\(\int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx\) [7606]

Optimal result

Integrand size = 40, antiderivative size = 21 \[ \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx=-1+\log (-2 (1+x)+4 x \log (e x)+\log (x \log (2))) \]

[Out]

ln(ln(x*ln(2))+4*x*ln(x*exp(1))-2*x-2)-1

Rubi [F]

\[ \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx=\int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx \]

[In]

Int[(1 + 2*x + 4*x*Log[E*x])/(-2*x - 2*x^2 + 4*x^2*Log[E*x] + x*Log[x*Log[2]]),x]

[Out]

6*Defer[Int][(-2 + 2*x + 4*x*Log[x] + Log[x*Log[2]])^(-1), x] + Defer[Int][1/(x*(-2 + 2*x + 4*x*Log[x] + Log[x
*Log[2]])), x] + 4*Defer[Int][Log[x]/(-2 + 2*x + 4*x*Log[x] + Log[x*Log[2]]), x]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx=\log (-2+2 x+4 x \log (x)+\log (x \log (2))) \]

[In]

Integrate[(1 + 2*x + 4*x*Log[E*x])/(-2*x - 2*x^2 + 4*x^2*Log[E*x] + x*Log[x*Log[2]]),x]

[Out]

Log[-2 + 2*x + 4*x*Log[x] + Log[x*Log[2]]]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
default	\(\ln \left (\ln \left (x \right )+\ln \left (\ln \left (2\right )\right )+4 x \ln \left (x \right )+2 x -2\right )\)	\(17\)
norman	\(\ln \left (\ln \left (x \ln \left (2\right )\right )+4 x \ln \left (x \,{\mathrm e}\right )-2 x -2\right )\)	\(20\)
parallelrisch	\(\ln \left (x \ln \left (x \,{\mathrm e}\right )-\frac {x}{2}+\frac {\ln \left (x \ln \left (2\right )\right )}{4}-\frac {1}{2}\right )\)	\(21\)
risch	\(\ln \left (1+4 x \right )+\ln \left (\ln \left (x \right )-\frac {i \left (2 i \ln \left (\ln \left (2\right )\right )+4 i x -4 i\right )}{2 \left (1+4 x \right )}\right )\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81 \[ \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx=\log \left (4 \, x + 1\right ) + \log \left (\frac {{\left (4 \, x + 1\right )} \log \left (x e\right ) - 2 \, x + \log \left (e^{\left (-1\right )} \log \left (2\right )\right ) - 2}{4 \, x + 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx=\log {\left (4 x + 1 \right )} + \log {\left (\log {\left (e x \right )} + \frac {- 2 x - 3 + \log {\left (\log {\left (2 \right )} \right )}}{4 x + 1} \right )} \]

[In]

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

[Out]

log(4*x + 1) + log(log(E*x) + (-2*x - 3 + log(log(2)))/(4*x + 1))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx=\log \left (4 \, x + 1\right ) + \log \left (\frac {{\left (4 \, x + 1\right )} \log \left (x\right ) + 2 \, x + \log \left (\log \left (2\right )\right ) - 2}{4 \, x + 1}\right ) \]

[In]

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

[Out]

log(4*x + 1) + log(((4*x + 1)*log(x) + 2*x + log(log(2)) - 2)/(4*x + 1))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx=\log \left (4 \, x \log \left (x\right ) + 2 \, x + \log \left (x\right ) + \log \left (\log \left (2\right )\right ) - 2\right ) \]

[In]

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

[Out]

log(4*x*log(x) + 2*x + log(x) + log(log(2)) - 2)

Mupad [B] (verification not implemented)

Time = 14.58 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx=\ln \left (2\,x+\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )+4\,x\,\ln \left (x\right )-2\right ) \]

[In]

int(-(2*x + 4*x*log(x*exp(1)) + 1)/(2*x - x*log(x*log(2)) + 2*x^2 - 4*x^2*log(x*exp(1))),x)

[Out]

log(2*x + log(log(2)) + log(x) + 4*x*log(x) - 2)