Integrand size = 69, antiderivative size = 31 \[ \int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx=2-2 x-x^2+\frac {1+x^2}{\log (5)}+(1-x) \log (\log (-2+x)) \] Output:
2+ln(ln(-2+x))*(1-x)-2*x-x^2+(x^2+1)/ln(5)
Time = 0.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx=-\frac {(-2+x)^2 (-2+\log (25))}{2 \log (5)}-\frac {(-2+x) (-4+3 \log (25))}{\log (5)}-\log (\log (-2+x))-(-2+x) \log (\log (-2+x)) \] Input:
Integrate[((1 - x)*Log[5] + (-4*x + 2*x^2 + (4 + 2*x - 2*x^2)*Log[5])*Log[ -2 + x] + (2 - x)*Log[5]*Log[-2 + x]*Log[Log[-2 + x]])/((-2 + x)*Log[5]*Lo g[-2 + x]),x]
Output:
-1/2*((-2 + x)^2*(-2 + Log[25]))/Log[5] - ((-2 + x)*(-4 + 3*Log[25]))/Log[ 5] - Log[Log[-2 + x]] - (-2 + x)*Log[Log[-2 + x]]
Time = 0.45 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {\left (2 x^2+\left (-2 x^2+2 x+4\right ) \log (5)-4 x\right ) \log (x-2)+(1-x) \log (5)+(2-x) \log (5) \log (x-2) \log (\log (x-2))}{(x-2) \log (5) \log (x-2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {\log (5) (1-x)-2 \left (-x^2+2 x-\left (-x^2+x+2\right ) \log (5)\right ) \log (x-2)+(2-x) \log (5) \log (x-2) \log (\log (x-2))}{(2-x) \log (x-2)}dx}{\log (5)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\log (5) (1-x)-2 \left (-x^2+2 x-\left (-x^2+x+2\right ) \log (5)\right ) \log (x-2)+(2-x) \log (5) \log (x-2) \log (\log (x-2))}{(2-x) \log (x-2)}dx}{\log (5)}\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle -\frac {\int \left (\frac {\log (5) (x-1)}{(x-2) \log (x-2)}+\log (5) \log (\log (x-2))+\log (25)+x (-2+\log (25))\right )dx}{\log (5)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{2} x^2 (2-\log (25))+x \log (25)-(2-x) \log (5) \log (\log (x-2))+\log (5) \log (\log (x-2))}{\log (5)}\) |
Input:
Int[((1 - x)*Log[5] + (-4*x + 2*x^2 + (4 + 2*x - 2*x^2)*Log[5])*Log[-2 + x ] + (2 - x)*Log[5]*Log[-2 + x]*Log[Log[-2 + x]])/((-2 + x)*Log[5]*Log[-2 + x]),x]
Output:
-((-1/2*(x^2*(2 - Log[25])) + x*Log[25] + Log[5]*Log[Log[-2 + x]] - (2 - x )*Log[5]*Log[Log[-2 + x]])/Log[5])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.40 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\ln \left (\ln \left (-2+x \right )\right )-2 x -\ln \left (\ln \left (-2+x \right )\right ) x -\frac {\left (\ln \left (5\right )-1\right ) x^{2}}{\ln \left (5\right )}\) | \(31\) |
risch | \(-\ln \left (\ln \left (-2+x \right )\right ) x -x^{2}-2 x +\frac {x^{2}}{\ln \left (5\right )}+\ln \left (\ln \left (-2+x \right )\right )\) | \(31\) |
parts | \(-\ln \left (\ln \left (-2+x \right )\right ) x -x^{2}-2 x +\frac {x^{2}}{\ln \left (5\right )}+\ln \left (\ln \left (-2+x \right )\right )\) | \(31\) |
default | \(\frac {\ln \left (5\right ) \left (-x^{2}-2 x +\ln \left (\ln \left (-2+x \right )\right )-\ln \left (\ln \left (-2+x \right )\right ) x \right )+x^{2}}{\ln \left (5\right )}\) | \(35\) |
parallelrisch | \(\frac {-x^{2} \ln \left (5\right )-\ln \left (5\right ) x \ln \left (\ln \left (-2+x \right )\right )-2 x \ln \left (5\right )+\ln \left (5\right ) \ln \left (\ln \left (-2+x \right )\right )+x^{2}-4 \ln \left (5\right )-4}{\ln \left (5\right )}\) | \(45\) |
Input:
int(((2-x)*ln(5)*ln(-2+x)*ln(ln(-2+x))+((-2*x^2+2*x+4)*ln(5)+2*x^2-4*x)*ln (-2+x)+(1-x)*ln(5))/(-2+x)/ln(5)/ln(-2+x),x,method=_RETURNVERBOSE)
Output:
ln(ln(-2+x))-2*x-ln(ln(-2+x))*x-(ln(5)-1)/ln(5)*x^2
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx=-\frac {{\left (x - 1\right )} \log \left (5\right ) \log \left (\log \left (x - 2\right )\right ) - x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (5\right )}{\log \left (5\right )} \] Input:
integrate(((2-x)*log(5)*log(-2+x)*log(log(-2+x))+((-2*x^2+2*x+4)*log(5)+2* x^2-4*x)*log(-2+x)+(1-x)*log(5))/(-2+x)/log(5)/log(-2+x),x, algorithm="fri cas")
Output:
-((x - 1)*log(5)*log(log(x - 2)) - x^2 + (x^2 + 2*x)*log(5))/log(5)
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx=\frac {x^{2} \cdot \left (1 - \log {\left (5 \right )}\right )}{\log {\left (5 \right )}} - 2 x + \left (1 - x\right ) \log {\left (\log {\left (x - 2 \right )} \right )} \] Input:
integrate(((2-x)*ln(5)*ln(-2+x)*ln(ln(-2+x))+((-2*x**2+2*x+4)*ln(5)+2*x**2 -4*x)*ln(-2+x)+(1-x)*ln(5))/(-2+x)/ln(5)/ln(-2+x),x)
Output:
x**2*(1 - log(5))/log(5) - 2*x + (1 - x)*log(log(x - 2))
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (30) = 60\).
Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.90 \[ \int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx=-\frac {x^{2} {\left (\log \left (5\right ) - 1\right )} - 6 \, \log \left (5\right ) \log \left (x - 2\right ) \log \left (\log \left (x - 2\right )\right ) + 4 \, {\left (\log \left (x - 2\right ) \log \left (\log \left (x - 2\right )\right ) - \log \left (x - 2\right )\right )} \log \left (5\right ) + 2 \, x \log \left (5\right ) + 4 \, \log \left (5\right ) \log \left (x - 2\right ) + {\left (x \log \left (5\right ) + 2 \, \log \left (5\right ) \log \left (x - 2\right )\right )} \log \left (\log \left (x - 2\right )\right ) - \log \left (5\right ) \log \left (\log \left (x - 2\right )\right )}{\log \left (5\right )} \] Input:
integrate(((2-x)*log(5)*log(-2+x)*log(log(-2+x))+((-2*x^2+2*x+4)*log(5)+2* x^2-4*x)*log(-2+x)+(1-x)*log(5))/(-2+x)/log(5)/log(-2+x),x, algorithm="max ima")
Output:
-(x^2*(log(5) - 1) - 6*log(5)*log(x - 2)*log(log(x - 2)) + 4*(log(x - 2)*l og(log(x - 2)) - log(x - 2))*log(5) + 2*x*log(5) + 4*log(5)*log(x - 2) + ( x*log(5) + 2*log(5)*log(x - 2))*log(log(x - 2)) - log(5)*log(log(x - 2)))/ log(5)
Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx=-\frac {x^{2} {\left (\log \left (5\right ) - 1\right )} + x \log \left (5\right ) \log \left (\log \left (x - 2\right )\right ) + 2 \, x \log \left (5\right ) - \log \left (5\right ) \log \left (\log \left (x - 2\right )\right )}{\log \left (5\right )} \] Input:
integrate(((2-x)*log(5)*log(-2+x)*log(log(-2+x))+((-2*x^2+2*x+4)*log(5)+2* x^2-4*x)*log(-2+x)+(1-x)*log(5))/(-2+x)/log(5)/log(-2+x),x, algorithm="gia c")
Output:
-(x^2*(log(5) - 1) + x*log(5)*log(log(x - 2)) + 2*x*log(5) - log(5)*log(lo g(x - 2)))/log(5)
Time = 3.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx=-2\,x-\ln \left (\ln \left (x-2\right )\right )-\ln \left (\ln \left (x-2\right )\right )\,\left (x-2\right )-\frac {x^2\,\left (\ln \left (25\right )-2\right )}{2\,\ln \left (5\right )} \] Input:
int(-(log(5)*(x - 1) - log(x - 2)*(log(5)*(2*x - 2*x^2 + 4) - 4*x + 2*x^2) + log(x - 2)*log(5)*log(log(x - 2))*(x - 2))/(log(x - 2)*log(5)*(x - 2)), x)
Output:
- 2*x - log(log(x - 2)) - log(log(x - 2))*(x - 2) - (x^2*(log(25) - 2))/(2 *log(5))
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx=\frac {-\mathrm {log}\left (\mathrm {log}\left (x -2\right )\right ) \mathrm {log}\left (5\right ) x +\mathrm {log}\left (\mathrm {log}\left (x -2\right )\right ) \mathrm {log}\left (5\right )-\mathrm {log}\left (5\right ) x^{2}-2 \,\mathrm {log}\left (5\right ) x +x^{2}}{\mathrm {log}\left (5\right )} \] Input:
int(((2-x)*log(5)*log(-2+x)*log(log(-2+x))+((-2*x^2+2*x+4)*log(5)+2*x^2-4* x)*log(-2+x)+(1-x)*log(5))/(-2+x)/log(5)/log(-2+x),x)
Output:
( - log(log(x - 2))*log(5)*x + log(log(x - 2))*log(5) - log(5)*x**2 - 2*lo g(5)*x + x**2)/log(5)