\(\int \frac {24 x+2 \log (4)+e^{e^x} (6+e^x (6 x+\log (4)))}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx\) [5595]

Optimal result

Integrand size = 52, antiderivative size = 17 \[ \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx=\log \left (\left (e^{e^x}+2 x\right ) (6 x+\log (4))\right ) \]

[Out]

ln((2*ln(2)+6*x)*(2*x+exp(exp(x))))

Rubi [F]

\[ \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx=\int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx \]

[In]

Int[(24*x + 2*Log[4] + E^E^x*(6 + E^x*(6*x + Log[4])))/(12*x^2 + 2*x*Log[4] + E^E^x*(6*x + Log[4])),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{\left (e^{e^x}+2 x\right ) (6 x+\log (4))} \, dx \\ & = \int \left (\frac {e^{e^x+x}}{e^{e^x}+2 x}+\frac {2 \left (3 e^{e^x}+12 x+\log (4)\right )}{\left (e^{e^x}+2 x\right ) (6 x+\log (4))}\right ) \, dx \\ & = 2 \int \frac {3 e^{e^x}+12 x+\log (4)}{\left (e^{e^x}+2 x\right ) (6 x+\log (4))} \, dx+\int \frac {e^{e^x+x}}{e^{e^x}+2 x} \, dx \\ & = 2 \int \left (\frac {1}{e^{e^x}+2 x}+\frac {3}{6 x+\log (4)}\right ) \, dx+\int \frac {e^{e^x+x}}{e^{e^x}+2 x} \, dx \\ & = \log (6 x+\log (4))+2 \int \frac {1}{e^{e^x}+2 x} \, dx+\int \frac {e^{e^x+x}}{e^{e^x}+2 x} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx=\int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx \]

[In]

Integrate[(24*x + 2*Log[4] + E^E^x*(6 + E^x*(6*x + Log[4])))/(12*x^2 + 2*x*Log[4] + E^E^x*(6*x + Log[4])),x]

[Out]

Integrate[(24*x + 2*Log[4] + E^E^x*(6 + E^x*(6*x + Log[4])))/(12*x^2 + 2*x*Log[4] + E^E^x*(6*x + Log[4])), x]

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
norman	\(\ln \left (2 x +{\mathrm e}^{{\mathrm e}^{x}}\right )+\ln \left (3 x +\ln \left (2\right )\right )\)	\(17\)
risch	\(\ln \left (2 x +{\mathrm e}^{{\mathrm e}^{x}}\right )+\ln \left (3 x +\ln \left (2\right )\right )\)	\(17\)
parallelrisch	\(\ln \left (x +\frac {{\mathrm e}^{{\mathrm e}^{x}}}{2}\right )+\ln \left (\frac {\ln \left (2\right )}{3}+x \right )\)	\(17\)

[In]

int((((2*ln(2)+6*x)*exp(x)+6)*exp(exp(x))+4*ln(2)+24*x)/((2*ln(2)+6*x)*exp(exp(x))+4*x*ln(2)+12*x^2),x,method=
_RETURNVERBOSE)

[Out]

ln(2*x+exp(exp(x)))+ln(3*x+ln(2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx=\log \left (3 \, x + \log \left (2\right )\right ) + \log \left (2 \, x + e^{\left (e^{x}\right )}\right ) \]

[In]

integrate((((2*log(2)+6*x)*exp(x)+6)*exp(exp(x))+4*log(2)+24*x)/((2*log(2)+6*x)*exp(exp(x))+4*x*log(2)+12*x^2)
,x, algorithm="fricas")

[Out]

log(3*x + log(2)) + log(2*x + e^(e^x))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx=\log {\left (2 x + e^{e^{x}} \right )} + \log {\left (3 x + \log {\left (2 \right )} \right )} \]

[In]

integrate((((2*ln(2)+6*x)*exp(x)+6)*exp(exp(x))+4*ln(2)+24*x)/((2*ln(2)+6*x)*exp(exp(x))+4*x*ln(2)+12*x**2),x)

[Out]

log(2*x + exp(exp(x))) + log(3*x + log(2))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx=\log \left (3 \, x + \log \left (2\right )\right ) + \log \left (2 \, x + e^{\left (e^{x}\right )}\right ) \]

[In]

integrate((((2*log(2)+6*x)*exp(x)+6)*exp(exp(x))+4*log(2)+24*x)/((2*log(2)+6*x)*exp(exp(x))+4*x*log(2)+12*x^2)
,x, algorithm="maxima")

[Out]

log(3*x + log(2)) + log(2*x + e^(e^x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx=-x + \log \left (2 \, x e^{x} + e^{\left (x + e^{x}\right )}\right ) + \log \left (3 \, x + \log \left (2\right )\right ) \]

[In]

integrate((((2*log(2)+6*x)*exp(x)+6)*exp(exp(x))+4*log(2)+24*x)/((2*log(2)+6*x)*exp(exp(x))+4*x*log(2)+12*x^2)
,x, algorithm="giac")

[Out]

-x + log(2*x*e^x + e^(x + e^x)) + log(3*x + log(2))

Mupad [B] (verification not implemented)

Time = 11.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {24 x+2 \log (4)+e^{e^x} \left (6+e^x (6 x+\log (4))\right )}{12 x^2+2 x \log (4)+e^{e^x} (6 x+\log (4))} \, dx=\ln \left (\frac {x\,\ln \left (2\right )}{3}+\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (6\,x+2\,\ln \left (2\right )\right )}{12}+x^2\right ) \]

[In]

int((24*x + 4*log(2) + exp(exp(x))*(exp(x)*(6*x + 2*log(2)) + 6))/(4*x*log(2) + exp(exp(x))*(6*x + 2*log(2)) +
12*x^2),x)

[Out]

log((x*log(2))/3 + (exp(exp(x))*(6*x + 2*log(2)))/12 + x^2)