\(\int \frac {-9+1048576 e^{5-5 e^x-5 x} (-1-5 x-5 e^x x)+\log (x)}{2 x^2} \, dx\) [2292]

Optimal result

Integrand size = 36, antiderivative size = 36 \[ \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx=\frac {4}{x}+\frac {e^{5+5 \left (-e^x-x+\log (16)\right )}+x-\log (x)}{2 x} \]

[Out]

1/2*(x-ln(x)+exp(-5*exp(x)+20*ln(2)-5*x+5))/x+4/x

Rubi [F]

\[ \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx=\int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx \]

[In]

Int[(-9 + 1048576*E^(5 - 5*E^x - 5*x)*(-1 - 5*x - 5*E^x*x) + Log[x])/(2*x^2),x]

[Out]

-1/2*1/x + (9 - Log[x])/(2*x) - 524288*Defer[Int][1/(E^(5*(-1 + E^x + x))*x^2), x] - 2621440*Defer[Int][E^(5 -
5*E^x - 4*x)/x, x] - 2621440*Defer[Int][1/(E^(5*(-1 + E^x + x))*x), x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {1048576 e^{-5 \left (-1+e^x+x\right )} (-1-5 x)}{x^2}-\frac {5242880 e^{5-5 e^x-4 x}}{x}+\frac {-9+\log (x)}{x^2}\right ) \, dx \\ & = \frac {1}{2} \int \frac {-9+\log (x)}{x^2} \, dx+524288 \int \frac {e^{-5 \left (-1+e^x+x\right )} (-1-5 x)}{x^2} \, dx-2621440 \int \frac {e^{5-5 e^x-4 x}}{x} \, dx \\ & = -\frac {1}{2 x}+\frac {9-\log (x)}{2 x}+524288 \int \left (-\frac {e^{-5 \left (-1+e^x+x\right )}}{x^2}-\frac {5 e^{-5 \left (-1+e^x+x\right )}}{x}\right ) \, dx-2621440 \int \frac {e^{5-5 e^x-4 x}}{x} \, dx \\ & = -\frac {1}{2 x}+\frac {9-\log (x)}{2 x}-524288 \int \frac {e^{-5 \left (-1+e^x+x\right )}}{x^2} \, dx-2621440 \int \frac {e^{5-5 e^x-4 x}}{x} \, dx-2621440 \int \frac {e^{-5 \left (-1+e^x+x\right )}}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx=\frac {8+1048576 e^{-5 \left (-1+e^x+x\right )}-\log (x)}{2 x} \]

[In]

Integrate[(-9 + 1048576*E^(5 - 5*E^x - 5*x)*(-1 - 5*x - 5*E^x*x) + Log[x])/(2*x^2),x]

[Out]

(8 + 1048576/E^(5*(-1 + E^x + x)) - Log[x])/(2*x)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.72

method	result	size
parallelrisch	\(-\frac {-8+\ln \left (x \right )-{\mathrm e}^{-5 \,{\mathrm e}^{x}+20 \ln \left (2\right )-5 x +5}}{2 x}\)	\(26\)
risch	\(-\frac {\ln \left (x \right )}{2 x}+\frac {4}{x}+\frac {524288 \,{\mathrm e}^{-5 \,{\mathrm e}^{x}+5-5 x}}{x}\)	\(29\)

[In]

int(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*ln(2)-5*x+5)+ln(x)-9)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/2/x*(-8+ln(x)-exp(-5*exp(x)+20*ln(2)-5*x+5))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx=\frac {e^{\left (-5 \, x - 5 \, e^{x} + 20 \, \log \left (2\right ) + 5\right )} - \log \left (x\right ) + 8}{2 \, x} \]

[In]

integrate(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*log(2)-5*x+5)+log(x)-9)/x^2,x, algorithm="fricas")

[Out]

1/2*(e^(-5*x - 5*e^x + 20*log(2) + 5) - log(x) + 8)/x

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx=\frac {524288 e^{- 5 x - 5 e^{x} + 5}}{x} - \frac {\log {\left (x \right )}}{2 x} + \frac {4}{x} \]

[In]

integrate(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*ln(2)-5*x+5)+ln(x)-9)/x**2,x)

[Out]

524288*exp(-5*x - 5*exp(x) + 5)/x - log(x)/(2*x) + 4/x

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx=-\frac {{\left ({\left (\log \left (x\right ) + 1\right )} e^{\left (5 \, x\right )} - 1048576 \, e^{\left (-5 \, e^{x} + 5\right )}\right )} e^{\left (-5 \, x\right )}}{2 \, x} + \frac {9}{2 \, x} \]

[In]

integrate(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*log(2)-5*x+5)+log(x)-9)/x^2,x, algorithm="maxima")

[Out]

-1/2*((log(x) + 1)*e^(5*x) - 1048576*e^(-5*e^x + 5))*e^(-5*x)/x + 9/2/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx=-\frac {{\left (e^{x} \log \left (x\right ) - 8 \, e^{x} - 1048576 \, e^{\left (-4 \, x - 5 \, e^{x} + 5\right )}\right )} e^{\left (-x\right )}}{2 \, x} \]

[In]

integrate(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*log(2)-5*x+5)+log(x)-9)/x^2,x, algorithm="giac")

[Out]

-1/2*(e^x*log(x) - 8*e^x - 1048576*e^(-4*x - 5*e^x + 5))*e^(-x)/x

Mupad [B] (verification not implemented)

Time = 8.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx=\frac {4}{x}-\frac {\ln \left (x\right )}{2\,x}+\frac {524288\,{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-5\,{\mathrm {e}}^x}}{x} \]

[In]

int(-((exp(20*log(2) - 5*x - 5*exp(x) + 5)*(5*x + 5*x*exp(x) + 1))/2 - log(x)/2 + 9/2)/x^2,x)

[Out]

4/x - log(x)/(2*x) + (524288*exp(-5*x)*exp(5)*exp(-5*exp(x)))/x