\(\int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log (625+50 x+x^2)}{25 x^2+x^3+(50 x+2 x^2) \log (5)+(25+x) \log ^2(5)+(100 x+4 x^2+(100+4 x) \log (5)) \log (16)+(100+4 x) \log ^2(16)+(-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)) \log (625+50 x+x^2)+(25+x) \log ^2(625+50 x+x^2)} \, dx\) [9]

Optimal result

Integrand size = 145, antiderivative size = 22 \[ \int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\right )} \, dx=2+\frac {x}{x+\log (5)+2 \log (16)-\log \left ((25+x)^2\right )} \] Output:

x/(8*ln(2)-ln((x+25)^2)+ln(5)+x)+2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 5.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\right )} \, dx=\frac {x}{x+\log (1280)-\log \left ((25+x)^2\right )} \] Input:

Integrate[(2*x + (25 + x)*Log[5] + (50 + 2*x)*Log[16] + (-25 - x)*Log[625 
+ 50*x + x^2])/(25*x^2 + x^3 + (50*x + 2*x^2)*Log[5] + (25 + x)*Log[5]^2 + 
 (100*x + 4*x^2 + (100 + 4*x)*Log[5])*Log[16] + (100 + 4*x)*Log[16]^2 + (- 
50*x - 2*x^2 + (-50 - 2*x)*Log[5] + (-100 - 4*x)*Log[16])*Log[625 + 50*x + 
 x^2] + (25 + x)*Log[625 + 50*x + x^2]^2),x]
 

Output:

x/(x + Log[1280] - Log[(25 + x)^2])
 

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.

\(\displaystyle \int \frac {(-x-25) \log \left (x^2+50 x+625\right )+2 x+(2 x+50) \log (16)+(x+25) \log (5)}{x^3+25 x^2+(x+25) \log ^2\left (x^2+50 x+625\right )+\left (-2 x^2-50 x+(-4 x-100) \log (16)+(-2 x-50) \log (5)\right ) \log \left (x^2+50 x+625\right )+\log (16) \left (4 x^2+100 x+(4 x+100) \log (5)\right )+\left (2 x^2+50 x\right ) \log (5)+(4 x+100) \log ^2(16)+(x+25) \log ^2(5)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {x (2+\log (1280))-(x+25) \log \left ((x+25)^2\right )+25 \log (1280)}{(x+25) \left (x-\log \left ((x+25)^2\right )+\log (1280)\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {1}{x-\log \left ((x+25)^2\right )+\log (1280)}-\frac {x (x+23)}{(x+25) \left (x-\log \left ((x+25)^2\right )+\log (1280)\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \int \frac {1}{\left (x-\log \left ((x+25)^2\right )+\log (1280)\right )^2}dx-\int \frac {x}{\left (x-\log \left ((x+25)^2\right )+\log (1280)\right )^2}dx-50 \int \frac {1}{(x+25) \left (x-\log \left ((x+25)^2\right )+\log (1280)\right )^2}dx+\int \frac {1}{x-\log \left ((x+25)^2\right )+\log (1280)}dx\)

Input:

Int[(2*x + (25 + x)*Log[5] + (50 + 2*x)*Log[16] + (-25 - x)*Log[625 + 50*x 
 + x^2])/(25*x^2 + x^3 + (50*x + 2*x^2)*Log[5] + (25 + x)*Log[5]^2 + (100* 
x + 4*x^2 + (100 + 4*x)*Log[5])*Log[16] + (100 + 4*x)*Log[16]^2 + (-50*x - 
 2*x^2 + (-50 - 2*x)*Log[5] + (-100 - 4*x)*Log[16])*Log[625 + 50*x + x^2] 
+ (25 + x)*Log[625 + 50*x + x^2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09

method	result	size
risch	\(\frac {x}{8 \ln \left (2\right )+\ln \left (5\right )+x -\ln \left (x^{2}+50 x +625\right )}\)	\(24\)
parallelrisch	\(\frac {x}{8 \ln \left (2\right )+\ln \left (5\right )+x -\ln \left (x^{2}+50 x +625\right )}\)	\(24\)
norman	\(\frac {\ln \left (x^{2}+50 x +625\right )-8 \ln \left (2\right )-\ln \left (5\right )}{8 \ln \left (2\right )+\ln \left (5\right )+x -\ln \left (x^{2}+50 x +625\right )}\)	\(41\)

Input:

int(((-x-25)*ln(x^2+50*x+625)+4*(2*x+50)*ln(2)+(x+25)*ln(5)+2*x)/((x+25)*l 
n(x^2+50*x+625)^2+(4*(-4*x-100)*ln(2)+(-2*x-50)*ln(5)-2*x^2-50*x)*ln(x^2+5 
0*x+625)+16*(4*x+100)*ln(2)^2+4*((4*x+100)*ln(5)+4*x^2+100*x)*ln(2)+(x+25) 
*ln(5)^2+(2*x^2+50*x)*ln(5)+x^3+25*x^2),x,method=_RETURNVERBOSE)
 

Output:

x/(8*ln(2)+ln(5)+x-ln(x^2+50*x+625))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\right )} \, dx=\frac {x}{x + \log \left (5\right ) + 8 \, \log \left (2\right ) - \log \left (x^{2} + 50 \, x + 625\right )} \] Input:

integrate(((-x-25)*log(x^2+50*x+625)+4*(2*x+50)*log(2)+(x+25)*log(5)+2*x)/ 
((x+25)*log(x^2+50*x+625)^2+(4*(-4*x-100)*log(2)+(-2*x-50)*log(5)-2*x^2-50 
*x)*log(x^2+50*x+625)+16*(4*x+100)*log(2)^2+4*((4*x+100)*log(5)+4*x^2+100* 
x)*log(2)+(x+25)*log(5)^2+(2*x^2+50*x)*log(5)+x^3+25*x^2),x, algorithm="fr 
icas")
 

Output:

x/(x + log(5) + 8*log(2) - log(x^2 + 50*x + 625))
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\right )} \, dx=- \frac {x}{- x + \log {\left (x^{2} + 50 x + 625 \right )} - 8 \log {\left (2 \right )} - \log {\left (5 \right )}} \] Input:

integrate(((-x-25)*ln(x**2+50*x+625)+4*(2*x+50)*ln(2)+(x+25)*ln(5)+2*x)/(( 
x+25)*ln(x**2+50*x+625)**2+(4*(-4*x-100)*ln(2)+(-2*x-50)*ln(5)-2*x**2-50*x 
)*ln(x**2+50*x+625)+16*(4*x+100)*ln(2)**2+4*((4*x+100)*ln(5)+4*x**2+100*x) 
*ln(2)+(x+25)*ln(5)**2+(2*x**2+50*x)*ln(5)+x**3+25*x**2),x)
 

Output:

-x/(-x + log(x**2 + 50*x + 625) - 8*log(2) - log(5))
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\right )} \, dx=\frac {x}{x + \log \left (5\right ) + 8 \, \log \left (2\right ) - 2 \, \log \left (x + 25\right )} \] Input:

integrate(((-x-25)*log(x^2+50*x+625)+4*(2*x+50)*log(2)+(x+25)*log(5)+2*x)/ 
((x+25)*log(x^2+50*x+625)^2+(4*(-4*x-100)*log(2)+(-2*x-50)*log(5)-2*x^2-50 
*x)*log(x^2+50*x+625)+16*(4*x+100)*log(2)^2+4*((4*x+100)*log(5)+4*x^2+100* 
x)*log(2)+(x+25)*log(5)^2+(2*x^2+50*x)*log(5)+x^3+25*x^2),x, algorithm="ma 
xima")
 

Output:

x/(x + log(5) + 8*log(2) - 2*log(x + 25))
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\right )} \, dx=\frac {x}{x + \log \left (5\right ) + 8 \, \log \left (2\right ) - \log \left (x^{2} + 50 \, x + 625\right )} \] Input:

integrate(((-x-25)*log(x^2+50*x+625)+4*(2*x+50)*log(2)+(x+25)*log(5)+2*x)/ 
((x+25)*log(x^2+50*x+625)^2+(4*(-4*x-100)*log(2)+(-2*x-50)*log(5)-2*x^2-50 
*x)*log(x^2+50*x+625)+16*(4*x+100)*log(2)^2+4*((4*x+100)*log(5)+4*x^2+100* 
x)*log(2)+(x+25)*log(5)^2+(2*x^2+50*x)*log(5)+x^3+25*x^2),x, algorithm="gi 
ac")
 

Output:

x/(x + log(5) + 8*log(2) - log(x^2 + 50*x + 625))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\right )} \, dx=\int \frac {2\,x+4\,\ln \left (2\right )\,\left (2\,x+50\right )+\ln \left (5\right )\,\left (x+25\right )-\ln \left (x^2+50\,x+625\right )\,\left (x+25\right )}{{\ln \left (5\right )}^2\,\left (x+25\right )+{\ln \left (x^2+50\,x+625\right )}^2\,\left (x+25\right )+\ln \left (5\right )\,\left (2\,x^2+50\,x\right )+4\,\ln \left (2\right )\,\left (100\,x+\ln \left (5\right )\,\left (4\,x+100\right )+4\,x^2\right )-\ln \left (x^2+50\,x+625\right )\,\left (50\,x+\ln \left (5\right )\,\left (2\,x+50\right )+4\,\ln \left (2\right )\,\left (4\,x+100\right )+2\,x^2\right )+16\,{\ln \left (2\right )}^2\,\left (4\,x+100\right )+25\,x^2+x^3} \,d x \] Input:

int((2*x + 4*log(2)*(2*x + 50) + log(5)*(x + 25) - log(50*x + x^2 + 625)*( 
x + 25))/(log(5)^2*(x + 25) + log(50*x + x^2 + 625)^2*(x + 25) + log(5)*(5 
0*x + 2*x^2) + 4*log(2)*(100*x + log(5)*(4*x + 100) + 4*x^2) - log(50*x + 
x^2 + 625)*(50*x + log(5)*(2*x + 50) + 4*log(2)*(4*x + 100) + 2*x^2) + 16* 
log(2)^2*(4*x + 100) + 25*x^2 + x^3),x)
 

Output:

int((2*x + 4*log(2)*(2*x + 50) + log(5)*(x + 25) - log(50*x + x^2 + 625)*( 
x + 25))/(log(5)^2*(x + 25) + log(50*x + x^2 + 625)^2*(x + 25) + log(5)*(5 
0*x + 2*x^2) + 4*log(2)*(100*x + log(5)*(4*x + 100) + 4*x^2) - log(50*x + 
x^2 + 625)*(50*x + log(5)*(2*x + 50) + 4*log(2)*(4*x + 100) + 2*x^2) + 16* 
log(2)^2*(4*x + 100) + 25*x^2 + x^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\right )} \, dx=\frac {-\mathrm {log}\left (x^{2}+50 x +625\right )+\mathrm {log}\left (5\right )+8 \,\mathrm {log}\left (2\right )}{\mathrm {log}\left (x^{2}+50 x +625\right )-\mathrm {log}\left (5\right )-8 \,\mathrm {log}\left (2\right )-x} \] Input:

int(((-x-25)*log(x^2+50*x+625)+4*(2*x+50)*log(2)+(x+25)*log(5)+2*x)/((x+25 
)*log(x^2+50*x+625)^2+(4*(-4*x-100)*log(2)+(-2*x-50)*log(5)-2*x^2-50*x)*lo 
g(x^2+50*x+625)+16*(4*x+100)*log(2)^2+4*((4*x+100)*log(5)+4*x^2+100*x)*log 
(2)+(x+25)*log(5)^2+(2*x^2+50*x)*log(5)+x^3+25*x^2),x)
 

Output:

( - log(x**2 + 50*x + 625) + log(5) + 8*log(2))/(log(x**2 + 50*x + 625) - 
log(5) - 8*log(2) - x)