\(\int \frac {-12+e^x (-3-3 x-x^2)+(e^{3 x} (6 x+2 x^2)+e^{2 x} (24 x+8 x^2)) \log (\frac {16+4 e^x}{3+x})+(e^{3 x} (6 x+2 x^2)+e^{2 x} (24 x+8 x^2)) \log (\frac {x}{\log (3)})}{(12 x+4 x^2+e^x (3 x+x^2)) \log (\frac {16+4 e^x}{3+x})+(12 x+4 x^2+e^x (3 x+x^2)) \log (\frac {x}{\log (3)})} \, dx\) [2195]

Optimal result

Integrand size = 168, antiderivative size = 30 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{2 x}-\log \left (\log \left (\frac {4 \left (4+e^x\right )}{3+x}\right )+\log \left (\frac {x}{\log (3)}\right )\right ) \] Output:

exp(x)^2-ln(ln(x/ln(3))+ln(4*(exp(x)+4)/(3+x)))
 

Mathematica [F]

\[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=\int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx \] Input:

Integrate[(-12 + E^x*(-3 - 3*x - x^2) + (E^(3*x)*(6*x + 2*x^2) + E^(2*x)*( 
24*x + 8*x^2))*Log[(16 + 4*E^x)/(3 + x)] + (E^(3*x)*(6*x + 2*x^2) + E^(2*x 
)*(24*x + 8*x^2))*Log[x/Log[3]])/((12*x + 4*x^2 + E^x*(3*x + x^2))*Log[(16 
 + 4*E^x)/(3 + x)] + (12*x + 4*x^2 + E^x*(3*x + x^2))*Log[x/Log[3]]),x]
 

Output:

Integrate[(-12 + E^x*(-3 - 3*x - x^2) + (E^(3*x)*(6*x + 2*x^2) + E^(2*x)*( 
24*x + 8*x^2))*Log[(16 + 4*E^x)/(3 + x)] + (E^(3*x)*(6*x + 2*x^2) + E^(2*x 
)*(24*x + 8*x^2))*Log[x/Log[3]])/((12*x + 4*x^2 + E^x*(3*x + x^2))*Log[(16 
 + 4*E^x)/(3 + x)] + (12*x + 4*x^2 + E^x*(3*x + x^2))*Log[x/Log[3]]), x]
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {e^x \left (-x^2-3 x-3\right )+\left (e^{3 x} \left (2 x^2+6 x\right )+e^{2 x} \left (8 x^2+24 x\right )\right ) \log \left (\frac {4 e^x+16}{x+3}\right )+\left (e^{3 x} \left (2 x^2+6 x\right )+e^{2 x} \left (8 x^2+24 x\right )\right ) \log \left (\frac {x}{\log (3)}\right )-12}{\left (4 x^2+e^x \left (x^2+3 x\right )+12 x\right ) \log \left (\frac {4 e^x+16}{x+3}\right )+\left (4 x^2+e^x \left (x^2+3 x\right )+12 x\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x \left (-x^2-3 x-3\right )+\left (e^{3 x} \left (2 x^2+6 x\right )+e^{2 x} \left (8 x^2+24 x\right )\right ) \log \left (\frac {4 e^x+16}{x+3}\right )+\left (e^{3 x} \left (2 x^2+6 x\right )+e^{2 x} \left (8 x^2+24 x\right )\right ) \log \left (\frac {x}{\log (3)}\right )-12}{\left (e^x+4\right ) x (x+3) \left (\log \left (\frac {4 \left (e^x+4\right )}{x+3}\right )+\log \left (\frac {x}{\log (3)}\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {-x^2-3 x-3}{x (x+3) \left (\log \left (\frac {e^x+4}{x+3}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}+2 e^{2 x}+\frac {4}{\left (e^x+4\right ) \left (\log \left (\frac {e^x+4}{x+3}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {1}{-\log \left (\frac {4+e^x}{x+3}\right )-\log \left (\frac {4 x}{\log (3)}\right )}dx+4 \int \frac {1}{\left (4+e^x\right ) \left (\log \left (\frac {4+e^x}{x+3}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}dx-\int \frac {1}{x \left (\log \left (\frac {4+e^x}{x+3}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}dx+\int \frac {1}{(x+3) \left (\log \left (\frac {4+e^x}{x+3}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}dx+e^{2 x}\)

Input:

Int[(-12 + E^x*(-3 - 3*x - x^2) + (E^(3*x)*(6*x + 2*x^2) + E^(2*x)*(24*x + 
 8*x^2))*Log[(16 + 4*E^x)/(3 + x)] + (E^(3*x)*(6*x + 2*x^2) + E^(2*x)*(24* 
x + 8*x^2))*Log[x/Log[3]])/((12*x + 4*x^2 + E^x*(3*x + x^2))*Log[(16 + 4*E 
^x)/(3 + x)] + (12*x + 4*x^2 + E^x*(3*x + x^2))*Log[x/Log[3]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 8.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
parallelrisch	\({\mathrm e}^{2 x}-\ln \left (\ln \left (\frac {4 \,{\mathrm e}^{x}+16}{3+x}\right )+\ln \left (\frac {x}{\ln \left (3\right )}\right )\right )\)	\(29\)
risch	\({\mathrm e}^{2 x}-\ln \left (\ln \left ({\mathrm e}^{x}+4\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{3+x}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+4\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+4\right )}{3+x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{3+x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+4\right )}{3+x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+4\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+4\right )}{3+x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+4\right )}{3+x}\right )}^{3}+4 i \ln \left (2\right )+2 i \ln \left (\frac {x}{\ln \left (3\right )}\right )-2 i \ln \left (3+x \right )\right )}{2}\right )\)	\(143\)

Input:

int((((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*ln((4*exp(x)+16)/(3+x))+ 
((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*ln(x/ln(3))+(-x^2-3*x-3)*exp( 
x)-12)/(((x^2+3*x)*exp(x)+4*x^2+12*x)*ln((4*exp(x)+16)/(3+x))+((x^2+3*x)*e 
xp(x)+4*x^2+12*x)*ln(x/ln(3))),x,method=_RETURNVERBOSE)
 

Output:

exp(x)^2-ln(ln(x/ln(3))+ln(4*(exp(x)+4)/(3+x)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{\left (2 \, x\right )} - \log \left (\log \left (\frac {4 \, {\left (e^{x} + 4\right )}}{x + 3}\right ) + \log \left (\frac {x}{\log \left (3\right )}\right )\right ) \] Input:

integrate((((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*log((4*exp(x)+16)/ 
(3+x))+((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*log(x/log(3))+(-x^2-3* 
x-3)*exp(x)-12)/(((x^2+3*x)*exp(x)+4*x^2+12*x)*log((4*exp(x)+16)/(3+x))+(( 
x^2+3*x)*exp(x)+4*x^2+12*x)*log(x/log(3))),x, algorithm="fricas")
 

Output:

e^(2*x) - log(log(4*(e^x + 4)/(x + 3)) + log(x/log(3)))
 

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{2 x} - \log {\left (\log {\left (\frac {x}{\log {\left (3 \right )}} \right )} + \log {\left (\frac {4 e^{x} + 16}{x + 3} \right )} \right )} \] Input:

integrate((((2*x**2+6*x)*exp(x)**3+(8*x**2+24*x)*exp(x)**2)*ln((4*exp(x)+1 
6)/(3+x))+((2*x**2+6*x)*exp(x)**3+(8*x**2+24*x)*exp(x)**2)*ln(x/ln(3))+(-x 
**2-3*x-3)*exp(x)-12)/(((x**2+3*x)*exp(x)+4*x**2+12*x)*ln((4*exp(x)+16)/(3 
+x))+((x**2+3*x)*exp(x)+4*x**2+12*x)*ln(x/ln(3))),x)
 

Output:

exp(2*x) - log(log(x/log(3)) + log((4*exp(x) + 16)/(x + 3)))
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{\left (2 \, x\right )} - \log \left (2 \, \log \left (2\right ) - \log \left (x + 3\right ) + \log \left (x\right ) + \log \left (e^{x} + 4\right ) - \log \left (\log \left (3\right )\right )\right ) \] Input:

integrate((((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*log((4*exp(x)+16)/ 
(3+x))+((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*log(x/log(3))+(-x^2-3* 
x-3)*exp(x)-12)/(((x^2+3*x)*exp(x)+4*x^2+12*x)*log((4*exp(x)+16)/(3+x))+(( 
x^2+3*x)*exp(x)+4*x^2+12*x)*log(x/log(3))),x, algorithm="maxima")
 

Output:

e^(2*x) - log(2*log(2) - log(x + 3) + log(x) + log(e^x + 4) - log(log(3)))
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{\left (2 \, x\right )} - \log \left (-\log \left (x + 3\right ) + \log \left (x\right ) + \log \left (4 \, e^{x} + 16\right ) - \log \left (\log \left (3\right )\right )\right ) \] Input:

integrate((((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*log((4*exp(x)+16)/ 
(3+x))+((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*log(x/log(3))+(-x^2-3* 
x-3)*exp(x)-12)/(((x^2+3*x)*exp(x)+4*x^2+12*x)*log((4*exp(x)+16)/(3+x))+(( 
x^2+3*x)*exp(x)+4*x^2+12*x)*log(x/log(3))),x, algorithm="giac")
 

Output:

e^(2*x) - log(-log(x + 3) + log(x) + log(4*e^x + 16) - log(log(3)))
 

Mupad [B] (verification not implemented)

Time = 3.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx={\mathrm {e}}^{2\,x}-\ln \left (\ln \left (\frac {4\,{\mathrm {e}}^x+16}{\ln \left (3\right )\,\left (x+3\right )}\right )+\ln \left (x\right )\right ) \] Input:

int(-(exp(x)*(3*x + x^2 + 3) - log((4*exp(x) + 16)/(x + 3))*(exp(3*x)*(6*x 
 + 2*x^2) + exp(2*x)*(24*x + 8*x^2)) - log(x/log(3))*(exp(3*x)*(6*x + 2*x^ 
2) + exp(2*x)*(24*x + 8*x^2)) + 12)/(log(x/log(3))*(12*x + exp(x)*(3*x + x 
^2) + 4*x^2) + log((4*exp(x) + 16)/(x + 3))*(12*x + exp(x)*(3*x + x^2) + 4 
*x^2)),x)
 

Output:

exp(2*x) - log(log((4*exp(x) + 16)/(log(3)*(x + 3))) + log(x))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{2 x}-\mathrm {log}\left (\mathrm {log}\left (\frac {4 e^{x}+16}{x +3}\right )+\mathrm {log}\left (\frac {x}{\mathrm {log}\left (3\right )}\right )\right ) \] Input:

int((((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*log((4*exp(x)+16)/(3+x)) 
+((2*x^2+6*x)*exp(x)^3+(8*x^2+24*x)*exp(x)^2)*log(x/log(3))+(-x^2-3*x-3)*e 
xp(x)-12)/(((x^2+3*x)*exp(x)+4*x^2+12*x)*log((4*exp(x)+16)/(3+x))+((x^2+3* 
x)*exp(x)+4*x^2+12*x)*log(x/log(3))),x)
 

Output:

e**(2*x) - log(log((4*e**x + 16)/(x + 3)) + log(x/log(3)))