3.8.99 \(\int \frac {16 x-4 e^x x+e^{2 x} x+(-112-128 x-16 x^2+e^{2 x} (-6-7 x-x^2)+e^x (52+60 x+8 x^2)) \log (\frac {-28-4 x+e^x (6+x)}{-4+e^x})}{(112 x^3+16 x^4+e^x (-52 x^3-8 x^4)+e^{2 x} (6 x^3+x^4)) \log (\frac {-28-4 x+e^x (6+x)}{-4+e^x})+(224 x^2+32 x^3+e^x (-104 x^2-16 x^3)+e^{2 x} (12 x^2+2 x^3)) \log (\frac {-28-4 x+e^x (6+x)}{-4+e^x}) \log (\frac {x}{\log (\frac {-28-4 x+e^x (6+x)}{-4+e^x})})+(112 x+16 x^2+e^x (-52 x-8 x^2)+e^{2 x} (6 x+x^2)) \log (\frac {-28-4 x+e^x (6+x)}{-4+e^x}) \log ^2(\frac {x}{\log (\frac {-28-4 x+e^x (6+x)}{-4+e^x})})} \, dx\) [799]

3.8.99.1 Optimal result
3.8.99.2 Mathematica [A] (verified)
3.8.99.3 Rubi [F]
3.8.99.4 Maple [A] (verified)
3.8.99.5 Fricas [A] (verification not implemented)
3.8.99.6 Sympy [A] (verification not implemented)
3.8.99.7 Maxima [A] (verification not implemented)
3.8.99.8 Giac [B] (verification not implemented)
3.8.99.9 Mupad [B] (verification not implemented)

3.8.99.1 Optimal result

Integrand size = 322, antiderivative size = 24 \[ \int \frac {16 x-4 e^x x+e^{2 x} x+\left (-112-128 x-16 x^2+e^{2 x} \left (-6-7 x-x^2\right )+e^x \left (52+60 x+8 x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}{\left (112 x^3+16 x^4+e^x \left (-52 x^3-8 x^4\right )+e^{2 x} \left (6 x^3+x^4\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )+\left (224 x^2+32 x^3+e^x \left (-104 x^2-16 x^3\right )+e^{2 x} \left (12 x^2+2 x^3\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log \left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )+\left (112 x+16 x^2+e^x \left (-52 x-8 x^2\right )+e^{2 x} \left (6 x+x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log ^2\left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )} \, dx=\frac {1}{x+\log \left (\frac {x}{\log \left (6+\frac {4}{4-e^x}+x\right )}\right )} \]

output
1/(x+ln(x/ln(6+x+4/(-exp(x)+4))))
 
3.8.99.2 Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {16 x-4 e^x x+e^{2 x} x+\left (-112-128 x-16 x^2+e^{2 x} \left (-6-7 x-x^2\right )+e^x \left (52+60 x+8 x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}{\left (112 x^3+16 x^4+e^x \left (-52 x^3-8 x^4\right )+e^{2 x} \left (6 x^3+x^4\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )+\left (224 x^2+32 x^3+e^x \left (-104 x^2-16 x^3\right )+e^{2 x} \left (12 x^2+2 x^3\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log \left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )+\left (112 x+16 x^2+e^x \left (-52 x-8 x^2\right )+e^{2 x} \left (6 x+x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log ^2\left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )} \, dx=\frac {1}{x+\log \left (\frac {x}{\log \left (\frac {e^x (6+x)-4 (7+x)}{-4+e^x}\right )}\right )} \]

input
Integrate[(16*x - 4*E^x*x + E^(2*x)*x + (-112 - 128*x - 16*x^2 + E^(2*x)*( 
-6 - 7*x - x^2) + E^x*(52 + 60*x + 8*x^2))*Log[(-28 - 4*x + E^x*(6 + x))/( 
-4 + E^x)])/((112*x^3 + 16*x^4 + E^x*(-52*x^3 - 8*x^4) + E^(2*x)*(6*x^3 + 
x^4))*Log[(-28 - 4*x + E^x*(6 + x))/(-4 + E^x)] + (224*x^2 + 32*x^3 + E^x* 
(-104*x^2 - 16*x^3) + E^(2*x)*(12*x^2 + 2*x^3))*Log[(-28 - 4*x + E^x*(6 + 
x))/(-4 + E^x)]*Log[x/Log[(-28 - 4*x + E^x*(6 + x))/(-4 + E^x)]] + (112*x 
+ 16*x^2 + E^x*(-52*x - 8*x^2) + E^(2*x)*(6*x + x^2))*Log[(-28 - 4*x + E^x 
*(6 + x))/(-4 + E^x)]*Log[x/Log[(-28 - 4*x + E^x*(6 + x))/(-4 + E^x)]]^2), 
x]
 
output
(x + Log[x/Log[(E^x*(6 + x) - 4*(7 + x))/(-4 + E^x)]])^(-1)
 
3.8.99.3 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.

\(\displaystyle \int \frac {\left (-16 x^2+e^{2 x} \left (-x^2-7 x-6\right )+e^x \left (8 x^2+60 x+52\right )-128 x-112\right ) \log \left (\frac {-4 x+e^x (x+6)-28}{e^x-4}\right )-4 e^x x+e^{2 x} x+16 x}{\left (16 x^2+e^x \left (-8 x^2-52 x\right )+e^{2 x} \left (x^2+6 x\right )+112 x\right ) \log \left (\frac {-4 x+e^x (x+6)-28}{e^x-4}\right ) \log ^2\left (\frac {x}{\log \left (\frac {-4 x+e^x (x+6)-28}{e^x-4}\right )}\right )+\left (16 x^4+112 x^3+e^x \left (-8 x^4-52 x^3\right )+e^{2 x} \left (x^4+6 x^3\right )\right ) \log \left (\frac {-4 x+e^x (x+6)-28}{e^x-4}\right )+\left (32 x^3+224 x^2+e^x \left (-16 x^3-104 x^2\right )+e^{2 x} \left (2 x^3+12 x^2\right )\right ) \log \left (\frac {-4 x+e^x (x+6)-28}{e^x-4}\right ) \log \left (\frac {x}{\log \left (\frac {-4 x+e^x (x+6)-28}{e^x-4}\right )}\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\frac {-4 e^x+e^{2 x}+16}{\left (e^x-4\right ) \left (e^x (x+6)-4 (x+7)\right ) \log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right )}-\frac {x+1}{x}}{\left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right )}\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {4 \left (x^2+13 x+43\right )}{(x+6) \left (e^x x-4 x+6 e^x-28\right ) \log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right ) \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right )}\right )\right )^2}+\frac {x^2 \left (-\log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right )\right )+x-7 x \log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right )-6 \log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right )}{x (x+6) \log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right ) \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right )}\right )\right )^2}-\frac {4}{\left (e^x-4\right ) \log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right ) \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{e^x-4}\right )}\right )\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \frac {1}{\left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right )}\right )\right )^2}dx-\int \frac {1}{x \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right )}\right )\right )^2}dx-4 \int \frac {1}{\left (-4+e^x\right ) \log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right ) \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right )}\right )\right )^2}dx+\int \frac {1}{(x+6) \log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right ) \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right )}\right )\right )^2}dx+28 \int \frac {1}{\left (e^x x-4 x+6 e^x-28\right ) \log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right ) \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right )}\right )\right )^2}dx+4 \int \frac {x}{\left (e^x x-4 x+6 e^x-28\right ) \log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right ) \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right )}\right )\right )^2}dx+4 \int \frac {1}{(x+6) \left (e^x x-4 x+6 e^x-28\right ) \log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right ) \left (x+\log \left (\frac {x}{\log \left (\frac {e^x (x+6)-4 (x+7)}{-4+e^x}\right )}\right )\right )^2}dx\)

input
Int[(16*x - 4*E^x*x + E^(2*x)*x + (-112 - 128*x - 16*x^2 + E^(2*x)*(-6 - 7 
*x - x^2) + E^x*(52 + 60*x + 8*x^2))*Log[(-28 - 4*x + E^x*(6 + x))/(-4 + E 
^x)])/((112*x^3 + 16*x^4 + E^x*(-52*x^3 - 8*x^4) + E^(2*x)*(6*x^3 + x^4))* 
Log[(-28 - 4*x + E^x*(6 + x))/(-4 + E^x)] + (224*x^2 + 32*x^3 + E^x*(-104* 
x^2 - 16*x^3) + E^(2*x)*(12*x^2 + 2*x^3))*Log[(-28 - 4*x + E^x*(6 + x))/(- 
4 + E^x)]*Log[x/Log[(-28 - 4*x + E^x*(6 + x))/(-4 + E^x)]] + (112*x + 16*x 
^2 + E^x*(-52*x - 8*x^2) + E^(2*x)*(6*x + x^2))*Log[(-28 - 4*x + E^x*(6 + 
x))/(-4 + E^x)]*Log[x/Log[(-28 - 4*x + E^x*(6 + x))/(-4 + E^x)]]^2),x]
 
output
$Aborted
 

3.8.99.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
3.8.99.4 Maple [A] (verified)

Time = 167.76 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method result size
parallelrisch \(\frac {1}{\ln \left (\frac {x}{\ln \left (\frac {\left (6+x \right ) {\mathrm e}^{x}-4 x -28}{{\mathrm e}^{x}-4}\right )}\right )+x}\) \(29\)
risch \(\text {Expression too large to display}\) \(2495\)

input
int((((-x^2-7*x-6)*exp(x)^2+(8*x^2+60*x+52)*exp(x)-16*x^2-128*x-112)*ln((( 
6+x)*exp(x)-4*x-28)/(exp(x)-4))+x*exp(x)^2-4*exp(x)*x+16*x)/(((x^2+6*x)*ex 
p(x)^2+(-8*x^2-52*x)*exp(x)+16*x^2+112*x)*ln(((6+x)*exp(x)-4*x-28)/(exp(x) 
-4))*ln(x/ln(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))^2+((2*x^3+12*x^2)*exp(x)^2 
+(-16*x^3-104*x^2)*exp(x)+32*x^3+224*x^2)*ln(((6+x)*exp(x)-4*x-28)/(exp(x) 
-4))*ln(x/ln(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))+((x^4+6*x^3)*exp(x)^2+(-8* 
x^4-52*x^3)*exp(x)+16*x^4+112*x^3)*ln(((6+x)*exp(x)-4*x-28)/(exp(x)-4))),x 
,method=_RETURNVERBOSE)
 
output
1/(ln(x/ln(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))+x)
 
3.8.99.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {16 x-4 e^x x+e^{2 x} x+\left (-112-128 x-16 x^2+e^{2 x} \left (-6-7 x-x^2\right )+e^x \left (52+60 x+8 x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}{\left (112 x^3+16 x^4+e^x \left (-52 x^3-8 x^4\right )+e^{2 x} \left (6 x^3+x^4\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )+\left (224 x^2+32 x^3+e^x \left (-104 x^2-16 x^3\right )+e^{2 x} \left (12 x^2+2 x^3\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log \left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )+\left (112 x+16 x^2+e^x \left (-52 x-8 x^2\right )+e^{2 x} \left (6 x+x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log ^2\left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )} \, dx=\frac {1}{x + \log \left (\frac {x}{\log \left (\frac {{\left (x + 6\right )} e^{x} - 4 \, x - 28}{e^{x} - 4}\right )}\right )} \]

input
integrate((((-x^2-7*x-6)*exp(x)^2+(8*x^2+60*x+52)*exp(x)-16*x^2-128*x-112) 
*log(((6+x)*exp(x)-4*x-28)/(exp(x)-4))+x*exp(x)^2-4*exp(x)*x+16*x)/(((x^2+ 
6*x)*exp(x)^2+(-8*x^2-52*x)*exp(x)+16*x^2+112*x)*log(((6+x)*exp(x)-4*x-28) 
/(exp(x)-4))*log(x/log(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))^2+((2*x^3+12*x^2 
)*exp(x)^2+(-16*x^3-104*x^2)*exp(x)+32*x^3+224*x^2)*log(((6+x)*exp(x)-4*x- 
28)/(exp(x)-4))*log(x/log(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))+((x^4+6*x^3)* 
exp(x)^2+(-8*x^4-52*x^3)*exp(x)+16*x^4+112*x^3)*log(((6+x)*exp(x)-4*x-28)/ 
(exp(x)-4))),x, algorithm=\
 
output
1/(x + log(x/log(((x + 6)*e^x - 4*x - 28)/(e^x - 4))))
 
3.8.99.6 Sympy [A] (verification not implemented)

Time = 2.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {16 x-4 e^x x+e^{2 x} x+\left (-112-128 x-16 x^2+e^{2 x} \left (-6-7 x-x^2\right )+e^x \left (52+60 x+8 x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}{\left (112 x^3+16 x^4+e^x \left (-52 x^3-8 x^4\right )+e^{2 x} \left (6 x^3+x^4\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )+\left (224 x^2+32 x^3+e^x \left (-104 x^2-16 x^3\right )+e^{2 x} \left (12 x^2+2 x^3\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log \left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )+\left (112 x+16 x^2+e^x \left (-52 x-8 x^2\right )+e^{2 x} \left (6 x+x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log ^2\left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )} \, dx=\frac {1}{x + \log {\left (\frac {x}{\log {\left (\frac {- 4 x + \left (x + 6\right ) e^{x} - 28}{e^{x} - 4} \right )}} \right )}} \]

input
integrate((((-x**2-7*x-6)*exp(x)**2+(8*x**2+60*x+52)*exp(x)-16*x**2-128*x- 
112)*ln(((6+x)*exp(x)-4*x-28)/(exp(x)-4))+x*exp(x)**2-4*exp(x)*x+16*x)/((( 
x**2+6*x)*exp(x)**2+(-8*x**2-52*x)*exp(x)+16*x**2+112*x)*ln(((6+x)*exp(x)- 
4*x-28)/(exp(x)-4))*ln(x/ln(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))**2+((2*x**3 
+12*x**2)*exp(x)**2+(-16*x**3-104*x**2)*exp(x)+32*x**3+224*x**2)*ln(((6+x) 
*exp(x)-4*x-28)/(exp(x)-4))*ln(x/ln(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))+((x 
**4+6*x**3)*exp(x)**2+(-8*x**4-52*x**3)*exp(x)+16*x**4+112*x**3)*ln(((6+x) 
*exp(x)-4*x-28)/(exp(x)-4))),x)
 
output
1/(x + log(x/log((-4*x + (x + 6)*exp(x) - 28)/(exp(x) - 4))))
 
3.8.99.7 Maxima [A] (verification not implemented)

Time = 7.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {16 x-4 e^x x+e^{2 x} x+\left (-112-128 x-16 x^2+e^{2 x} \left (-6-7 x-x^2\right )+e^x \left (52+60 x+8 x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}{\left (112 x^3+16 x^4+e^x \left (-52 x^3-8 x^4\right )+e^{2 x} \left (6 x^3+x^4\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )+\left (224 x^2+32 x^3+e^x \left (-104 x^2-16 x^3\right )+e^{2 x} \left (12 x^2+2 x^3\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log \left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )+\left (112 x+16 x^2+e^x \left (-52 x-8 x^2\right )+e^{2 x} \left (6 x+x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log ^2\left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )} \, dx=\frac {1}{x + \log \left (x\right ) - \log \left (\log \left ({\left (x + 6\right )} e^{x} - 4 \, x - 28\right ) - \log \left (e^{x} - 4\right )\right )} \]

input
integrate((((-x^2-7*x-6)*exp(x)^2+(8*x^2+60*x+52)*exp(x)-16*x^2-128*x-112) 
*log(((6+x)*exp(x)-4*x-28)/(exp(x)-4))+x*exp(x)^2-4*exp(x)*x+16*x)/(((x^2+ 
6*x)*exp(x)^2+(-8*x^2-52*x)*exp(x)+16*x^2+112*x)*log(((6+x)*exp(x)-4*x-28) 
/(exp(x)-4))*log(x/log(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))^2+((2*x^3+12*x^2 
)*exp(x)^2+(-16*x^3-104*x^2)*exp(x)+32*x^3+224*x^2)*log(((6+x)*exp(x)-4*x- 
28)/(exp(x)-4))*log(x/log(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))+((x^4+6*x^3)* 
exp(x)^2+(-8*x^4-52*x^3)*exp(x)+16*x^4+112*x^3)*log(((6+x)*exp(x)-4*x-28)/ 
(exp(x)-4))),x, algorithm=\
 
output
1/(x + log(x) - log(log((x + 6)*e^x - 4*x - 28) - log(e^x - 4)))
 
3.8.99.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3523 vs. \(2 (21) = 42\).

Time = 205.73 (sec) , antiderivative size = 3523, normalized size of antiderivative = 146.79 \[ \int \frac {16 x-4 e^x x+e^{2 x} x+\left (-112-128 x-16 x^2+e^{2 x} \left (-6-7 x-x^2\right )+e^x \left (52+60 x+8 x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}{\left (112 x^3+16 x^4+e^x \left (-52 x^3-8 x^4\right )+e^{2 x} \left (6 x^3+x^4\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )+\left (224 x^2+32 x^3+e^x \left (-104 x^2-16 x^3\right )+e^{2 x} \left (12 x^2+2 x^3\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log \left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )+\left (112 x+16 x^2+e^x \left (-52 x-8 x^2\right )+e^{2 x} \left (6 x+x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log ^2\left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )} \, dx=\text {Too large to display} \]

input
integrate((((-x^2-7*x-6)*exp(x)^2+(8*x^2+60*x+52)*exp(x)-16*x^2-128*x-112) 
*log(((6+x)*exp(x)-4*x-28)/(exp(x)-4))+x*exp(x)^2-4*exp(x)*x+16*x)/(((x^2+ 
6*x)*exp(x)^2+(-8*x^2-52*x)*exp(x)+16*x^2+112*x)*log(((6+x)*exp(x)-4*x-28) 
/(exp(x)-4))*log(x/log(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))^2+((2*x^3+12*x^2 
)*exp(x)^2+(-16*x^3-104*x^2)*exp(x)+32*x^3+224*x^2)*log(((6+x)*exp(x)-4*x- 
28)/(exp(x)-4))*log(x/log(((6+x)*exp(x)-4*x-28)/(exp(x)-4)))+((x^4+6*x^3)* 
exp(x)^2+(-8*x^4-52*x^3)*exp(x)+16*x^4+112*x^3)*log(((6+x)*exp(x)-4*x-28)/ 
(exp(x)-4))),x, algorithm=\
 
output
(x^2*e^(2*x)*log(x*e^x - 4*x + 6*e^x - 28)*log((x*e^x - 4*x + 6*e^x - 28)/ 
(e^x - 4)) - 8*x^2*e^x*log(x*e^x - 4*x + 6*e^x - 28)*log((x*e^x - 4*x + 6* 
e^x - 28)/(e^x - 4)) - x^2*e^(2*x)*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4 
))*log(e^x - 4) + 8*x^2*e^x*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4))*log( 
e^x - 4) + 16*x^2*log(x*e^x - 4*x + 6*e^x - 28)*log((x*e^x - 4*x + 6*e^x - 
 28)/(e^x - 4)) + 7*x*e^(2*x)*log(x*e^x - 4*x + 6*e^x - 28)*log((x*e^x - 4 
*x + 6*e^x - 28)/(e^x - 4)) - 60*x*e^x*log(x*e^x - 4*x + 6*e^x - 28)*log(( 
x*e^x - 4*x + 6*e^x - 28)/(e^x - 4)) - 16*x^2*log((x*e^x - 4*x + 6*e^x - 2 
8)/(e^x - 4))*log(e^x - 4) - 7*x*e^(2*x)*log((x*e^x - 4*x + 6*e^x - 28)/(e 
^x - 4))*log(e^x - 4) + 60*x*e^x*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4)) 
*log(e^x - 4) - x*e^(2*x)*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4)) + 4*x* 
e^x*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4)) + 128*x*log(x*e^x - 4*x + 6* 
e^x - 28)*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4)) + 6*e^(2*x)*log(x*e^x 
- 4*x + 6*e^x - 28)*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4)) - 52*e^x*log 
(x*e^x - 4*x + 6*e^x - 28)*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4)) - 128 
*x*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4))*log(e^x - 4) - 6*e^(2*x)*log( 
(x*e^x - 4*x + 6*e^x - 28)/(e^x - 4))*log(e^x - 4) + 52*e^x*log((x*e^x - 4 
*x + 6*e^x - 28)/(e^x - 4))*log(e^x - 4) - 16*x*log((x*e^x - 4*x + 6*e^x - 
 28)/(e^x - 4)) + 112*log(x*e^x - 4*x + 6*e^x - 28)*log((x*e^x - 4*x + 6*e 
^x - 28)/(e^x - 4)) - 112*log((x*e^x - 4*x + 6*e^x - 28)/(e^x - 4))*log...
 
3.8.99.9 Mupad [B] (verification not implemented)

Time = 10.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {16 x-4 e^x x+e^{2 x} x+\left (-112-128 x-16 x^2+e^{2 x} \left (-6-7 x-x^2\right )+e^x \left (52+60 x+8 x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}{\left (112 x^3+16 x^4+e^x \left (-52 x^3-8 x^4\right )+e^{2 x} \left (6 x^3+x^4\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )+\left (224 x^2+32 x^3+e^x \left (-104 x^2-16 x^3\right )+e^{2 x} \left (12 x^2+2 x^3\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log \left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )+\left (112 x+16 x^2+e^x \left (-52 x-8 x^2\right )+e^{2 x} \left (6 x+x^2\right )\right ) \log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right ) \log ^2\left (\frac {x}{\log \left (\frac {-28-4 x+e^x (6+x)}{-4+e^x}\right )}\right )} \, dx=\frac {1}{x+\ln \left (\frac {x}{\ln \left (-\frac {4\,x-{\mathrm {e}}^x\,\left (x+6\right )+28}{{\mathrm {e}}^x-4}\right )}\right )} \]

input
int((16*x + x*exp(2*x) - 4*x*exp(x) - log(-(4*x - exp(x)*(x + 6) + 28)/(ex 
p(x) - 4))*(128*x + exp(2*x)*(7*x + x^2 + 6) - exp(x)*(60*x + 8*x^2 + 52) 
+ 16*x^2 + 112))/(log(-(4*x - exp(x)*(x + 6) + 28)/(exp(x) - 4))*(exp(2*x) 
*(6*x^3 + x^4) - exp(x)*(52*x^3 + 8*x^4) + 112*x^3 + 16*x^4) + log(x/log(- 
(4*x - exp(x)*(x + 6) + 28)/(exp(x) - 4)))^2*log(-(4*x - exp(x)*(x + 6) + 
28)/(exp(x) - 4))*(112*x + exp(2*x)*(6*x + x^2) - exp(x)*(52*x + 8*x^2) + 
16*x^2) + log(x/log(-(4*x - exp(x)*(x + 6) + 28)/(exp(x) - 4)))*log(-(4*x 
- exp(x)*(x + 6) + 28)/(exp(x) - 4))*(exp(2*x)*(12*x^2 + 2*x^3) - exp(x)*( 
104*x^2 + 16*x^3) + 224*x^2 + 32*x^3)),x)
 
output
1/(x + log(x/log(-(4*x - exp(x)*(x + 6) + 28)/(exp(x) - 4))))