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

Optimal. Leaf size=20 \[ -x+\log \left (25 \log \left (\frac {16}{25} e^{-2 x} x^2\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6742, 6684} \begin {gather*} \log \left (\log \left (\frac {16}{25} e^{-2 x} x^2\right )\right )-x \end {gather*}

Antiderivative was successfully verified.

[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 6684

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 18, normalized size = 0.90 \begin {gather*} -x+\log \left (\log \left (\frac {16}{25} e^{-2 x} x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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))]]

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 15, normalized size = 0.75 \begin {gather*} -x + \log \left (\log \left (\frac {16}{25} \, x^{2} e^{\left (-2 \, x\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________

giac [A]  time = 0.17, size = 17, normalized size = 0.85 \begin {gather*} -x + \log \left (2 \, x - \log \left (\frac {16}{25} \, x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

maple [A]  time = 0.11, size = 16, normalized size = 0.80

method	result	size
norman	\(-x +\ln \left (\ln \left (\frac {16 x^{2} {\mathrm e}^{-2 x}}{25}\right )\right )\)	\(16\)
default	\(-x +\ln \left (-\ln \left (\frac {16 x^{2} {\mathrm e}^{2 x} {\mathrm e}^{-4 x}}{25}\right )\right )\)	\(24\)
risch	\(-x +\ln \left (\ln \left ({\mathrm e}^{x}\right )+\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-2 x}\right )-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-2 x}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-2 x}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-2 x}\right )^{3}+8 i \ln \relax (2)-4 i \ln \relax (5)+4 i \ln \relax (x )\right )}{4}\right )\)	\(214\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 19, normalized size = 0.95 \begin {gather*} -x + \log \left (-x - \log \relax (5) + 2 \, \log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

mupad [B]  time = 2.64, size = 15, normalized size = 0.75 \begin {gather*} \ln \left (\ln \left (\frac {16\,x^2}{25}\right )-2\,x\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 15, normalized size = 0.75 \begin {gather*} - x + \log {\left (\log {\left (\frac {16 x^{2} e^{- 2 x}}{25} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________