Integrand size = 238, antiderivative size = 32 \[ \int \frac {64+208 x+33 x^2+2 x^3+\left (-16-50 x-4 x^2\right ) \log (5)+(1+3 x) \log ^2(5)+e^x \left (-80-266 x-70 x^2-10 x^3-x^4+\left (26+82 x+10 x^2+x^3\right ) \log (5)+(-2-6 x) \log ^2(5)\right )+e^{2 x} \left (25+85 x+31 x^2+3 x^3+\left (-10-32 x-6 x^2\right ) \log (5)+(1+3 x) \log ^2(5)\right )}{64 x+16 x^2+x^3+\left (-16 x-2 x^2\right ) \log (5)+x \log ^2(5)+e^x \left (-80 x-26 x^2-2 x^3+\left (26 x+4 x^2\right ) \log (5)-2 x \log ^2(5)\right )+e^{2 x} \left (25 x+10 x^2+x^3+\left (-10 x-2 x^2\right ) \log (5)+x \log ^2(5)\right )} \, dx=3 x+\frac {x}{-\frac {3}{x}+\frac {\left (-1+e^x\right ) (5+x-\log (5))}{x}}+\log (x) \] Output:
3*x+x/((-1+exp(x))/x*(5-ln(5)+x)-3/x)+ln(x)
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {64+208 x+33 x^2+2 x^3+\left (-16-50 x-4 x^2\right ) \log (5)+(1+3 x) \log ^2(5)+e^x \left (-80-266 x-70 x^2-10 x^3-x^4+\left (26+82 x+10 x^2+x^3\right ) \log (5)+(-2-6 x) \log ^2(5)\right )+e^{2 x} \left (25+85 x+31 x^2+3 x^3+\left (-10-32 x-6 x^2\right ) \log (5)+(1+3 x) \log ^2(5)\right )}{64 x+16 x^2+x^3+\left (-16 x-2 x^2\right ) \log (5)+x \log ^2(5)+e^x \left (-80 x-26 x^2-2 x^3+\left (26 x+4 x^2\right ) \log (5)-2 x \log ^2(5)\right )+e^{2 x} \left (25 x+10 x^2+x^3+\left (-10 x-2 x^2\right ) \log (5)+x \log ^2(5)\right )} \, dx=x \left (3+\frac {x}{-8-x+e^x (5+x-\log (5))+\log (5)}\right )+\log (x) \] Input:
Integrate[(64 + 208*x + 33*x^2 + 2*x^3 + (-16 - 50*x - 4*x^2)*Log[5] + (1 + 3*x)*Log[5]^2 + E^x*(-80 - 266*x - 70*x^2 - 10*x^3 - x^4 + (26 + 82*x + 10*x^2 + x^3)*Log[5] + (-2 - 6*x)*Log[5]^2) + E^(2*x)*(25 + 85*x + 31*x^2 + 3*x^3 + (-10 - 32*x - 6*x^2)*Log[5] + (1 + 3*x)*Log[5]^2))/(64*x + 16*x^ 2 + x^3 + (-16*x - 2*x^2)*Log[5] + x*Log[5]^2 + E^x*(-80*x - 26*x^2 - 2*x^ 3 + (26*x + 4*x^2)*Log[5] - 2*x*Log[5]^2) + E^(2*x)*(25*x + 10*x^2 + x^3 + (-10*x - 2*x^2)*Log[5] + x*Log[5]^2)),x]
Output:
x*(3 + x/(-8 - x + E^x*(5 + x - Log[5]) + Log[5])) + Log[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.
| \(\displaystyle \int \frac {2 x^3+33 x^2+\left (-4 x^2-50 x-16\right ) \log (5)+e^{2 x} \left (3 x^3+31 x^2+\left (-6 x^2-32 x-10\right ) \log (5)+85 x+(3 x+1) \log ^2(5)+25\right )+e^x \left (-x^4-10 x^3-70 x^2+\left (x^3+10 x^2+82 x+26\right ) \log (5)-266 x+(-6 x-2) \log ^2(5)-80\right )+208 x+(3 x+1) \log ^2(5)+64}{x^3+16 x^2+\left (-2 x^2-16 x\right ) \log (5)+e^x \left (-2 x^3-26 x^2+\left (4 x^2+26 x\right ) \log (5)-80 x-2 x \log ^2(5)\right )+e^{2 x} \left (x^3+10 x^2+\left (-2 x^2-10 x\right ) \log (5)+25 x+x \log ^2(5)\right )+64 x+x \log ^2(5)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {2 x^3+33 x^2+\left (-4 x^2-50 x-16\right ) \log (5)+e^{2 x} \left (3 x^3+31 x^2+\left (-6 x^2-32 x-10\right ) \log (5)+85 x+(3 x+1) \log ^2(5)+25\right )+e^x \left (-x^4-10 x^3-70 x^2+\left (x^3+10 x^2+82 x+26\right ) \log (5)-266 x+(-6 x-2) \log ^2(5)-80\right )+208 x+(3 x+1) \log ^2(5)+64}{x^3+16 x^2+\left (-2 x^2-16 x\right ) \log (5)+e^x \left (-2 x^3-26 x^2+\left (4 x^2+26 x\right ) \log (5)-80 x-2 x \log ^2(5)\right )+e^{2 x} \left (x^3+10 x^2+\left (-2 x^2-10 x\right ) \log (5)+25 x+x \log ^2(5)\right )+x \left (64+\log ^2(5)\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 x^3+x^2 (33-4 \log (5))-e^x \left (x^4-x^3 (\log (5)-10)-10 x^2 (\log (5)-7)+x \left (266+6 \log ^2(5)-82 \log (5)\right )+2 \left (40+\log ^2(5)-13 \log (5)\right )\right )+x \left (208+3 \log ^2(5)-50 \log (5)\right )+e^{2 x} (3 x+1) (x+5-\log (5))^2+(\log (5)-8)^2}{x \left (x-e^x (x+5-\log (5))+8 \left (1-\frac {\log (5)}{8}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x^2 \left (-x^2-x (13-2 \log (5))-43-\log ^2(5)+13 \log (5)\right )}{\left (-e^x x+x-5 e^x \left (1-\frac {\log (5)}{5}\right )+8 \left (1-\frac {\log (5)}{8}\right )\right )^2 (x+5-\log (5))}+\frac {x \left (x^2+x (4-\log (5))-10+\log (25)\right )}{\left (-e^x x+x-5 e^x \left (1-\frac {\log (5)}{5}\right )+8 \left (1-\frac {\log (5)}{8}\right )\right ) (x+5-\log (5))}+\frac {3 x+1}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int \frac {x^3}{\left (-e^x x+x+8 \left (1-\frac {\log (5)}{8}\right )-5 e^x \left (1-\frac {\log (5)}{5}\right )\right )^2}dx-(8-\log (5)) \int \frac {x^2}{\left (-e^x x+x+8 \left (1-\frac {\log (5)}{8}\right )-5 e^x \left (1-\frac {\log (5)}{5}\right )\right )^2}dx+\int \frac {x^2}{-e^x x+x+8 \left (1-\frac {\log (5)}{8}\right )-5 e^x \left (1-\frac {\log (5)}{5}\right )}dx+(5-\log (5)) \left (3+2 \log ^2(5)-\log (5) \log (25)\right ) \int \frac {1}{\left (-e^x x+x+8 \left (1-\frac {\log (5)}{8}\right )-5 e^x \left (1-\frac {\log (5)}{5}\right )\right )^2}dx-\left (3+2 \log ^2(5)-\log (5) \log (25)\right ) \int \frac {x}{\left (-e^x x+x+8 \left (1-\frac {\log (5)}{8}\right )-5 e^x \left (1-\frac {\log (5)}{5}\right )\right )^2}dx-(5-\log (5))^2 \left (3+2 \log ^2(5)-\log (5) \log (25)\right ) \int \frac {1}{\left (-e^x x+x+8 \left (1-\frac {\log (5)}{8}\right )-5 e^x \left (1-\frac {\log (5)}{5}\right )\right )^2 (x-\log (5)+5)}dx+(5-\log (5)) \int \frac {1}{e^x x-x-8 \left (1-\frac {\log (5)}{8}\right )+5 e^x \left (1-\frac {\log (5)}{5}\right )}dx+\int \frac {x}{e^x x-x-8 \left (1-\frac {\log (5)}{8}\right )+5 e^x \left (1-\frac {\log (5)}{5}\right )}dx+(5-\log (5))^2 \int \frac {1}{\left (-e^x x+x+8 \left (1-\frac {\log (5)}{8}\right )-5 e^x \left (1-\frac {\log (5)}{5}\right )\right ) (x-\log (5)+5)}dx+3 x+\log (x)\) |
Input:
Int[(64 + 208*x + 33*x^2 + 2*x^3 + (-16 - 50*x - 4*x^2)*Log[5] + (1 + 3*x) *Log[5]^2 + E^x*(-80 - 266*x - 70*x^2 - 10*x^3 - x^4 + (26 + 82*x + 10*x^2 + x^3)*Log[5] + (-2 - 6*x)*Log[5]^2) + E^(2*x)*(25 + 85*x + 31*x^2 + 3*x^ 3 + (-10 - 32*x - 6*x^2)*Log[5] + (1 + 3*x)*Log[5]^2))/(64*x + 16*x^2 + x^ 3 + (-16*x - 2*x^2)*Log[5] + x*Log[5]^2 + E^x*(-80*x - 26*x^2 - 2*x^3 + (2 6*x + 4*x^2)*Log[5] - 2*x*Log[5]^2) + E^(2*x)*(25*x + 10*x^2 + x^3 + (-10* x - 2*x^2)*Log[5] + x*Log[5]^2)),x]
Output:
$Aborted
Time = 1.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
| method | result | size |
| risch | \(3 x +\ln \left (x \right )-\frac {x^{2}}{{\mathrm e}^{x} \ln \left (5\right )-{\mathrm e}^{x} x -\ln \left (5\right )-5 \,{\mathrm e}^{x}+x +8}\) | \(35\) |
| norman | \(\frac {9 x +\left (3 \ln \left (5\right )^{2}-30 \ln \left (5\right )+75\right ) {\mathrm e}^{x}+2 x^{2}-3 \,{\mathrm e}^{x} x^{2}-3 \ln \left (5\right )^{2}+39 \ln \left (5\right )-120}{{\mathrm e}^{x} \ln \left (5\right )-{\mathrm e}^{x} x -\ln \left (5\right )-5 \,{\mathrm e}^{x}+x +8}+\ln \left (x \right )\) | \(70\) |
| parallelrisch | \(\frac {3 \ln \left (5\right )^{2} {\mathrm e}^{x}+\ln \left (5\right ) {\mathrm e}^{x} \ln \left (x \right )-120-x \,{\mathrm e}^{x} \ln \left (x \right )-3 \,{\mathrm e}^{x} x^{2}-3 \ln \left (5\right )^{2}-\ln \left (5\right ) \ln \left (x \right )-30 \,{\mathrm e}^{x} \ln \left (5\right )+x \ln \left (x \right )-5 \,{\mathrm e}^{x} \ln \left (x \right )+2 x^{2}+39 \ln \left (5\right )+8 \ln \left (x \right )+9 x +75 \,{\mathrm e}^{x}}{{\mathrm e}^{x} \ln \left (5\right )-{\mathrm e}^{x} x -\ln \left (5\right )-5 \,{\mathrm e}^{x}+x +8}\) | \(104\) |
Input:
int((((1+3*x)*ln(5)^2+(-6*x^2-32*x-10)*ln(5)+3*x^3+31*x^2+85*x+25)*exp(x)^ 2+((-6*x-2)*ln(5)^2+(x^3+10*x^2+82*x+26)*ln(5)-x^4-10*x^3-70*x^2-266*x-80) *exp(x)+(1+3*x)*ln(5)^2+(-4*x^2-50*x-16)*ln(5)+2*x^3+33*x^2+208*x+64)/((x* ln(5)^2+(-2*x^2-10*x)*ln(5)+x^3+10*x^2+25*x)*exp(x)^2+(-2*x*ln(5)^2+(4*x^2 +26*x)*ln(5)-2*x^3-26*x^2-80*x)*exp(x)+x*ln(5)^2+(-2*x^2-16*x)*ln(5)+x^3+1 6*x^2+64*x),x,method=_RETURNVERBOSE)
Output:
3*x+ln(x)-x^2/(exp(x)*ln(5)-exp(x)*x-ln(5)-5*exp(x)+x+8)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (31) = 62\).
Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.25 \[ \int \frac {64+208 x+33 x^2+2 x^3+\left (-16-50 x-4 x^2\right ) \log (5)+(1+3 x) \log ^2(5)+e^x \left (-80-266 x-70 x^2-10 x^3-x^4+\left (26+82 x+10 x^2+x^3\right ) \log (5)+(-2-6 x) \log ^2(5)\right )+e^{2 x} \left (25+85 x+31 x^2+3 x^3+\left (-10-32 x-6 x^2\right ) \log (5)+(1+3 x) \log ^2(5)\right )}{64 x+16 x^2+x^3+\left (-16 x-2 x^2\right ) \log (5)+x \log ^2(5)+e^x \left (-80 x-26 x^2-2 x^3+\left (26 x+4 x^2\right ) \log (5)-2 x \log ^2(5)\right )+e^{2 x} \left (25 x+10 x^2+x^3+\left (-10 x-2 x^2\right ) \log (5)+x \log ^2(5)\right )} \, dx=-\frac {2 \, x^{2} - 3 \, {\left (x^{2} - x \log \left (5\right ) + 5 \, x\right )} e^{x} - 3 \, x \log \left (5\right ) - {\left ({\left (x - \log \left (5\right ) + 5\right )} e^{x} - x + \log \left (5\right ) - 8\right )} \log \left (x\right ) + 24 \, x}{{\left (x - \log \left (5\right ) + 5\right )} e^{x} - x + \log \left (5\right ) - 8} \] Input:
integrate((((1+3*x)*log(5)^2+(-6*x^2-32*x-10)*log(5)+3*x^3+31*x^2+85*x+25) *exp(x)^2+((-6*x-2)*log(5)^2+(x^3+10*x^2+82*x+26)*log(5)-x^4-10*x^3-70*x^2 -266*x-80)*exp(x)+(1+3*x)*log(5)^2+(-4*x^2-50*x-16)*log(5)+2*x^3+33*x^2+20 8*x+64)/((x*log(5)^2+(-2*x^2-10*x)*log(5)+x^3+10*x^2+25*x)*exp(x)^2+(-2*x* log(5)^2+(4*x^2+26*x)*log(5)-2*x^3-26*x^2-80*x)*exp(x)+x*log(5)^2+(-2*x^2- 16*x)*log(5)+x^3+16*x^2+64*x),x, algorithm="fricas")
Output:
-(2*x^2 - 3*(x^2 - x*log(5) + 5*x)*e^x - 3*x*log(5) - ((x - log(5) + 5)*e^ x - x + log(5) - 8)*log(x) + 24*x)/((x - log(5) + 5)*e^x - x + log(5) - 8)
Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {64+208 x+33 x^2+2 x^3+\left (-16-50 x-4 x^2\right ) \log (5)+(1+3 x) \log ^2(5)+e^x \left (-80-266 x-70 x^2-10 x^3-x^4+\left (26+82 x+10 x^2+x^3\right ) \log (5)+(-2-6 x) \log ^2(5)\right )+e^{2 x} \left (25+85 x+31 x^2+3 x^3+\left (-10-32 x-6 x^2\right ) \log (5)+(1+3 x) \log ^2(5)\right )}{64 x+16 x^2+x^3+\left (-16 x-2 x^2\right ) \log (5)+x \log ^2(5)+e^x \left (-80 x-26 x^2-2 x^3+\left (26 x+4 x^2\right ) \log (5)-2 x \log ^2(5)\right )+e^{2 x} \left (25 x+10 x^2+x^3+\left (-10 x-2 x^2\right ) \log (5)+x \log ^2(5)\right )} \, dx=\frac {x^{2}}{- x + \left (x - \log {\left (5 \right )} + 5\right ) e^{x} - 8 + \log {\left (5 \right )}} + 3 x + \log {\left (x \right )} \] Input:
integrate((((1+3*x)*ln(5)**2+(-6*x**2-32*x-10)*ln(5)+3*x**3+31*x**2+85*x+2 5)*exp(x)**2+((-6*x-2)*ln(5)**2+(x**3+10*x**2+82*x+26)*ln(5)-x**4-10*x**3- 70*x**2-266*x-80)*exp(x)+(1+3*x)*ln(5)**2+(-4*x**2-50*x-16)*ln(5)+2*x**3+3 3*x**2+208*x+64)/((x*ln(5)**2+(-2*x**2-10*x)*ln(5)+x**3+10*x**2+25*x)*exp( x)**2+(-2*x*ln(5)**2+(4*x**2+26*x)*ln(5)-2*x**3-26*x**2-80*x)*exp(x)+x*ln( 5)**2+(-2*x**2-16*x)*ln(5)+x**3+16*x**2+64*x),x)
Output:
x**2/(-x + (x - log(5) + 5)*exp(x) - 8 + log(5)) + 3*x + log(x)
Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {64+208 x+33 x^2+2 x^3+\left (-16-50 x-4 x^2\right ) \log (5)+(1+3 x) \log ^2(5)+e^x \left (-80-266 x-70 x^2-10 x^3-x^4+\left (26+82 x+10 x^2+x^3\right ) \log (5)+(-2-6 x) \log ^2(5)\right )+e^{2 x} \left (25+85 x+31 x^2+3 x^3+\left (-10-32 x-6 x^2\right ) \log (5)+(1+3 x) \log ^2(5)\right )}{64 x+16 x^2+x^3+\left (-16 x-2 x^2\right ) \log (5)+x \log ^2(5)+e^x \left (-80 x-26 x^2-2 x^3+\left (26 x+4 x^2\right ) \log (5)-2 x \log ^2(5)\right )+e^{2 x} \left (25 x+10 x^2+x^3+\left (-10 x-2 x^2\right ) \log (5)+x \log ^2(5)\right )} \, dx=-\frac {2 \, x^{2} - 3 \, x {\left (\log \left (5\right ) - 8\right )} - 3 \, {\left (x^{2} - x {\left (\log \left (5\right ) - 5\right )}\right )} e^{x}}{{\left (x - \log \left (5\right ) + 5\right )} e^{x} - x + \log \left (5\right ) - 8} + \log \left (x\right ) \] Input:
integrate((((1+3*x)*log(5)^2+(-6*x^2-32*x-10)*log(5)+3*x^3+31*x^2+85*x+25) *exp(x)^2+((-6*x-2)*log(5)^2+(x^3+10*x^2+82*x+26)*log(5)-x^4-10*x^3-70*x^2 -266*x-80)*exp(x)+(1+3*x)*log(5)^2+(-4*x^2-50*x-16)*log(5)+2*x^3+33*x^2+20 8*x+64)/((x*log(5)^2+(-2*x^2-10*x)*log(5)+x^3+10*x^2+25*x)*exp(x)^2+(-2*x* log(5)^2+(4*x^2+26*x)*log(5)-2*x^3-26*x^2-80*x)*exp(x)+x*log(5)^2+(-2*x^2- 16*x)*log(5)+x^3+16*x^2+64*x),x, algorithm="maxima")
Output:
-(2*x^2 - 3*x*(log(5) - 8) - 3*(x^2 - x*(log(5) - 5))*e^x)/((x - log(5) + 5)*e^x - x + log(5) - 8) + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (31) = 62\).
Time = 0.16 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.84 \[ \int \frac {64+208 x+33 x^2+2 x^3+\left (-16-50 x-4 x^2\right ) \log (5)+(1+3 x) \log ^2(5)+e^x \left (-80-266 x-70 x^2-10 x^3-x^4+\left (26+82 x+10 x^2+x^3\right ) \log (5)+(-2-6 x) \log ^2(5)\right )+e^{2 x} \left (25+85 x+31 x^2+3 x^3+\left (-10-32 x-6 x^2\right ) \log (5)+(1+3 x) \log ^2(5)\right )}{64 x+16 x^2+x^3+\left (-16 x-2 x^2\right ) \log (5)+x \log ^2(5)+e^x \left (-80 x-26 x^2-2 x^3+\left (26 x+4 x^2\right ) \log (5)-2 x \log ^2(5)\right )+e^{2 x} \left (25 x+10 x^2+x^3+\left (-10 x-2 x^2\right ) \log (5)+x \log ^2(5)\right )} \, dx=\frac {3 \, x^{2} e^{x} - 3 \, x e^{x} \log \left (5\right ) + x e^{x} \log \left (x\right ) - e^{x} \log \left (5\right ) \log \left (x\right ) - 2 \, x^{2} + 15 \, x e^{x} + 3 \, x \log \left (5\right ) - x \log \left (x\right ) + 5 \, e^{x} \log \left (x\right ) + \log \left (5\right ) \log \left (x\right ) - 24 \, x - 8 \, \log \left (x\right )}{x e^{x} - e^{x} \log \left (5\right ) - x + 5 \, e^{x} + \log \left (5\right ) - 8} \] Input:
integrate((((1+3*x)*log(5)^2+(-6*x^2-32*x-10)*log(5)+3*x^3+31*x^2+85*x+25) *exp(x)^2+((-6*x-2)*log(5)^2+(x^3+10*x^2+82*x+26)*log(5)-x^4-10*x^3-70*x^2 -266*x-80)*exp(x)+(1+3*x)*log(5)^2+(-4*x^2-50*x-16)*log(5)+2*x^3+33*x^2+20 8*x+64)/((x*log(5)^2+(-2*x^2-10*x)*log(5)+x^3+10*x^2+25*x)*exp(x)^2+(-2*x* log(5)^2+(4*x^2+26*x)*log(5)-2*x^3-26*x^2-80*x)*exp(x)+x*log(5)^2+(-2*x^2- 16*x)*log(5)+x^3+16*x^2+64*x),x, algorithm="giac")
Output:
(3*x^2*e^x - 3*x*e^x*log(5) + x*e^x*log(x) - e^x*log(5)*log(x) - 2*x^2 + 1 5*x*e^x + 3*x*log(5) - x*log(x) + 5*e^x*log(x) + log(5)*log(x) - 24*x - 8* log(x))/(x*e^x - e^x*log(5) - x + 5*e^x + log(5) - 8)
Time = 2.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.84 \[ \int \frac {64+208 x+33 x^2+2 x^3+\left (-16-50 x-4 x^2\right ) \log (5)+(1+3 x) \log ^2(5)+e^x \left (-80-266 x-70 x^2-10 x^3-x^4+\left (26+82 x+10 x^2+x^3\right ) \log (5)+(-2-6 x) \log ^2(5)\right )+e^{2 x} \left (25+85 x+31 x^2+3 x^3+\left (-10-32 x-6 x^2\right ) \log (5)+(1+3 x) \log ^2(5)\right )}{64 x+16 x^2+x^3+\left (-16 x-2 x^2\right ) \log (5)+x \log ^2(5)+e^x \left (-80 x-26 x^2-2 x^3+\left (26 x+4 x^2\right ) \log (5)-2 x \log ^2(5)\right )+e^{2 x} \left (25 x+10 x^2+x^3+\left (-10 x-2 x^2\right ) \log (5)+x \log ^2(5)\right )} \, dx=\frac {24\,x+8\,\ln \left (x\right )-3\,x^2\,{\mathrm {e}}^x-5\,{\mathrm {e}}^x\,\ln \left (x\right )-3\,x\,\ln \left (5\right )-\ln \left (5\right )\,\ln \left (x\right )-15\,x\,{\mathrm {e}}^x+x\,\ln \left (x\right )+2\,x^2+3\,x\,{\mathrm {e}}^x\,\ln \left (5\right )+{\mathrm {e}}^x\,\ln \left (5\right )\,\ln \left (x\right )-x\,{\mathrm {e}}^x\,\ln \left (x\right )}{x-\ln \left (5\right )-5\,{\mathrm {e}}^x+{\mathrm {e}}^x\,\ln \left (5\right )-x\,{\mathrm {e}}^x+8} \] Input:
int((208*x - exp(x)*(266*x - log(5)*(82*x + 10*x^2 + x^3 + 26) + log(5)^2* (6*x + 2) + 70*x^2 + 10*x^3 + x^4 + 80) - log(5)*(50*x + 4*x^2 + 16) + log (5)^2*(3*x + 1) + 33*x^2 + 2*x^3 + exp(2*x)*(85*x - log(5)*(32*x + 6*x^2 + 10) + log(5)^2*(3*x + 1) + 31*x^2 + 3*x^3 + 25) + 64)/(64*x - log(5)*(16* x + 2*x^2) + x*log(5)^2 + exp(2*x)*(25*x - log(5)*(10*x + 2*x^2) + x*log(5 )^2 + 10*x^2 + x^3) - exp(x)*(80*x - log(5)*(26*x + 4*x^2) + 2*x*log(5)^2 + 26*x^2 + 2*x^3) + 16*x^2 + x^3),x)
Output:
(24*x + 8*log(x) - 3*x^2*exp(x) - 5*exp(x)*log(x) - 3*x*log(5) - log(5)*lo g(x) - 15*x*exp(x) + x*log(x) + 2*x^2 + 3*x*exp(x)*log(5) + exp(x)*log(5)* log(x) - x*exp(x)*log(x))/(x - log(5) - 5*exp(x) + exp(x)*log(5) - x*exp(x ) + 8)
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.12 \[ \int \frac {64+208 x+33 x^2+2 x^3+\left (-16-50 x-4 x^2\right ) \log (5)+(1+3 x) \log ^2(5)+e^x \left (-80-266 x-70 x^2-10 x^3-x^4+\left (26+82 x+10 x^2+x^3\right ) \log (5)+(-2-6 x) \log ^2(5)\right )+e^{2 x} \left (25+85 x+31 x^2+3 x^3+\left (-10-32 x-6 x^2\right ) \log (5)+(1+3 x) \log ^2(5)\right )}{64 x+16 x^2+x^3+\left (-16 x-2 x^2\right ) \log (5)+x \log ^2(5)+e^x \left (-80 x-26 x^2-2 x^3+\left (26 x+4 x^2\right ) \log (5)-2 x \log ^2(5)\right )+e^{2 x} \left (25 x+10 x^2+x^3+\left (-10 x-2 x^2\right ) \log (5)+x \log ^2(5)\right )} \, dx=\frac {e^{x} \mathrm {log}\left (x \right ) \mathrm {log}\left (5\right )-e^{x} \mathrm {log}\left (x \right ) x -5 e^{x} \mathrm {log}\left (x \right )+3 e^{x} \mathrm {log}\left (5\right ) x -3 e^{x} x^{2}-15 e^{x} x -\mathrm {log}\left (x \right ) \mathrm {log}\left (5\right )+\mathrm {log}\left (x \right ) x +8 \,\mathrm {log}\left (x \right )-3 \,\mathrm {log}\left (5\right ) x +2 x^{2}+24 x}{e^{x} \mathrm {log}\left (5\right )-e^{x} x -5 e^{x}-\mathrm {log}\left (5\right )+x +8} \] Input:
int((((1+3*x)*log(5)^2+(-6*x^2-32*x-10)*log(5)+3*x^3+31*x^2+85*x+25)*exp(x )^2+((-6*x-2)*log(5)^2+(x^3+10*x^2+82*x+26)*log(5)-x^4-10*x^3-70*x^2-266*x -80)*exp(x)+(1+3*x)*log(5)^2+(-4*x^2-50*x-16)*log(5)+2*x^3+33*x^2+208*x+64 )/((x*log(5)^2+(-2*x^2-10*x)*log(5)+x^3+10*x^2+25*x)*exp(x)^2+(-2*x*log(5) ^2+(4*x^2+26*x)*log(5)-2*x^3-26*x^2-80*x)*exp(x)+x*log(5)^2+(-2*x^2-16*x)* log(5)+x^3+16*x^2+64*x),x)
Output:
(e**x*log(x)*log(5) - e**x*log(x)*x - 5*e**x*log(x) + 3*e**x*log(5)*x - 3* e**x*x**2 - 15*e**x*x - log(x)*log(5) + log(x)*x + 8*log(x) - 3*log(5)*x + 2*x**2 + 24*x)/(e**x*log(5) - e**x*x - 5*e**x - log(5) + x + 8)