Integrand size = 53, antiderivative size = 30 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=e^{3+4 e^{2 x} x^2}+\frac {1}{5} (1+x+\log (4)) \log \left (\frac {5}{x}\right ) \] Output:
exp(4*exp(x)^2*x^2+3)+1/5*(2*ln(2)+x+1)*ln(5/x)
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\frac {1}{5} \left (5 e^{3+4 e^{2 x} x^2}+x \log \left (\frac {5}{x}\right )-(1+\log (4)) \log (x)\right ) \] Input:
Integrate[(-1 - x + E^(3 + 2*x + 4*E^(2*x)*x^2)*(40*x^2 + 40*x^3) - Log[4] + x*Log[5/x])/(5*x),x]
Output:
(5*E^(3 + 4*E^(2*x)*x^2) + x*Log[5/x] - (1 + Log[4])*Log[x])/5
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.
\(\displaystyle \int \frac {e^{4 e^{2 x} x^2+2 x+3} \left (40 x^3+40 x^2\right )-x+x \log \left (\frac {5}{x}\right )-1-\log (4)}{5 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {-\log \left (\frac {5}{x}\right ) x+x-40 e^{4 e^{2 x} x^2+2 x+3} \left (x^3+x^2\right )+\log (4)+1}{x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {-\log \left (\frac {5}{x}\right ) x+x-40 e^{4 e^{2 x} x^2+2 x+3} \left (x^3+x^2\right )+\log (4)+1}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {-\log \left (\frac {5}{x}\right ) x+x+\log (4)+1}{x}-40 e^{4 e^{2 x} x^2+2 x+3} x (x+1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (40 \int e^{4 e^{2 x} x^2+2 x+3} xdx+40 \int e^{4 e^{2 x} x^2+2 x+3} x^2dx+x \log \left (\frac {5}{x}\right )-(1+\log (4)) \log (x)\right )\) |
Input:
Int[(-1 - x + E^(3 + 2*x + 4*E^(2*x)*x^2)*(40*x^2 + 40*x^3) - Log[4] + x*L og[5/x])/(5*x),x]
Output:
$Aborted
Time = 5.41 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10
method | result | size |
default | \(-\frac {\left (1+2 \ln \left (2\right )\right ) \ln \left (x \right )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}+\frac {x \ln \left (\frac {5}{x}\right )}{5}\) | \(33\) |
parts | \(-\frac {\left (1+2 \ln \left (2\right )\right ) \ln \left (x \right )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}+\frac {x \ln \left (\frac {5}{x}\right )}{5}\) | \(33\) |
risch | \(-\frac {x \ln \left (x \right )}{5}+\frac {x \ln \left (5\right )}{5}-\frac {2 \ln \left (2\right ) \ln \left (x \right )}{5}-\frac {\ln \left (x \right )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}\) | \(34\) |
parallelrisch | \(\frac {2 \ln \left (\frac {5}{x}\right ) \ln \left (2\right )}{5}+\frac {x \ln \left (\frac {5}{x}\right )}{5}+\frac {\ln \left (\frac {5}{x}\right )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}\) | \(41\) |
Input:
int(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*ln(5/x)-2*ln(2)- x-1)/x,x,method=_RETURNVERBOSE)
Output:
-1/5*(1+2*ln(2))*ln(x)+exp(4*exp(x)^2*x^2+3)+1/5*x*ln(5/x)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\frac {1}{5} \, {\left ({\left (x + 2 \, \log \left (2\right ) + 1\right )} e^{\left (2 \, x\right )} \log \left (\frac {5}{x}\right ) + 5 \, e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 2 \, x + 3\right )}\right )} e^{\left (-2 \, x\right )} \] Input:
integrate(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*log(5/x)-2 *log(2)-x-1)/x,x, algorithm="fricas")
Output:
1/5*((x + 2*log(2) + 1)*e^(2*x)*log(5/x) + 5*e^(4*x^2*e^(2*x) + 2*x + 3))* e^(-2*x)
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\frac {x \log {\left (\frac {5}{x} \right )}}{5} + e^{4 x^{2} e^{2 x} + 3} + \left (- \frac {2 \log {\left (2 \right )}}{5} - \frac {1}{5}\right ) \log {\left (x \right )} \] Input:
integrate(1/5*((40*x**3+40*x**2)*exp(x)**2*exp(4*exp(x)**2*x**2+3)+x*ln(5/ x)-2*ln(2)-x-1)/x,x)
Output:
x*log(5/x)/5 + exp(4*x**2*exp(2*x) + 3) + (-2*log(2)/5 - 1/5)*log(x)
Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=-\frac {2}{5} \, \log \left (2\right ) \log \left (x\right ) + \frac {1}{5} \, x \log \left (\frac {5}{x}\right ) + e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 3\right )} - \frac {1}{5} \, \log \left (x\right ) \] Input:
integrate(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*log(5/x)-2 *log(2)-x-1)/x,x, algorithm="maxima")
Output:
-2/5*log(2)*log(x) + 1/5*x*log(5/x) + e^(4*x^2*e^(2*x) + 3) - 1/5*log(x)
\[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\int { \frac {40 \, {\left (x^{3} + x^{2}\right )} e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 2 \, x + 3\right )} + x \log \left (\frac {5}{x}\right ) - x - 2 \, \log \left (2\right ) - 1}{5 \, x} \,d x } \] Input:
integrate(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*log(5/x)-2 *log(2)-x-1)/x,x, algorithm="giac")
Output:
integrate(1/5*(40*(x^3 + x^2)*e^(4*x^2*e^(2*x) + 2*x + 3) + x*log(5/x) - x - 2*log(2) - 1)/x, x)
Time = 3.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx={\mathrm {e}}^{4\,x^2\,{\mathrm {e}}^{2\,x}+3}-\ln \left (x\right )\,\left (\frac {\ln \left (4\right )}{5}+\frac {1}{5}\right )+\frac {x\,\ln \left (\frac {5}{x}\right )}{5} \] Input:
int(-(x/5 + (2*log(2))/5 - (x*log(5/x))/5 - (exp(4*x^2*exp(2*x) + 3)*exp(2 *x)*(40*x^2 + 40*x^3))/5 + 1/5)/x,x)
Output:
exp(4*x^2*exp(2*x) + 3) - log(x)*(log(4)/5 + 1/5) + (x*log(5/x))/5
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=e^{4 e^{2 x} x^{2}} e^{3}+\frac {\mathrm {log}\left (\frac {5}{x}\right ) x}{5}-\frac {2 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )}{5}-\frac {\mathrm {log}\left (x \right )}{5} \] Input:
int(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*log(5/x)-2*log(2 )-x-1)/x,x)
Output:
(5*e**(4*e**(2*x)*x**2)*e**3 + log(5/x)*x - 2*log(x)*log(2) - log(x))/5