Integrand size = 96, antiderivative size = 33 \[ \int \frac {100-20 e^3-15 x+8 x^2+e^x (10+5 x)+\left (50-10 e^3+5 e^x-5 x+2 x^2\right ) \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right )}{50-10 e^3+5 e^x-5 x+2 x^2} \, dx=x+x \left (2+\log \left (x+\frac {5 \left (5-e^3-x+\frac {1}{2} \left (e^x+x\right )\right )}{x}\right )\right ) \] Output:
x*(ln(5/x*(5+1/2*exp(x)-1/2*x-exp(3))+x)+2)+x
Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {100-20 e^3-15 x+8 x^2+e^x (10+5 x)+\left (50-10 e^3+5 e^x-5 x+2 x^2\right ) \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right )}{50-10 e^3+5 e^x-5 x+2 x^2} \, dx=3 x+x \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right ) \] Input:
Integrate[(100 - 20*E^3 - 15*x + 8*x^2 + E^x*(10 + 5*x) + (50 - 10*E^3 + 5 *E^x - 5*x + 2*x^2)*Log[(50 - 10*E^3 + 5*E^x - 5*x + 2*x^2)/(2*x)])/(50 - 10*E^3 + 5*E^x - 5*x + 2*x^2),x]
Output:
3*x + x*Log[(50 - 10*E^3 + 5*E^x - 5*x + 2*x^2)/(2*x)]
Time = 1.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {8 x^2+\left (2 x^2-5 x+5 e^x-10 e^3+50\right ) \log \left (\frac {2 x^2-5 x+5 e^x-10 e^3+50}{2 x}\right )-15 x+e^x (5 x+10)-20 e^3+100}{2 x^2-5 x+5 e^x-10 e^3+50} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {8 x^2+\left (2 x^2-5 x+5 e^x-10 e^3+50\right ) \log \left (\frac {2 x^2-5 x+5 e^x-10 e^3+50}{2 x}\right )-15 x+e^x (5 x+10)+100 \left (1-\frac {e^3}{5}\right )}{2 x^2-5 x+5 e^x+50 \left (1-\frac {e^3}{5}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (-2 x^2+9 x-5 \left (11-2 e^3\right )\right ) x}{2 x^2-5 x+5 e^x+50 \left (1-\frac {e^3}{5}\right )}+\log \left (\frac {2 x^2-5 x+5 e^x+50 \left (1-\frac {e^3}{5}\right )}{2 x}\right )+x+2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x \log \left (\frac {2 x^2-5 x+5 e^x+10 \left (5-e^3\right )}{2 x}\right )+3 x\) |
Input:
Int[(100 - 20*E^3 - 15*x + 8*x^2 + E^x*(10 + 5*x) + (50 - 10*E^3 + 5*E^x - 5*x + 2*x^2)*Log[(50 - 10*E^3 + 5*E^x - 5*x + 2*x^2)/(2*x)])/(50 - 10*E^3 + 5*E^x - 5*x + 2*x^2),x]
Output:
3*x + x*Log[(5*E^x + 10*(5 - E^3) - 5*x + 2*x^2)/(2*x)]
Time = 0.85 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94
method | result | size |
norman | \(\ln \left (\frac {5 \,{\mathrm e}^{x}-10 \,{\mathrm e}^{3}+2 x^{2}-5 x +50}{2 x}\right ) x +3 x\) | \(31\) |
parallelrisch | \(15+\ln \left (\frac {5 \,{\mathrm e}^{x}-10 \,{\mathrm e}^{3}+2 x^{2}-5 x +50}{2 x}\right ) x +3 x\) | \(32\) |
risch | \(x \ln \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )-x \ln \left (x \right )-i \pi x {\operatorname {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )}^{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )\right ) \operatorname {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )}^{3}}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+i \pi x +x \ln \left (5\right )+3 x\) | \(240\) |
Input:
int(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*ln(1/2*(5*exp(x)-10*exp(3)+2*x^2-5* x+50)/x)+(5*x+10)*exp(x)-20*exp(3)+8*x^2-15*x+100)/(5*exp(x)-10*exp(3)+2*x ^2-5*x+50),x,method=_RETURNVERBOSE)
Output:
ln(1/2*(5*exp(x)-10*exp(3)+2*x^2-5*x+50)/x)*x+3*x
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {100-20 e^3-15 x+8 x^2+e^x (10+5 x)+\left (50-10 e^3+5 e^x-5 x+2 x^2\right ) \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right )}{50-10 e^3+5 e^x-5 x+2 x^2} \, dx=x \log \left (\frac {2 \, x^{2} - 5 \, x - 10 \, e^{3} + 5 \, e^{x} + 50}{2 \, x}\right ) + 3 \, x \] Input:
integrate(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*log(1/2*(5*exp(x)-10*exp(3)+2 *x^2-5*x+50)/x)+(5*x+10)*exp(x)-20*exp(3)+8*x^2-15*x+100)/(5*exp(x)-10*exp (3)+2*x^2-5*x+50),x, algorithm="fricas")
Output:
x*log(1/2*(2*x^2 - 5*x - 10*e^3 + 5*e^x + 50)/x) + 3*x
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {100-20 e^3-15 x+8 x^2+e^x (10+5 x)+\left (50-10 e^3+5 e^x-5 x+2 x^2\right ) \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right )}{50-10 e^3+5 e^x-5 x+2 x^2} \, dx=x \log {\left (\frac {x^{2} - \frac {5 x}{2} + \frac {5 e^{x}}{2} - 5 e^{3} + 25}{x} \right )} + 3 x \] Input:
integrate(((5*exp(x)-10*exp(3)+2*x**2-5*x+50)*ln(1/2*(5*exp(x)-10*exp(3)+2 *x**2-5*x+50)/x)+(5*x+10)*exp(x)-20*exp(3)+8*x**2-15*x+100)/(5*exp(x)-10*e xp(3)+2*x**2-5*x+50),x)
Output:
x*log((x**2 - 5*x/2 + 5*exp(x)/2 - 5*exp(3) + 25)/x) + 3*x
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {100-20 e^3-15 x+8 x^2+e^x (10+5 x)+\left (50-10 e^3+5 e^x-5 x+2 x^2\right ) \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right )}{50-10 e^3+5 e^x-5 x+2 x^2} \, dx=-x {\left (\log \left (2\right ) - 3\right )} + x \log \left (2 \, x^{2} - 5 \, x - 10 \, e^{3} + 5 \, e^{x} + 50\right ) - x \log \left (x\right ) \] Input:
integrate(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*log(1/2*(5*exp(x)-10*exp(3)+2 *x^2-5*x+50)/x)+(5*x+10)*exp(x)-20*exp(3)+8*x^2-15*x+100)/(5*exp(x)-10*exp (3)+2*x^2-5*x+50),x, algorithm="maxima")
Output:
-x*(log(2) - 3) + x*log(2*x^2 - 5*x - 10*e^3 + 5*e^x + 50) - x*log(x)
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {100-20 e^3-15 x+8 x^2+e^x (10+5 x)+\left (50-10 e^3+5 e^x-5 x+2 x^2\right ) \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right )}{50-10 e^3+5 e^x-5 x+2 x^2} \, dx=x \log \left (\frac {2 \, x^{2} - 5 \, x - 10 \, e^{3} + 5 \, e^{x} + 50}{2 \, x}\right ) + 3 \, x \] Input:
integrate(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*log(1/2*(5*exp(x)-10*exp(3)+2 *x^2-5*x+50)/x)+(5*x+10)*exp(x)-20*exp(3)+8*x^2-15*x+100)/(5*exp(x)-10*exp (3)+2*x^2-5*x+50),x, algorithm="giac")
Output:
x*log(1/2*(2*x^2 - 5*x - 10*e^3 + 5*e^x + 50)/x) + 3*x
Time = 3.98 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {100-20 e^3-15 x+8 x^2+e^x (10+5 x)+\left (50-10 e^3+5 e^x-5 x+2 x^2\right ) \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right )}{50-10 e^3+5 e^x-5 x+2 x^2} \, dx=x\,\left (\ln \left (\frac {\frac {5\,{\mathrm {e}}^x}{2}-5\,{\mathrm {e}}^3-\frac {5\,x}{2}+x^2+25}{x}\right )+3\right ) \] Input:
int((exp(x)*(5*x + 10) - 20*exp(3) - 15*x + log(((5*exp(x))/2 - 5*exp(3) - (5*x)/2 + x^2 + 25)/x)*(5*exp(x) - 10*exp(3) - 5*x + 2*x^2 + 50) + 8*x^2 + 100)/(5*exp(x) - 10*exp(3) - 5*x + 2*x^2 + 50),x)
Output:
x*(log(((5*exp(x))/2 - 5*exp(3) - (5*x)/2 + x^2 + 25)/x) + 3)
\[ \int \frac {100-20 e^3-15 x+8 x^2+e^x (10+5 x)+\left (50-10 e^3+5 e^x-5 x+2 x^2\right ) \log \left (\frac {50-10 e^3+5 e^x-5 x+2 x^2}{2 x}\right )}{50-10 e^3+5 e^x-5 x+2 x^2} \, dx=-10 \left (\int \frac {\mathrm {log}\left (\frac {5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}{2 x}\right )}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right ) e^{3}+50 \left (\int \frac {\mathrm {log}\left (\frac {5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}{2 x}\right )}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right )+5 \left (\int \frac {e^{x} \mathrm {log}\left (\frac {5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}{2 x}\right )}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right )+5 \left (\int \frac {e^{x} x}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right )+2 \left (\int \frac {\mathrm {log}\left (\frac {5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}{2 x}\right ) x^{2}}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right )-5 \left (\int \frac {\mathrm {log}\left (\frac {5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}{2 x}\right ) x}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right )+13 \left (\int \frac {x}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right )+20 \left (\int \frac {1}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right ) e^{3}-110 \left (\int \frac {1}{5 e^{x}-10 e^{3}+2 x^{2}-5 x +50}d x \right )-2 \,\mathrm {log}\left (5 e^{x}-10 e^{3}+2 x^{2}-5 x +50\right )+4 x \] Input:
int(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*log(1/2*(5*exp(x)-10*exp(3)+2*x^2-5 *x+50)/x)+(5*x+10)*exp(x)-20*exp(3)+8*x^2-15*x+100)/(5*exp(x)-10*exp(3)+2* x^2-5*x+50),x)
Output:
- 10*int(log((5*e**x - 10*e**3 + 2*x**2 - 5*x + 50)/(2*x))/(5*e**x - 10*e **3 + 2*x**2 - 5*x + 50),x)*e**3 + 50*int(log((5*e**x - 10*e**3 + 2*x**2 - 5*x + 50)/(2*x))/(5*e**x - 10*e**3 + 2*x**2 - 5*x + 50),x) + 5*int((e**x* log((5*e**x - 10*e**3 + 2*x**2 - 5*x + 50)/(2*x)))/(5*e**x - 10*e**3 + 2*x **2 - 5*x + 50),x) + 5*int((e**x*x)/(5*e**x - 10*e**3 + 2*x**2 - 5*x + 50) ,x) + 2*int((log((5*e**x - 10*e**3 + 2*x**2 - 5*x + 50)/(2*x))*x**2)/(5*e* *x - 10*e**3 + 2*x**2 - 5*x + 50),x) - 5*int((log((5*e**x - 10*e**3 + 2*x* *2 - 5*x + 50)/(2*x))*x)/(5*e**x - 10*e**3 + 2*x**2 - 5*x + 50),x) + 13*in t(x/(5*e**x - 10*e**3 + 2*x**2 - 5*x + 50),x) + 20*int(1/(5*e**x - 10*e**3 + 2*x**2 - 5*x + 50),x)*e**3 - 110*int(1/(5*e**x - 10*e**3 + 2*x**2 - 5*x + 50),x) - 2*log(5*e**x - 10*e**3 + 2*x**2 - 5*x + 50) + 4*x