\(\int \frac {e^x (-784-170 x-34 x^2-2 x^3+(196+28 x+x^2) \log (4))+e^x (112+12 x+2 x^2+(-28-2 x) \log (4)) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx\) [842]

Optimal result

Integrand size = 98, antiderivative size = 24 \[ \int \frac {e^x \left (-784-170 x-34 x^2-2 x^3+\left (196+28 x+x^2\right ) \log (4)\right )+e^x \left (112+12 x+2 x^2+(-28-2 x) \log (4)\right ) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx=e^x \left (-4+\log (4)+\frac {2 x^2}{-14-x+\log (3 x)}\right ) \] Output:

(4*x^2/(2*ln(3*x)-2*x-28)-4+2*ln(2))*exp(x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 5.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-784-170 x-34 x^2-2 x^3+\left (196+28 x+x^2\right ) \log (4)\right )+e^x \left (112+12 x+2 x^2+(-28-2 x) \log (4)\right ) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx=e^x \left (-4+\log (4)-\frac {2 x^2}{14+x-\log (3 x)}\right ) \] Input:

Integrate[(E^x*(-784 - 170*x - 34*x^2 - 2*x^3 + (196 + 28*x + x^2)*Log[4]) 
 + E^x*(112 + 12*x + 2*x^2 + (-28 - 2*x)*Log[4])*Log[3*x] + E^x*(-4 + Log[ 
4])*Log[3*x]^2)/(196 + 28*x + x^2 + (-28 - 2*x)*Log[3*x] + Log[3*x]^2),x]
 

Output:

E^x*(-4 + Log[4] - (2*x^2)/(14 + 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.

\(\displaystyle \int \frac {e^x \left (2 x^2+12 x+(-2 x-28) \log (4)+112\right ) \log (3 x)+e^x \left (-2 x^3-34 x^2+\left (x^2+28 x+196\right ) \log (4)-170 x-784\right )+e^x (\log (4)-4) \log ^2(3 x)}{x^2+28 x+\log ^2(3 x)+(-2 x-28) \log (3 x)+196} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x \left (2 x^2+12 x+(-2 x-28) \log (4)+112\right ) \log (3 x)+e^x \left (-2 x^3-34 x^2+\left (x^2+28 x+196\right ) \log (4)-170 x-784\right )+e^x (\log (4)-4) \log ^2(3 x)}{(x-\log (3 x)+14)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {2 e^x x^3}{(x-\log (3 x)+14)^2}+\frac {2 e^x x^2 \log (3 x)}{(x-\log (3 x)+14)^2}-\frac {34 e^x x^2 \left (1-\frac {\log (2)}{17}\right )}{(x-\log (3 x)+14)^2}+\frac {e^x (\log (4)-4) \log ^2(3 x)}{(x-\log (3 x)+14)^2}+\frac {12 e^x x \left (1-\frac {\log (2)}{3}\right ) \log (3 x)}{(x-\log (3 x)+14)^2}-\frac {170 e^x x \left (1-\frac {28 \log (2)}{85}\right )}{(x-\log (3 x)+14)^2}+\frac {112 e^x \left (1-\frac {\log (2)}{2}\right ) \log (3 x)}{(x-\log (3 x)+14)^2}-\frac {784 e^x \left (1-\frac {\log (2)}{2}\right )}{(x-\log (3 x)+14)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -(4-\log (4)) \int \frac {e^x x^2}{(x-\log (3 x)+14)^2}dx-2 (17-\log (2)) \int \frac {e^x x^2}{(x-\log (3 x)+14)^2}dx+4 (3-\log (2)) \int \frac {e^x x^2}{(x-\log (3 x)+14)^2}dx+28 \int \frac {e^x x^2}{(x-\log (3 x)+14)^2}dx-2 \int \frac {e^x x^2}{x-\log (3 x)+14}dx-196 (4-\log (4)) \int \frac {e^x}{(x-\log (3 x)+14)^2}dx+392 (2-\log (2)) \int \frac {e^x}{(x-\log (3 x)+14)^2}dx-28 (4-\log (4)) \int \frac {e^x x}{(x-\log (3 x)+14)^2}dx+56 (3-\log (2)) \int \frac {e^x x}{(x-\log (3 x)+14)^2}dx+56 (2-\log (2)) \int \frac {e^x x}{(x-\log (3 x)+14)^2}dx-2 (85-28 \log (2)) \int \frac {e^x x}{(x-\log (3 x)+14)^2}dx+28 (4-\log (4)) \int \frac {e^x}{x-\log (3 x)+14}dx+2 (4-\log (4)) \int \frac {e^x x}{x-\log (3 x)+14}dx-4 (3-\log (2)) \int \frac {e^x x}{x-\log (3 x)+14}dx+56 (2-\log (2)) \int \frac {e^x}{-x+\log (3 x)-14}dx-e^x (4-\log (4))\)

Input:

Int[(E^x*(-784 - 170*x - 34*x^2 - 2*x^3 + (196 + 28*x + x^2)*Log[4]) + E^x 
*(112 + 12*x + 2*x^2 + (-28 - 2*x)*Log[4])*Log[3*x] + E^x*(-4 + Log[4])*Lo 
g[3*x]^2)/(196 + 28*x + x^2 + (-28 - 2*x)*Log[3*x] + Log[3*x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25

method	result	size
risch	\(2 \,{\mathrm e}^{x} \ln \left (2\right )-4 \,{\mathrm e}^{x}-\frac {2 x^{2} {\mathrm e}^{x}}{x -\ln \left (3 x \right )+14}\)	\(30\)
parallelrisch	\(\frac {2 x \ln \left (2\right ) {\mathrm e}^{x}-2 \ln \left (2\right ) {\mathrm e}^{x} \ln \left (3 x \right )-2 \,{\mathrm e}^{x} x^{2}+28 \,{\mathrm e}^{x} \ln \left (2\right )-4 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x} \ln \left (3 x \right )-56 \,{\mathrm e}^{x}}{x -\ln \left (3 x \right )+14}\)	\(61\)

Input:

int(((2*ln(2)-4)*exp(x)*ln(3*x)^2+(2*(-2*x-28)*ln(2)+2*x^2+12*x+112)*exp(x 
)*ln(3*x)+(2*(x^2+28*x+196)*ln(2)-2*x^3-34*x^2-170*x-784)*exp(x))/(ln(3*x) 
^2+(-2*x-28)*ln(3*x)+x^2+28*x+196),x,method=_RETURNVERBOSE)
 

Output:

2*exp(x)*ln(2)-4*exp(x)-2*x^2*exp(x)/(x-ln(3*x)+14)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {e^x \left (-784-170 x-34 x^2-2 x^3+\left (196+28 x+x^2\right ) \log (4)\right )+e^x \left (112+12 x+2 x^2+(-28-2 x) \log (4)\right ) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx=-\frac {2 \, {\left ({\left (\log \left (2\right ) - 2\right )} e^{x} \log \left (3 \, x\right ) + {\left (x^{2} - {\left (x + 14\right )} \log \left (2\right ) + 2 \, x + 28\right )} e^{x}\right )}}{x - \log \left (3 \, x\right ) + 14} \] Input:

integrate(((2*log(2)-4)*exp(x)*log(3*x)^2+(2*(-2*x-28)*log(2)+2*x^2+12*x+1 
12)*exp(x)*log(3*x)+(2*(x^2+28*x+196)*log(2)-2*x^3-34*x^2-170*x-784)*exp(x 
))/(log(3*x)^2+(-2*x-28)*log(3*x)+x^2+28*x+196),x, algorithm="fricas")
 

Output:

-2*((log(2) - 2)*e^x*log(3*x) + (x^2 - (x + 14)*log(2) + 2*x + 28)*e^x)/(x 
 - log(3*x) + 14)
 

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {e^x \left (-784-170 x-34 x^2-2 x^3+\left (196+28 x+x^2\right ) \log (4)\right )+e^x \left (112+12 x+2 x^2+(-28-2 x) \log (4)\right ) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx=\frac {\left (- 2 x^{2} - 4 x + 2 x \log {\left (2 \right )} - 2 \log {\left (2 \right )} \log {\left (3 x \right )} + 4 \log {\left (3 x \right )} - 56 + 28 \log {\left (2 \right )}\right ) e^{x}}{x - \log {\left (3 x \right )} + 14} \] Input:

integrate(((2*ln(2)-4)*exp(x)*ln(3*x)**2+(2*(-2*x-28)*ln(2)+2*x**2+12*x+11 
2)*exp(x)*ln(3*x)+(2*(x**2+28*x+196)*ln(2)-2*x**3-34*x**2-170*x-784)*exp(x 
))/(ln(3*x)**2+(-2*x-28)*ln(3*x)+x**2+28*x+196),x)
 

Output:

(-2*x**2 - 4*x + 2*x*log(2) - 2*log(2)*log(3*x) + 4*log(3*x) - 56 + 28*log 
(2))*exp(x)/(x - log(3*x) + 14)
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {e^x \left (-784-170 x-34 x^2-2 x^3+\left (196+28 x+x^2\right ) \log (4)\right )+e^x \left (112+12 x+2 x^2+(-28-2 x) \log (4)\right ) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx=-\frac {2 \, {\left (x^{2} - x {\left (\log \left (2\right ) - 2\right )} + {\left (\log \left (3\right ) - 14\right )} \log \left (2\right ) + {\left (\log \left (2\right ) - 2\right )} \log \left (x\right ) - 2 \, \log \left (3\right ) + 28\right )} e^{x}}{x - \log \left (3\right ) - \log \left (x\right ) + 14} \] Input:

integrate(((2*log(2)-4)*exp(x)*log(3*x)^2+(2*(-2*x-28)*log(2)+2*x^2+12*x+1 
12)*exp(x)*log(3*x)+(2*(x^2+28*x+196)*log(2)-2*x^3-34*x^2-170*x-784)*exp(x 
))/(log(3*x)^2+(-2*x-28)*log(3*x)+x^2+28*x+196),x, algorithm="maxima")
 

Output:

-2*(x^2 - x*(log(2) - 2) + (log(3) - 14)*log(2) + (log(2) - 2)*log(x) - 2* 
log(3) + 28)*e^x/(x - log(3) - log(x) + 14)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).

Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.46 \[ \int \frac {e^x \left (-784-170 x-34 x^2-2 x^3+\left (196+28 x+x^2\right ) \log (4)\right )+e^x \left (112+12 x+2 x^2+(-28-2 x) \log (4)\right ) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx=-\frac {2 \, {\left (x^{2} e^{x} - x e^{x} \log \left (2\right ) + e^{x} \log \left (2\right ) \log \left (3 \, x\right ) + 2 \, x e^{x} - 14 \, e^{x} \log \left (2\right ) - 2 \, e^{x} \log \left (3 \, x\right ) + 28 \, e^{x}\right )}}{x - \log \left (3 \, x\right ) + 14} \] Input:

integrate(((2*log(2)-4)*exp(x)*log(3*x)^2+(2*(-2*x-28)*log(2)+2*x^2+12*x+1 
12)*exp(x)*log(3*x)+(2*(x^2+28*x+196)*log(2)-2*x^3-34*x^2-170*x-784)*exp(x 
))/(log(3*x)^2+(-2*x-28)*log(3*x)+x^2+28*x+196),x, algorithm="giac")
 

Output:

-2*(x^2*e^x - x*e^x*log(2) + e^x*log(2)*log(3*x) + 2*x*e^x - 14*e^x*log(2) 
 - 2*e^x*log(3*x) + 28*e^x)/(x - log(3*x) + 14)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x \left (-784-170 x-34 x^2-2 x^3+\left (196+28 x+x^2\right ) \log (4)\right )+e^x \left (112+12 x+2 x^2+(-28-2 x) \log (4)\right ) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx=\int \frac {{\mathrm {e}}^x\,\left (2\,\ln \left (2\right )-4\right )\,{\ln \left (3\,x\right )}^2+{\mathrm {e}}^x\,\left (12\,x-2\,\ln \left (2\right )\,\left (2\,x+28\right )+2\,x^2+112\right )\,\ln \left (3\,x\right )-{\mathrm {e}}^x\,\left (170\,x+34\,x^2+2\,x^3-2\,\ln \left (2\right )\,\left (x^2+28\,x+196\right )+784\right )}{28\,x+{\ln \left (3\,x\right )}^2+x^2-\ln \left (3\,x\right )\,\left (2\,x+28\right )+196} \,d x \] Input:

int((log(3*x)^2*exp(x)*(2*log(2) - 4) - exp(x)*(170*x + 34*x^2 + 2*x^3 - 2 
*log(2)*(28*x + x^2 + 196) + 784) + log(3*x)*exp(x)*(12*x - 2*log(2)*(2*x 
+ 28) + 2*x^2 + 112))/(28*x + log(3*x)^2 + x^2 - log(3*x)*(2*x + 28) + 196 
),x)
 

Output:

int((log(3*x)^2*exp(x)*(2*log(2) - 4) - exp(x)*(170*x + 34*x^2 + 2*x^3 - 2 
*log(2)*(28*x + x^2 + 196) + 784) + log(3*x)*exp(x)*(12*x - 2*log(2)*(2*x 
+ 28) + 2*x^2 + 112))/(28*x + log(3*x)^2 + x^2 - log(3*x)*(2*x + 28) + 196 
), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {e^x \left (-784-170 x-34 x^2-2 x^3+\left (196+28 x+x^2\right ) \log (4)\right )+e^x \left (112+12 x+2 x^2+(-28-2 x) \log (4)\right ) \log (3 x)+e^x (-4+\log (4)) \log ^2(3 x)}{196+28 x+x^2+(-28-2 x) \log (3 x)+\log ^2(3 x)} \, dx=\frac {2 e^{x} \left (\mathrm {log}\left (3 x \right ) \mathrm {log}\left (2\right )-2 \,\mathrm {log}\left (3 x \right )-\mathrm {log}\left (2\right ) x -14 \,\mathrm {log}\left (2\right )+x^{2}+2 x +28\right )}{\mathrm {log}\left (3 x \right )-x -14} \] Input:

int(((2*log(2)-4)*exp(x)*log(3*x)^2+(2*(-2*x-28)*log(2)+2*x^2+12*x+112)*ex 
p(x)*log(3*x)+(2*(x^2+28*x+196)*log(2)-2*x^3-34*x^2-170*x-784)*exp(x))/(lo 
g(3*x)^2+(-2*x-28)*log(3*x)+x^2+28*x+196),x)
 

Output:

(2*e**x*(log(3*x)*log(2) - 2*log(3*x) - log(2)*x - 14*log(2) + x**2 + 2*x 
+ 28))/(log(3*x) - x - 14)