\(\int \frac {(-5 x-4 x^2) \log (x)+(-16-20 x-8 x^2) \log (x) \log (x^2)+(4+5 x+2 x^2) \log ^2(x^2)+(-4 x-5 x^2-2 x^3) \log ^4(x^2)+(4+5 x+2 x^2+(-8 x-10 x^2-4 x^3) \log ^2(x^2)) \log (4+5 x+2 x^2)+(-4 x-5 x^2-2 x^3) \log ^2(4+5 x+2 x^2)}{(4 x+5 x^2+2 x^3) \log ^4(x^2)+(8 x+10 x^2+4 x^3) \log ^2(x^2) \log (4+5 x+2 x^2)+(4 x+5 x^2+2 x^3) \log ^2(4+5 x+2 x^2)} \, dx\) [2020]

Optimal result

Integrand size = 224, antiderivative size = 27 \[ \int \frac {\left (-5 x-4 x^2\right ) \log (x)+\left (-16-20 x-8 x^2\right ) \log (x) \log \left (x^2\right )+\left (4+5 x+2 x^2\right ) \log ^2\left (x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^4\left (x^2\right )+\left (4+5 x+2 x^2+\left (-8 x-10 x^2-4 x^3\right ) \log ^2\left (x^2\right )\right ) \log \left (4+5 x+2 x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )}{\left (4 x+5 x^2+2 x^3\right ) \log ^4\left (x^2\right )+\left (8 x+10 x^2+4 x^3\right ) \log ^2\left (x^2\right ) \log \left (4+5 x+2 x^2\right )+\left (4 x+5 x^2+2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )} \, dx=-x+\frac {\log (x)}{\log ^2\left (x^2\right )+\log (4+x+x (4+2 x))} \] Output:

ln(x)/(ln(x^2)^2+ln(x+4+x*(4+2*x)))-x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-5 x-4 x^2\right ) \log (x)+\left (-16-20 x-8 x^2\right ) \log (x) \log \left (x^2\right )+\left (4+5 x+2 x^2\right ) \log ^2\left (x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^4\left (x^2\right )+\left (4+5 x+2 x^2+\left (-8 x-10 x^2-4 x^3\right ) \log ^2\left (x^2\right )\right ) \log \left (4+5 x+2 x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )}{\left (4 x+5 x^2+2 x^3\right ) \log ^4\left (x^2\right )+\left (8 x+10 x^2+4 x^3\right ) \log ^2\left (x^2\right ) \log \left (4+5 x+2 x^2\right )+\left (4 x+5 x^2+2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )} \, dx=-x+\frac {\log (x)}{\log ^2\left (x^2\right )+\log \left (4+5 x+2 x^2\right )} \] Input:

Integrate[((-5*x - 4*x^2)*Log[x] + (-16 - 20*x - 8*x^2)*Log[x]*Log[x^2] + 
(4 + 5*x + 2*x^2)*Log[x^2]^2 + (-4*x - 5*x^2 - 2*x^3)*Log[x^2]^4 + (4 + 5* 
x + 2*x^2 + (-8*x - 10*x^2 - 4*x^3)*Log[x^2]^2)*Log[4 + 5*x + 2*x^2] + (-4 
*x - 5*x^2 - 2*x^3)*Log[4 + 5*x + 2*x^2]^2)/((4*x + 5*x^2 + 2*x^3)*Log[x^2 
]^4 + (8*x + 10*x^2 + 4*x^3)*Log[x^2]^2*Log[4 + 5*x + 2*x^2] + (4*x + 5*x^ 
2 + 2*x^3)*Log[4 + 5*x + 2*x^2]^2),x]
 

Output:

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

\(\displaystyle \int \frac {\left (2 x^2+5 x+4\right ) \log ^2\left (x^2\right )+\left (-8 x^2-20 x-16\right ) \log (x) \log \left (x^2\right )+\left (-4 x^2-5 x\right ) \log (x)+\left (-2 x^3-5 x^2-4 x\right ) \log ^4\left (x^2\right )+\left (-2 x^3-5 x^2-4 x\right ) \log ^2\left (2 x^2+5 x+4\right )+\left (2 x^2+\left (-4 x^3-10 x^2-8 x\right ) \log ^2\left (x^2\right )+5 x+4\right ) \log \left (2 x^2+5 x+4\right )}{\left (2 x^3+5 x^2+4 x\right ) \log ^4\left (x^2\right )+\left (4 x^3+10 x^2+8 x\right ) \log \left (2 x^2+5 x+4\right ) \log ^2\left (x^2\right )+\left (2 x^3+5 x^2+4 x\right ) \log ^2\left (2 x^2+5 x+4\right )} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {\log (x) \left (4 x^2+8 x^2 \log \left (x^2\right )+20 x \log \left (x^2\right )+16 \log \left (x^2\right )+5 x\right )}{x \left (2 x^2+5 x+4\right ) \left (\log ^2\left (x^2\right )+\log \left (2 x^2+5 x+4\right )\right )^2}+\frac {1}{x \left (\log ^2\left (x^2\right )+\log \left (2 x^2+5 x+4\right )\right )}-1\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (-\frac {\log (x) \left (4 x^2+8 x^2 \log \left (x^2\right )+20 x \log \left (x^2\right )+16 \log \left (x^2\right )+5 x\right )}{x \left (2 x^2+5 x+4\right ) \left (\log ^2\left (x^2\right )+\log \left (2 x^2+5 x+4\right )\right )^2}+\frac {1}{x \left (\log ^2\left (x^2\right )+\log \left (2 x^2+5 x+4\right )\right )}-1\right )dx\)

Input:

Int[((-5*x - 4*x^2)*Log[x] + (-16 - 20*x - 8*x^2)*Log[x]*Log[x^2] + (4 + 5 
*x + 2*x^2)*Log[x^2]^2 + (-4*x - 5*x^2 - 2*x^3)*Log[x^2]^4 + (4 + 5*x + 2* 
x^2 + (-8*x - 10*x^2 - 4*x^3)*Log[x^2]^2)*Log[4 + 5*x + 2*x^2] + (-4*x - 5 
*x^2 - 2*x^3)*Log[4 + 5*x + 2*x^2]^2)/((4*x + 5*x^2 + 2*x^3)*Log[x^2]^4 + 
(8*x + 10*x^2 + 4*x^3)*Log[x^2]^2*Log[4 + 5*x + 2*x^2] + (4*x + 5*x^2 + 2* 
x^3)*Log[4 + 5*x + 2*x^2]^2),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.26 (sec) , antiderivative size = 182, normalized size of antiderivative = 6.74

\[-x +\frac {4 \ln \left (x \right )}{-\pi ^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}+4 \pi ^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}-6 \pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}+4 \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}-\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}-8 i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+16 i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-8 i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+16 \ln \left (x \right )^{2}+4 \ln \left (2 x^{2}+5 x +4\right )}\]

Input:

int(((-2*x^3-5*x^2-4*x)*ln(2*x^2+5*x+4)^2+((-4*x^3-10*x^2-8*x)*ln(x^2)^2+2 
*x^2+5*x+4)*ln(2*x^2+5*x+4)+(-2*x^3-5*x^2-4*x)*ln(x^2)^4+(2*x^2+5*x+4)*ln( 
x^2)^2+(-8*x^2-20*x-16)*ln(x)*ln(x^2)+(-4*x^2-5*x)*ln(x))/((2*x^3+5*x^2+4* 
x)*ln(2*x^2+5*x+4)^2+(4*x^3+10*x^2+8*x)*ln(x^2)^2*ln(2*x^2+5*x+4)+(2*x^3+5 
*x^2+4*x)*ln(x^2)^4),x)
 

Output:

-x+4*ln(x)/(-Pi^2*csgn(I*x)^4*csgn(I*x^2)^2+4*Pi^2*csgn(I*x)^3*csgn(I*x^2) 
^3-6*Pi^2*csgn(I*x)^2*csgn(I*x^2)^4+4*Pi^2*csgn(I*x)*csgn(I*x^2)^5-Pi^2*cs 
gn(I*x^2)^6-8*I*ln(x)*Pi*csgn(I*x)^2*csgn(I*x^2)+16*I*ln(x)*Pi*csgn(I*x)*c 
sgn(I*x^2)^2-8*I*ln(x)*Pi*csgn(I*x^2)^3+16*ln(x)^2+4*ln(2*x^2+5*x+4))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {\left (-5 x-4 x^2\right ) \log (x)+\left (-16-20 x-8 x^2\right ) \log (x) \log \left (x^2\right )+\left (4+5 x+2 x^2\right ) \log ^2\left (x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^4\left (x^2\right )+\left (4+5 x+2 x^2+\left (-8 x-10 x^2-4 x^3\right ) \log ^2\left (x^2\right )\right ) \log \left (4+5 x+2 x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )}{\left (4 x+5 x^2+2 x^3\right ) \log ^4\left (x^2\right )+\left (8 x+10 x^2+4 x^3\right ) \log ^2\left (x^2\right ) \log \left (4+5 x+2 x^2\right )+\left (4 x+5 x^2+2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )} \, dx=-\frac {4 \, x \log \left (x\right )^{2} + x \log \left (2 \, x^{2} + 5 \, x + 4\right ) - \log \left (x\right )}{4 \, \log \left (x\right )^{2} + \log \left (2 \, x^{2} + 5 \, x + 4\right )} \] Input:

integrate(((-2*x^3-5*x^2-4*x)*log(2*x^2+5*x+4)^2+((-4*x^3-10*x^2-8*x)*log( 
x^2)^2+2*x^2+5*x+4)*log(2*x^2+5*x+4)+(-2*x^3-5*x^2-4*x)*log(x^2)^4+(2*x^2+ 
5*x+4)*log(x^2)^2+(-8*x^2-20*x-16)*log(x)*log(x^2)+(-4*x^2-5*x)*log(x))/(( 
2*x^3+5*x^2+4*x)*log(2*x^2+5*x+4)^2+(4*x^3+10*x^2+8*x)*log(x^2)^2*log(2*x^ 
2+5*x+4)+(2*x^3+5*x^2+4*x)*log(x^2)^4),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {\left (-5 x-4 x^2\right ) \log (x)+\left (-16-20 x-8 x^2\right ) \log (x) \log \left (x^2\right )+\left (4+5 x+2 x^2\right ) \log ^2\left (x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^4\left (x^2\right )+\left (4+5 x+2 x^2+\left (-8 x-10 x^2-4 x^3\right ) \log ^2\left (x^2\right )\right ) \log \left (4+5 x+2 x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )}{\left (4 x+5 x^2+2 x^3\right ) \log ^4\left (x^2\right )+\left (8 x+10 x^2+4 x^3\right ) \log ^2\left (x^2\right ) \log \left (4+5 x+2 x^2\right )+\left (4 x+5 x^2+2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )} \, dx=- x + \frac {\log {\left (x \right )}}{4 \log {\left (x \right )}^{2} + \log {\left (2 x^{2} + 5 x + 4 \right )}} \] Input:

integrate(((-2*x**3-5*x**2-4*x)*ln(2*x**2+5*x+4)**2+((-4*x**3-10*x**2-8*x) 
*ln(x**2)**2+2*x**2+5*x+4)*ln(2*x**2+5*x+4)+(-2*x**3-5*x**2-4*x)*ln(x**2)* 
*4+(2*x**2+5*x+4)*ln(x**2)**2+(-8*x**2-20*x-16)*ln(x)*ln(x**2)+(-4*x**2-5* 
x)*ln(x))/((2*x**3+5*x**2+4*x)*ln(2*x**2+5*x+4)**2+(4*x**3+10*x**2+8*x)*ln 
(x**2)**2*ln(2*x**2+5*x+4)+(2*x**3+5*x**2+4*x)*ln(x**2)**4),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {\left (-5 x-4 x^2\right ) \log (x)+\left (-16-20 x-8 x^2\right ) \log (x) \log \left (x^2\right )+\left (4+5 x+2 x^2\right ) \log ^2\left (x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^4\left (x^2\right )+\left (4+5 x+2 x^2+\left (-8 x-10 x^2-4 x^3\right ) \log ^2\left (x^2\right )\right ) \log \left (4+5 x+2 x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )}{\left (4 x+5 x^2+2 x^3\right ) \log ^4\left (x^2\right )+\left (8 x+10 x^2+4 x^3\right ) \log ^2\left (x^2\right ) \log \left (4+5 x+2 x^2\right )+\left (4 x+5 x^2+2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )} \, dx=-\frac {4 \, x \log \left (x\right )^{2} + x \log \left (2 \, x^{2} + 5 \, x + 4\right ) - \log \left (x\right )}{4 \, \log \left (x\right )^{2} + \log \left (2 \, x^{2} + 5 \, x + 4\right )} \] Input:

integrate(((-2*x^3-5*x^2-4*x)*log(2*x^2+5*x+4)^2+((-4*x^3-10*x^2-8*x)*log( 
x^2)^2+2*x^2+5*x+4)*log(2*x^2+5*x+4)+(-2*x^3-5*x^2-4*x)*log(x^2)^4+(2*x^2+ 
5*x+4)*log(x^2)^2+(-8*x^2-20*x-16)*log(x)*log(x^2)+(-4*x^2-5*x)*log(x))/(( 
2*x^3+5*x^2+4*x)*log(2*x^2+5*x+4)^2+(4*x^3+10*x^2+8*x)*log(x^2)^2*log(2*x^ 
2+5*x+4)+(2*x^3+5*x^2+4*x)*log(x^2)^4),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-5 x-4 x^2\right ) \log (x)+\left (-16-20 x-8 x^2\right ) \log (x) \log \left (x^2\right )+\left (4+5 x+2 x^2\right ) \log ^2\left (x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^4\left (x^2\right )+\left (4+5 x+2 x^2+\left (-8 x-10 x^2-4 x^3\right ) \log ^2\left (x^2\right )\right ) \log \left (4+5 x+2 x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )}{\left (4 x+5 x^2+2 x^3\right ) \log ^4\left (x^2\right )+\left (8 x+10 x^2+4 x^3\right ) \log ^2\left (x^2\right ) \log \left (4+5 x+2 x^2\right )+\left (4 x+5 x^2+2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )} \, dx=-x + \frac {\log \left (x\right )}{4 \, \log \left (x\right )^{2} + \log \left (2 \, x^{2} + 5 \, x + 4\right )} \] Input:

integrate(((-2*x^3-5*x^2-4*x)*log(2*x^2+5*x+4)^2+((-4*x^3-10*x^2-8*x)*log( 
x^2)^2+2*x^2+5*x+4)*log(2*x^2+5*x+4)+(-2*x^3-5*x^2-4*x)*log(x^2)^4+(2*x^2+ 
5*x+4)*log(x^2)^2+(-8*x^2-20*x-16)*log(x)*log(x^2)+(-4*x^2-5*x)*log(x))/(( 
2*x^3+5*x^2+4*x)*log(2*x^2+5*x+4)^2+(4*x^3+10*x^2+8*x)*log(x^2)^2*log(2*x^ 
2+5*x+4)+(2*x^3+5*x^2+4*x)*log(x^2)^4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-5 x-4 x^2\right ) \log (x)+\left (-16-20 x-8 x^2\right ) \log (x) \log \left (x^2\right )+\left (4+5 x+2 x^2\right ) \log ^2\left (x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^4\left (x^2\right )+\left (4+5 x+2 x^2+\left (-8 x-10 x^2-4 x^3\right ) \log ^2\left (x^2\right )\right ) \log \left (4+5 x+2 x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )}{\left (4 x+5 x^2+2 x^3\right ) \log ^4\left (x^2\right )+\left (8 x+10 x^2+4 x^3\right ) \log ^2\left (x^2\right ) \log \left (4+5 x+2 x^2\right )+\left (4 x+5 x^2+2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )} \, dx=\int -\frac {{\ln \left (2\,x^2+5\,x+4\right )}^2\,\left (2\,x^3+5\,x^2+4\,x\right )-\ln \left (2\,x^2+5\,x+4\right )\,\left (5\,x-{\ln \left (x^2\right )}^2\,\left (4\,x^3+10\,x^2+8\,x\right )+2\,x^2+4\right )-{\ln \left (x^2\right )}^2\,\left (2\,x^2+5\,x+4\right )+{\ln \left (x^2\right )}^4\,\left (2\,x^3+5\,x^2+4\,x\right )+\ln \left (x\right )\,\left (4\,x^2+5\,x\right )+\ln \left (x^2\right )\,\ln \left (x\right )\,\left (8\,x^2+20\,x+16\right )}{\left (2\,x^3+5\,x^2+4\,x\right )\,{\ln \left (x^2\right )}^4+\left (4\,x^3+10\,x^2+8\,x\right )\,{\ln \left (x^2\right )}^2\,\ln \left (2\,x^2+5\,x+4\right )+\left (2\,x^3+5\,x^2+4\,x\right )\,{\ln \left (2\,x^2+5\,x+4\right )}^2} \,d x \] Input:

int(-(log(5*x + 2*x^2 + 4)^2*(4*x + 5*x^2 + 2*x^3) - log(5*x + 2*x^2 + 4)* 
(5*x - log(x^2)^2*(8*x + 10*x^2 + 4*x^3) + 2*x^2 + 4) - log(x^2)^2*(5*x + 
2*x^2 + 4) + log(x^2)^4*(4*x + 5*x^2 + 2*x^3) + log(x)*(5*x + 4*x^2) + log 
(x^2)*log(x)*(20*x + 8*x^2 + 16))/(log(5*x + 2*x^2 + 4)^2*(4*x + 5*x^2 + 2 
*x^3) + log(x^2)^4*(4*x + 5*x^2 + 2*x^3) + log(x^2)^2*log(5*x + 2*x^2 + 4) 
*(8*x + 10*x^2 + 4*x^3)),x)
 

Output:

int(-(log(5*x + 2*x^2 + 4)^2*(4*x + 5*x^2 + 2*x^3) - log(5*x + 2*x^2 + 4)* 
(5*x - log(x^2)^2*(8*x + 10*x^2 + 4*x^3) + 2*x^2 + 4) - log(x^2)^2*(5*x + 
2*x^2 + 4) + log(x^2)^4*(4*x + 5*x^2 + 2*x^3) + log(x)*(5*x + 4*x^2) + log 
(x^2)*log(x)*(20*x + 8*x^2 + 16))/(log(5*x + 2*x^2 + 4)^2*(4*x + 5*x^2 + 2 
*x^3) + log(x^2)^4*(4*x + 5*x^2 + 2*x^3) + log(x^2)^2*log(5*x + 2*x^2 + 4) 
*(8*x + 10*x^2 + 4*x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {\left (-5 x-4 x^2\right ) \log (x)+\left (-16-20 x-8 x^2\right ) \log (x) \log \left (x^2\right )+\left (4+5 x+2 x^2\right ) \log ^2\left (x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^4\left (x^2\right )+\left (4+5 x+2 x^2+\left (-8 x-10 x^2-4 x^3\right ) \log ^2\left (x^2\right )\right ) \log \left (4+5 x+2 x^2\right )+\left (-4 x-5 x^2-2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )}{\left (4 x+5 x^2+2 x^3\right ) \log ^4\left (x^2\right )+\left (8 x+10 x^2+4 x^3\right ) \log ^2\left (x^2\right ) \log \left (4+5 x+2 x^2\right )+\left (4 x+5 x^2+2 x^3\right ) \log ^2\left (4+5 x+2 x^2\right )} \, dx=\frac {-\mathrm {log}\left (x^{2}\right )^{2} x -\mathrm {log}\left (2 x^{2}+5 x +4\right ) x +\mathrm {log}\left (x \right )}{\mathrm {log}\left (x^{2}\right )^{2}+\mathrm {log}\left (2 x^{2}+5 x +4\right )} \] Input:

int(((-2*x^3-5*x^2-4*x)*log(2*x^2+5*x+4)^2+((-4*x^3-10*x^2-8*x)*log(x^2)^2 
+2*x^2+5*x+4)*log(2*x^2+5*x+4)+(-2*x^3-5*x^2-4*x)*log(x^2)^4+(2*x^2+5*x+4) 
*log(x^2)^2+(-8*x^2-20*x-16)*log(x)*log(x^2)+(-4*x^2-5*x)*log(x))/((2*x^3+ 
5*x^2+4*x)*log(2*x^2+5*x+4)^2+(4*x^3+10*x^2+8*x)*log(x^2)^2*log(2*x^2+5*x+ 
4)+(2*x^3+5*x^2+4*x)*log(x^2)^4),x)
 

Output:

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