\(\int \frac {60+49 x-4 x^2-3 x^3+(-60-17 x+4 x^2+x^3+e^x (-45 x-9 x^2+5 x^3+x^4)) \log (15 x+8 x^2+x^3) \log (\log (15 x+8 x^2+x^3))+(-15 x-8 x^2-x^3) \log (15 x+8 x^2+x^3) \log (\log (15 x+8 x^2+x^3)) \log (\frac {\log (\log (15 x+8 x^2+x^3))}{x})}{(15 x+8 x^2+x^3) \log (15 x+8 x^2+x^3) \log (\log (15 x+8 x^2+x^3))} \, dx\) [1507]

Optimal result

Integrand size = 184, antiderivative size = 27 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=(4-x) \left (-e^x+\log \left (\frac {\log (\log (x (3+x) (5+x)))}{x}\right )\right ) \] Output:

(ln(ln(ln((3+x)*(5+x)*x))/x)-exp(x))*(4-x)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=e^x (-4+x)-4 \log (x)+4 \log \left (\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )\right )-x \log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right ) \] Input:

Integrate[(60 + 49*x - 4*x^2 - 3*x^3 + (-60 - 17*x + 4*x^2 + x^3 + E^x*(-4 
5*x - 9*x^2 + 5*x^3 + x^4))*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + 
 x^3]] + (-15*x - 8*x^2 - x^3)*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^ 
2 + x^3]]*Log[Log[Log[15*x + 8*x^2 + x^3]]/x])/((15*x + 8*x^2 + x^3)*Log[1 
5*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^3]]),x]
 

Output:

E^x*(-4 + x) - 4*Log[x] + 4*Log[Log[Log[x*(15 + 8*x + x^2)]]] - x*Log[Log[ 
Log[x*(15 + 8*x + x^2)]]/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.

\(\displaystyle \int \frac {-3 x^3-4 x^2+\left (-x^3-8 x^2-15 x\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right ) \log \left (\frac {\log \left (\log \left (x^3+8 x^2+15 x\right )\right )}{x}\right )+\left (x^3+4 x^2+e^x \left (x^4+5 x^3-9 x^2-45 x\right )-17 x-60\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right )+49 x+60}{\left (x^3+8 x^2+15 x\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {-3 x^3-4 x^2+\left (-x^3-8 x^2-15 x\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right ) \log \left (\frac {\log \left (\log \left (x^3+8 x^2+15 x\right )\right )}{x}\right )+\left (x^3+4 x^2+e^x \left (x^4+5 x^3-9 x^2-45 x\right )-17 x-60\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right )+49 x+60}{x \left (x^2+8 x+15\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right )}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {x^2}{x^2+8 x+15}+\frac {4 x}{x^2+8 x+15}-\frac {17}{x^2+8 x+15}-\frac {60}{\left (x^2+8 x+15\right ) x}-\frac {3 x^2}{(x+3) (x+5) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}-\frac {4 x}{(x+3) (x+5) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}-\log \left (\frac {\log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}{x}\right )+\frac {49}{(x+3) (x+5) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}+\frac {60}{(x+3) (x+5) x \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}+e^x (x-3)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \int \frac {1}{x \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}dx+4 \int \frac {1}{(x+3) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}dx+4 \int \frac {1}{(x+5) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}dx-x \log \left (\frac {\log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}{x}\right )-e^x (3-x)-e^x-4 \log (x)\)

Input:

Int[(60 + 49*x - 4*x^2 - 3*x^3 + (-60 - 17*x + 4*x^2 + x^3 + E^x*(-45*x - 
9*x^2 + 5*x^3 + x^4))*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^3]] 
 + (-15*x - 8*x^2 - x^3)*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^ 
3]]*Log[Log[Log[15*x + 8*x^2 + x^3]]/x])/((15*x + 8*x^2 + x^3)*Log[15*x + 
8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^3]]),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.25 (sec) , antiderivative size = 733, normalized size of antiderivative = 27.15

\[\text {Expression too large to display}\]

Input:

int(((-x^3-8*x^2-15*x)*ln(x^3+8*x^2+15*x)*ln(ln(x^3+8*x^2+15*x))*ln(ln(ln( 
x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-17*x-60)*ln(x 
^3+8*x^2+15*x)*ln(ln(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/(x^3+8*x^2+15*x 
)/ln(x^3+8*x^2+15*x)/ln(ln(x^3+8*x^2+15*x)),x)
 

Output:

-x*ln(ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*( 
x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x+15)))))+x 
*ln(x)+1/2*I*Pi*x*csgn(I/x)*csgn(I*ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I 
*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+1 
5))+csgn(I*(x^2+8*x+15)))))*csgn(I/x*ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn 
(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x 
+15))+csgn(I*(x^2+8*x+15)))))-1/2*I*Pi*x*csgn(I/x)*csgn(I/x*ln(ln(x)+ln(x^ 
2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x+15))+csgn(I* 
x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x+15)))))^2-1/2*I*Pi*x*csgn(I*l 
n(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x 
+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x+15)))))*csgn(I/x 
*ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8 
*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x+15)))))^2+1/2* 
I*Pi*x*csgn(I/x*ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(- 
csgn(I*x*(x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x 
+15)))))^3-4*ln(x)+exp(x)*x-4*exp(x)+4*ln(ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi 
*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^ 
2+8*x+15))+csgn(I*(x^2+8*x+15)))))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx={\left (x - 4\right )} e^{x} - {\left (x - 4\right )} \log \left (\frac {\log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )}{x}\right ) \] Input:

integrate(((-x^3-8*x^2-15*x)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))* 
log(log(log(x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-1 
7*x-60)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/ 
(x^3+8*x^2+15*x)/log(x^3+8*x^2+15*x)/log(log(x^3+8*x^2+15*x)),x, algorithm 
="fricas")
 

Output:

(x - 4)*e^x - (x - 4)*log(log(log(x^3 + 8*x^2 + 15*x))/x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=\text {Timed out} \] Input:

integrate(((-x**3-8*x**2-15*x)*ln(x**3+8*x**2+15*x)*ln(ln(x**3+8*x**2+15*x 
))*ln(ln(ln(x**3+8*x**2+15*x))/x)+((x**4+5*x**3-9*x**2-45*x)*exp(x)+x**3+4 
*x**2-17*x-60)*ln(x**3+8*x**2+15*x)*ln(ln(x**3+8*x**2+15*x))-3*x**3-4*x**2 
+49*x+60)/(x**3+8*x**2+15*x)/ln(x**3+8*x**2+15*x)/ln(ln(x**3+8*x**2+15*x)) 
,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx={\left (x - 4\right )} e^{x} + {\left (x - 4\right )} \log \left (x\right ) - {\left (x - 4\right )} \log \left (\log \left (\log \left (x + 5\right ) + \log \left (x + 3\right ) + \log \left (x\right )\right )\right ) \] Input:

integrate(((-x^3-8*x^2-15*x)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))* 
log(log(log(x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-1 
7*x-60)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/ 
(x^3+8*x^2+15*x)/log(x^3+8*x^2+15*x)/log(log(x^3+8*x^2+15*x)),x, algorithm 
="maxima")
 

Output:

(x - 4)*e^x + (x - 4)*log(x) - (x - 4)*log(log(log(x + 5) + log(x + 3) + l 
og(x)))
 

Giac [F]

\[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=\int { -\frac {{\left (x^{3} + 8 \, x^{2} + 15 \, x\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\frac {\log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )}{x}\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right ) + 3 \, x^{3} - {\left (x^{3} + 4 \, x^{2} + {\left (x^{4} + 5 \, x^{3} - 9 \, x^{2} - 45 \, x\right )} e^{x} - 17 \, x - 60\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right ) + 4 \, x^{2} - 49 \, x - 60}{{\left (x^{3} + 8 \, x^{2} + 15 \, x\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )} \,d x } \] Input:

integrate(((-x^3-8*x^2-15*x)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))* 
log(log(log(x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-1 
7*x-60)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/ 
(x^3+8*x^2+15*x)/log(x^3+8*x^2+15*x)/log(log(x^3+8*x^2+15*x)),x, algorithm 
="giac")
 

Output:

integrate(-((x^3 + 8*x^2 + 15*x)*log(x^3 + 8*x^2 + 15*x)*log(log(log(x^3 + 
 8*x^2 + 15*x))/x)*log(log(x^3 + 8*x^2 + 15*x)) + 3*x^3 - (x^3 + 4*x^2 + ( 
x^4 + 5*x^3 - 9*x^2 - 45*x)*e^x - 17*x - 60)*log(x^3 + 8*x^2 + 15*x)*log(l 
og(x^3 + 8*x^2 + 15*x)) + 4*x^2 - 49*x - 60)/((x^3 + 8*x^2 + 15*x)*log(x^3 
 + 8*x^2 + 15*x)*log(log(x^3 + 8*x^2 + 15*x))), x)
 

Mupad [B] (verification not implemented)

Time = 2.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.81 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=4\,\ln \left (\ln \left (\ln \left (x^3+8\,x^2+15\,x\right )\right )\right )-4\,\ln \left (x\right )+{\mathrm {e}}^x\,\left (x-4\right )-\frac {\ln \left (\frac {\ln \left (\ln \left (x^3+8\,x^2+15\,x\right )\right )}{x}\right )\,\left (x^4+8\,x^3+15\,x^2\right )}{x\,\left (x^2+8\,x+15\right )} \] Input:

int(-(4*x^2 - 49*x + 3*x^3 + log(log(15*x + 8*x^2 + x^3))*log(15*x + 8*x^2 
 + x^3)*(17*x + exp(x)*(45*x + 9*x^2 - 5*x^3 - x^4) - 4*x^2 - x^3 + 60) + 
log(log(15*x + 8*x^2 + x^3))*log(15*x + 8*x^2 + x^3)*log(log(log(15*x + 8* 
x^2 + x^3))/x)*(15*x + 8*x^2 + x^3) - 60)/(log(log(15*x + 8*x^2 + x^3))*lo 
g(15*x + 8*x^2 + x^3)*(15*x + 8*x^2 + x^3)),x)
 

Output:

4*log(log(log(15*x + 8*x^2 + x^3))) - 4*log(x) + exp(x)*(x - 4) - (log(log 
(log(15*x + 8*x^2 + x^3))/x)*(15*x^2 + 8*x^3 + x^4))/(x*(8*x + x^2 + 15))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=e^{x} x -4 e^{x}+4 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x^{3}+8 x^{2}+15 x \right )\right )\right )-\mathrm {log}\left (\frac {\mathrm {log}\left (\mathrm {log}\left (x^{3}+8 x^{2}+15 x \right )\right )}{x}\right ) x -4 \,\mathrm {log}\left (x \right ) \] Input:

int(((-x^3-8*x^2-15*x)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))*log(lo 
g(log(x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-17*x-60 
)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/(x^3+8 
*x^2+15*x)/log(x^3+8*x^2+15*x)/log(log(x^3+8*x^2+15*x)),x)
 

Output:

e**x*x - 4*e**x + 4*log(log(log(x**3 + 8*x**2 + 15*x))) - log(log(log(x**3 
 + 8*x**2 + 15*x))/x)*x - 4*log(x)