\(\int \frac {2-2 x-x \log (\frac {16}{25} e^{-2 x} x^2)}{x \log (\frac {16}{25} e^{-2 x} x^2)} \, dx\) [4007]

Optimal result

Integrand size = 40, antiderivative size = 20 \[ \int \frac {2-2 x-x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx=-x+\log \left (25 \log \left (\frac {16}{25} e^{-2 x} x^2\right )\right ) \]

[Out]

ln(25*ln(16/25*x^2*exp(x)^2/exp(2*x)^2))-x

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6874, 6816} \[ \int \frac {2-2 x-x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx=\log \left (\log \left (\frac {16}{25} e^{-2 x} x^2\right )\right )-x \]

[In]

Int[(2 - 2*x - x*Log[(16*x^2)/(25*E^(2*x))])/(x*Log[(16*x^2)/(25*E^(2*x))]),x]

[Out]

-x + Log[Log[(16*x^2)/(25*E^(2*x))]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-1-\frac {2 (-1+x)}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}\right ) \, dx \\ & = -x-2 \int \frac {-1+x}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx \\ & = -x+\log \left (\log \left (\frac {16}{25} e^{-2 x} x^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {2-2 x-x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx=-x+\log \left (\log \left (\frac {16}{25} e^{-2 x} x^2\right )\right ) \]

[In]

Integrate[(2 - 2*x - x*Log[(16*x^2)/(25*E^(2*x))])/(x*Log[(16*x^2)/(25*E^(2*x))]),x]

[Out]

-x + Log[Log[(16*x^2)/(25*E^(2*x))]]

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
norman	\(-x +\ln \left (\ln \left (\frac {16 x^{2} {\mathrm e}^{-2 x}}{25}\right )\right )\)	\(16\)
default	\(-x +\ln \left (\ln \left (\frac {16 x^{2} {\mathrm e}^{2 x} {\mathrm e}^{-4 x}}{25}\right )\right )\)	\(22\)
parallelrisch	\(-x +\ln \left (\ln \left (\frac {16 x^{2} {\mathrm e}^{2 x} {\mathrm e}^{-4 x}}{25}\right )\right )\)	\(22\)
parts	\(-x +\ln \left (\ln \left (\frac {16 x^{2} {\mathrm e}^{2 x} {\mathrm e}^{-4 x}}{25}\right )\right )\)	\(22\)
risch	\(-x +\ln \left (\ln \left ({\mathrm e}^{x}\right )+\frac {i \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (i x^{2} {\mathrm e}^{-2 x}\right )-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{2} {\mathrm e}^{-2 x}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (i x^{2} {\mathrm e}^{-2 x}\right )^{2}-\pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+2 \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2} {\mathrm e}^{-2 x}\right )^{3}+8 i \ln \left (2\right )+4 i \ln \left (x \right )-4 i \ln \left (5\right )\right )}{4}\right )\)	\(214\)

[In]

int((-x*ln(16/25*x^2*exp(x)^2/exp(2*x)^2)-2*x+2)/x/ln(16/25*x^2*exp(x)^2/exp(2*x)^2),x,method=_RETURNVERBOSE)

[Out]

-x+ln(ln(16/25*x^2/exp(x)^2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {2-2 x-x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx=-x + \log \left (\log \left (\frac {16}{25} \, x^{2} e^{\left (-2 \, x\right )}\right )\right ) \]

[In]

integrate((-x*log(16/25*x^2*exp(x)^2/exp(2*x)^2)-2*x+2)/x/log(16/25*x^2*exp(x)^2/exp(2*x)^2),x, algorithm="fri
cas")

[Out]

-x + log(log(16/25*x^2*e^(-2*x)))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {2-2 x-x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx=- x + \log {\left (\log {\left (\frac {16 x^{2} e^{- 2 x}}{25} \right )} \right )} \]

[In]

integrate((-x*ln(16/25*x**2*exp(x)**2/exp(2*x)**2)-2*x+2)/x/ln(16/25*x**2*exp(x)**2/exp(2*x)**2),x)

[Out]

-x + log(log(16*x**2*exp(-2*x)/25))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {2-2 x-x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx=-x + \log \left (-x - \log \left (5\right ) + 2 \, \log \left (2\right ) + \log \left (x\right )\right ) \]

[In]

integrate((-x*log(16/25*x^2*exp(x)^2/exp(2*x)^2)-2*x+2)/x/log(16/25*x^2*exp(x)^2/exp(2*x)^2),x, algorithm="max
ima")

[Out]

-x + log(-x - log(5) + 2*log(2) + log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2-2 x-x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx=-x + \log \left (2 \, x - \log \left (\frac {16}{25} \, x^{2}\right )\right ) \]

[In]

integrate((-x*log(16/25*x^2*exp(x)^2/exp(2*x)^2)-2*x+2)/x/log(16/25*x^2*exp(x)^2/exp(2*x)^2),x, algorithm="gia
c")

[Out]

-x + log(2*x - log(16/25*x^2))

Mupad [B] (verification not implemented)

Time = 9.37 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {2-2 x-x \log \left (\frac {16}{25} e^{-2 x} x^2\right )}{x \log \left (\frac {16}{25} e^{-2 x} x^2\right )} \, dx=\ln \left (\ln \left (\frac {16\,x^2}{25}\right )-2\,x\right )-x \]

[In]

int(-(2*x + x*log((16*x^2*exp(-2*x))/25) - 2)/(x*log((16*x^2*exp(-2*x))/25)),x)

[Out]

log(log((16*x^2)/25) - 2*x) - x