\(\int \frac {e^{\frac {-3 x^2+(15 x+9 x^2) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} (75 x^2+45 x^3+(750 x^2+225 x^3+(-150 x^2-45 x^3) \log (x)) \log (5-\log (x)) \log (\log (5-\log (x)))+(-625-2625 x-2475 x^2-675 x^3+(125+525 x+495 x^2+135 x^3) \log (x)) \log (5-\log (x)) \log ^2(\log (5-\log (x))))}{(-125-150 x-45 x^2+(25+30 x+9 x^2) \log (x)) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx\) [50]

Optimal result

Integrand size = 192, antiderivative size = 31 \[ \int \frac {e^{\frac {-3 x^2+\left (15 x+9 x^2\right ) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} \left (75 x^2+45 x^3+\left (750 x^2+225 x^3+\left (-150 x^2-45 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))+\left (-625-2625 x-2475 x^2-675 x^3+\left (125+525 x+495 x^2+135 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))\right )}{\left (-125-150 x-45 x^2+\left (25+30 x+9 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx=5 e^{3 \left (x-\frac {x^2}{(5+3 x) \log (\log (5-\log (x)))}\right )} x \] Output:

5*exp(3*x-3*x^2/ln(ln(5-ln(x)))/(5+3*x))*x
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-3 x^2+\left (15 x+9 x^2\right ) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} \left (75 x^2+45 x^3+\left (750 x^2+225 x^3+\left (-150 x^2-45 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))+\left (-625-2625 x-2475 x^2-675 x^3+\left (125+525 x+495 x^2+135 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))\right )}{\left (-125-150 x-45 x^2+\left (25+30 x+9 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx=5 e^{3 x-\frac {3 x^2}{(5+3 x) \log (\log (5-\log (x)))}} x \] Input:

Integrate[(E^((-3*x^2 + (15*x + 9*x^2)*Log[Log[5 - Log[x]]])/((5 + 3*x)*Lo 
g[Log[5 - Log[x]]]))*(75*x^2 + 45*x^3 + (750*x^2 + 225*x^3 + (-150*x^2 - 4 
5*x^3)*Log[x])*Log[5 - Log[x]]*Log[Log[5 - Log[x]]] + (-625 - 2625*x - 247 
5*x^2 - 675*x^3 + (125 + 525*x + 495*x^2 + 135*x^3)*Log[x])*Log[5 - Log[x] 
]*Log[Log[5 - Log[x]]]^2))/((-125 - 150*x - 45*x^2 + (25 + 30*x + 9*x^2)*L 
og[x])*Log[5 - Log[x]]*Log[Log[5 - Log[x]]]^2),x]
 

Output:

5*E^(3*x - (3*x^2)/((5 + 3*x)*Log[Log[5 - Log[x]]]))*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.

\(\displaystyle \int \frac {\left (45 x^3+75 x^2+\left (-675 x^3-2475 x^2+\left (135 x^3+495 x^2+525 x+125\right ) \log (x)-2625 x-625\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))+\left (225 x^3+750 x^2+\left (-45 x^3-150 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))\right ) \exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )}{\left (-45 x^2+\left (9 x^2+30 x+25\right ) \log (x)-150 x-125\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (-45 x^3-75 x^2-\left (-675 x^3-2475 x^2+\left (135 x^3+495 x^2+525 x+125\right ) \log (x)-2625 x-625\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))-\left (225 x^3+750 x^2+\left (-45 x^3-150 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))\right ) \exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )}{(3 x+5)^2 (5-\log (x)) \log (5-\log (x)) \log ^2(\log (5-\log (x)))}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {15 x^2 \exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )}{(3 x+5) (\log (x)-5) \log (5-\log (x)) \log ^2(\log (5-\log (x)))}-\frac {15 (3 x+10) x^2 \exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )}{(3 x+5)^2 \log (\log (5-\log (x)))}+5 (3 x+1) \exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {25}{3} \int \frac {\exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )}{(\log (x)-5) \log (5-\log (x)) \log ^2(\log (5-\log (x)))}dx+5 \int \frac {\exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right ) x}{(\log (x)-5) \log (5-\log (x)) \log ^2(\log (5-\log (x)))}dx+\frac {125}{3} \int \frac {\exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )}{(3 x+5) (\log (x)-5) \log (5-\log (x)) \log ^2(\log (5-\log (x)))}dx+5 \int \exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )dx+15 \int \exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right ) xdx-5 \int \frac {\exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right ) x}{\log (\log (5-\log (x)))}dx-\frac {625}{3} \int \frac {\exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )}{(3 x+5)^2 \log (\log (5-\log (x)))}dx+\frac {125}{3} \int \frac {\exp \left (\frac {\left (9 x^2+15 x\right ) \log (\log (5-\log (x)))-3 x^2}{(3 x+5) \log (\log (5-\log (x)))}\right )}{(3 x+5) \log (\log (5-\log (x)))}dx\)

Input:

Int[(E^((-3*x^2 + (15*x + 9*x^2)*Log[Log[5 - Log[x]]])/((5 + 3*x)*Log[Log[ 
5 - Log[x]]]))*(75*x^2 + 45*x^3 + (750*x^2 + 225*x^3 + (-150*x^2 - 45*x^3) 
*Log[x])*Log[5 - Log[x]]*Log[Log[5 - Log[x]]] + (-625 - 2625*x - 2475*x^2 
- 675*x^3 + (125 + 525*x + 495*x^2 + 135*x^3)*Log[x])*Log[5 - Log[x]]*Log[ 
Log[5 - Log[x]]]^2))/((-125 - 150*x - 45*x^2 + (25 + 30*x + 9*x^2)*Log[x]) 
*Log[5 - Log[x]]*Log[Log[5 - Log[x]]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61

\[5 x \,{\mathrm e}^{\frac {3 x \left (3 \ln \left (\ln \left (5-\ln \left (x \right )\right )\right ) x +5 \ln \left (\ln \left (5-\ln \left (x \right )\right )\right )-x \right )}{\left (3 x +5\right ) \ln \left (\ln \left (5-\ln \left (x \right )\right )\right )}}\]

Input:

int((((135*x^3+495*x^2+525*x+125)*ln(x)-675*x^3-2475*x^2-2625*x-625)*ln(5- 
ln(x))*ln(ln(5-ln(x)))^2+((-45*x^3-150*x^2)*ln(x)+225*x^3+750*x^2)*ln(5-ln 
(x))*ln(ln(5-ln(x)))+45*x^3+75*x^2)*exp(((9*x^2+15*x)*ln(ln(5-ln(x)))-3*x^ 
2)/(3*x+5)/ln(ln(5-ln(x))))/((9*x^2+30*x+25)*ln(x)-45*x^2-150*x-125)/ln(5- 
ln(x))/ln(ln(5-ln(x)))^2,x)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {-3 x^2+\left (15 x+9 x^2\right ) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} \left (75 x^2+45 x^3+\left (750 x^2+225 x^3+\left (-150 x^2-45 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))+\left (-625-2625 x-2475 x^2-675 x^3+\left (125+525 x+495 x^2+135 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))\right )}{\left (-125-150 x-45 x^2+\left (25+30 x+9 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx=5 \, x e^{\left (-\frac {3 \, {\left (x^{2} - {\left (3 \, x^{2} + 5 \, x\right )} \log \left (\log \left (-\log \left (x\right ) + 5\right )\right )\right )}}{{\left (3 \, x + 5\right )} \log \left (\log \left (-\log \left (x\right ) + 5\right )\right )}\right )} \] Input:

integrate((((135*x^3+495*x^2+525*x+125)*log(x)-675*x^3-2475*x^2-2625*x-625 
)*log(5-log(x))*log(log(5-log(x)))^2+((-45*x^3-150*x^2)*log(x)+225*x^3+750 
*x^2)*log(5-log(x))*log(log(5-log(x)))+45*x^3+75*x^2)*exp(((9*x^2+15*x)*lo 
g(log(5-log(x)))-3*x^2)/(3*x+5)/log(log(5-log(x))))/((9*x^2+30*x+25)*log(x 
)-45*x^2-150*x-125)/log(5-log(x))/log(log(5-log(x)))^2,x, algorithm="frica 
s")
 

Output:

5*x*e^(-3*(x^2 - (3*x^2 + 5*x)*log(log(-log(x) + 5)))/((3*x + 5)*log(log(- 
log(x) + 5))))
 

Sympy [A] (verification not implemented)

Time = 19.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {-3 x^2+\left (15 x+9 x^2\right ) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} \left (75 x^2+45 x^3+\left (750 x^2+225 x^3+\left (-150 x^2-45 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))+\left (-625-2625 x-2475 x^2-675 x^3+\left (125+525 x+495 x^2+135 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))\right )}{\left (-125-150 x-45 x^2+\left (25+30 x+9 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx=5 x e^{\frac {- 3 x^{2} + \left (9 x^{2} + 15 x\right ) \log {\left (\log {\left (5 - \log {\left (x \right )} \right )} \right )}}{\left (3 x + 5\right ) \log {\left (\log {\left (5 - \log {\left (x \right )} \right )} \right )}}} \] Input:

integrate((((135*x**3+495*x**2+525*x+125)*ln(x)-675*x**3-2475*x**2-2625*x- 
625)*ln(5-ln(x))*ln(ln(5-ln(x)))**2+((-45*x**3-150*x**2)*ln(x)+225*x**3+75 
0*x**2)*ln(5-ln(x))*ln(ln(5-ln(x)))+45*x**3+75*x**2)*exp(((9*x**2+15*x)*ln 
(ln(5-ln(x)))-3*x**2)/(3*x+5)/ln(ln(5-ln(x))))/((9*x**2+30*x+25)*ln(x)-45* 
x**2-150*x-125)/ln(5-ln(x))/ln(ln(5-ln(x)))**2,x)
 

Output:

5*x*exp((-3*x**2 + (9*x**2 + 15*x)*log(log(5 - log(x))))/((3*x + 5)*log(lo 
g(5 - log(x)))))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-3 x^2+\left (15 x+9 x^2\right ) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} \left (75 x^2+45 x^3+\left (750 x^2+225 x^3+\left (-150 x^2-45 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))+\left (-625-2625 x-2475 x^2-675 x^3+\left (125+525 x+495 x^2+135 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))\right )}{\left (-125-150 x-45 x^2+\left (25+30 x+9 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((((135*x^3+495*x^2+525*x+125)*log(x)-675*x^3-2475*x^2-2625*x-625 
)*log(5-log(x))*log(log(5-log(x)))^2+((-45*x^3-150*x^2)*log(x)+225*x^3+750 
*x^2)*log(5-log(x))*log(log(5-log(x)))+45*x^3+75*x^2)*exp(((9*x^2+15*x)*lo 
g(log(5-log(x)))-3*x^2)/(3*x+5)/log(log(5-log(x))))/((9*x^2+30*x+25)*log(x 
)-45*x^2-150*x-125)/log(5-log(x))/log(log(5-log(x)))^2,x, algorithm="maxim 
a")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [F]

\[ \int \frac {e^{\frac {-3 x^2+\left (15 x+9 x^2\right ) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} \left (75 x^2+45 x^3+\left (750 x^2+225 x^3+\left (-150 x^2-45 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))+\left (-625-2625 x-2475 x^2-675 x^3+\left (125+525 x+495 x^2+135 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))\right )}{\left (-125-150 x-45 x^2+\left (25+30 x+9 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx=\int { \frac {5 \, {\left ({\left (135 \, x^{3} + 495 \, x^{2} - {\left (27 \, x^{3} + 99 \, x^{2} + 105 \, x + 25\right )} \log \left (x\right ) + 525 \, x + 125\right )} \log \left (-\log \left (x\right ) + 5\right ) \log \left (\log \left (-\log \left (x\right ) + 5\right )\right )^{2} - 9 \, x^{3} - 3 \, {\left (15 \, x^{3} + 50 \, x^{2} - {\left (3 \, x^{3} + 10 \, x^{2}\right )} \log \left (x\right )\right )} \log \left (-\log \left (x\right ) + 5\right ) \log \left (\log \left (-\log \left (x\right ) + 5\right )\right ) - 15 \, x^{2}\right )} e^{\left (-\frac {3 \, {\left (x^{2} - {\left (3 \, x^{2} + 5 \, x\right )} \log \left (\log \left (-\log \left (x\right ) + 5\right )\right )\right )}}{{\left (3 \, x + 5\right )} \log \left (\log \left (-\log \left (x\right ) + 5\right )\right )}\right )}}{{\left (45 \, x^{2} - {\left (9 \, x^{2} + 30 \, x + 25\right )} \log \left (x\right ) + 150 \, x + 125\right )} \log \left (-\log \left (x\right ) + 5\right ) \log \left (\log \left (-\log \left (x\right ) + 5\right )\right )^{2}} \,d x } \] Input:

integrate((((135*x^3+495*x^2+525*x+125)*log(x)-675*x^3-2475*x^2-2625*x-625 
)*log(5-log(x))*log(log(5-log(x)))^2+((-45*x^3-150*x^2)*log(x)+225*x^3+750 
*x^2)*log(5-log(x))*log(log(5-log(x)))+45*x^3+75*x^2)*exp(((9*x^2+15*x)*lo 
g(log(5-log(x)))-3*x^2)/(3*x+5)/log(log(5-log(x))))/((9*x^2+30*x+25)*log(x 
)-45*x^2-150*x-125)/log(5-log(x))/log(log(5-log(x)))^2,x, algorithm="giac" 
)
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 3.51 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {-3 x^2+\left (15 x+9 x^2\right ) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} \left (75 x^2+45 x^3+\left (750 x^2+225 x^3+\left (-150 x^2-45 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))+\left (-625-2625 x-2475 x^2-675 x^3+\left (125+525 x+495 x^2+135 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))\right )}{\left (-125-150 x-45 x^2+\left (25+30 x+9 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx=5\,x\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{-\frac {3\,x^2}{5\,\ln \left (\ln \left (5-\ln \left (x\right )\right )\right )+3\,x\,\ln \left (\ln \left (5-\ln \left (x\right )\right )\right )}} \] Input:

int(-(exp((log(log(5 - log(x)))*(15*x + 9*x^2) - 3*x^2)/(log(log(5 - log(x 
)))*(3*x + 5)))*(75*x^2 + 45*x^3 + log(5 - log(x))*log(log(5 - log(x)))*(7 
50*x^2 - log(x)*(150*x^2 + 45*x^3) + 225*x^3) - log(5 - log(x))*log(log(5 
- log(x)))^2*(2625*x + 2475*x^2 + 675*x^3 - log(x)*(525*x + 495*x^2 + 135* 
x^3 + 125) + 625)))/(log(5 - log(x))*log(log(5 - log(x)))^2*(150*x - log(x 
)*(30*x + 9*x^2 + 25) + 45*x^2 + 125)),x)
 

Output:

5*x*exp(3*x)*exp(-(3*x^2)/(5*log(log(5 - log(x))) + 3*x*log(log(5 - log(x) 
))))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {-3 x^2+\left (15 x+9 x^2\right ) \log (\log (5-\log (x)))}{(5+3 x) \log (\log (5-\log (x)))}} \left (75 x^2+45 x^3+\left (750 x^2+225 x^3+\left (-150 x^2-45 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log (\log (5-\log (x)))+\left (-625-2625 x-2475 x^2-675 x^3+\left (125+525 x+495 x^2+135 x^3\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))\right )}{\left (-125-150 x-45 x^2+\left (25+30 x+9 x^2\right ) \log (x)\right ) \log (5-\log (x)) \log ^2(\log (5-\log (x)))} \, dx=\frac {5 e^{3 x} x}{e^{\frac {3 x^{2}}{3 \,\mathrm {log}\left (\mathrm {log}\left (-\mathrm {log}\left (x \right )+5\right )\right ) x +5 \,\mathrm {log}\left (\mathrm {log}\left (-\mathrm {log}\left (x \right )+5\right )\right )}}} \] Input:

int((((135*x^3+495*x^2+525*x+125)*log(x)-675*x^3-2475*x^2-2625*x-625)*log( 
5-log(x))*log(log(5-log(x)))^2+((-45*x^3-150*x^2)*log(x)+225*x^3+750*x^2)* 
log(5-log(x))*log(log(5-log(x)))+45*x^3+75*x^2)*exp(((9*x^2+15*x)*log(log( 
5-log(x)))-3*x^2)/(3*x+5)/log(log(5-log(x))))/((9*x^2+30*x+25)*log(x)-45*x 
^2-150*x-125)/log(5-log(x))/log(log(5-log(x)))^2,x)
 

Output:

(5*e**(3*x)*x)/e**((3*x**2)/(3*log(log( - log(x) + 5))*x + 5*log(log( - lo 
g(x) + 5))))