2 e^{50}\right ) \log \left (4 x+2 x \log \left (\frac {10 x+10 e^{25}+20}{x+e^{25}}\right )\right )+\left (-2 x^3-8 x^2+e^{25} \left (-4 x^2-12 x-8\right )-8 x+e^{50} (-2 x-4)\right ) \log \left (\frac {10 x+10 e^{25}+20}{x+e^{25}}\right )-18 x+e^{50} (-4 x-8)}{2 x^2+\left (x^2+2 x+e^{25} (2 x+2)+e^{50}\right ) \log \left (\frac {10 x+10 e^{25}+20}{x+e^{25}}\right )+4 x+e^{25} (4 x+4)+2 e^{50}} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-4 x^3-16 x^2+e^{25} \left (-8 x^2-24 x-16\right )+\left (2 x^2+\left (x^2+2 x+e^{25} (2 x+2)+e^{50}\right ) \log \left (\frac {10 x+10 e^{25}+20}{x+e^{25}}\right )+4 x+e^{25} (4 x+4)+2 e^{50}\right ) \log \left (4 x+2 x \log \left (\frac {10 x+10 e^{25}+20}{x+e^{25}}\right )\right )+\left (-2 x^3-8 x^2+e^{25} \left (-4 x^2-12 x-8\right )-8 x+e^{50} (-2 x-4)\right ) \log \left (\frac {10 x+10 e^{25}+20}{x+e^{25}}\right )-18 x+e^{50} (-4 x-8)}{\left (x^2+2 \left (1+e^{25}\right ) x+e^{25} \left (2+e^{25}\right )\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (-\frac {4 x^3}{\left (x+e^{25}\right ) \left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}-\frac {16 x^2}{\left (x+e^{25}\right ) \left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}-\frac {18 x}{\left (x+e^{25}\right ) \left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}+\log \left (2 x \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )\right )-\frac {2 (x+2) \log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )}{\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2}-\frac {8 e^{25} (x+1) (x+2)}{\left (x+e^{25}\right ) \left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}-\frac {4 e^{50} (x+2)}{\left (x+e^{25}\right ) \left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 8 \left (1+e^{25}\right ) \int \frac {1}{\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2}dx-8 e^{25} \int \frac {1}{\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2}dx-16 \int \frac {1}{\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2}dx-4 \int \frac {x}{\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2}dx+4 \int \frac {x+2}{\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2}dx-4 e^{25} \left (2-3 e^{25}+e^{50}\right ) \int \frac {1}{\left (x+e^{25}\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx-2 e^{50} \left (2-e^{25}\right ) \int \frac {1}{\left (x+e^{25}\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx+2 e^{75} \int \frac {1}{\left (x+e^{25}\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx-8 e^{50} \int \frac {1}{\left (x+e^{25}\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx+8 e^{25} \int \frac {1}{\left (x+e^{25}\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx-2 \left (2+e^{25}\right )^3 \int \frac {1}{\left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx+8 \left (2+e^{25}\right )^2 \int \frac {1}{\left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx-8 \left (2+e^{25}\right ) \int \frac {1}{\left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx+4 e^{50} \left (1+e^{25}\right ) \int \frac {1}{\left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx-2 e^{75} \int \frac {1}{\left (x+e^{25}+2\right ) \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )}dx-x^2-5 x+x \log \left (2 x \left (\log \left (\frac {10 \left (x+e^{25}+2\right )}{x+e^{25}}\right )+2\right )\right )\) |
Input:
Int[(E^50*(-8 - 4*x) - 18*x - 16*x^2 - 4*x^3 + E^25*(-16 - 24*x - 8*x^2) + (E^50*(-4 - 2*x) - 8*x - 8*x^2 - 2*x^3 + E^25*(-8 - 12*x - 4*x^2))*Log[(2 0 + 10*E^25 + 10*x)/(E^25 + x)] + (2*E^50 + 4*x + 2*x^2 + E^25*(4 + 4*x) + (E^50 + 2*x + x^2 + E^25*(2 + 2*x))*Log[(20 + 10*E^25 + 10*x)/(E^25 + x)] )*Log[4*x + 2*x*Log[(20 + 10*E^25 + 10*x)/(E^25 + x)]])/(2*E^50 + 4*x + 2* x^2 + E^25*(4 + 4*x) + (E^50 + 2*x + x^2 + E^25*(2 + 2*x))*Log[(20 + 10*E^ 25 + 10*x)/(E^25 + x)]),x]
Output:
$Aborted
Time = 5.80 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67
| method | result | size |
| parallelrisch | \(6 \,{\mathrm e}^{50}+24-x^{2}+\ln \left (2 x \ln \left (\frac {10 \,{\mathrm e}^{25}+10 x +20}{{\mathrm e}^{25}+x}\right )+4 x \right ) x +32 \,{\mathrm e}^{25}-5 x\) | \(45\) |
| default | \(\frac {\frac {4 \left ({\mathrm e}^{25}+x +2\right ) {\mathrm e}^{25}}{{\mathrm e}^{25}+x}-4 \,{\mathrm e}^{25}-\frac {8 \left ({\mathrm e}^{25}+x +2\right )}{{\mathrm e}^{25}+x}+4}{\left (\frac {{\mathrm e}^{25}+x +2}{{\mathrm e}^{25}+x}-1\right )^{2}}+x \ln \left (2\right )+x \ln \left (x \ln \left (\frac {10 \,{\mathrm e}^{25}+10 x +20}{{\mathrm e}^{25}+x}\right )+2 x \right )-\frac {2}{\frac {{\mathrm e}^{25}+x +2}{{\mathrm e}^{25}+x}-1}\) | \(98\) |
Input:
int((((exp(25)^2+(2+2*x)*exp(25)+x^2+2*x)*ln((10*exp(25)+10*x+20)/(exp(25) +x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x^2+4*x)*ln(2*x*ln((10*exp(25)+10*x+20) /(exp(25)+x))+4*x)+((-2*x-4)*exp(25)^2+(-4*x^2-12*x-8)*exp(25)-2*x^3-8*x^2 -8*x)*ln((10*exp(25)+10*x+20)/(exp(25)+x))+(-4*x-8)*exp(25)^2+(-8*x^2-24*x -16)*exp(25)-4*x^3-16*x^2-18*x)/((exp(25)^2+(2+2*x)*exp(25)+x^2+2*x)*ln((1 0*exp(25)+10*x+20)/(exp(25)+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x^2+4*x),x,m ethod=_RETURNVERBOSE)
Output:
6*exp(25)^2+24-x^2+ln(2*x*ln(10*(exp(25)+x+2)/(exp(25)+x))+4*x)*x+32*exp(2 5)-5*x
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx=-x^{2} + x \log \left (2 \, x \log \left (\frac {10 \, {\left (x + e^{25} + 2\right )}}{x + e^{25}}\right ) + 4 \, x\right ) - 5 \, x \] Input:
integrate((((exp(25)^2+(2+2*x)*exp(25)+x^2+2*x)*log((10*exp(25)+10*x+20)/( exp(25)+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x^2+4*x)*log(2*x*log((10*exp(25) +10*x+20)/(exp(25)+x))+4*x)+((-2*x-4)*exp(25)^2+(-4*x^2-12*x-8)*exp(25)-2* x^3-8*x^2-8*x)*log((10*exp(25)+10*x+20)/(exp(25)+x))+(-4*x-8)*exp(25)^2+(- 8*x^2-24*x-16)*exp(25)-4*x^3-16*x^2-18*x)/((exp(25)^2+(2+2*x)*exp(25)+x^2+ 2*x)*log((10*exp(25)+10*x+20)/(exp(25)+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x ^2+4*x),x, algorithm="fricas")
Output:
-x^2 + x*log(2*x*log(10*(x + e^25 + 2)/(x + e^25)) + 4*x) - 5*x
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (20) = 40\).
Time = 1.02 (sec) , antiderivative size = 139, normalized size of antiderivative = 5.15 \[ \int \frac {e^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx=- x^{2} - 5 x + \left (x + \frac {1}{6} + \frac {e^{25}}{6}\right ) \log {\left (2 x \log {\left (\frac {10 x + 20 + 10 e^{25}}{x + e^{25}} \right )} + 4 x \right )} - \frac {\left (1 + e\right ) \left (- e^{3} - e + 1 + e^{2} + e^{4}\right ) \left (- e^{15} - e^{5} + 1 + e^{10} + e^{20}\right ) \log {\left (x \right )}}{6} - \frac {\left (1 + e\right ) \left (- e^{3} - e + 1 + e^{2} + e^{4}\right ) \left (- e^{15} - e^{5} + 1 + e^{10} + e^{20}\right ) \log {\left (\log {\left (\frac {10 x + 20 + 10 e^{25}}{x + e^{25}} \right )} + 2 \right )}}{6} \] Input:
integrate((((exp(25)**2+(2+2*x)*exp(25)+x**2+2*x)*ln((10*exp(25)+10*x+20)/ (exp(25)+x))+2*exp(25)**2+(4+4*x)*exp(25)+2*x**2+4*x)*ln(2*x*ln((10*exp(25 )+10*x+20)/(exp(25)+x))+4*x)+((-2*x-4)*exp(25)**2+(-4*x**2-12*x-8)*exp(25) -2*x**3-8*x**2-8*x)*ln((10*exp(25)+10*x+20)/(exp(25)+x))+(-4*x-8)*exp(25)* *2+(-8*x**2-24*x-16)*exp(25)-4*x**3-16*x**2-18*x)/((exp(25)**2+(2+2*x)*exp (25)+x**2+2*x)*ln((10*exp(25)+10*x+20)/(exp(25)+x))+2*exp(25)**2+(4+4*x)*e xp(25)+2*x**2+4*x),x)
Output:
-x**2 - 5*x + (x + 1/6 + exp(25)/6)*log(2*x*log((10*x + 20 + 10*exp(25))/( x + exp(25))) + 4*x) - (1 + E)*(-exp(3) - E + 1 + exp(2) + exp(4))*(-exp(1 5) - exp(5) + 1 + exp(10) + exp(20))*log(x)/6 - (1 + E)*(-exp(3) - E + 1 + exp(2) + exp(4))*(-exp(15) - exp(5) + 1 + exp(10) + exp(20))*log(log((10* x + 20 + 10*exp(25))/(x + exp(25))) + 2)/6
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (25) = 50\).
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.52 \[ \int \frac {e^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx=-x^{2} + x {\left (\log \left (2\right ) - 5\right )} + x \log \left (x\right ) + {\left (x - 4 \, e^{50} - 8 \, e^{25}\right )} \log \left (\log \left (5\right ) + \log \left (2\right ) + \log \left (x + e^{25} + 2\right ) - \log \left (x + e^{25}\right ) + 2\right ) + 4 \, e^{50} \log \left (\log \left (5\right ) + \log \left (2\right ) + \log \left (x + e^{25} + 2\right ) - \log \left (x + e^{25}\right ) + 2\right ) + 8 \, e^{25} \log \left (\log \left (5\right ) + \log \left (2\right ) + \log \left (x + e^{25} + 2\right ) - \log \left (x + e^{25}\right ) + 2\right ) \] Input:
integrate((((exp(25)^2+(2+2*x)*exp(25)+x^2+2*x)*log((10*exp(25)+10*x+20)/( exp(25)+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x^2+4*x)*log(2*x*log((10*exp(25) +10*x+20)/(exp(25)+x))+4*x)+((-2*x-4)*exp(25)^2+(-4*x^2-12*x-8)*exp(25)-2* x^3-8*x^2-8*x)*log((10*exp(25)+10*x+20)/(exp(25)+x))+(-4*x-8)*exp(25)^2+(- 8*x^2-24*x-16)*exp(25)-4*x^3-16*x^2-18*x)/((exp(25)^2+(2+2*x)*exp(25)+x^2+ 2*x)*log((10*exp(25)+10*x+20)/(exp(25)+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x ^2+4*x),x, algorithm="maxima")
Output:
-x^2 + x*(log(2) - 5) + x*log(x) + (x - 4*e^50 - 8*e^25)*log(log(5) + log( 2) + log(x + e^25 + 2) - log(x + e^25) + 2) + 4*e^50*log(log(5) + log(2) + log(x + e^25 + 2) - log(x + e^25) + 2) + 8*e^25*log(log(5) + log(2) + log (x + e^25 + 2) - log(x + e^25) + 2)
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (25) = 50\).
Time = 0.59 (sec) , antiderivative size = 254, normalized size of antiderivative = 9.41 \[ \int \frac {e^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx=-e^{50} \log \left (2\right ) \log \left (x + e^{25} + 2\right ) - e^{25} \log \left (2\right ) \log \left (x + e^{25} + 2\right ) - e^{50} \log \left (2\right ) \log \left (x + e^{25}\right ) + e^{25} \log \left (2\right ) \log \left (x + e^{25}\right ) + e^{50} \log \left (2\right ) \log \left (-x - e^{25}\right ) - e^{25} \log \left (2\right ) \log \left (-x - e^{25}\right ) + e^{50} \log \left (2\right ) \log \left (-x - e^{25} - 2\right ) + e^{25} \log \left (2\right ) \log \left (-x - e^{25} - 2\right ) - x^{2} + x \log \left (2\right ) + x \log \left (x \log \left (\frac {10 \, {\left (x + e^{25} + 2\right )}}{x + e^{25}}\right ) + 2 \, x\right ) - 2 \, e^{75} \log \left (x + e^{25} + 2\right ) - 2 \, e^{50} \log \left (x + e^{25} + 2\right ) - 2 \, e^{75} \log \left (x + e^{25}\right ) + 6 \, e^{50} \log \left (x + e^{25}\right ) - 4 \, e^{25} \log \left (x + e^{25}\right ) + 2 \, e^{75} \log \left (-x - e^{25}\right ) - 6 \, e^{50} \log \left (-x - e^{25}\right ) + 4 \, e^{25} \log \left (-x - e^{25}\right ) + 2 \, e^{75} \log \left (-x - e^{25} - 2\right ) + 2 \, e^{50} \log \left (-x - e^{25} - 2\right ) - 5 \, x \] Input:
integrate((((exp(25)^2+(2+2*x)*exp(25)+x^2+2*x)*log((10*exp(25)+10*x+20)/( exp(25)+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x^2+4*x)*log(2*x*log((10*exp(25) +10*x+20)/(exp(25)+x))+4*x)+((-2*x-4)*exp(25)^2+(-4*x^2-12*x-8)*exp(25)-2* x^3-8*x^2-8*x)*log((10*exp(25)+10*x+20)/(exp(25)+x))+(-4*x-8)*exp(25)^2+(- 8*x^2-24*x-16)*exp(25)-4*x^3-16*x^2-18*x)/((exp(25)^2+(2+2*x)*exp(25)+x^2+ 2*x)*log((10*exp(25)+10*x+20)/(exp(25)+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x ^2+4*x),x, algorithm="giac")
Output:
-e^50*log(2)*log(x + e^25 + 2) - e^25*log(2)*log(x + e^25 + 2) - e^50*log( 2)*log(x + e^25) + e^25*log(2)*log(x + e^25) + e^50*log(2)*log(-x - e^25) - e^25*log(2)*log(-x - e^25) + e^50*log(2)*log(-x - e^25 - 2) + e^25*log(2 )*log(-x - e^25 - 2) - x^2 + x*log(2) + x*log(x*log(10*(x + e^25 + 2)/(x + e^25)) + 2*x) - 2*e^75*log(x + e^25 + 2) - 2*e^50*log(x + e^25 + 2) - 2*e ^75*log(x + e^25) + 6*e^50*log(x + e^25) - 4*e^25*log(x + e^25) + 2*e^75*l og(-x - e^25) - 6*e^50*log(-x - e^25) + 4*e^25*log(-x - e^25) + 2*e^75*log (-x - e^25 - 2) + 2*e^50*log(-x - e^25 - 2) - 5*x
Time = 2.58 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx=-x\,\left (x-\ln \left (4\,x+2\,x\,\ln \left (\frac {10\,x+10\,{\mathrm {e}}^{25}+20}{x+{\mathrm {e}}^{25}}\right )\right )+5\right ) \] Input:
int(-(18*x + exp(25)*(24*x + 8*x^2 + 16) + log((10*x + 10*exp(25) + 20)/(x + exp(25)))*(8*x + exp(25)*(12*x + 4*x^2 + 8) + 8*x^2 + 2*x^3 + exp(50)*( 2*x + 4)) - log(4*x + 2*x*log((10*x + 10*exp(25) + 20)/(x + exp(25))))*(4* x + 2*exp(50) + log((10*x + 10*exp(25) + 20)/(x + exp(25)))*(2*x + exp(50) + x^2 + exp(25)*(2*x + 2)) + 2*x^2 + exp(25)*(4*x + 4)) + 16*x^2 + 4*x^3 + exp(50)*(4*x + 8))/(4*x + 2*exp(50) + log((10*x + 10*exp(25) + 20)/(x + exp(25)))*(2*x + exp(50) + x^2 + exp(25)*(2*x + 2)) + 2*x^2 + exp(25)*(4*x + 4)),x)
Output:
-x*(x - log(4*x + 2*x*log((10*x + 10*exp(25) + 20)/(x + exp(25)))) + 5)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx=x \left (\mathrm {log}\left (2 \,\mathrm {log}\left (\frac {10 e^{25}+10 x +20}{e^{25}+x}\right ) x +4 x \right )-x -5\right ) \] Input:
int((((exp(25)^2+(2+2*x)*exp(25)+x^2+2*x)*log((10*exp(25)+10*x+20)/(exp(25 )+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x^2+4*x)*log(2*x*log((10*exp(25)+10*x+ 20)/(exp(25)+x))+4*x)+((-2*x-4)*exp(25)^2+(-4*x^2-12*x-8)*exp(25)-2*x^3-8* x^2-8*x)*log((10*exp(25)+10*x+20)/(exp(25)+x))+(-4*x-8)*exp(25)^2+(-8*x^2- 24*x-16)*exp(25)-4*x^3-16*x^2-18*x)/((exp(25)^2+(2+2*x)*exp(25)+x^2+2*x)*l og((10*exp(25)+10*x+20)/(exp(25)+x))+2*exp(25)^2+(4+4*x)*exp(25)+2*x^2+4*x ),x)
Output:
x*(log(2*log((10*e**25 + 10*x + 20)/(e**25 + x))*x + 4*x) - x - 5)