Integrand size = 88, antiderivative size = 22 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=-3+\log (4-x)+\frac {1}{e^{10} (1-\log (-1+x))} \] Output:
-3+ln(4-x)+1/exp(5)^2/(1-ln(-1+x))
Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\frac {e^{10} \log (4-x)-\frac {1}{-1+\log (-1+x)}}{e^{10}} \] Input:
Integrate[(-4 + E^10*(-1 + x) + x + E^10*(2 - 2*x)*Log[-1 + x] + E^10*(-1 + x)*Log[-1 + x]^2)/(E^10*(4 - 5*x + x^2) + E^10*(-8 + 10*x - 2*x^2)*Log[- 1 + x] + E^10*(4 - 5*x + x^2)*Log[-1 + x]^2),x]
Output:
(E^10*Log[4 - x] - (-1 + Log[-1 + x])^(-1))/E^10
Time = 0.58 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {7292, 27, 25, 7279, 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.
\(\displaystyle \int \frac {e^{10} (x-1)+x+e^{10} (x-1) \log ^2(x-1)+e^{10} (2-2 x) \log (x-1)-4}{e^{10} \left (x^2-5 x+4\right )+e^{10} \left (x^2-5 x+4\right ) \log ^2(x-1)+e^{10} \left (-2 x^2+10 x-8\right ) \log (x-1)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{10} (x-1)+x+e^{10} (x-1) \log ^2(x-1)+e^{10} (2-2 x) \log (x-1)-4}{e^{10} \left (x^2-5 x+4\right ) (1-\log (x-1))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^{10} (1-x) \log ^2(x-1)-2 e^{10} (1-x) \log (x-1)+e^{10} (1-x)-x+4}{\left (x^2-5 x+4\right ) (1-\log (x-1))^2}dx}{e^{10}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{10} (1-x) \log ^2(x-1)-2 e^{10} (1-x) \log (x-1)+e^{10} (1-x)-x+4}{\left (x^2-5 x+4\right ) (1-\log (x-1))^2}dx}{e^{10}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle -\frac {\int \left (-\frac {1}{(x-1) (\log (x-1)-1)^2}-\frac {e^{10}}{x-4}\right )dx}{e^{10}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-e^{10} \log (4-x)-\frac {1}{1-\log (x-1)}}{e^{10}}\) |
Input:
Int[(-4 + E^10*(-1 + x) + x + E^10*(2 - 2*x)*Log[-1 + x] + E^10*(-1 + x)*L og[-1 + x]^2)/(E^10*(4 - 5*x + x^2) + E^10*(-8 + 10*x - 2*x^2)*Log[-1 + x] + E^10*(4 - 5*x + x^2)*Log[-1 + x]^2),x]
Output:
-((-(E^10*Log[4 - x]) - (1 - Log[-1 + x])^(-1))/E^10)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.55 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-\frac {{\mathrm e}^{-10}}{\ln \left (-1+x \right )-1}+\ln \left (x -4\right )\) | \(18\) |
norman | \(-\frac {{\mathrm e}^{-10}}{\ln \left (-1+x \right )-1}+\ln \left (x -4\right )\) | \(20\) |
derivativedivides | \({\mathrm e}^{-10} \left (-\frac {1}{\ln \left (-1+x \right )-1}+\ln \left (x -4\right ) {\mathrm e}^{10}\right )\) | \(26\) |
default | \({\mathrm e}^{-10} \left (-\frac {1}{\ln \left (-1+x \right )-1}+\ln \left (x -4\right ) {\mathrm e}^{10}\right )\) | \(26\) |
parallelrisch | \(\frac {\left ({\mathrm e}^{10} \ln \left (x -4\right ) \ln \left (-1+x \right )-1-\ln \left (x -4\right ) {\mathrm e}^{10}\right ) {\mathrm e}^{-10}}{\ln \left (-1+x \right )-1}\) | \(39\) |
Input:
int(((-1+x)*exp(5)^2*ln(-1+x)^2+(2-2*x)*exp(5)^2*ln(-1+x)+(-1+x)*exp(5)^2+ x-4)/((x^2-5*x+4)*exp(5)^2*ln(-1+x)^2+(-2*x^2+10*x-8)*exp(5)^2*ln(-1+x)+(x ^2-5*x+4)*exp(5)^2),x,method=_RETURNVERBOSE)
Output:
-exp(-10)/(ln(-1+x)-1)+ln(x-4)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\frac {e^{10} \log \left (x - 1\right ) \log \left (x - 4\right ) - e^{10} \log \left (x - 4\right ) - 1}{e^{10} \log \left (x - 1\right ) - e^{10}} \] Input:
integrate(((-1+x)*exp(5)^2*log(-1+x)^2+(2-2*x)*exp(5)^2*log(-1+x)+(-1+x)*e xp(5)^2+x-4)/((x^2-5*x+4)*exp(5)^2*log(-1+x)^2+(-2*x^2+10*x-8)*exp(5)^2*lo g(-1+x)+(x^2-5*x+4)*exp(5)^2),x, algorithm="fricas")
Output:
(e^10*log(x - 1)*log(x - 4) - e^10*log(x - 4) - 1)/(e^10*log(x - 1) - e^10 )
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\log {\left (x - 4 \right )} - \frac {1}{e^{10} \log {\left (x - 1 \right )} - e^{10}} \] Input:
integrate(((-1+x)*exp(5)**2*ln(-1+x)**2+(2-2*x)*exp(5)**2*ln(-1+x)+(-1+x)* exp(5)**2+x-4)/((x**2-5*x+4)*exp(5)**2*ln(-1+x)**2+(-2*x**2+10*x-8)*exp(5) **2*ln(-1+x)+(x**2-5*x+4)*exp(5)**2),x)
Output:
log(x - 4) - 1/(exp(10)*log(x - 1) - exp(10))
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=-\frac {1}{e^{10} \log \left (x - 1\right ) - e^{10}} + \log \left (x - 4\right ) \] Input:
integrate(((-1+x)*exp(5)^2*log(-1+x)^2+(2-2*x)*exp(5)^2*log(-1+x)+(-1+x)*e xp(5)^2+x-4)/((x^2-5*x+4)*exp(5)^2*log(-1+x)^2+(-2*x^2+10*x-8)*exp(5)^2*lo g(-1+x)+(x^2-5*x+4)*exp(5)^2),x, algorithm="maxima")
Output:
-1/(e^10*log(x - 1) - e^10) + log(x - 4)
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\frac {e^{10} \log \left (x - 1\right ) \log \left (x - 4\right ) - e^{10} \log \left (x - 4\right ) - 1}{e^{10} \log \left (x - 1\right ) - e^{10}} \] Input:
integrate(((-1+x)*exp(5)^2*log(-1+x)^2+(2-2*x)*exp(5)^2*log(-1+x)+(-1+x)*e xp(5)^2+x-4)/((x^2-5*x+4)*exp(5)^2*log(-1+x)^2+(-2*x^2+10*x-8)*exp(5)^2*lo g(-1+x)+(x^2-5*x+4)*exp(5)^2),x, algorithm="giac")
Output:
(e^10*log(x - 1)*log(x - 4) - e^10*log(x - 4) - 1)/(e^10*log(x - 1) - e^10 )
Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\ln \left (x-4\right )-\frac {{\mathrm {e}}^{-10}}{\ln \left (x-1\right )-1} \] Input:
int((x + exp(10)*(x - 1) - log(x - 1)*exp(10)*(2*x - 2) + log(x - 1)^2*exp (10)*(x - 1) - 4)/(exp(10)*(x^2 - 5*x + 4) - log(x - 1)*exp(10)*(2*x^2 - 1 0*x + 8) + log(x - 1)^2*exp(10)*(x^2 - 5*x + 4)),x)
Output:
log(x - 4) - exp(-10)/(log(x - 1) - 1)
Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\frac {\mathrm {log}\left (x -4\right ) \mathrm {log}\left (x -1\right ) e^{10}-\mathrm {log}\left (x -4\right ) e^{10}-\mathrm {log}\left (x -1\right )}{e^{10} \left (\mathrm {log}\left (x -1\right )-1\right )} \] Input:
int(((-1+x)*exp(5)^2*log(-1+x)^2+(2-2*x)*exp(5)^2*log(-1+x)+(-1+x)*exp(5)^ 2+x-4)/((x^2-5*x+4)*exp(5)^2*log(-1+x)^2+(-2*x^2+10*x-8)*exp(5)^2*log(-1+x )+(x^2-5*x+4)*exp(5)^2),x)
Output:
(log(x - 4)*log(x - 1)*e**10 - log(x - 4)*e**10 - log(x - 1))/(e**10*(log( x - 1) - 1))