Integrand size = 70, antiderivative size = 28 \[ \int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} \left (4 e^{e^2+e^3}+\left (-4 e^{e^2+e^3}-x\right ) \log (5 x)\right )}{4 x^2} \, dx=\frac {e^{-\frac {1}{4} e^{-e^2-e^3} x} \log (5 x)}{x} \] Output:
ln(5*x)/exp(1/4*x/exp(exp(3)+exp(2)))/x
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} \left (4 e^{e^2+e^3}+\left (-4 e^{e^2+e^3}-x\right ) \log (5 x)\right )}{4 x^2} \, dx=\frac {e^{-\frac {1}{4} e^{-e^2 (1+e)} x} \log (5 x)}{x} \] Input:
Integrate[(E^(-E^2 - E^3 - (E^(-E^2 - E^3)*x)/4)*(4*E^(E^2 + E^3) + (-4*E^ (E^2 + E^3) - x)*Log[5*x]))/(4*x^2),x]
Output:
Log[5*x]/(E^(x/(4*E^(E^2*(1 + E))))*x)
Time = 1.89 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {27, 7293, 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.
\(\displaystyle \int \frac {e^{-\frac {1}{4} e^{-e^2-e^3} x-e^3-e^2} \left (\left (-x-4 e^{e^2+e^3}\right ) \log (5 x)+4 e^{e^2+e^3}\right )}{4 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {e^{-\frac {1}{4} e^{-e^2 (1+e)} x-e^2 (1+e)} \left (4 e^{e^2+e^3}-\left (x+4 e^{e^2+e^3}\right ) \log (5 x)\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (\frac {4 \exp \left (-\frac {1}{4} e^{-e^2 (1+e)} x-e^2 (1+e)+e^3+e^2\right )}{x^2}-\frac {e^{-\frac {1}{4} e^{-e^2 (1+e)} x-e^2 (1+e)} \left (x+4 e^{e^2+e^3}\right ) \log (5 x)}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{-\frac {1}{4} e^{-e^2 (1+e)} x} \log (5 x)}{x}\) |
Input:
Int[(E^(-E^2 - E^3 - (E^(-E^2 - E^3)*x)/4)*(4*E^(E^2 + E^3) + (-4*E^(E^2 + E^3) - x)*Log[5*x]))/(4*x^2),x]
Output:
Log[5*x]/(E^(x/(4*E^(E^2*(1 + E))))*x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.48 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
norman | \(\frac {\ln \left (5 x \right ) {\mathrm e}^{-\frac {x \,{\mathrm e}^{-{\mathrm e}^{3}-{\mathrm e}^{2}}}{4}}}{x}\) | \(23\) |
risch | \(\frac {\ln \left (5 x \right ) {\mathrm e}^{-\frac {x \,{\mathrm e}^{-{\mathrm e}^{3}-{\mathrm e}^{2}}}{4}}}{x}\) | \(23\) |
parallelrisch | \(\frac {\ln \left (5 x \right ) {\mathrm e}^{-\frac {x \,{\mathrm e}^{-{\mathrm e}^{3}-{\mathrm e}^{2}}}{4}}}{x}\) | \(23\) |
Input:
int(1/4*((-4*exp(exp(3)+exp(2))-x)*ln(5*x)+4*exp(exp(3)+exp(2)))/x^2/exp(e xp(3)+exp(2))/exp(1/4*x/exp(exp(3)+exp(2))),x,method=_RETURNVERBOSE)
Output:
ln(5*x)/exp(1/4*x/exp(exp(3)+exp(2)))/x
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} \left (4 e^{e^2+e^3}+\left (-4 e^{e^2+e^3}-x\right ) \log (5 x)\right )}{4 x^2} \, dx=\frac {e^{\left (-\frac {1}{4} \, {\left (4 \, {\left (e^{3} + e^{2}\right )} e^{\left (e^{3} + e^{2}\right )} + x\right )} e^{\left (-e^{3} - e^{2}\right )} + e^{3} + e^{2}\right )} \log \left (5 \, x\right )}{x} \] Input:
integrate(1/4*((-4*exp(exp(3)+exp(2))-x)*log(5*x)+4*exp(exp(3)+exp(2)))/x^ 2/exp(exp(3)+exp(2))/exp(1/4*x/exp(exp(3)+exp(2))),x, algorithm="fricas")
Output:
e^(-1/4*(4*(e^3 + e^2)*e^(e^3 + e^2) + x)*e^(-e^3 - e^2) + e^3 + e^2)*log( 5*x)/x
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} \left (4 e^{e^2+e^3}+\left (-4 e^{e^2+e^3}-x\right ) \log (5 x)\right )}{4 x^2} \, dx=\frac {e^{- \frac {x}{4 e^{e^{2} + e^{3}}}} \log {\left (5 x \right )}}{x} \] Input:
integrate(1/4*((-4*exp(exp(3)+exp(2))-x)*ln(5*x)+4*exp(exp(3)+exp(2)))/x** 2/exp(exp(3)+exp(2))/exp(1/4*x/exp(exp(3)+exp(2))),x)
Output:
exp(-x*exp(-exp(3) - exp(2))/4)*log(5*x)/x
\[ \int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} \left (4 e^{e^2+e^3}+\left (-4 e^{e^2+e^3}-x\right ) \log (5 x)\right )}{4 x^2} \, dx=\int { -\frac {{\left ({\left (x + 4 \, e^{\left (e^{3} + e^{2}\right )}\right )} \log \left (5 \, x\right ) - 4 \, e^{\left (e^{3} + e^{2}\right )}\right )} e^{\left (-\frac {1}{4} \, x e^{\left (-e^{3} - e^{2}\right )} - e^{3} - e^{2}\right )}}{4 \, x^{2}} \,d x } \] Input:
integrate(1/4*((-4*exp(exp(3)+exp(2))-x)*log(5*x)+4*exp(exp(3)+exp(2)))/x^ 2/exp(exp(3)+exp(2))/exp(1/4*x/exp(exp(3)+exp(2))),x, algorithm="maxima")
Output:
-1/4*e^(-e^3 - e^2)*gamma(-1, 1/4*x*e^(-e^3 - e^2)) + e^(-1/4*x*e^(-e^3 - e^2))*log(x)/x - 1/4*integrate((4*(log(5) + 1)*e^(e^3 + e^2) + x*log(5))*e ^(-1/4*x*e^(-e^3 - e^2) - e^3 - e^2)/x^2, x)
\[ \int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} \left (4 e^{e^2+e^3}+\left (-4 e^{e^2+e^3}-x\right ) \log (5 x)\right )}{4 x^2} \, dx=\int { -\frac {{\left ({\left (x + 4 \, e^{\left (e^{3} + e^{2}\right )}\right )} \log \left (5 \, x\right ) - 4 \, e^{\left (e^{3} + e^{2}\right )}\right )} e^{\left (-\frac {1}{4} \, x e^{\left (-e^{3} - e^{2}\right )} - e^{3} - e^{2}\right )}}{4 \, x^{2}} \,d x } \] Input:
integrate(1/4*((-4*exp(exp(3)+exp(2))-x)*log(5*x)+4*exp(exp(3)+exp(2)))/x^ 2/exp(exp(3)+exp(2))/exp(1/4*x/exp(exp(3)+exp(2))),x, algorithm="giac")
Output:
integrate(-1/4*((x + 4*e^(e^3 + e^2))*log(5*x) - 4*e^(e^3 + e^2))*e^(-1/4* x*e^(-e^3 - e^2) - e^3 - e^2)/x^2, x)
Timed out. \[ \int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} \left (4 e^{e^2+e^3}+\left (-4 e^{e^2+e^3}-x\right ) \log (5 x)\right )}{4 x^2} \, dx=\int \frac {{\mathrm {e}}^{-{\mathrm {e}}^2-{\mathrm {e}}^3}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{-{\mathrm {e}}^2-{\mathrm {e}}^3}}{4}}\,\left ({\mathrm {e}}^{{\mathrm {e}}^2+{\mathrm {e}}^3}-\frac {\ln \left (5\,x\right )\,\left (x+4\,{\mathrm {e}}^{{\mathrm {e}}^2+{\mathrm {e}}^3}\right )}{4}\right )}{x^2} \,d x \] Input:
int((exp(- exp(2) - exp(3))*exp(-(x*exp(- exp(2) - exp(3)))/4)*(exp(exp(2) + exp(3)) - (log(5*x)*(x + 4*exp(exp(2) + exp(3))))/4))/x^2,x)
Output:
int((exp(- exp(2) - exp(3))*exp(-(x*exp(- exp(2) - exp(3)))/4)*(exp(exp(2) + exp(3)) - (log(5*x)*(x + 4*exp(exp(2) + exp(3))))/4))/x^2, x)
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} \left (4 e^{e^2+e^3}+\left (-4 e^{e^2+e^3}-x\right ) \log (5 x)\right )}{4 x^2} \, dx=\frac {\mathrm {log}\left (5 x \right )}{e^{\frac {x}{4 e^{e^{3}+e^{2}}}} x} \] Input:
int(1/4*((-4*exp(exp(3)+exp(2))-x)*log(5*x)+4*exp(exp(3)+exp(2)))/x^2/exp( exp(3)+exp(2))/exp(1/4*x/exp(exp(3)+exp(2))),x)
Output:
log(5*x)/(e**(x/(4*e**(e**3 + e**2)))*x)