Integrand size = 64, antiderivative size = 31 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=5+\frac {e^{\frac {e^{-x} x}{15}} \left (-e^{3+3 x}+\log (x)\right )}{\log (2)} \]
Time = 2.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {e^{\frac {e^{-x} x}{15}} \left (-15 e^{3+3 x}+15 \log (x)\right )}{15 \log (2)} \]
Integrate[(E^(-x + x/(15*E^x))*(15*E^x + E^(3 + 3*x)*(-x - 45*E^x*x + x^2) + (x - x^2)*Log[x]))/(15*x*Log[2]),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.
\(\displaystyle \int \frac {e^{\frac {e^{-x} x}{15}-x} \left (e^{3 x+3} \left (x^2-45 e^x x-x\right )+\left (x-x^2\right ) \log (x)+15 e^x\right )}{15 x \log (2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {e^{\frac {e^{-x} x}{15}-x} \left (-e^{3 x+3} \left (-x^2+45 e^x x+x\right )+15 e^x+\left (x-x^2\right ) \log (x)\right )}{x}dx}{15 \log (2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (e^{\frac {e^{-x} x}{15}+2 x+3} (x-1)-e^{\frac {e^{-x} x}{15}-x} \log (x) (x-1)-45 e^{\frac {e^{-x} x}{15}+3 x+3}+\frac {15 e^{\frac {e^{-x} x}{15}}}{x}\right )dx}{15 \log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\int e^{\frac {e^{-x} x}{15}+2 x+3}dx-45 \int e^{\frac {e^{-x} x}{15}+3 x+3}dx+15 \int \frac {e^{\frac {e^{-x} x}{15}}}{x}dx+\int e^{\frac {e^{-x} x}{15}+2 x+3} xdx-\int \frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x}dx}{x}dx+\int \frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} xdx}{x}dx+\log (x) \int e^{\frac {e^{-x} x}{15}-x}dx-\log (x) \int e^{\frac {e^{-x} x}{15}-x} xdx}{15 \log (2)}\) |
Int[(E^(-x + x/(15*E^x))*(15*E^x + E^(3 + 3*x)*(-x - 45*E^x*x + x^2) + (x - x^2)*Log[x]))/(15*x*Log[2]),x]
3.16.79.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.42 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {\left (-15 \,{\mathrm e}^{3 x +3}+15 \ln \left (x \right )\right ) {\mathrm e}^{\frac {x \,{\mathrm e}^{-x}}{15}}}{15 \ln \left (2\right )}\) | \(28\) |
parallelrisch | \(\frac {15 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-x}}{15}} \ln \left (x \right )-15 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-x}}{15}} {\mathrm e}^{3 x +3}}{15 \ln \left (2\right )}\) | \(36\) |
int(1/15*((-x^2+x)*ln(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x))*exp(1/ 15*x/exp(x))/x/ln(2)/exp(x),x,method=_RETURNVERBOSE)
Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {{\left (e^{x} \log \left (x\right ) - e^{\left (4 \, x + 3\right )}\right )} e^{\left (-\frac {1}{15} \, {\left (15 \, x e^{x} - x\right )} e^{\left (-x\right )}\right )}}{\log \left (2\right )} \]
integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x)) *exp(1/15*x/exp(x))/x/log(2)/exp(x),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {\left (- e^{3} e^{3 x} + \log {\left (x \right )}\right ) e^{\frac {x e^{- x}}{15}}}{\log {\left (2 \right )}} \]
integrate(1/15*((-x**2+x)*ln(x)+(-45*exp(x)*x+x**2-x)*exp(3*x+3)+15*exp(x) )*exp(1/15*x/exp(x))/x/ln(2)/exp(x),x)
\[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int { \frac {{\left ({\left (x^{2} - 45 \, x e^{x} - x\right )} e^{\left (3 \, x + 3\right )} - {\left (x^{2} - x\right )} \log \left (x\right ) + 15 \, e^{x}\right )} e^{\left (\frac {1}{15} \, x e^{\left (-x\right )} - x\right )}}{15 \, x \log \left (2\right )} \,d x } \]
integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x)) *exp(1/15*x/exp(x))/x/log(2)/exp(x),x, algorithm=\
-1/15*(15*e^(1/15*x*e^(-x) + 3*x + 3) - integrate(-((x^2 - x)*log(x) - 15* e^x)*e^(1/15*x*e^(-x) - x)/x, x))/log(2)
\[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int { \frac {{\left ({\left (x^{2} - 45 \, x e^{x} - x\right )} e^{\left (3 \, x + 3\right )} - {\left (x^{2} - x\right )} \log \left (x\right ) + 15 \, e^{x}\right )} e^{\left (\frac {1}{15} \, x e^{\left (-x\right )} - x\right )}}{15 \, x \log \left (2\right )} \,d x } \]
integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x)) *exp(1/15*x/exp(x))/x/log(2)/exp(x),x, algorithm=\
integrate(1/15*((x^2 - 45*x*e^x - x)*e^(3*x + 3) - (x^2 - x)*log(x) + 15*e ^x)*e^(1/15*x*e^(-x) - x)/(x*log(2)), x)
Timed out. \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int \frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-x}}{15}-x}\,\left (15\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (x-x^2\right )-{\mathrm {e}}^{3\,x+3}\,\left (x+45\,x\,{\mathrm {e}}^x-x^2\right )\right )}{15\,x\,\ln \left (2\right )} \,d x \]
int((exp(-x)*exp((x*exp(-x))/15)*(15*exp(x) + log(x)*(x - x^2) - exp(3*x + 3)*(x + 45*x*exp(x) - x^2)))/(15*x*log(2)),x)