\(\int \frac {e^{-5 x} (e^{2 e^{-5 x}} (-2 e^{5 x}-10 x)-e^{5 x} x^3 \log (4))}{x^3 \log (4)} \, dx\) [2509]

Optimal result

Integrand size = 47, antiderivative size = 22 \[ \int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx=-5-x+\frac {e^{2 e^{-5 x}}}{x^2 \log (4)} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx=\int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx \]

[In]

Int[(E^(2/E^(5*x))*(-2*E^(5*x) - 10*x) - E^(5*x)*x^3*Log[4])/(E^(5*x)*x^3*Log[4]),x]

[Out]

-x - (2*Defer[Int][E^(2/E^(5*x))/x^3, x])/Log[4] - (10*Defer[Int][E^(2/E^(5*x) - 5*x)/x^2, x])/Log[4]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3} \, dx}{\log (4)} \\ & = \frac {\int \left (-\frac {10 e^{2 e^{-5 x}-5 x}}{x^2}-\frac {2 e^{2 e^{-5 x}}+x^3 \log (4)}{x^3}\right ) \, dx}{\log (4)} \\ & = -\frac {\int \frac {2 e^{2 e^{-5 x}}+x^3 \log (4)}{x^3} \, dx}{\log (4)}-\frac {10 \int \frac {e^{2 e^{-5 x}-5 x}}{x^2} \, dx}{\log (4)} \\ & = -\frac {\int \left (\frac {2 e^{2 e^{-5 x}}}{x^3}+\log (4)\right ) \, dx}{\log (4)}-\frac {10 \int \frac {e^{2 e^{-5 x}-5 x}}{x^2} \, dx}{\log (4)} \\ & = -x-\frac {2 \int \frac {e^{2 e^{-5 x}}}{x^3} \, dx}{\log (4)}-\frac {10 \int \frac {e^{2 e^{-5 x}-5 x}}{x^2} \, dx}{\log (4)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx=-\frac {-\frac {e^{2 e^{-5 x}}}{x^2}+x \log (4)}{\log (4)} \]

[In]

Integrate[(E^(2/E^(5*x))*(-2*E^(5*x) - 10*x) - E^(5*x)*x^3*Log[4])/(E^(5*x)*x^3*Log[4]),x]

[Out]

-((-(E^(2/E^(5*x))/x^2) + x*Log[4])/Log[4])

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
risch	\(-x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{-5 x}}}{2 x^{2} \ln \left (2\right )}\)	\(21\)
parallelrisch	\(\frac {-2 x^{3} \ln \left (2\right )+{\mathrm e}^{2 \,{\mathrm e}^{-5 x}}}{2 \ln \left (2\right ) x^{2}}\)	\(27\)

[In]

int(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x^3*ln(2)*exp(5*x))/x^3/ln(2)/exp(5*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx=-\frac {2 \, x^{3} \log \left (2\right ) - e^{\left (2 \, e^{\left (-5 \, x\right )}\right )}}{2 \, x^{2} \log \left (2\right )} \]

[In]

integrate(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x^3*log(2)*exp(5*x))/x^3/log(2)/exp(5*x),x, algorithm="fri
cas")

[Out]

-1/2*(2*x^3*log(2) - e^(2*e^(-5*x)))/(x^2*log(2))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx=- x + \frac {e^{2 e^{- 5 x}}}{2 x^{2} \log {\left (2 \right )}} \]

[In]

integrate(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x**3*ln(2)*exp(5*x))/x**3/ln(2)/exp(5*x),x)

[Out]

-x + exp(2*exp(-5*x))/(2*x**2*log(2))

Maxima [F]

\[ \int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx=\int { -\frac {{\left (x^{3} e^{\left (5 \, x\right )} \log \left (2\right ) + {\left (5 \, x + e^{\left (5 \, x\right )}\right )} e^{\left (2 \, e^{\left (-5 \, x\right )}\right )}\right )} e^{\left (-5 \, x\right )}}{x^{3} \log \left (2\right )} \,d x } \]

[In]

integrate(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x^3*log(2)*exp(5*x))/x^3/log(2)/exp(5*x),x, algorithm="max
ima")

[Out]

-(x*log(2) + integrate((5*x + e^(5*x))*e^(-5*x + 2*e^(-5*x))/x^3, x))/log(2)

Giac [F]

\[ \int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx=\int { -\frac {{\left (x^{3} e^{\left (5 \, x\right )} \log \left (2\right ) + {\left (5 \, x + e^{\left (5 \, x\right )}\right )} e^{\left (2 \, e^{\left (-5 \, x\right )}\right )}\right )} e^{\left (-5 \, x\right )}}{x^{3} \log \left (2\right )} \,d x } \]

[In]

integrate(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x^3*log(2)*exp(5*x))/x^3/log(2)/exp(5*x),x, algorithm="gia
c")

[Out]

integrate(-(x^3*e^(5*x)*log(2) + (5*x + e^(5*x))*e^(2*e^(-5*x)))*e^(-5*x)/(x^3*log(2)), x)

Mupad [B] (verification not implemented)

Time = 10.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx=\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{-5\,x}}}{2\,x^2\,\ln \left (2\right )}-x \]

[In]

int(-(exp(-5*x)*((exp(2*exp(-5*x))*(10*x + 2*exp(5*x)))/2 + x^3*exp(5*x)*log(2)))/(x^3*log(2)),x)

[Out]

exp(2*exp(-5*x))/(2*x^2*log(2)) - x