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

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

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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} $$\begin{array}{c}\log (\log \left(\frac{16}{25}{e}^{-2x}{x}^2\right))-x\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-1-\frac{2(-1+x)}{x\log \left(\frac{16}{25}{e}^{-2x}{x}^2\right)})dx\\ & =-x-2\int \frac{-1+x}{x\log \left(\frac{16}{25}{e}^{-2x}{x}^2\right)}dx\\ & =-x+\log (\log \left(\frac{16}{25}{e}^{-2x}{x}^2\right))\end{array}\end{array}$$

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

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))]]

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

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)))

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

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))

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

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

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))

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maxima [A]  time = 0.49, size = 19, normalized size = 0.95 $$\begin{array}{c}-x+\log (-x-\log (5)+2\log (2)+\log (x))\end{array}$$

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))

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

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

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

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))

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