\(\int \frac {-192 x^2-64 x^3-12 e^2 x^4+e (-96 x^3-12 x^4)+e^x (48 x^2+16 x^3-x^4+3 e^2 x^4+e (24 x^3+3 x^4))+(-96 x^2-16 x^3-24 e x^3+e^x (24 x^2+4 x^3+6 e x^3)) \log (4-e^x)+(-12 x^2+3 e^x x^2) \log ^2(4-e^x)}{-64-32 e x-4 e^2 x^2+e^x (16+8 e x+e^2 x^2)+(-32-8 e x+e^x (8+2 e x)) \log (4-e^x)+(-4+e^x) \log ^2(4-e^x)} \, dx\) [452]

Optimal result

Integrand size = 220, antiderivative size = 23 \[ \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx=x^3+\frac {x^4}{4+e x+\log \left (4-e^x\right )} \] Output:

x^4/(ln(-exp(x)+4)+4+x*exp(1))+x^3
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx=x^3+\frac {x^4}{4+e x+\log \left (4-e^x\right )} \] Input:

Integrate[(-192*x^2 - 64*x^3 - 12*E^2*x^4 + E*(-96*x^3 - 12*x^4) + E^x*(48 
*x^2 + 16*x^3 - x^4 + 3*E^2*x^4 + E*(24*x^3 + 3*x^4)) + (-96*x^2 - 16*x^3 
- 24*E*x^3 + E^x*(24*x^2 + 4*x^3 + 6*E*x^3))*Log[4 - E^x] + (-12*x^2 + 3*E 
^x*x^2)*Log[4 - E^x]^2)/(-64 - 32*E*x - 4*E^2*x^2 + E^x*(16 + 8*E*x + E^2* 
x^2) + (-32 - 8*E*x + E^x*(8 + 2*E*x))*Log[4 - E^x] + (-4 + E^x)*Log[4 - E 
^x]^2),x]
 

Output:

x^3 + x^4/(4 + E*x + Log[4 - E^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.

\(\displaystyle \int \frac {-12 e^2 x^4-64 x^3-192 x^2+\left (3 e^x x^2-12 x^2\right ) \log ^2\left (4-e^x\right )+e \left (-12 x^4-96 x^3\right )+\left (-24 e x^3-16 x^3-96 x^2+e^x \left (6 e x^3+4 x^3+24 x^2\right )\right ) \log \left (4-e^x\right )+e^x \left (3 e^2 x^4-x^4+16 x^3+48 x^2+e \left (3 x^4+24 x^3\right )\right )}{-4 e^2 x^2+e^x \left (e^2 x^2+8 e x+16\right )-32 e x+\left (e^x-4\right ) \log ^2\left (4-e^x\right )+\left (-8 e x+e^x (2 e x+8)-32\right ) \log \left (4-e^x\right )-64} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {x^2 \left (-3 e^{x+2} x^2+12 e^2 x^2-e^x \left (-x^2+16 x+48\right )-3 e^{x+1} (x+8) x+12 e (x+8) x+64 (x+3)-3 \left (e^x-4\right ) \log ^2\left (4-e^x\right )-2 \left (e^x-4\right ) ((2+3 e) x+12) \log \left (4-e^x\right )\right )}{\left (4-e^x\right ) \left (e x+\log \left (4-e^x\right )+4\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^2 \left (-\left ((1-3 e (1+e)) x^2\right )+16 \left (1+\frac {3 e}{2}\right ) x+3 \log ^2\left (4-e^x\right )+4 \left (1+\frac {3 e}{2}\right ) x \log \left (4-e^x\right )+24 \log \left (4-e^x\right )+48\right )}{\left (e x+\log \left (4-e^x\right )+4\right )^2}-\frac {4 x^4}{\left (e^x-4\right ) \left (e x+\log \left (4-e^x\right )+4\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(-192*x^2 - 64*x^3 - 12*E^2*x^4 + E*(-96*x^3 - 12*x^4) + E^x*(48*x^2 + 
 16*x^3 - x^4 + 3*E^2*x^4 + E*(24*x^3 + 3*x^4)) + (-96*x^2 - 16*x^3 - 24*E 
*x^3 + E^x*(24*x^2 + 4*x^3 + 6*E*x^3))*Log[4 - E^x] + (-12*x^2 + 3*E^x*x^2 
)*Log[4 - E^x]^2)/(-64 - 32*E*x - 4*E^2*x^2 + E^x*(16 + 8*E*x + E^2*x^2) + 
 (-32 - 8*E*x + E^x*(8 + 2*E*x))*Log[4 - E^x] + (-4 + E^x)*Log[4 - E^x]^2) 
,x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {x^{4}}{\ln \left (-{\mathrm e}^{x}+4\right )+4+x \,{\mathrm e}}+x^{3}\)	\(24\)
parallelrisch	\(\frac {x^{4} {\mathrm e}+x^{4}+x^{3} \ln \left (-{\mathrm e}^{x}+4\right )+4 x^{3}}{\ln \left (-{\mathrm e}^{x}+4\right )+4+x \,{\mathrm e}}\)	\(43\)

Input:

int(((3*exp(x)*x^2-12*x^2)*ln(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24*x^2)*ex 
p(x)-24*x^3*exp(1)-16*x^3-96*x^2)*ln(-exp(x)+4)+(3*x^4*exp(1)^2+(3*x^4+24* 
x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-96*x^3)*exp 
(1)-64*x^3-192*x^2)/((exp(x)-4)*ln(-exp(x)+4)^2+((2*x*exp(1)+8)*exp(x)-8*x 
*exp(1)-32)*ln(-exp(x)+4)+(x^2*exp(1)^2+8*x*exp(1)+16)*exp(x)-4*x^2*exp(1) 
^2-32*x*exp(1)-64),x,method=_RETURNVERBOSE)
 

Output:

x^4/(ln(-exp(x)+4)+4+x*exp(1))+x^3
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx=\frac {x^{4} e + x^{4} + x^{3} \log \left (-e^{x} + 4\right ) + 4 \, x^{3}}{x e + \log \left (-e^{x} + 4\right ) + 4} \] Input:

integrate(((3*exp(x)*x^2-12*x^2)*log(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24* 
x^2)*exp(x)-24*x^3*exp(1)-16*x^3-96*x^2)*log(-exp(x)+4)+(3*x^4*exp(1)^2+(3 
*x^4+24*x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-96* 
x^3)*exp(1)-64*x^3-192*x^2)/((exp(x)-4)*log(-exp(x)+4)^2+((2*exp(1)*x+8)*e 
xp(x)-8*exp(1)*x-32)*log(-exp(x)+4)+(x^2*exp(1)^2+8*exp(1)*x+16)*exp(x)-4* 
x^2*exp(1)^2-32*exp(1)*x-64),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx=\frac {x^{4}}{e x + \log {\left (4 - e^{x} \right )} + 4} + x^{3} \] Input:

integrate(((3*exp(x)*x**2-12*x**2)*ln(-exp(x)+4)**2+((6*x**3*exp(1)+4*x**3 
+24*x**2)*exp(x)-24*x**3*exp(1)-16*x**3-96*x**2)*ln(-exp(x)+4)+(3*x**4*exp 
(1)**2+(3*x**4+24*x**3)*exp(1)-x**4+16*x**3+48*x**2)*exp(x)-12*x**4*exp(1) 
**2+(-12*x**4-96*x**3)*exp(1)-64*x**3-192*x**2)/((exp(x)-4)*ln(-exp(x)+4)* 
*2+((2*exp(1)*x+8)*exp(x)-8*exp(1)*x-32)*ln(-exp(x)+4)+(x**2*exp(1)**2+8*e 
xp(1)*x+16)*exp(x)-4*x**2*exp(1)**2-32*exp(1)*x-64),x)
 

Output:

x**4/(E*x + log(4 - exp(x)) + 4) + x**3
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx=\frac {x^{4} {\left (e + 1\right )} + x^{3} \log \left (-e^{x} + 4\right ) + 4 \, x^{3}}{x e + \log \left (-e^{x} + 4\right ) + 4} \] Input:

integrate(((3*exp(x)*x^2-12*x^2)*log(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24* 
x^2)*exp(x)-24*x^3*exp(1)-16*x^3-96*x^2)*log(-exp(x)+4)+(3*x^4*exp(1)^2+(3 
*x^4+24*x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-96* 
x^3)*exp(1)-64*x^3-192*x^2)/((exp(x)-4)*log(-exp(x)+4)^2+((2*exp(1)*x+8)*e 
xp(x)-8*exp(1)*x-32)*log(-exp(x)+4)+(x^2*exp(1)^2+8*exp(1)*x+16)*exp(x)-4* 
x^2*exp(1)^2-32*exp(1)*x-64),x, algorithm="maxima")
 

Output:

(x^4*(e + 1) + x^3*log(-e^x + 4) + 4*x^3)/(x*e + log(-e^x + 4) + 4)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx=\frac {x^{4} e + x^{4} + x^{3} \log \left (-e^{x} + 4\right ) + 4 \, x^{3}}{x e + \log \left (-e^{x} + 4\right ) + 4} \] Input:

integrate(((3*exp(x)*x^2-12*x^2)*log(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24* 
x^2)*exp(x)-24*x^3*exp(1)-16*x^3-96*x^2)*log(-exp(x)+4)+(3*x^4*exp(1)^2+(3 
*x^4+24*x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-96* 
x^3)*exp(1)-64*x^3-192*x^2)/((exp(x)-4)*log(-exp(x)+4)^2+((2*exp(1)*x+8)*e 
xp(x)-8*exp(1)*x-32)*log(-exp(x)+4)+(x^2*exp(1)^2+8*exp(1)*x+16)*exp(x)-4* 
x^2*exp(1)^2-32*exp(1)*x-64),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 139, normalized size of antiderivative = 6.04 \[ \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx=\frac {\frac {x^2\,\left (64\,x+x^2\,{\mathrm {e}}^x-3\,x^2\,{\mathrm {e}}^{x+1}+12\,x^2\,\mathrm {e}-16\,x\,{\mathrm {e}}^x\right )}{{\mathrm {e}}^{x+1}-4\,\mathrm {e}+{\mathrm {e}}^x}-\frac {4\,x^3\,\ln \left (4-{\mathrm {e}}^x\right )\,\left ({\mathrm {e}}^x-4\right )}{{\mathrm {e}}^{x+1}-4\,\mathrm {e}+{\mathrm {e}}^x}}{\ln \left (4-{\mathrm {e}}^x\right )+x\,\mathrm {e}+4}+\frac {x^3\,\left (3\,\mathrm {e}+15\right )}{3\,\left (\mathrm {e}+1\right )}-\frac {16\,x^3}{\left ({\mathrm {e}}^x-\frac {4\,\mathrm {e}}{\mathrm {e}+1}\right )\,{\left (\mathrm {e}+1\right )}^2} \] Input:

int((log(4 - exp(x))*(24*x^3*exp(1) - exp(x)*(6*x^3*exp(1) + 24*x^2 + 4*x^ 
3) + 96*x^2 + 16*x^3) - log(4 - exp(x))^2*(3*x^2*exp(x) - 12*x^2) - exp(x) 
*(exp(1)*(24*x^3 + 3*x^4) + 3*x^4*exp(2) + 48*x^2 + 16*x^3 - x^4) + exp(1) 
*(96*x^3 + 12*x^4) + 12*x^4*exp(2) + 192*x^2 + 64*x^3)/(32*x*exp(1) - log( 
4 - exp(x))^2*(exp(x) - 4) - exp(x)*(8*x*exp(1) + x^2*exp(2) + 16) + 4*x^2 
*exp(2) + log(4 - exp(x))*(8*x*exp(1) - exp(x)*(2*x*exp(1) + 8) + 32) + 64 
),x)
 

Output:

((x^2*(64*x + x^2*exp(x) - 3*x^2*exp(x + 1) + 12*x^2*exp(1) - 16*x*exp(x)) 
)/(exp(x + 1) - 4*exp(1) + exp(x)) - (4*x^3*log(4 - exp(x))*(exp(x) - 4))/ 
(exp(x + 1) - 4*exp(1) + exp(x)))/(log(4 - exp(x)) + x*exp(1) + 4) + (x^3* 
(3*exp(1) + 15))/(3*(exp(1) + 1)) - (16*x^3)/((exp(x) - (4*exp(1))/(exp(1) 
 + 1))*(exp(1) + 1)^2)
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {-192 x^2-64 x^3-12 e^2 x^4+e \left (-96 x^3-12 x^4\right )+e^x \left (48 x^2+16 x^3-x^4+3 e^2 x^4+e \left (24 x^3+3 x^4\right )\right )+\left (-96 x^2-16 x^3-24 e x^3+e^x \left (24 x^2+4 x^3+6 e x^3\right )\right ) \log \left (4-e^x\right )+\left (-12 x^2+3 e^x x^2\right ) \log ^2\left (4-e^x\right )}{-64-32 e x-4 e^2 x^2+e^x \left (16+8 e x+e^2 x^2\right )+\left (-32-8 e x+e^x (8+2 e x)\right ) \log \left (4-e^x\right )+\left (-4+e^x\right ) \log ^2\left (4-e^x\right )} \, dx=\frac {x^{3} \left (\mathrm {log}\left (-e^{x}+4\right )+e x +x +4\right )}{\mathrm {log}\left (-e^{x}+4\right )+e x +4} \] Input:

int(((3*exp(x)*x^2-12*x^2)*log(-exp(x)+4)^2+((6*x^3*exp(1)+4*x^3+24*x^2)*e 
xp(x)-24*x^3*exp(1)-16*x^3-96*x^2)*log(-exp(x)+4)+(3*x^4*exp(1)^2+(3*x^4+2 
4*x^3)*exp(1)-x^4+16*x^3+48*x^2)*exp(x)-12*x^4*exp(1)^2+(-12*x^4-96*x^3)*e 
xp(1)-64*x^3-192*x^2)/((exp(x)-4)*log(-exp(x)+4)^2+((2*exp(1)*x+8)*exp(x)- 
8*exp(1)*x-32)*log(-exp(x)+4)+(x^2*exp(1)^2+8*exp(1)*x+16)*exp(x)-4*x^2*ex 
p(1)^2-32*exp(1)*x-64),x)
 

Output:

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