\(\int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx\) [3975]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 38, antiderivative size = 17 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (e^x+4 x (1-x \log (2 x))\right ) \]

[Out]

ln(x*(4-4*x*ln(2*x))+exp(x))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6816} \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (-4 x^2 \log (2 x)+4 x+e^x\right ) \]

[In]

Int[(-4 - E^x + 4*x + 8*x*Log[2*x])/(-E^x - 4*x + 4*x^2*Log[2*x]),x]

[Out]

Log[E^x + 4*x - 4*x^2*Log[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]]

Rubi steps \begin{align*} \text {integral}& = \log \left (e^x+4 x-4 x^2 \log (2 x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (e^x+4 x-4 x^2 \log (2 x)\right ) \]

[In]

Integrate[(-4 - E^x + 4*x + 8*x*Log[2*x])/(-E^x - 4*x + 4*x^2*Log[2*x]),x]

[Out]

Log[E^x + 4*x - 4*x^2*Log[2*x]]

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method result size
parallelrisch \(\ln \left (x^{2} \ln \left (2 x \right )-x -\frac {{\mathrm e}^{x}}{4}\right )\) \(18\)
derivativedivides \(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\) \(19\)
default \(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\) \(19\)
norman \(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\) \(19\)
risch \(2 \ln \left (x \right )+\ln \left (\ln \left (2 x \right )-\frac {4 x +{\mathrm e}^{x}}{4 x^{2}}\right )\) \(23\)

[In]

int((8*x*ln(2*x)-exp(x)+4*x-4)/(4*x^2*ln(2*x)-exp(x)-4*x),x,method=_RETURNVERBOSE)

[Out]

ln(x^2*ln(2*x)-x-1/4*exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=2 \, \log \left (2 \, x\right ) + \log \left (\frac {4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}}{x^{2}}\right ) \]

[In]

integrate((8*x*log(2*x)-exp(x)+4*x-4)/(4*x^2*log(2*x)-exp(x)-4*x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log {\left (- 4 x^{2} \log {\left (2 x \right )} + 4 x + e^{x} \right )} \]

[In]

integrate((8*x*ln(2*x)-exp(x)+4*x-4)/(4*x**2*ln(2*x)-exp(x)-4*x),x)

[Out]

log(-4*x**2*log(2*x) + 4*x + exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}\right ) \]

[In]

integrate((8*x*log(2*x)-exp(x)+4*x-4)/(4*x^2*log(2*x)-exp(x)-4*x),x, algorithm="maxima")

[Out]

log(4*x^2*log(2*x) - 4*x - e^x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}\right ) \]

[In]

integrate((8*x*log(2*x)-exp(x)+4*x-4)/(4*x^2*log(2*x)-exp(x)-4*x),x, algorithm="giac")

[Out]

log(4*x^2*log(2*x) - 4*x - e^x)

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\ln \left (4\,x^2\,\ln \left (2\,x\right )-{\mathrm {e}}^x-4\,x\right ) \]

[In]

int(-(4*x - exp(x) + 8*x*log(2*x) - 4)/(4*x + exp(x) - 4*x^2*log(2*x)),x)

[Out]

log(4*x^2*log(2*x) - exp(x) - 4*x)