Integrand size = 69, antiderivative size = 27 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x+\frac {4+2 \left (-2-\left (-3+e^3\right )^2+x\right )}{\log \left (-\frac {1}{5}+x\right )} \] Output:
x+(-2*(exp(3)-3)^2+2*x)/ln(x-1/5)
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x-\frac {2 \left (9-6 e^3+e^6-x\right )}{\log \left (-\frac {1}{5}+x\right )} \] Input:
Integrate[(90 - 60*E^3 + 10*E^6 - 10*x + (-2 + 10*x)*Log[(-1 + 5*x)/5] + ( -1 + 5*x)*Log[(-1 + 5*x)/5]^2)/((-1 + 5*x)*Log[(-1 + 5*x)/5]^2),x]
Output:
x - (2*(9 - 6*E^3 + E^6 - x))/Log[-1/5 + x]
Time = 1.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {7292, 7239, 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.
\(\displaystyle \int \frac {-10 x+(5 x-1) \log ^2\left (\frac {1}{5} (5 x-1)\right )+(10 x-2) \log \left (\frac {1}{5} (5 x-1)\right )+10 e^6-60 e^3+90}{(5 x-1) \log ^2\left (\frac {1}{5} (5 x-1)\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {10 x-\left ((5 x-1) \log ^2\left (\frac {1}{5} (5 x-1)\right )\right )-(10 x-2) \log \left (\frac {1}{5} (5 x-1)\right )-90 \left (1+\frac {1}{9} e^3 \left (e^3-6\right )\right )}{(1-5 x) \log ^2\left (x-\frac {1}{5}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {10 \left (-x+e^6-6 e^3+9\right )}{(5 x-1) \log ^2\left (x-\frac {1}{5}\right )}+\frac {2}{\log \left (x-\frac {1}{5}\right )}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x-\frac {2 (1-5 x)}{5 \log \left (x-\frac {1}{5}\right )}-\frac {2 \left (44-30 e^3+5 e^6\right )}{5 \log \left (x-\frac {1}{5}\right )}\) |
Input:
Int[(90 - 60*E^3 + 10*E^6 - 10*x + (-2 + 10*x)*Log[(-1 + 5*x)/5] + (-1 + 5 *x)*Log[(-1 + 5*x)/5]^2)/((-1 + 5*x)*Log[(-1 + 5*x)/5]^2),x]
Output:
x - (2*(44 - 30*E^3 + 5*E^6))/(5*Log[-1/5 + x]) - (2*(1 - 5*x))/(5*Log[-1/ 5 + x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
risch | \(x -\frac {2 \left ({\mathrm e}^{6}-6 \,{\mathrm e}^{3}-x +9\right )}{\ln \left (x -\frac {1}{5}\right )}\) | \(22\) |
norman | \(\frac {\ln \left (x -\frac {1}{5}\right ) x +12 \,{\mathrm e}^{3}-2 \,{\mathrm e}^{6}+2 x -18}{\ln \left (x -\frac {1}{5}\right )}\) | \(29\) |
parallelrisch | \(-\frac {2 \,{\mathrm e}^{6}-\ln \left (x -\frac {1}{5}\right ) x +18-12 \,{\mathrm e}^{3}-2 x}{\ln \left (x -\frac {1}{5}\right )}\) | \(31\) |
parts | \(x -\frac {2 \,{\mathrm e}^{6}}{\ln \left (x -\frac {1}{5}\right )}+\frac {12 \,{\mathrm e}^{3}}{\ln \left (x -\frac {1}{5}\right )}+\frac {2 x -\frac {2}{5}}{\ln \left (x -\frac {1}{5}\right )}-\frac {88}{5 \ln \left (x -\frac {1}{5}\right )}\) | \(44\) |
derivativedivides | \(x -\frac {1}{5}-\frac {2 \,{\mathrm e}^{6}}{\ln \left (x -\frac {1}{5}\right )}+\frac {12 \,{\mathrm e}^{3}}{\ln \left (x -\frac {1}{5}\right )}-\frac {88}{5 \ln \left (x -\frac {1}{5}\right )}+\frac {2 x -\frac {2}{5}}{\ln \left (x -\frac {1}{5}\right )}\) | \(45\) |
default | \(x -\frac {1}{5}-\frac {2 \,{\mathrm e}^{6}}{\ln \left (x -\frac {1}{5}\right )}+\frac {12 \,{\mathrm e}^{3}}{\ln \left (x -\frac {1}{5}\right )}-\frac {88}{5 \ln \left (x -\frac {1}{5}\right )}+\frac {2 x -\frac {2}{5}}{\ln \left (x -\frac {1}{5}\right )}\) | \(45\) |
Input:
int(((5*x-1)*ln(x-1/5)^2+(10*x-2)*ln(x-1/5)+10*exp(3)^2-60*exp(3)-10*x+90) /(5*x-1)/ln(x-1/5)^2,x,method=_RETURNVERBOSE)
Output:
x-2*(exp(6)-6*exp(3)-x+9)/ln(x-1/5)
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=\frac {x \log \left (x - \frac {1}{5}\right ) + 2 \, x - 2 \, e^{6} + 12 \, e^{3} - 18}{\log \left (x - \frac {1}{5}\right )} \] Input:
integrate(((5*x-1)*log(x-1/5)^2+(10*x-2)*log(x-1/5)+10*exp(3)^2-60*exp(3)- 10*x+90)/(5*x-1)/log(x-1/5)^2,x, algorithm="fricas")
Output:
(x*log(x - 1/5) + 2*x - 2*e^6 + 12*e^3 - 18)/log(x - 1/5)
Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x + \frac {2 x - 2 e^{6} - 18 + 12 e^{3}}{\log {\left (x - \frac {1}{5} \right )}} \] Input:
integrate(((5*x-1)*ln(x-1/5)**2+(10*x-2)*ln(x-1/5)+10*exp(3)**2-60*exp(3)- 10*x+90)/(5*x-1)/ln(x-1/5)**2,x)
Output:
x + (2*x - 2*exp(6) - 18 + 12*exp(3))/log(x - 1/5)
Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (20) = 40\).
Time = 0.17 (sec) , antiderivative size = 171, normalized size of antiderivative = 6.33 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=-\frac {2}{5} \, {\left (\log \left (5\right ) - \log \left (5 \, x - 1\right )\right )} \log \left (-\log \left (5\right ) + \log \left (5 \, x - 1\right )\right ) - \frac {2}{5} \, \log \left (x - \frac {1}{5}\right ) \log \left (-\log \left (5\right ) + \log \left (5 \, x - 1\right )\right ) - \frac {\log \left (x - \frac {1}{5}\right )^{2}}{5 \, {\left (\log \left (5\right ) - \log \left (5 \, x - 1\right )\right )}} + \frac {x {\left (\log \left (5\right ) - 2\right )} - x \log \left (5 \, x - 1\right )}{\log \left (5\right ) - \log \left (5 \, x - 1\right )} + \frac {2 \, e^{6}}{\log \left (5\right ) - \log \left (5 \, x - 1\right )} - \frac {12 \, e^{3}}{\log \left (5\right ) - \log \left (5 \, x - 1\right )} - \frac {2 \, \log \left (x - \frac {1}{5}\right )}{5 \, {\left (\log \left (5\right ) - \log \left (5 \, x - 1\right )\right )}} + \frac {18}{\log \left (5\right ) - \log \left (5 \, x - 1\right )} - \frac {1}{5} \, \log \left (5 \, x - 1\right ) \] Input:
integrate(((5*x-1)*log(x-1/5)^2+(10*x-2)*log(x-1/5)+10*exp(3)^2-60*exp(3)- 10*x+90)/(5*x-1)/log(x-1/5)^2,x, algorithm="maxima")
Output:
-2/5*(log(5) - log(5*x - 1))*log(-log(5) + log(5*x - 1)) - 2/5*log(x - 1/5 )*log(-log(5) + log(5*x - 1)) - 1/5*log(x - 1/5)^2/(log(5) - log(5*x - 1)) + (x*(log(5) - 2) - x*log(5*x - 1))/(log(5) - log(5*x - 1)) + 2*e^6/(log( 5) - log(5*x - 1)) - 12*e^3/(log(5) - log(5*x - 1)) - 2/5*log(x - 1/5)/(lo g(5) - log(5*x - 1)) + 18/(log(5) - log(5*x - 1)) - 1/5*log(5*x - 1)
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=\frac {x \log \left (x - \frac {1}{5}\right ) + 2 \, x - 2 \, e^{6} + 12 \, e^{3} - 18}{\log \left (x - \frac {1}{5}\right )} \] Input:
integrate(((5*x-1)*log(x-1/5)^2+(10*x-2)*log(x-1/5)+10*exp(3)^2-60*exp(3)- 10*x+90)/(5*x-1)/log(x-1/5)^2,x, algorithm="giac")
Output:
(x*log(x - 1/5) + 2*x - 2*e^6 + 12*e^3 - 18)/log(x - 1/5)
Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x+\frac {2\,x+12\,{\mathrm {e}}^3-2\,{\mathrm {e}}^6-18}{\ln \left (x-\frac {1}{5}\right )} \] Input:
int((10*exp(6) - 60*exp(3) - 10*x + log(x - 1/5)^2*(5*x - 1) + log(x - 1/5 )*(10*x - 2) + 90)/(log(x - 1/5)^2*(5*x - 1)),x)
Output:
x + (2*x + 12*exp(3) - 2*exp(6) - 18)/log(x - 1/5)
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=\frac {\mathrm {log}\left (x -\frac {1}{5}\right ) x -2 e^{6}+12 e^{3}+2 x -18}{\mathrm {log}\left (x -\frac {1}{5}\right )} \] Input:
int(((5*x-1)*log(x-1/5)^2+(10*x-2)*log(x-1/5)+10*exp(3)^2-60*exp(3)-10*x+9 0)/(5*x-1)/log(x-1/5)^2,x)
Output:
(log((5*x - 1)/5)*x - 2*e**6 + 12*e**3 + 2*x - 18)/log((5*x - 1)/5)