\(\int \frac {164-36 e^4+32 x+e^x (48+4 x-2 x^2+e^4 (-12+2 x))+(16-4 e^4+4 x) \log (4-e^4+x)}{-256+e^{2 x} (-4+e^4-x)-192 x-48 x^2-4 x^3+e^4 (64+32 x+4 x^2)+e^x (-64-32 x-4 x^2+e^4 (16+4 x))+(-128+e^x (-16+4 e^4-4 x)-64 x-8 x^2+e^4 (32+8 x)) \log (4-e^4+x)+(-16+4 e^4-4 x) \log ^2(4-e^4+x)} \, dx\) [2906]

Optimal result

Integrand size = 187, antiderivative size = 27 \[ \int \frac {164-36 e^4+32 x+e^x \left (48+4 x-2 x^2+e^4 (-12+2 x)\right )+\left (16-4 e^4+4 x\right ) \log \left (4-e^4+x\right )}{-256+e^{2 x} \left (-4+e^4-x\right )-192 x-48 x^2-4 x^3+e^4 \left (64+32 x+4 x^2\right )+e^x \left (-64-32 x-4 x^2+e^4 (16+4 x)\right )+\left (-128+e^x \left (-16+4 e^4-4 x\right )-64 x-8 x^2+e^4 (32+8 x)\right ) \log \left (4-e^4+x\right )+\left (-16+4 e^4-4 x\right ) \log ^2\left (4-e^4+x\right )} \, dx=\frac {5-x}{4+\frac {e^x}{2}+x+\log \left (4-e^4+x\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {164-36 e^4+32 x+e^x \left (48+4 x-2 x^2+e^4 (-12+2 x)\right )+\left (16-4 e^4+4 x\right ) \log \left (4-e^4+x\right )}{-256+e^{2 x} \left (-4+e^4-x\right )-192 x-48 x^2-4 x^3+e^4 \left (64+32 x+4 x^2\right )+e^x \left (-64-32 x-4 x^2+e^4 (16+4 x)\right )+\left (-128+e^x \left (-16+4 e^4-4 x\right )-64 x-8 x^2+e^4 (32+8 x)\right ) \log \left (4-e^4+x\right )+\left (-16+4 e^4-4 x\right ) \log ^2\left (4-e^4+x\right )} \, dx=\frac {2 (5-x)}{8+e^x+2 x+2 \log \left (4-e^4+x\right )} \] Input:

Integrate[(164 - 36*E^4 + 32*x + E^x*(48 + 4*x - 2*x^2 + E^4*(-12 + 2*x)) 
+ (16 - 4*E^4 + 4*x)*Log[4 - E^4 + x])/(-256 + E^(2*x)*(-4 + E^4 - x) - 19 
2*x - 48*x^2 - 4*x^3 + E^4*(64 + 32*x + 4*x^2) + E^x*(-64 - 32*x - 4*x^2 + 
 E^4*(16 + 4*x)) + (-128 + E^x*(-16 + 4*E^4 - 4*x) - 64*x - 8*x^2 + E^4*(3 
2 + 8*x))*Log[4 - E^4 + x] + (-16 + 4*E^4 - 4*x)*Log[4 - E^4 + x]^2),x]
 

Output:

(2*(5 - x))/(8 + E^x + 2*x + 2*Log[4 - E^4 + 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.

\(\displaystyle \int \frac {e^x \left (-2 x^2+4 x+e^4 (2 x-12)+48\right )+32 x+\left (4 x-4 e^4+16\right ) \log \left (x-e^4+4\right )-36 e^4+164}{-4 x^3-48 x^2+e^4 \left (4 x^2+32 x+64\right )+e^x \left (-4 x^2-32 x+e^4 (4 x+16)-64\right )+\left (-8 x^2-64 x+e^x \left (-4 x+4 e^4-16\right )+e^4 (8 x+32)-128\right ) \log \left (x-e^4+4\right )-192 x+e^{2 x} \left (-x+e^4-4\right )+\left (-4 x+4 e^4-16\right ) \log ^2\left (x-e^4+4\right )-256} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {2 \left (-e^x \left (-x^2+2 x+24\right )-e^{x+4} (x-6)-16 x-\left (2 x-2 e^4+8\right ) \log \left (x-e^4+4\right )-82 \left (1-\frac {9 e^4}{41}\right )\right )}{\left (x-e^4+4\right ) \left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int -\frac {-e^{x+4} (6-x)+16 x+e^x \left (-x^2+2 x+24\right )+2 \left (x-e^4+4\right ) \log \left (x-e^4+4\right )+2 \left (41-9 e^4\right )}{\left (x-e^4+4\right ) \left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -2 \int \frac {-e^{x+4} (6-x)+16 x+e^x \left (-x^2+2 x+24\right )+2 \left (x-e^4+4\right ) \log \left (x-e^4+4\right )+2 \left (41-9 e^4\right )}{\left (x-e^4+4\right ) \left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -2 \int \left (\frac {2 (5-x) \left (-x^2-\log \left (x-e^4+4\right ) x-7 \left (1-\frac {e^4}{7}\right ) x-4 \left (1-\frac {e^4}{4}\right ) \log \left (x-e^4+4\right )-11 \left (1-\frac {3 e^4}{11}\right )\right )}{\left (x-e^4+4\right ) \left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}-\frac {x-6}{2 x+e^x+2 \log \left (x-e^4+4\right )+8}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \left (2 \int \frac {x^2}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx-2 \left (7-e^4\right ) \left (9-e^4\right ) \int \frac {1}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx+2 \left (4-e^4\right ) \left (9-e^4\right ) \int \frac {1}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx+2 \left (11-3 e^4\right ) \int \frac {1}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx-2 \left (4-e^4\right ) \left (7-e^4\right ) \left (9-e^4\right ) \int \frac {1}{\left (-x+e^4-4\right ) \left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx+2 \left (4-e^4\right )^2 \left (9-e^4\right ) \int \frac {1}{\left (-x+e^4-4\right ) \left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx+2 \left (11-3 e^4\right ) \left (9-e^4\right ) \int \frac {1}{\left (-x+e^4-4\right ) \left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx-2 \left (9-e^4\right ) \int \frac {x}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx+2 \left (7-e^4\right ) \int \frac {x}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx-2 \left (9-e^4\right ) \int \frac {\log \left (x-e^4+4\right )}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx+2 \left (4-e^4\right ) \int \frac {\log \left (x-e^4+4\right )}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx+2 \int \frac {x \log \left (x-e^4+4\right )}{\left (2 x+e^x+2 \log \left (x-e^4+4\right )+8\right )^2}dx+6 \int \frac {1}{2 x+e^x+2 \log \left (x-e^4+4\right )+8}dx-\int \frac {x}{2 x+e^x+2 \log \left (x-e^4+4\right )+8}dx\right )\)

Input:

Int[(164 - 36*E^4 + 32*x + E^x*(48 + 4*x - 2*x^2 + E^4*(-12 + 2*x)) + (16 
- 4*E^4 + 4*x)*Log[4 - E^4 + x])/(-256 + E^(2*x)*(-4 + E^4 - x) - 192*x - 
48*x^2 - 4*x^3 + E^4*(64 + 32*x + 4*x^2) + E^x*(-64 - 32*x - 4*x^2 + E^4*( 
16 + 4*x)) + (-128 + E^x*(-16 + 4*E^4 - 4*x) - 64*x - 8*x^2 + E^4*(32 + 8* 
x))*Log[4 - E^4 + x] + (-16 + 4*E^4 - 4*x)*Log[4 - E^4 + x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {2 \left (-5+x \right )}{{\mathrm e}^{x}+2 \ln \left (4-{\mathrm e}^{4}+x \right )+2 x +8}\)	\(25\)
parallelrisch	\(-\frac {4 x -20}{2 \left ({\mathrm e}^{x}+2 \ln \left (4-{\mathrm e}^{4}+x \right )+2 x +8\right )}\)	\(27\)

Input:

int(((-4*exp(4)+4*x+16)*ln(4-exp(4)+x)+((2*x-12)*exp(4)-2*x^2+4*x+48)*exp( 
x)-36*exp(4)+32*x+164)/((4*exp(4)-16-4*x)*ln(4-exp(4)+x)^2+((4*exp(4)-16-4 
*x)*exp(x)+(8*x+32)*exp(4)-8*x^2-64*x-128)*ln(4-exp(4)+x)+(exp(4)-x-4)*exp 
(x)^2+((4*x+16)*exp(4)-4*x^2-32*x-64)*exp(x)+(4*x^2+32*x+64)*exp(4)-4*x^3- 
48*x^2-192*x-256),x,method=_RETURNVERBOSE)
 

Output:

-2*(-5+x)/(exp(x)+2*ln(4-exp(4)+x)+2*x+8)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {164-36 e^4+32 x+e^x \left (48+4 x-2 x^2+e^4 (-12+2 x)\right )+\left (16-4 e^4+4 x\right ) \log \left (4-e^4+x\right )}{-256+e^{2 x} \left (-4+e^4-x\right )-192 x-48 x^2-4 x^3+e^4 \left (64+32 x+4 x^2\right )+e^x \left (-64-32 x-4 x^2+e^4 (16+4 x)\right )+\left (-128+e^x \left (-16+4 e^4-4 x\right )-64 x-8 x^2+e^4 (32+8 x)\right ) \log \left (4-e^4+x\right )+\left (-16+4 e^4-4 x\right ) \log ^2\left (4-e^4+x\right )} \, dx=-\frac {2 \, {\left (x - 5\right )}}{2 \, x + e^{x} + 2 \, \log \left (x - e^{4} + 4\right ) + 8} \] Input:

integrate(((-4*exp(4)+4*x+16)*log(4-exp(4)+x)+((2*x-12)*exp(4)-2*x^2+4*x+4 
8)*exp(x)-36*exp(4)+32*x+164)/((4*exp(4)-16-4*x)*log(4-exp(4)+x)^2+((4*exp 
(4)-16-4*x)*exp(x)+(8*x+32)*exp(4)-8*x^2-64*x-128)*log(4-exp(4)+x)+(exp(4) 
-x-4)*exp(x)^2+((4*x+16)*exp(4)-4*x^2-32*x-64)*exp(x)+(4*x^2+32*x+64)*exp( 
4)-4*x^3-48*x^2-192*x-256),x, algorithm="fricas")
 

Output:

-2*(x - 5)/(2*x + e^x + 2*log(x - e^4 + 4) + 8)
 

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {164-36 e^4+32 x+e^x \left (48+4 x-2 x^2+e^4 (-12+2 x)\right )+\left (16-4 e^4+4 x\right ) \log \left (4-e^4+x\right )}{-256+e^{2 x} \left (-4+e^4-x\right )-192 x-48 x^2-4 x^3+e^4 \left (64+32 x+4 x^2\right )+e^x \left (-64-32 x-4 x^2+e^4 (16+4 x)\right )+\left (-128+e^x \left (-16+4 e^4-4 x\right )-64 x-8 x^2+e^4 (32+8 x)\right ) \log \left (4-e^4+x\right )+\left (-16+4 e^4-4 x\right ) \log ^2\left (4-e^4+x\right )} \, dx=\frac {10 - 2 x}{2 x + e^{x} + 2 \log {\left (x - e^{4} + 4 \right )} + 8} \] Input:

integrate(((-4*exp(4)+4*x+16)*ln(4-exp(4)+x)+((2*x-12)*exp(4)-2*x**2+4*x+4 
8)*exp(x)-36*exp(4)+32*x+164)/((4*exp(4)-16-4*x)*ln(4-exp(4)+x)**2+((4*exp 
(4)-16-4*x)*exp(x)+(8*x+32)*exp(4)-8*x**2-64*x-128)*ln(4-exp(4)+x)+(exp(4) 
-x-4)*exp(x)**2+((4*x+16)*exp(4)-4*x**2-32*x-64)*exp(x)+(4*x**2+32*x+64)*e 
xp(4)-4*x**3-48*x**2-192*x-256),x)
 

Output:

(10 - 2*x)/(2*x + exp(x) + 2*log(x - exp(4) + 4) + 8)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {164-36 e^4+32 x+e^x \left (48+4 x-2 x^2+e^4 (-12+2 x)\right )+\left (16-4 e^4+4 x\right ) \log \left (4-e^4+x\right )}{-256+e^{2 x} \left (-4+e^4-x\right )-192 x-48 x^2-4 x^3+e^4 \left (64+32 x+4 x^2\right )+e^x \left (-64-32 x-4 x^2+e^4 (16+4 x)\right )+\left (-128+e^x \left (-16+4 e^4-4 x\right )-64 x-8 x^2+e^4 (32+8 x)\right ) \log \left (4-e^4+x\right )+\left (-16+4 e^4-4 x\right ) \log ^2\left (4-e^4+x\right )} \, dx=-\frac {2 \, {\left (x - 5\right )}}{2 \, x + e^{x} + 2 \, \log \left (x - e^{4} + 4\right ) + 8} \] Input:

integrate(((-4*exp(4)+4*x+16)*log(4-exp(4)+x)+((2*x-12)*exp(4)-2*x^2+4*x+4 
8)*exp(x)-36*exp(4)+32*x+164)/((4*exp(4)-16-4*x)*log(4-exp(4)+x)^2+((4*exp 
(4)-16-4*x)*exp(x)+(8*x+32)*exp(4)-8*x^2-64*x-128)*log(4-exp(4)+x)+(exp(4) 
-x-4)*exp(x)^2+((4*x+16)*exp(4)-4*x^2-32*x-64)*exp(x)+(4*x^2+32*x+64)*exp( 
4)-4*x^3-48*x^2-192*x-256),x, algorithm="maxima")
 

Output:

-2*(x - 5)/(2*x + e^x + 2*log(x - e^4 + 4) + 8)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {164-36 e^4+32 x+e^x \left (48+4 x-2 x^2+e^4 (-12+2 x)\right )+\left (16-4 e^4+4 x\right ) \log \left (4-e^4+x\right )}{-256+e^{2 x} \left (-4+e^4-x\right )-192 x-48 x^2-4 x^3+e^4 \left (64+32 x+4 x^2\right )+e^x \left (-64-32 x-4 x^2+e^4 (16+4 x)\right )+\left (-128+e^x \left (-16+4 e^4-4 x\right )-64 x-8 x^2+e^4 (32+8 x)\right ) \log \left (4-e^4+x\right )+\left (-16+4 e^4-4 x\right ) \log ^2\left (4-e^4+x\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((-4*exp(4)+4*x+16)*log(4-exp(4)+x)+((2*x-12)*exp(4)-2*x^2+4*x+4 
8)*exp(x)-36*exp(4)+32*x+164)/((4*exp(4)-16-4*x)*log(4-exp(4)+x)^2+((4*exp 
(4)-16-4*x)*exp(x)+(8*x+32)*exp(4)-8*x^2-64*x-128)*log(4-exp(4)+x)+(exp(4) 
-x-4)*exp(x)^2+((4*x+16)*exp(4)-4*x^2-32*x-64)*exp(x)+(4*x^2+32*x+64)*exp( 
4)-4*x^3-48*x^2-192*x-256),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Not invertible Error: Bad Argument 
Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {164-36 e^4+32 x+e^x \left (48+4 x-2 x^2+e^4 (-12+2 x)\right )+\left (16-4 e^4+4 x\right ) \log \left (4-e^4+x\right )}{-256+e^{2 x} \left (-4+e^4-x\right )-192 x-48 x^2-4 x^3+e^4 \left (64+32 x+4 x^2\right )+e^x \left (-64-32 x-4 x^2+e^4 (16+4 x)\right )+\left (-128+e^x \left (-16+4 e^4-4 x\right )-64 x-8 x^2+e^4 (32+8 x)\right ) \log \left (4-e^4+x\right )+\left (-16+4 e^4-4 x\right ) \log ^2\left (4-e^4+x\right )} \, dx=\int -\frac {32\,x-36\,{\mathrm {e}}^4+\ln \left (x-{\mathrm {e}}^4+4\right )\,\left (4\,x-4\,{\mathrm {e}}^4+16\right )+{\mathrm {e}}^x\,\left (4\,x-2\,x^2+{\mathrm {e}}^4\,\left (2\,x-12\right )+48\right )+164}{192\,x+\ln \left (x-{\mathrm {e}}^4+4\right )\,\left (64\,x+{\mathrm {e}}^x\,\left (4\,x-4\,{\mathrm {e}}^4+16\right )+8\,x^2-{\mathrm {e}}^4\,\left (8\,x+32\right )+128\right )-{\mathrm {e}}^4\,\left (4\,x^2+32\,x+64\right )+{\mathrm {e}}^{2\,x}\,\left (x-{\mathrm {e}}^4+4\right )+{\mathrm {e}}^x\,\left (32\,x+4\,x^2-{\mathrm {e}}^4\,\left (4\,x+16\right )+64\right )+{\ln \left (x-{\mathrm {e}}^4+4\right )}^2\,\left (4\,x-4\,{\mathrm {e}}^4+16\right )+48\,x^2+4\,x^3+256} \,d x \] Input:

int(-(32*x - 36*exp(4) + log(x - exp(4) + 4)*(4*x - 4*exp(4) + 16) + exp(x 
)*(4*x - 2*x^2 + exp(4)*(2*x - 12) + 48) + 164)/(192*x + log(x - exp(4) + 
4)*(64*x + exp(x)*(4*x - 4*exp(4) + 16) + 8*x^2 - exp(4)*(8*x + 32) + 128) 
 - exp(4)*(32*x + 4*x^2 + 64) + exp(2*x)*(x - exp(4) + 4) + exp(x)*(32*x + 
 4*x^2 - exp(4)*(4*x + 16) + 64) + log(x - exp(4) + 4)^2*(4*x - 4*exp(4) + 
 16) + 48*x^2 + 4*x^3 + 256),x)
 

Output:

int(-(32*x - 36*exp(4) + log(x - exp(4) + 4)*(4*x - 4*exp(4) + 16) + exp(x 
)*(4*x - 2*x^2 + exp(4)*(2*x - 12) + 48) + 164)/(192*x + log(x - exp(4) + 
4)*(64*x + exp(x)*(4*x - 4*exp(4) + 16) + 8*x^2 - exp(4)*(8*x + 32) + 128) 
 - exp(4)*(32*x + 4*x^2 + 64) + exp(2*x)*(x - exp(4) + 4) + exp(x)*(32*x + 
 4*x^2 - exp(4)*(4*x + 16) + 64) + log(x - exp(4) + 4)^2*(4*x - 4*exp(4) + 
 16) + 48*x^2 + 4*x^3 + 256), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {164-36 e^4+32 x+e^x \left (48+4 x-2 x^2+e^4 (-12+2 x)\right )+\left (16-4 e^4+4 x\right ) \log \left (4-e^4+x\right )}{-256+e^{2 x} \left (-4+e^4-x\right )-192 x-48 x^2-4 x^3+e^4 \left (64+32 x+4 x^2\right )+e^x \left (-64-32 x-4 x^2+e^4 (16+4 x)\right )+\left (-128+e^x \left (-16+4 e^4-4 x\right )-64 x-8 x^2+e^4 (32+8 x)\right ) \log \left (4-e^4+x\right )+\left (-16+4 e^4-4 x\right ) \log ^2\left (4-e^4+x\right )} \, dx=\frac {-2 x +10}{e^{x}+2 \,\mathrm {log}\left (-e^{4}+x +4\right )+2 x +8} \] Input:

int(((-4*exp(4)+4*x+16)*log(4-exp(4)+x)+((2*x-12)*exp(4)-2*x^2+4*x+48)*exp 
(x)-36*exp(4)+32*x+164)/((4*exp(4)-16-4*x)*log(4-exp(4)+x)^2+((4*exp(4)-16 
-4*x)*exp(x)+(8*x+32)*exp(4)-8*x^2-64*x-128)*log(4-exp(4)+x)+(exp(4)-x-4)* 
exp(x)^2+((4*x+16)*exp(4)-4*x^2-32*x-64)*exp(x)+(4*x^2+32*x+64)*exp(4)-4*x 
^3-48*x^2-192*x-256),x)
 

Output:

(2*( - x + 5))/(e**x + 2*log( - e**4 + x + 4) + 2*x + 8)