Integrand size = 79, antiderivative size = 31 \[ \int \frac {16^{-e^x-x} e^{16^{-e^x-x} (5-x)^{e^x+x}} (5-x)^{e^x+x} \left (e^x+x+\left (-5+e^x (-5+x)+x\right ) \log \left (\frac {5-x}{16}\right )\right )}{-5+x} \, dx=e^{4^{-e^x-x} \left (1+\frac {1-x}{4}\right )^{e^x+x}} \] Output:
exp(exp((exp(x)+x)*ln(5/16-1/16*x)))
Time = 8.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {16^{-e^x-x} e^{16^{-e^x-x} (5-x)^{e^x+x}} (5-x)^{e^x+x} \left (e^x+x+\left (-5+e^x (-5+x)+x\right ) \log \left (\frac {5-x}{16}\right )\right )}{-5+x} \, dx=e^{16^{-e^x-x} (5-x)^{e^x+x}} \] Input:
Integrate[(16^(-E^x - x)*E^(16^(-E^x - x)*(5 - x)^(E^x + x))*(5 - x)^(E^x + x)*(E^x + x + (-5 + E^x*(-5 + x) + x)*Log[(5 - x)/16]))/(-5 + x),x]
Output:
E^(16^(-E^x - x)*(5 - x)^(E^x + x))
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.
\(\displaystyle \int \frac {16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}} (5-x)^{x+e^x} \left (x+e^x+\left (e^x (x-5)+x-5\right ) \log \left (\frac {5-x}{16}\right )\right )}{x-5} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}+x} (5-x)^{x+e^x-1} \left (x \log \left (\frac {5-x}{16}\right )-5 \log \left (\frac {5-x}{16}\right )+1\right )-16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}} (5-x)^{x+e^x-1} \left (x+x \log \left (\frac {5-x}{16}\right )-5 \log \left (\frac {5-x}{16}\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}+x} (5-x)^{x+e^x-1}dx-\int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}} (5-x)^{x+e^x-1} xdx-5 \int \frac {\int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}} (5-x)^{x+e^x-1}dx}{x-5}dx-5 \int \frac {\int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}+x} (5-x)^{x+e^x-1}dx}{x-5}dx+\int \frac {\int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}} (5-x)^{x+e^x-1} xdx}{x-5}dx+\int \frac {\int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}+x} (5-x)^{x+e^x-1} xdx}{x-5}dx+5 \log \left (\frac {5}{16}-\frac {x}{16}\right ) \int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}} (5-x)^{x+e^x-1}dx+5 \log \left (\frac {5}{16}-\frac {x}{16}\right ) \int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}+x} (5-x)^{x+e^x-1}dx-\log \left (\frac {5}{16}-\frac {x}{16}\right ) \int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}} (5-x)^{x+e^x-1} xdx-\log \left (\frac {5}{16}-\frac {x}{16}\right ) \int 16^{-x-e^x} e^{16^{-x-e^x} (5-x)^{x+e^x}+x} (5-x)^{x+e^x-1} xdx\) |
Input:
Int[(16^(-E^x - x)*E^(16^(-E^x - x)*(5 - x)^(E^x + x))*(5 - x)^(E^x + x)*( E^x + x + (-5 + E^x*(-5 + x) + x)*Log[(5 - x)/16]))/(-5 + x),x]
Output:
$Aborted
Time = 2.43 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.39
method | result | size |
risch | \({\mathrm e}^{\left (\frac {5}{16}-\frac {x}{16}\right )^{{\mathrm e}^{x}+x}}\) | \(12\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\left ({\mathrm e}^{x}+x \right ) \ln \left (\frac {5}{16}-\frac {x}{16}\right )}}\) | \(14\) |
Input:
int((((-5+x)*exp(x)+x-5)*ln(5/16-1/16*x)+exp(x)+x)*exp((exp(x)+x)*ln(5/16- 1/16*x))*exp(exp((exp(x)+x)*ln(5/16-1/16*x)))/(-5+x),x,method=_RETURNVERBO SE)
Output:
exp((5/16-1/16*x)^(exp(x)+x))
Time = 0.10 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.35 \[ \int \frac {16^{-e^x-x} e^{16^{-e^x-x} (5-x)^{e^x+x}} (5-x)^{e^x+x} \left (e^x+x+\left (-5+e^x (-5+x)+x\right ) \log \left (\frac {5-x}{16}\right )\right )}{-5+x} \, dx=e^{\left ({\left (-\frac {1}{16} \, x + \frac {5}{16}\right )}^{x + e^{x}}\right )} \] Input:
integrate((((-5+x)*exp(x)+x-5)*log(5/16-1/16*x)+exp(x)+x)*exp((exp(x)+x)*l og(5/16-1/16*x))*exp(exp((exp(x)+x)*log(5/16-1/16*x)))/(-5+x),x, algorithm ="fricas")
Output:
e^((-1/16*x + 5/16)^(x + e^x))
Time = 16.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.48 \[ \int \frac {16^{-e^x-x} e^{16^{-e^x-x} (5-x)^{e^x+x}} (5-x)^{e^x+x} \left (e^x+x+\left (-5+e^x (-5+x)+x\right ) \log \left (\frac {5-x}{16}\right )\right )}{-5+x} \, dx=e^{e^{\left (x + e^{x}\right ) \log {\left (\frac {5}{16} - \frac {x}{16} \right )}}} \] Input:
integrate((((-5+x)*exp(x)+x-5)*ln(5/16-1/16*x)+exp(x)+x)*exp((exp(x)+x)*ln (5/16-1/16*x))*exp(exp((exp(x)+x)*ln(5/16-1/16*x)))/(-5+x),x)
Output:
exp(exp((x + exp(x))*log(5/16 - x/16)))
Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (11) = 22\).
Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {16^{-e^x-x} e^{16^{-e^x-x} (5-x)^{e^x+x}} (5-x)^{e^x+x} \left (e^x+x+\left (-5+e^x (-5+x)+x\right ) \log \left (\frac {5-x}{16}\right )\right )}{-5+x} \, dx=e^{\left (e^{\left (-4 \, x \log \left (2\right ) - 4 \, e^{x} \log \left (2\right ) + x \log \left (-x + 5\right ) + e^{x} \log \left (-x + 5\right )\right )}\right )} \] Input:
integrate((((-5+x)*exp(x)+x-5)*log(5/16-1/16*x)+exp(x)+x)*exp((exp(x)+x)*l og(5/16-1/16*x))*exp(exp((exp(x)+x)*log(5/16-1/16*x)))/(-5+x),x, algorithm ="maxima")
Output:
e^(e^(-4*x*log(2) - 4*e^x*log(2) + x*log(-x + 5) + e^x*log(-x + 5)))
\[ \int \frac {16^{-e^x-x} e^{16^{-e^x-x} (5-x)^{e^x+x}} (5-x)^{e^x+x} \left (e^x+x+\left (-5+e^x (-5+x)+x\right ) \log \left (\frac {5-x}{16}\right )\right )}{-5+x} \, dx=\int { \frac {{\left ({\left ({\left (x - 5\right )} e^{x} + x - 5\right )} \log \left (-\frac {1}{16} \, x + \frac {5}{16}\right ) + x + e^{x}\right )} {\left (-\frac {1}{16} \, x + \frac {5}{16}\right )}^{x + e^{x}} e^{\left ({\left (-\frac {1}{16} \, x + \frac {5}{16}\right )}^{x + e^{x}}\right )}}{x - 5} \,d x } \] Input:
integrate((((-5+x)*exp(x)+x-5)*log(5/16-1/16*x)+exp(x)+x)*exp((exp(x)+x)*l og(5/16-1/16*x))*exp(exp((exp(x)+x)*log(5/16-1/16*x)))/(-5+x),x, algorithm ="giac")
Output:
integrate((((x - 5)*e^x + x - 5)*log(-1/16*x + 5/16) + x + e^x)*(-1/16*x + 5/16)^(x + e^x)*e^((-1/16*x + 5/16)^(x + e^x))/(x - 5), x)
Time = 2.63 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.35 \[ \int \frac {16^{-e^x-x} e^{16^{-e^x-x} (5-x)^{e^x+x}} (5-x)^{e^x+x} \left (e^x+x+\left (-5+e^x (-5+x)+x\right ) \log \left (\frac {5-x}{16}\right )\right )}{-5+x} \, dx={\mathrm {e}}^{{\left (\frac {5}{16}-\frac {x}{16}\right )}^{x+{\mathrm {e}}^x}} \] Input:
int((exp(log(5/16 - x/16)*(x + exp(x)))*exp(exp(log(5/16 - x/16)*(x + exp( x))))*(x + exp(x) + log(5/16 - x/16)*(x + exp(x)*(x - 5) - 5)))/(x - 5),x)
Output:
exp((5/16 - x/16)^(x + exp(x)))
\[ \int \frac {16^{-e^x-x} e^{16^{-e^x-x} (5-x)^{e^x+x}} (5-x)^{e^x+x} \left (e^x+x+\left (-5+e^x (-5+x)+x\right ) \log \left (\frac {5-x}{16}\right )\right )}{-5+x} \, dx=\int \frac {\left (\left (\left (-5+x \right ) {\mathrm e}^{x}+x -5\right ) \mathrm {log}\left (\frac {5}{16}-\frac {x}{16}\right )+{\mathrm e}^{x}+x \right ) {\mathrm e}^{\left ({\mathrm e}^{x}+x \right ) \mathrm {log}\left (\frac {5}{16}-\frac {x}{16}\right )} {\mathrm e}^{{\mathrm e}^{\left ({\mathrm e}^{x}+x \right ) \mathrm {log}\left (\frac {5}{16}-\frac {x}{16}\right )}}}{-5+x}d x \] Input:
int((((-5+x)*exp(x)+x-5)*log(5/16-1/16*x)+exp(x)+x)*exp((exp(x)+x)*log(5/1 6-1/16*x))*exp(exp((exp(x)+x)*log(5/16-1/16*x)))/(-5+x),x)
Output:
int((((-5+x)*exp(x)+x-5)*log(5/16-1/16*x)+exp(x)+x)*exp((exp(x)+x)*log(5/1 6-1/16*x))*exp(exp((exp(x)+x)*log(5/16-1/16*x)))/(-5+x),x)