Integrand size = 64, antiderivative size = 31 \[ \int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5+2 e^3+10 \log (x)} \, dx=x \left (-4+x-\log \left (-1+\frac {e^3}{5}-x+\frac {1}{2} (3+2 x)+\log (x)\right )\right ) \] Output:
(x-4-ln(ln(x)+1/5*exp(3)+1/2))*x
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5+2 e^3+10 \log (x)} \, dx=-4 x+x^2-x \log \left (\frac {1}{2}+\frac {e^3}{5}+\log (x)\right ) \] Input:
Integrate[(-30 + 10*x + E^3*(-8 + 4*x) + (-40 + 20*x)*Log[x] + (-5 - 2*E^3 - 10*Log[x])*Log[(5 + 2*E^3 + 10*Log[x])/10])/(5 + 2*E^3 + 10*Log[x]),x]
Output:
-4*x + x^2 - x*Log[1/2 + E^3/5 + Log[x]]
Time = 0.95 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, 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.
\(\displaystyle \int \frac {10 x+e^3 (4 x-8)+(20 x-40) \log (x)+\left (-10 \log (x)-2 e^3-5\right ) \log \left (\frac {1}{10} \left (10 \log (x)+2 e^3+5\right )\right )-30}{10 \log (x)+2 e^3+5} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {10 x+e^3 (4 x-8)+(20 x-40) \log (x)+\left (-10 \log (x)-2 e^3-5\right ) \log \left (\frac {1}{10} \left (10 \log (x)+2 e^3+5\right )\right )-30}{10 \log (x)+5 \left (1+\frac {2 e^3}{5}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (5 \left (1+\frac {2 e^3}{5}\right ) x+10 x \log (x)-20 \log (x)-15 \left (1+\frac {4 e^3}{15}\right )\right )}{10 \log (x)+5 \left (1+\frac {2 e^3}{5}\right )}-\log \left (\frac {1}{10} \left (10 \log (x)+5 \left (1+\frac {2 e^3}{5}\right )\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2-4 x-x \log \left (\frac {1}{10} \left (10 \log (x)+2 e^3+5\right )\right )\) |
Input:
Int[(-30 + 10*x + E^3*(-8 + 4*x) + (-40 + 20*x)*Log[x] + (-5 - 2*E^3 - 10* Log[x])*Log[(5 + 2*E^3 + 10*Log[x])/10])/(5 + 2*E^3 + 10*Log[x]),x]
Output:
-4*x + x^2 - x*Log[(5 + 2*E^3 + 10*Log[x])/10]
Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65
method | result | size |
norman | \(x^{2}-4 x -x \ln \left (\ln \left (x \right )+\frac {{\mathrm e}^{3}}{5}+\frac {1}{2}\right )\) | \(20\) |
risch | \(x^{2}-4 x -x \ln \left (\ln \left (x \right )+\frac {{\mathrm e}^{3}}{5}+\frac {1}{2}\right )\) | \(20\) |
parallelrisch | \(x^{2}-4 x -x \ln \left (\ln \left (x \right )+\frac {{\mathrm e}^{3}}{5}+\frac {1}{2}\right )\) | \(20\) |
default | \(-4 x +x^{2}-\ln \left (10 \ln \left (x \right )+2 \,{\mathrm e}^{3}+5\right ) x +x \ln \left (10\right )\) | \(26\) |
Input:
int(((-10*ln(x)-2*exp(3)-5)*ln(ln(x)+1/5*exp(3)+1/2)+(20*x-40)*ln(x)+(4*x- 8)*exp(3)+10*x-30)/(10*ln(x)+2*exp(3)+5),x,method=_RETURNVERBOSE)
Output:
x^2-4*x-x*ln(ln(x)+1/5*exp(3)+1/2)
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5+2 e^3+10 \log (x)} \, dx=x^{2} - x \log \left (\frac {1}{5} \, e^{3} + \log \left (x\right ) + \frac {1}{2}\right ) - 4 \, x \] Input:
integrate(((-10*log(x)-2*exp(3)-5)*log(log(x)+1/5*exp(3)+1/2)+(20*x-40)*lo g(x)+(4*x-8)*exp(3)+10*x-30)/(10*log(x)+2*exp(3)+5),x, algorithm="fricas")
Output:
x^2 - x*log(1/5*e^3 + log(x) + 1/2) - 4*x
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5+2 e^3+10 \log (x)} \, dx=x^{2} - x \log {\left (\log {\left (x \right )} + \frac {1}{2} + \frac {e^{3}}{5} \right )} - 4 x \] Input:
integrate(((-10*ln(x)-2*exp(3)-5)*ln(ln(x)+1/5*exp(3)+1/2)+(20*x-40)*ln(x) +(4*x-8)*exp(3)+10*x-30)/(10*ln(x)+2*exp(3)+5),x)
Output:
x**2 - x*log(log(x) + 1/2 + exp(3)/5) - 4*x
\[ \int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5+2 e^3+10 \log (x)} \, dx=\int { \frac {4 \, {\left (x - 2\right )} e^{3} + 20 \, {\left (x - 2\right )} \log \left (x\right ) - {\left (2 \, e^{3} + 10 \, \log \left (x\right ) + 5\right )} \log \left (\frac {1}{5} \, e^{3} + \log \left (x\right ) + \frac {1}{2}\right ) + 10 \, x - 30}{2 \, e^{3} + 10 \, \log \left (x\right ) + 5} \,d x } \] Input:
integrate(((-10*log(x)-2*exp(3)-5)*log(log(x)+1/5*exp(3)+1/2)+(20*x-40)*lo g(x)+(4*x-8)*exp(3)+10*x-30)/(10*log(x)+2*exp(3)+5),x, algorithm="maxima")
Output:
4*e^(-1/5*e^3 - 1/2)*exp_integral_e(1, -1/5*e^3 - log(x) - 1/2)*log(x) - 2 *e^(-2/5*e^3 - 1)*exp_integral_e(1, -2/5*e^3 - 2*log(x) - 1)*log(x) + x*(l og(5) + log(2)) - 4*e^(-1/5*e^3 - 1/2)*exp_integral_e(2, -1/5*e^3 - log(x) - 1/2) + e^(-2/5*e^3 - 1)*exp_integral_e(2, -2/5*e^3 - 2*log(x) - 1) + 4/ 5*e^(-1/5*e^3 + 5/2)*exp_integral_e(1, -1/5*e^3 - log(x) - 1/2) + 3*e^(-1/ 5*e^3 - 1/2)*exp_integral_e(1, -1/5*e^3 - log(x) - 1/2) - 2/5*e^(-2/5*e^3 + 2)*exp_integral_e(1, -2/5*e^3 - 2*log(x) - 1) - e^(-2/5*e^3 - 1)*exp_int egral_e(1, -2/5*e^3 - 2*log(x) - 1) - x*log(2*e^3 + 10*log(x) + 5) + 10*in tegrate(1/(2*e^3 + 10*log(x) + 5), x)
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5+2 e^3+10 \log (x)} \, dx=x^{2} + x \log \left (10\right ) - x \log \left (2 \, e^{3} + 10 \, \log \left (x\right ) + 5\right ) - 4 \, x \] Input:
integrate(((-10*log(x)-2*exp(3)-5)*log(log(x)+1/5*exp(3)+1/2)+(20*x-40)*lo g(x)+(4*x-8)*exp(3)+10*x-30)/(10*log(x)+2*exp(3)+5),x, algorithm="giac")
Output:
x^2 + x*log(10) - x*log(2*e^3 + 10*log(x) + 5) - 4*x
Time = 2.96 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5+2 e^3+10 \log (x)} \, dx=-x\,\left (\ln \left (\frac {{\mathrm {e}}^3}{5}+\ln \left (x\right )+\frac {1}{2}\right )-x+4\right ) \] Input:
int((10*x - log(exp(3)/5 + log(x) + 1/2)*(2*exp(3) + 10*log(x) + 5) + log( x)*(20*x - 40) + exp(3)*(4*x - 8) - 30)/(2*exp(3) + 10*log(x) + 5),x)
Output:
-x*(log(exp(3)/5 + log(x) + 1/2) - x + 4)
Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5+2 e^3+10 \log (x)} \, dx=x \left (-\mathrm {log}\left (\mathrm {log}\left (x \right )+\frac {e^{3}}{5}+\frac {1}{2}\right )+x -4\right ) \] Input:
int(((-10*log(x)-2*exp(3)-5)*log(log(x)+1/5*exp(3)+1/2)+(20*x-40)*log(x)+( 4*x-8)*exp(3)+10*x-30)/(10*log(x)+2*exp(3)+5),x)
Output:
x*( - log((10*log(x) + 2*e**3 + 5)/10) + x - 4)