\(\int \frac {(-2 x^2+e^2 (250 x-250 x^2+2 x^3-2 x^4+e (50 x-50 x^2))) \log (x)+(e^2 (125-250 x+126 x^2-2 x^3+x^4+e (25-50 x+25 x^2))+(125+25 e+x^2) \log (\frac {1}{25} (125+25 e+x^2))) \log (e^2 (1-2 x+x^2)+\log (\frac {1}{25} (125+25 e+x^2)))}{(e^2 (125 x-250 x^2+126 x^3-2 x^4+x^5+e (25 x-50 x^2+25 x^3))+(125 x+25 e x+x^3) \log (\frac {1}{25} (125+25 e+x^2))) \log ^2(e^2 (1-2 x+x^2)+\log (\frac {1}{25} (125+25 e+x^2)))} \, dx\) [2194]

Optimal result

Integrand size = 229, antiderivative size = 27 \[ \int \frac {\left (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx=\frac {\log (x)}{\log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \] Output:

ln(x)/ln((-1+x)^2*exp(2)+ln(exp(1)+1/25*x^2+5))
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx=\frac {\log (x)}{\log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \] Input:

Integrate[((-2*x^2 + E^2*(250*x - 250*x^2 + 2*x^3 - 2*x^4 + E*(50*x - 50*x 
^2)))*Log[x] + (E^2*(125 - 250*x + 126*x^2 - 2*x^3 + x^4 + E*(25 - 50*x + 
25*x^2)) + (125 + 25*E + x^2)*Log[(125 + 25*E + x^2)/25])*Log[E^2*(1 - 2*x 
 + x^2) + Log[(125 + 25*E + x^2)/25]])/((E^2*(125*x - 250*x^2 + 126*x^3 - 
2*x^4 + x^5 + E*(25*x - 50*x^2 + 25*x^3)) + (125*x + 25*E*x + x^3)*Log[(12 
5 + 25*E + x^2)/25])*Log[E^2*(1 - 2*x + x^2) + Log[(125 + 25*E + x^2)/25]] 
^2),x]
 

Output:

Log[x]/Log[E^2*(-1 + x)^2 + Log[5 + E + x^2/25]]
 

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.

\(\displaystyle \int \frac {\left (e^2 \left (-2 x^4+2 x^3-250 x^2+e \left (50 x-50 x^2\right )+250 x\right )-2 x^2\right ) \log (x)+\left (\left (x^2+25 e+125\right ) \log \left (\frac {1}{25} \left (x^2+25 e+125\right )\right )+e^2 \left (x^4-2 x^3+126 x^2+e \left (25 x^2-50 x+25\right )-250 x+125\right )\right ) \log \left (e^2 \left (x^2-2 x+1\right )+\log \left (\frac {1}{25} \left (x^2+25 e+125\right )\right )\right )}{\left (\left (x^3+25 e x+125 x\right ) \log \left (\frac {1}{25} \left (x^2+25 e+125\right )\right )+e^2 \left (x^5-2 x^4+126 x^3-250 x^2+e \left (25 x^3-50 x^2+25 x\right )+125 x\right )\right ) \log ^2\left (e^2 \left (x^2-2 x+1\right )+\log \left (\frac {1}{25} \left (x^2+25 e+125\right )\right )\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\frac {\left (-2 x^2-2 e^2 (x-1) \left (x^2+25 e+125\right ) x\right ) \log (x)}{\left (x^2+25 e+125\right ) \left (\log \left (\frac {x^2}{25}+e+5\right )+e^2 (x-1)^2\right )}+\log \left (\log \left (\frac {x^2}{25}+e+5\right )+e^2 (x-1)^2\right )}{x \log ^2\left (\log \left (\frac {x^2}{25}+e+5\right )+e^2 (x-1)^2\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {1}{x \log \left (\log \left (\frac {x^2}{25}+e+5\right )+e^2 (x-1)^2\right )}+\frac {2 \left (-e^2 x^3+e^2 x^2-\left (1+125 e^2+25 e^3\right ) x+25 e^2 (5+e)\right ) \log (x)}{\left (x^2+25 e+125\right ) \left (e^2 x^2+\log \left (\frac {x^2}{25}+e+5\right )-2 e^2 x+e^2\right ) \log ^2\left (\log \left (\frac {x^2}{25}+e+5\right )+e^2 (x-1)^2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 e^2 \int \frac {\log (x)}{\left (e^2 x^2-2 e^2 x+\log \left (\frac {x^2}{25}+e+5\right )+e^2\right ) \log ^2\left (e^2 (x-1)^2+\log \left (\frac {x^2}{25}+e+5\right )\right )}dx+\int \frac {\log (x)}{\left (5 i \sqrt {5+e}-x\right ) \left (e^2 x^2-2 e^2 x+\log \left (\frac {x^2}{25}+e+5\right )+e^2\right ) \log ^2\left (e^2 (x-1)^2+\log \left (\frac {x^2}{25}+e+5\right )\right )}dx-2 e^2 \int \frac {x \log (x)}{\left (e^2 x^2-2 e^2 x+\log \left (\frac {x^2}{25}+e+5\right )+e^2\right ) \log ^2\left (e^2 (x-1)^2+\log \left (\frac {x^2}{25}+e+5\right )\right )}dx-\int \frac {\log (x)}{\left (x+5 i \sqrt {5+e}\right ) \left (e^2 x^2-2 e^2 x+\log \left (\frac {x^2}{25}+e+5\right )+e^2\right ) \log ^2\left (e^2 (x-1)^2+\log \left (\frac {x^2}{25}+e+5\right )\right )}dx+\int \frac {1}{x \log \left (e^2 (x-1)^2+\log \left (\frac {x^2}{25}+e+5\right )\right )}dx\)

Input:

Int[((-2*x^2 + E^2*(250*x - 250*x^2 + 2*x^3 - 2*x^4 + E*(50*x - 50*x^2)))* 
Log[x] + (E^2*(125 - 250*x + 126*x^2 - 2*x^3 + x^4 + E*(25 - 50*x + 25*x^2 
)) + (125 + 25*E + x^2)*Log[(125 + 25*E + x^2)/25])*Log[E^2*(1 - 2*x + x^2 
) + Log[(125 + 25*E + x^2)/25]])/((E^2*(125*x - 250*x^2 + 126*x^3 - 2*x^4 
+ x^5 + E*(25*x - 50*x^2 + 25*x^3)) + (125*x + 25*E*x + x^3)*Log[(125 + 25 
*E + x^2)/25])*Log[E^2*(1 - 2*x + x^2) + Log[(125 + 25*E + x^2)/25]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 8.74 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07

\[\frac {\ln \left (x \right )}{\ln \left (\ln \left ({\mathrm e}+\frac {x^{2}}{25}+5\right )+\left (x^{2}-2 x +1\right ) {\mathrm e}^{2}\right )}\]

Input:

int((((25*exp(1)+x^2+125)*ln(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*exp(1)+x 
^4-2*x^3+126*x^2-250*x+125)*exp(2))*ln(ln(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*e 
xp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2)-2*x^2)*ln 
(x))/((25*x*exp(1)+x^3+125*x)*ln(exp(1)+1/25*x^2+5)+((25*x^3-50*x^2+25*x)* 
exp(1)+x^5-2*x^4+126*x^3-250*x^2+125*x)*exp(2))/ln(ln(exp(1)+1/25*x^2+5)+( 
x^2-2*x+1)*exp(2))^2,x)
 

Output:

ln(x)/ln(ln(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx=\frac {\log \left (x\right )}{\log \left ({\left (x^{2} - 2 \, x + 1\right )} e^{2} + \log \left (\frac {1}{25} \, x^{2} + e + 5\right )\right )} \] Input:

integrate((((25*exp(1)+x^2+125)*log(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*e 
xp(1)+x^4-2*x^3+126*x^2-250*x+125)*exp(2))*log(log(exp(1)+1/25*x^2+5)+(x^2 
-2*x+1)*exp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2)- 
2*x^2)*log(x))/((25*exp(1)*x+x^3+125*x)*log(exp(1)+1/25*x^2+5)+((25*x^3-50 
*x^2+25*x)*exp(1)+x^5-2*x^4+126*x^3-250*x^2+125*x)*exp(2))/log(log(exp(1)+ 
1/25*x^2+5)+(x^2-2*x+1)*exp(2))^2,x, algorithm="fricas")
 

Output:

log(x)/log((x^2 - 2*x + 1)*e^2 + log(1/25*x^2 + e + 5))
 

Sympy [A] (verification not implemented)

Time = 4.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx=\frac {\log {\left (x \right )}}{\log {\left (\left (x^{2} - 2 x + 1\right ) e^{2} + \log {\left (\frac {x^{2}}{25} + e + 5 \right )} \right )}} \] Input:

integrate((((25*exp(1)+x**2+125)*ln(exp(1)+1/25*x**2+5)+((25*x**2-50*x+25) 
*exp(1)+x**4-2*x**3+126*x**2-250*x+125)*exp(2))*ln(ln(exp(1)+1/25*x**2+5)+ 
(x**2-2*x+1)*exp(2))+(((-50*x**2+50*x)*exp(1)-2*x**4+2*x**3-250*x**2+250*x 
)*exp(2)-2*x**2)*ln(x))/((25*exp(1)*x+x**3+125*x)*ln(exp(1)+1/25*x**2+5)+( 
(25*x**3-50*x**2+25*x)*exp(1)+x**5-2*x**4+126*x**3-250*x**2+125*x)*exp(2)) 
/ln(ln(exp(1)+1/25*x**2+5)+(x**2-2*x+1)*exp(2))**2,x)
 

Output:

log(x)/log((x**2 - 2*x + 1)*exp(2) + log(x**2/25 + E + 5))
 

Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx=\frac {\log \left (x\right )}{\log \left (x^{2} e^{2} - 2 \, x e^{2} + e^{2} - 2 \, \log \left (5\right ) + \log \left (x^{2} + 25 \, e + 125\right )\right )} \] Input:

integrate((((25*exp(1)+x^2+125)*log(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*e 
xp(1)+x^4-2*x^3+126*x^2-250*x+125)*exp(2))*log(log(exp(1)+1/25*x^2+5)+(x^2 
-2*x+1)*exp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2)- 
2*x^2)*log(x))/((25*exp(1)*x+x^3+125*x)*log(exp(1)+1/25*x^2+5)+((25*x^3-50 
*x^2+25*x)*exp(1)+x^5-2*x^4+126*x^3-250*x^2+125*x)*exp(2))/log(log(exp(1)+ 
1/25*x^2+5)+(x^2-2*x+1)*exp(2))^2,x, algorithm="maxima")
 

Output:

log(x)/log(x^2*e^2 - 2*x*e^2 + e^2 - 2*log(5) + log(x^2 + 25*e + 125))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1134 vs. \(2 (25) = 50\).

Time = 1.13 (sec) , antiderivative size = 1134, normalized size of antiderivative = 42.00 \[ \int \frac {\left (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx=\text {Too large to display} \] Input:

integrate((((25*exp(1)+x^2+125)*log(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*e 
xp(1)+x^4-2*x^3+126*x^2-250*x+125)*exp(2))*log(log(exp(1)+1/25*x^2+5)+(x^2 
-2*x+1)*exp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2)- 
2*x^2)*log(x))/((25*exp(1)*x+x^3+125*x)*log(exp(1)+1/25*x^2+5)+((25*x^3-50 
*x^2+25*x)*exp(1)+x^5-2*x^4+126*x^3-250*x^2+125*x)*exp(2))/log(log(exp(1)+ 
1/25*x^2+5)+(x^2-2*x+1)*exp(2))^2,x, algorithm="giac")
 

Output:

(x^5*e^4*log(x) - 3*x^4*e^4*log(x) + x^3*e^2*log(1/25*x^2 + e + 5)*log(x) 
+ 25*x^3*e^5*log(x) + 128*x^3*e^4*log(x) + x^3*e^2*log(x) - x^2*e^2*log(1/ 
25*x^2 + e + 5)*log(x) - 75*x^2*e^5*log(x) - 376*x^2*e^4*log(x) - 2*x^2*e^ 
2*log(x) + 25*x*e^3*log(1/25*x^2 + e + 5)*log(x) + 125*x*e^2*log(1/25*x^2 
+ e + 5)*log(x) + 75*x*e^5*log(x) + 375*x*e^4*log(x) + x*e^2*log(x) - 2*x* 
log(5)*log(x) + x*log(x^2 + 25*e + 125)*log(x) - 25*e^3*log(1/25*x^2 + e + 
 5)*log(x) - 125*e^2*log(1/25*x^2 + e + 5)*log(x) - 25*e^5*log(x) - 125*e^ 
4*log(x))/(x^5*e^4*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 
3*x^4*e^4*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 2*x^3*e^2 
*log(5)*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + x^3*e^2*log 
(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5))*log(x^2 + 25*e + 125) + 
25*x^3*e^5*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + 128*x^3* 
e^4*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + x^3*e^2*log(x^2 
*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + 2*x^2*e^2*log(5)*log(x^2*e 
^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - x^2*e^2*log(x^2*e^2 - 2*x*e^ 
2 + e^2 + log(1/25*x^2 + e + 5))*log(x^2 + 25*e + 125) - 75*x^2*e^5*log(x^ 
2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 376*x^2*e^4*log(x^2*e^2 - 
 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 2*x^2*e^2*log(x^2*e^2 - 2*x*e^2 
+ e^2 + log(1/25*x^2 + e + 5)) - 50*x*e^3*log(5)*log(x^2*e^2 - 2*x*e^2 + e 
^2 + log(1/25*x^2 + e + 5)) - 250*x*e^2*log(5)*log(x^2*e^2 - 2*x*e^2 + ...
 

Mupad [B] (verification not implemented)

Time = 5.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx=\frac {\ln \left (x\right )}{\ln \left (\ln \left (\frac {x^2}{25}+\mathrm {e}+5\right )+{\mathrm {e}}^2\,\left (x^2-2\,x+1\right )\right )} \] Input:

int((log(x)*(exp(2)*(250*x + exp(1)*(50*x - 50*x^2) - 250*x^2 + 2*x^3 - 2* 
x^4) - 2*x^2) + log(log(exp(1) + x^2/25 + 5) + exp(2)*(x^2 - 2*x + 1))*(ex 
p(2)*(exp(1)*(25*x^2 - 50*x + 25) - 250*x + 126*x^2 - 2*x^3 + x^4 + 125) + 
 log(exp(1) + x^2/25 + 5)*(25*exp(1) + x^2 + 125)))/(log(log(exp(1) + x^2/ 
25 + 5) + exp(2)*(x^2 - 2*x + 1))^2*(log(exp(1) + x^2/25 + 5)*(125*x + 25* 
x*exp(1) + x^3) + exp(2)*(125*x + exp(1)*(25*x - 50*x^2 + 25*x^3) - 250*x^ 
2 + 126*x^3 - 2*x^4 + x^5))),x)
 

Output:

log(x)/log(log(exp(1) + x^2/25 + 5) + exp(2)*(x^2 - 2*x + 1))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx=\frac {\mathrm {log}\left (x \right )}{\mathrm {log}\left (\mathrm {log}\left (\frac {x^{2}}{25}+e +5\right )+e^{2} x^{2}-2 e^{2} x +e^{2}\right )} \] Input:

int((((25*exp(1)+x^2+125)*log(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*exp(1)+ 
x^4-2*x^3+126*x^2-250*x+125)*exp(2))*log(log(exp(1)+1/25*x^2+5)+(x^2-2*x+1 
)*exp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2)-2*x^2) 
*log(x))/((25*exp(1)*x+x^3+125*x)*log(exp(1)+1/25*x^2+5)+((25*x^3-50*x^2+2 
5*x)*exp(1)+x^5-2*x^4+126*x^3-250*x^2+125*x)*exp(2))/log(log(exp(1)+1/25*x 
^2+5)+(x^2-2*x+1)*exp(2))^2,x)
 

Output:

log(x)/log(log((25*e + x**2 + 125)/25) + e**2*x**2 - 2*e**2*x + e**2)