\(\int \frac {(-20 x-10 x^2+10 x^3+e^4 (-20-10 x+10 x^2)+(-10 e^4-10 x) \log (4)) \log (\frac {5+x}{x})+(-10 x-2 x^2-5 x^3-x^4+e^4 (5 x-9 x^2-2 x^3)+(-5 x-x^2) \log (4)) \log ^2(\frac {5+x}{x})}{5 x^3+x^4+e^8 (5 x+x^2)+e^4 (10 x^2+2 x^3)} \, dx\) [161]

Optimal result

Integrand size = 148, antiderivative size = 28 \[ \int \frac {\left (-20 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx=\frac {\left (2+x-x^2+\log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{e^4+x} \] Output:

(2*ln(2)-x^2+x+2)/(x+exp(4))*ln(1/x*(5+x))^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 3.97 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-20 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx=\frac {\left (2+x-x^2+\log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{e^4+x} \] Input:

Integrate[((-20*x - 10*x^2 + 10*x^3 + E^4*(-20 - 10*x + 10*x^2) + (-10*E^4 
 - 10*x)*Log[4])*Log[(5 + x)/x] + (-10*x - 2*x^2 - 5*x^3 - x^4 + E^4*(5*x 
- 9*x^2 - 2*x^3) + (-5*x - x^2)*Log[4])*Log[(5 + x)/x]^2)/(5*x^3 + x^4 + E 
^8*(5*x + x^2) + E^4*(10*x^2 + 2*x^3)),x]
 

Output:

((2 + x - x^2 + Log[4])*Log[(5 + x)/x]^2)/(E^4 + 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.

\(\displaystyle \int \frac {\left (10 x^3-10 x^2+e^4 \left (10 x^2-10 x-20\right )-20 x+\left (-10 x-10 e^4\right ) \log (4)\right ) \log \left (\frac {x+5}{x}\right )+\left (-x^4-5 x^3-2 x^2+\left (-x^2-5 x\right ) \log (4)+e^4 \left (-2 x^3-9 x^2+5 x\right )-10 x\right ) \log ^2\left (\frac {x+5}{x}\right )}{x^4+5 x^3+e^8 \left (x^2+5 x\right )+e^4 \left (2 x^3+10 x^2\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (10 x^3-10 x^2+e^4 \left (10 x^2-10 x-20\right )-20 x+\left (-10 x-10 e^4\right ) \log (4)\right ) \log \left (\frac {x+5}{x}\right )+\left (-x^4-5 x^3-2 x^2+\left (-x^2-5 x\right ) \log (4)+e^4 \left (-2 x^3-9 x^2+5 x\right )-10 x\right ) \log ^2\left (\frac {x+5}{x}\right )}{x \left (x^3+\left (5+2 e^4\right ) x^2+e^4 \left (10+e^4\right ) x+5 e^8\right )}dx\)

\(\Big \downarrow \) 2463

\(\displaystyle \int \left (\frac {\left (10 x^3-10 x^2+e^4 \left (10 x^2-10 x-20\right )-20 x+\left (-10 x-10 e^4\right ) \log (4)\right ) \log \left (\frac {x+5}{x}\right )+\left (-x^4-5 x^3-2 x^2+\left (-x^2-5 x\right ) \log (4)+e^4 \left (-2 x^3-9 x^2+5 x\right )-10 x\right ) \log ^2\left (\frac {x+5}{x}\right )}{\left (e^4-5\right )^2 x (x+5)}-\frac {\left (10 x^3-10 x^2+e^4 \left (10 x^2-10 x-20\right )-20 x+\left (-10 x-10 e^4\right ) \log (4)\right ) \log \left (\frac {x+5}{x}\right )+\left (-x^4-5 x^3-2 x^2+\left (-x^2-5 x\right ) \log (4)+e^4 \left (-2 x^3-9 x^2+5 x\right )-10 x\right ) \log ^2\left (\frac {x+5}{x}\right )}{\left (e^4-5\right )^2 x \left (x+e^4\right )}-\frac {\left (10 x^3-10 x^2+e^4 \left (10 x^2-10 x-20\right )-20 x+\left (-10 x-10 e^4\right ) \log (4)\right ) \log \left (\frac {x+5}{x}\right )+\left (-x^4-5 x^3-2 x^2+\left (-x^2-5 x\right ) \log (4)+e^4 \left (-2 x^3-9 x^2+5 x\right )-10 x\right ) \log ^2\left (\frac {x+5}{x}\right )}{\left (e^4-5\right ) x \left (x+e^4\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (5+e^4\right ) \log ^2\left (\frac {x+5}{x}\right ) x^2}{2 \left (5-e^4\right )^2}-\frac {\log ^2\left (\frac {x+5}{x}\right ) x^2}{2 \left (5-e^4\right )}-\frac {e^4 \log ^2\left (\frac {x+5}{x}\right ) x^2}{\left (5-e^4\right )^2}+\frac {5 \log \left (1+\frac {5}{x}\right ) x^2}{\left (5-e^4\right )^2}+\frac {5 \log (x) x^2}{\left (5-e^4\right )^2}-\frac {5 \log (x+5) x^2}{\left (5-e^4\right )^2}-\frac {5 \left (\log (x)-\log (x+5)+\log \left (\frac {x+5}{x}\right )\right ) x^2}{\left (5-e^4\right )^2}-\frac {10 \left (6-e^4\right ) \log \left (1+\frac {5}{x}\right ) x}{\left (5-e^4\right )^2}-\frac {10 \log (x) x}{5-e^4}-\frac {10 \log (x) x}{\left (5-e^4\right )^2}+\frac {5 \left (5+e^4\right ) \log \left (\frac {x+5}{x}\right ) x}{\left (5-e^4\right )^2}-\frac {5 \log \left (\frac {x+5}{x}\right ) x}{5-e^4}-\frac {50 \log \left (\frac {x+5}{x}\right ) x}{3 \left (5-e^4\right )^2}+\frac {10 \left (\log (x)-\log (x+5)+\log \left (\frac {x+5}{x}\right )\right ) x}{5-e^4}+\frac {10 \left (\log (x)-\log (x+5)+\log \left (\frac {x+5}{x}\right )\right ) x}{\left (5-e^4\right )^2}-\frac {5 (x+5) \log ^2\left (1+\frac {5}{x}\right )}{5-e^4}+\frac {(x+5) \left (2+4 e^4-e^8+\log (4)\right ) \log ^2\left (1+\frac {5}{x}\right )}{\left (5-e^4\right )^2}+\frac {5 (2+\log (4)) \log ^2(x)}{e^4 \left (5-e^4\right )}-\frac {5 (2+\log (4)) \log ^2(x)}{\left (5-e^4\right )^2}-\frac {(x+5) \left (2-e^4+\log (4)\right ) \log ^2\left (\frac {x+5}{x}\right )}{\left (5-e^4\right )^2}-\frac {25 \left (5+e^4\right ) \log ^2\left (\frac {x+5}{x}\right )}{2 \left (5-e^4\right )^2}+\frac {25 \log ^2\left (\frac {x+5}{x}\right )}{2 \left (5-e^4\right )}+\frac {125 \log ^2\left (\frac {x+5}{x}\right )}{3 \left (5-e^4\right )^2}-\frac {10 (2+\log (4)) \log (5) \log (x)}{e^4 \left (5-e^4\right )}+\frac {10 (2+\log (4)) \log (5) \log (x)}{\left (5-e^4\right )^2}+\frac {25 \left (5+e^4\right ) \log (x)}{\left (5-e^4\right )^2}-\frac {25 \log (x)}{5-e^4}-\frac {50 e^4 \log (x)}{\left (5-e^4\right )^2}+\frac {10 (x+5) \log (x+5)}{5-e^4}+\frac {10 (x+5) \log (x+5)}{\left (5-e^4\right )^2}+\frac {2 (28-\log (4)) \log \left (1+\frac {5}{x}\right ) \log (x+5)}{5-e^4}-\frac {50 \left (6-e^4\right ) \log (x+5)}{\left (5-e^4\right )^2}-\frac {10 e^4 (x+5) \log \left (\frac {x+5}{x}\right )}{\left (5-e^4\right )^2}+\frac {50 (x+5) \log \left (\frac {x+5}{x}\right )}{3 \left (5-e^4\right )^2}+\frac {25 \left (5+e^4\right ) \log \left (\frac {x+5}{x}\right )}{\left (5-e^4\right )^2}-\frac {25 \log \left (\frac {x+5}{x}\right )}{5-e^4}-\frac {250 \log \left (\frac {x+5}{x}\right )}{3 \left (5-e^4\right )^2}-\frac {10 (2+\log (4)) \log (x) \left (\log (x)-\log (x+5)+\log \left (\frac {x+5}{x}\right )\right )}{e^4 \left (5-e^4\right )}+\frac {10 (2+\log (4)) \log (x) \left (\log (x)-\log (x+5)+\log \left (\frac {x+5}{x}\right )\right )}{\left (5-e^4\right )^2}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \left (\log (x)-\log (x+5)+\log \left (\frac {x+5}{x}\right )\right ) \log \left (x+e^4\right )}{e^4 \left (5-e^4\right )}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log (x+5) \log \left (-\frac {x+e^4}{5-e^4}\right )}{e^4 \left (5-e^4\right )}-\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log (x) \log \left (\frac {x}{e^4}+1\right )}{e^4 \left (5-e^4\right )}+\frac {2 (28-\log (4)) \log (x+5) \log \left (1-\frac {5}{x+5}\right )}{5-e^4}-\frac {50 e^4 \log \left (\frac {x+5}{x}\right ) \log \left (1-\frac {x}{x+5}\right )}{\left (5-e^4\right )^2}+\frac {250 \log \left (\frac {x+5}{x}\right ) \log \left (1-\frac {x}{x+5}\right )}{3 \left (5-e^4\right )^2}+\frac {10 \left (2+4 e^4-e^8+\log (4)\right ) \operatorname {PolyLog}\left (2,-\frac {5}{x}\right )}{\left (5-e^4\right )^2}-\frac {10 \left (2-e^4+\log (4)\right ) \operatorname {PolyLog}\left (2,-\frac {5}{x}\right )}{\left (5-e^4\right )^2}-\frac {25 \left (5+e^4\right ) \operatorname {PolyLog}\left (2,-\frac {5}{x}\right )}{\left (5-e^4\right )^2}-\frac {25 \operatorname {PolyLog}\left (2,-\frac {5}{x}\right )}{5-e^4}+\frac {250 \operatorname {PolyLog}\left (2,-\frac {5}{x}\right )}{3 \left (5-e^4\right )^2}+\frac {10 (2+\log (4)) \operatorname {PolyLog}\left (2,-\frac {x}{5}\right )}{e^4 \left (5-e^4\right )}-\frac {10 (2+\log (4)) \operatorname {PolyLog}\left (2,-\frac {x}{5}\right )}{\left (5-e^4\right )^2}-\frac {10 \left (2-e^4-e^8+\log (4)\right ) \operatorname {PolyLog}\left (2,-\frac {x}{e^4}\right )}{e^4 \left (5-e^4\right )}-\frac {2 (28-\log (4)) \operatorname {PolyLog}\left (2,\frac {5}{x+5}\right )}{5-e^4}+\frac {50 e^4 \operatorname {PolyLog}\left (2,\frac {x}{x+5}\right )}{\left (5-e^4\right )^2}-\frac {250 \operatorname {PolyLog}\left (2,\frac {x}{x+5}\right )}{3 \left (5-e^4\right )^2}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \operatorname {PolyLog}\left (2,\frac {x+5}{5-e^4}\right )}{e^4 \left (5-e^4\right )}-\frac {2 e^4 (2+\log (4)) \operatorname {PolyLog}\left (2,1-\frac {x+5}{x}\right )}{\left (5-e^4\right )^2}-\left (2-e^4-e^8+\log (4)\right ) \int \frac {\log ^2\left (1+\frac {5}{x}\right )}{\left (x+e^4\right )^2}dx\)

Input:

Int[((-20*x - 10*x^2 + 10*x^3 + E^4*(-20 - 10*x + 10*x^2) + (-10*E^4 - 10* 
x)*Log[4])*Log[(5 + x)/x] + (-10*x - 2*x^2 - 5*x^3 - x^4 + E^4*(5*x - 9*x^ 
2 - 2*x^3) + (-5*x - x^2)*Log[4])*Log[(5 + x)/x]^2)/(5*x^3 + x^4 + E^8*(5* 
x + x^2) + E^4*(10*x^2 + 2*x^3)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 17.54 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89

method	result	size
norman	\(\frac {x \ln \left (\frac {5+x}{x}\right )^{2}+\left (2+2 \ln \left (2\right )\right ) \ln \left (\frac {5+x}{x}\right )^{2}-\ln \left (\frac {5+x}{x}\right )^{2} x^{2}}{x +{\mathrm e}^{4}}\)	\(53\)
parallelrisch	\(\frac {-\ln \left (\frac {5+x}{x}\right )^{2} x^{2}+2 \ln \left (\frac {5+x}{x}\right )^{2} \ln \left (2\right )+x \ln \left (\frac {5+x}{x}\right )^{2}+2 \ln \left (\frac {5+x}{x}\right )^{2}}{x +{\mathrm e}^{4}}\)	\(62\)

Input:

int(((2*(-x^2-5*x)*ln(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2-10*x)*l 
n(1/x*(5+x))^2+(2*(-10*exp(4)-10*x)*ln(2)+(10*x^2-10*x-20)*exp(4)+10*x^3-1 
0*x^2-20*x)*ln(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*exp(4)+x^4+5 
*x^3),x,method=_RETURNVERBOSE)
 

Output:

(x*ln(1/x*(5+x))^2+(2+2*ln(2))*ln(1/x*(5+x))^2-ln(1/x*(5+x))^2*x^2)/(x+exp 
(4))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\left (-20 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx=-\frac {{\left (x^{2} - x - 2 \, \log \left (2\right ) - 2\right )} \log \left (\frac {x + 5}{x}\right )^{2}}{x + e^{4}} \] Input:

integrate(((2*(-x^2-5*x)*log(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2- 
10*x)*log(1/x*(5+x))^2+(2*(-10*exp(4)-10*x)*log(2)+(10*x^2-10*x-20)*exp(4) 
+10*x^3-10*x^2-20*x)*log(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*ex 
p(4)+x^4+5*x^3),x, algorithm="fricas")
 

Output:

-(x^2 - x - 2*log(2) - 2)*log((x + 5)/x)^2/(x + e^4)
 

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-20 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx=\frac {\left (- x^{2} + x + 2 \log {\left (2 \right )} + 2\right ) \log {\left (\frac {x + 5}{x} \right )}^{2}}{x + e^{4}} \] Input:

integrate(((2*(-x**2-5*x)*ln(2)+(-2*x**3-9*x**2+5*x)*exp(4)-x**4-5*x**3-2* 
x**2-10*x)*ln(1/x*(5+x))**2+(2*(-10*exp(4)-10*x)*ln(2)+(10*x**2-10*x-20)*e 
xp(4)+10*x**3-10*x**2-20*x)*ln(1/x*(5+x)))/((x**2+5*x)*exp(4)**2+(2*x**3+1 
0*x**2)*exp(4)+x**4+5*x**3),x)
 

Output:

(-x**2 + x + 2*log(2) + 2)*log((x + 5)/x)**2/(x + exp(4))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (30) = 60\).

Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.32 \[ \int \frac {\left (-20 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx=-\frac {{\left (x^{2} - x - 2 \, \log \left (2\right ) - 2\right )} \log \left (x + 5\right )^{2} - 2 \, {\left (x^{2} - x - 2 \, \log \left (2\right ) - 2\right )} \log \left (x + 5\right ) \log \left (x\right ) + {\left (x^{2} - x - 2 \, \log \left (2\right ) - 2\right )} \log \left (x\right )^{2}}{x + e^{4}} \] Input:

integrate(((2*(-x^2-5*x)*log(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2- 
10*x)*log(1/x*(5+x))^2+(2*(-10*exp(4)-10*x)*log(2)+(10*x^2-10*x-20)*exp(4) 
+10*x^3-10*x^2-20*x)*log(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*ex 
p(4)+x^4+5*x^3),x, algorithm="maxima")
 

Output:

-((x^2 - x - 2*log(2) - 2)*log(x + 5)^2 - 2*(x^2 - x - 2*log(2) - 2)*log(x 
 + 5)*log(x) + (x^2 - x - 2*log(2) - 2)*log(x)^2)/(x + e^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (30) = 60\).

Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 5.07 \[ \int \frac {\left (-20 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx=\frac {\frac {2 \, {\left (x + 5\right )}^{2} \log \left (2\right ) \log \left (\frac {x + 5}{x}\right )^{2}}{x^{2}} - \frac {4 \, {\left (x + 5\right )} \log \left (2\right ) \log \left (\frac {x + 5}{x}\right )^{2}}{x} + 2 \, \log \left (2\right ) \log \left (\frac {x + 5}{x}\right )^{2} + \frac {2 \, {\left (x + 5\right )}^{2} \log \left (\frac {x + 5}{x}\right )^{2}}{x^{2}} + \frac {{\left (x + 5\right )} \log \left (\frac {x + 5}{x}\right )^{2}}{x} - 28 \, \log \left (\frac {x + 5}{x}\right )^{2}}{\frac {{\left (x + 5\right )}^{2} e^{4}}{x^{2}} - \frac {2 \, {\left (x + 5\right )} e^{4}}{x} + \frac {5 \, {\left (x + 5\right )}}{x} + e^{4} - 5} \] Input:

integrate(((2*(-x^2-5*x)*log(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2- 
10*x)*log(1/x*(5+x))^2+(2*(-10*exp(4)-10*x)*log(2)+(10*x^2-10*x-20)*exp(4) 
+10*x^3-10*x^2-20*x)*log(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*ex 
p(4)+x^4+5*x^3),x, algorithm="giac")
 

Output:

(2*(x + 5)^2*log(2)*log((x + 5)/x)^2/x^2 - 4*(x + 5)*log(2)*log((x + 5)/x) 
^2/x + 2*log(2)*log((x + 5)/x)^2 + 2*(x + 5)^2*log((x + 5)/x)^2/x^2 + (x + 
 5)*log((x + 5)/x)^2/x - 28*log((x + 5)/x)^2)/((x + 5)^2*e^4/x^2 - 2*(x + 
5)*e^4/x + 5*(x + 5)/x + e^4 - 5)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-20 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx=\int -\frac {\left (10\,x+{\mathrm {e}}^4\,\left (2\,x^3+9\,x^2-5\,x\right )+2\,x^2+5\,x^3+x^4+2\,\ln \left (2\right )\,\left (x^2+5\,x\right )\right )\,{\ln \left (\frac {x+5}{x}\right )}^2+\left (20\,x+{\mathrm {e}}^4\,\left (-10\,x^2+10\,x+20\right )+2\,\ln \left (2\right )\,\left (10\,x+10\,{\mathrm {e}}^4\right )+10\,x^2-10\,x^3\right )\,\ln \left (\frac {x+5}{x}\right )}{{\mathrm {e}}^4\,\left (2\,x^3+10\,x^2\right )+{\mathrm {e}}^8\,\left (x^2+5\,x\right )+5\,x^3+x^4} \,d x \] Input:

int(-(log((x + 5)/x)*(20*x + exp(4)*(10*x - 10*x^2 + 20) + 2*log(2)*(10*x 
+ 10*exp(4)) + 10*x^2 - 10*x^3) + log((x + 5)/x)^2*(10*x + exp(4)*(9*x^2 - 
 5*x + 2*x^3) + 2*x^2 + 5*x^3 + x^4 + 2*log(2)*(5*x + x^2)))/(exp(4)*(10*x 
^2 + 2*x^3) + exp(8)*(5*x + x^2) + 5*x^3 + x^4),x)
 

Output:

int(-(log((x + 5)/x)*(20*x + exp(4)*(10*x - 10*x^2 + 20) + 2*log(2)*(10*x 
+ 10*exp(4)) + 10*x^2 - 10*x^3) + log((x + 5)/x)^2*(10*x + exp(4)*(9*x^2 - 
 5*x + 2*x^3) + 2*x^2 + 5*x^3 + x^4 + 2*log(2)*(5*x + x^2)))/(exp(4)*(10*x 
^2 + 2*x^3) + exp(8)*(5*x + x^2) + 5*x^3 + x^4), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\left (-20 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx=\frac {\mathrm {log}\left (\frac {x +5}{x}\right )^{2} \left (2 \,\mathrm {log}\left (2\right )-x^{2}+x +2\right )}{e^{4}+x} \] Input:

int(((2*(-x^2-5*x)*log(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2-10*x)* 
log(1/x*(5+x))^2+(2*(-10*exp(4)-10*x)*log(2)+(10*x^2-10*x-20)*exp(4)+10*x^ 
3-10*x^2-20*x)*log(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*exp(4)+x 
^4+5*x^3),x)
 

Output:

(log((x + 5)/x)**2*(2*log(2) - x**2 + x + 2))/(e**4 + x)