Integrand size = 50, antiderivative size = 22 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x \left (-1+5^{-x} \left (x \left (-1+(-2+\log (5))^2\right )\right )^x\right ) \] Output:
x*(exp(x*ln(1/5*x*((ln(5)-2)^2-1)))-1)
Time = 1.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=-x+\frac {5^{-x} \left (x \left (3-4 \log (5)+\log ^2(5)\right )\right )^{1+x}}{3-4 \log (5)+\log ^2(5)} \] Input:
Integrate[-1 + ((3*x - 4*x*Log[5] + x*Log[5]^2)^x*(1 + x + x*Log[(3*x - 4* x*Log[5] + x*Log[5]^2)/5]))/5^x,x]
Output:
-x + (x*(3 - 4*Log[5] + Log[5]^2))^(1 + x)/(5^x*(3 - 4*Log[5] + Log[5]^2))
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.
\(\displaystyle \int \left (5^{-x} \left (3 x+x \log ^2(5)-4 x \log (5)\right )^x \left (x+x \log \left (\frac {1}{5} \left (3 x+x \log ^2(5)-4 x \log (5)\right )\right )+1\right )-1\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int (x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^xdx+\frac {\log \left (\frac {1}{5} x (1-\log (5)) (3-\log (5))\right ) \int (x (1-\log (5)))^{x+1} \left (\frac {1}{5} (3-\log (5))\right )^xdx}{1-\log (5)}+\frac {\int (x (1-\log (5)))^{x+1} \left (\frac {1}{5} (3-\log (5))\right )^xdx}{1-\log (5)}-\frac {\int \frac {\int (x (1-\log (5)))^{x+1} \left (\frac {1}{5} (3-\log (5))\right )^xdx}{x}dx}{1-\log (5)}-x\) |
Input:
Int[-1 + ((3*x - 4*x*Log[5] + x*Log[5]^2)^x*(1 + x + x*Log[(3*x - 4*x*Log[ 5] + x*Log[5]^2)/5]))/5^x,x]
Output:
$Aborted
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \({\mathrm e}^{x \ln \left (\frac {x \left (\ln \left (5\right )^{2}-4 \ln \left (5\right )+3\right )}{5}\right )} x -x\) | \(24\) |
risch | \(-x +\left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )^{x} x\) | \(25\) |
default | \(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\) | \(27\) |
norman | \(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\) | \(27\) |
parts | \(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\) | \(27\) |
Input:
int((x*ln(1/5*x*ln(5)^2-4/5*x*ln(5)+3/5*x)+x+1)*exp(x*ln(1/5*x*ln(5)^2-4/5 *x*ln(5)+3/5*x))-1,x,method=_RETURNVERBOSE)
Output:
exp(x*ln(1/5*x*(ln(5)^2-4*ln(5)+3)))*x-x
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx={\left (\frac {1}{5} \, x \log \left (5\right )^{2} - \frac {4}{5} \, x \log \left (5\right ) + \frac {3}{5} \, x\right )}^{x} x - x \] Input:
integrate((x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x)+x+1)*exp(x*log(1/5*x*l og(5)^2-4/5*x*log(5)+3/5*x))-1,x, algorithm="fricas")
Output:
(1/5*x*log(5)^2 - 4/5*x*log(5) + 3/5*x)^x*x - x
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x e^{x \log {\left (- \frac {4 x \log {\left (5 \right )}}{5} + \frac {x \log {\left (5 \right )}^{2}}{5} + \frac {3 x}{5} \right )}} - x \] Input:
integrate((x*ln(1/5*x*ln(5)**2-4/5*x*ln(5)+3/5*x)+x+1)*exp(x*ln(1/5*x*ln(5 )**2-4/5*x*ln(5)+3/5*x))-1,x)
Output:
x*exp(x*log(-4*x*log(5)/5 + x*log(5)**2/5 + 3*x/5)) - x
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x e^{\left (-x \log \left (5\right ) + x \log \left (x\right ) + x \log \left (\log \left (5\right ) - 1\right ) + x \log \left (\log \left (5\right ) - 3\right )\right )} - x \] Input:
integrate((x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x)+x+1)*exp(x*log(1/5*x*l og(5)^2-4/5*x*log(5)+3/5*x))-1,x, algorithm="maxima")
Output:
x*e^(-x*log(5) + x*log(x) + x*log(log(5) - 1) + x*log(log(5) - 3)) - x
Timed out. \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=\text {Timed out} \] Input:
integrate((x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x)+x+1)*exp(x*log(1/5*x*l og(5)^2-4/5*x*log(5)+3/5*x))-1,x, algorithm="giac")
Output:
Timed out
Time = 3.52 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x\,\left ({\left (\frac {3\,x}{5}-\frac {4\,x\,\ln \left (5\right )}{5}+\frac {x\,{\ln \left (5\right )}^2}{5}\right )}^x-1\right ) \] Input:
int(exp(x*log((3*x)/5 - (4*x*log(5))/5 + (x*log(5)^2)/5))*(x + x*log((3*x) /5 - (4*x*log(5))/5 + (x*log(5)^2)/5) + 1) - 1,x)
Output:
x*(((3*x)/5 - (4*x*log(5))/5 + (x*log(5)^2)/5)^x - 1)
\[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=\int \frac {\left (\mathrm {log}\left (5\right )^{2} x -4 \,\mathrm {log}\left (5\right ) x +3 x \right )^{x}}{5^{x}}d x +\int \frac {\left (\mathrm {log}\left (5\right )^{2} x -4 \,\mathrm {log}\left (5\right ) x +3 x \right )^{x} \mathrm {log}\left (\frac {\mathrm {log}\left (5\right )^{2} x}{5}-\frac {4 \,\mathrm {log}\left (5\right ) x}{5}+\frac {3 x}{5}\right ) x}{5^{x}}d x +\int \frac {\left (\mathrm {log}\left (5\right )^{2} x -4 \,\mathrm {log}\left (5\right ) x +3 x \right )^{x} x}{5^{x}}d x -x \] Input:
int((x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x)+x+1)*exp(x*log(1/5*x*log(5)^ 2-4/5*x*log(5)+3/5*x))-1,x)
Output:
int((log(5)**2*x - 4*log(5)*x + 3*x)**x/5**x,x) + int(((log(5)**2*x - 4*lo g(5)*x + 3*x)**x*log((log(5)**2*x - 4*log(5)*x + 3*x)/5)*x)/5**x,x) + int( ((log(5)**2*x - 4*log(5)*x + 3*x)**x*x)/5**x,x) - x