\(\int \frac {e^{4-x} (4+(-4-2 x) \log (\frac {4}{x^2}))}{5 x^3 \log ^2(\frac {4}{x^2})} \, dx\) [8975]

Optimal result

Integrand size = 36, antiderivative size = 22 \[ \int \frac {e^{4-x} \left (4+(-4-2 x) \log \left (\frac {4}{x^2}\right )\right )}{5 x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx=\frac {2 e^{4-x}}{5 x^2 \log \left (\frac {4}{x^2}\right )} \]

[Out]

2/5/x^2/exp(x-4)/ln(4/x^2)

Rubi [F]

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

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{4-x} \left (4+(-4-2 x) \log \left (\frac {4}{x^2}\right )\right )}{x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx \\ & = \frac {1}{5} \int \left (\frac {4 e^{4-x}}{x^3 \log ^2\left (\frac {4}{x^2}\right )}-\frac {2 e^{4-x} (2+x)}{x^3 \log \left (\frac {4}{x^2}\right )}\right ) \, dx \\ & = -\left (\frac {2}{5} \int \frac {e^{4-x} (2+x)}{x^3 \log \left (\frac {4}{x^2}\right )} \, dx\right )+\frac {4}{5} \int \frac {e^{4-x}}{x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx \\ & = -\left (\frac {2}{5} \int \left (\frac {2 e^{4-x}}{x^3 \log \left (\frac {4}{x^2}\right )}+\frac {e^{4-x}}{x^2 \log \left (\frac {4}{x^2}\right )}\right ) \, dx\right )+\frac {4}{5} \int \frac {e^{4-x}}{x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx \\ & = -\left (\frac {2}{5} \int \frac {e^{4-x}}{x^2 \log \left (\frac {4}{x^2}\right )} \, dx\right )+\frac {4}{5} \int \frac {e^{4-x}}{x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx-\frac {4}{5} \int \frac {e^{4-x}}{x^3 \log \left (\frac {4}{x^2}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4-x} \left (4+(-4-2 x) \log \left (\frac {4}{x^2}\right )\right )}{5 x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx=\frac {2 e^{4-x}}{5 x^2 \log \left (\frac {4}{x^2}\right )} \]

[In]

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

[Out]

(2*E^(4 - x))/(5*x^2*Log[4/x^2])

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
parallelrisch	\(\frac {2 \,{\mathrm e}^{-x +4}}{5 x^{2} \ln \left (\frac {4}{x^{2}}\right )}\)	\(20\)
risch	\(-\frac {4 i {\mathrm e}^{-x +4}}{5 x^{2} \left (4 i \ln \left (x \right )-4 i \ln \left (2\right )+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}\right )}\)	\(70\)

[In]

int(1/5*((-2*x-4)*ln(4/x^2)+4)/x^3/exp(x-4)/ln(4/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

2/5/x^2/exp(x-4)/ln(4/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{4-x} \left (4+(-4-2 x) \log \left (\frac {4}{x^2}\right )\right )}{5 x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx=\frac {2 \, e^{\left (-x + 4\right )}}{5 \, x^{2} \log \left (\frac {4}{x^{2}}\right )} \]

[In]

integrate(1/5*((-2*x-4)*log(4/x^2)+4)/x^3/exp(x-4)/log(4/x^2)^2,x, algorithm="fricas")

[Out]

2/5*e^(-x + 4)/(x^2*log(4/x^2))

Sympy [A] (verification not implemented)

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

[In]

integrate(1/5*((-2*x-4)*ln(4/x**2)+4)/x**3/exp(x-4)/ln(4/x**2)**2,x)

[Out]

2*exp(4 - x)/(5*x**2*log(4/x**2))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{4-x} \left (4+(-4-2 x) \log \left (\frac {4}{x^2}\right )\right )}{5 x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx=\frac {e^{\left (-x + 4\right )}}{5 \, {\left (x^{2} \log \left (2\right ) - x^{2} \log \left (x\right )\right )}} \]

[In]

integrate(1/5*((-2*x-4)*log(4/x^2)+4)/x^3/exp(x-4)/log(4/x^2)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{4-x} \left (4+(-4-2 x) \log \left (\frac {4}{x^2}\right )\right )}{5 x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx=\frac {2 \, e^{\left (-x + 4\right )}}{5 \, x^{2} \log \left (\frac {4}{x^{2}}\right )} \]

[In]

integrate(1/5*((-2*x-4)*log(4/x^2)+4)/x^3/exp(x-4)/log(4/x^2)^2,x, algorithm="giac")

[Out]

2/5*e^(-x + 4)/(x^2*log(4/x^2))

Mupad [B] (verification not implemented)

Time = 13.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{4-x} \left (4+(-4-2 x) \log \left (\frac {4}{x^2}\right )\right )}{5 x^3 \log ^2\left (\frac {4}{x^2}\right )} \, dx=\frac {2\,{\mathrm {e}}^{4-x}}{5\,x^2\,\ln \left (\frac {4}{x^2}\right )} \]

[In]

int(-(exp(4 - x)*((log(4/x^2)*(2*x + 4))/5 - 4/5))/(x^3*log(4/x^2)^2),x)

[Out]

(2*exp(4 - x))/(5*x^2*log(4/x^2))