\(\int \frac {(160+160 x+40 x^2+e^x (-8 x^2-4 x^3)) \log (2 x) \log (x^2)+(40+40 x+10 x^2+e^x (-2 x^2-x^3)+e^x (-4 x^2-3 x^3-x^4) \log (2 x)) \log ^2(x^2)}{4 x+4 x^2+x^3} \, dx\) [1237]

Optimal result

Integrand size = 106, antiderivative size = 26 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=\left (10-\frac {e^x x^2}{2+x}\right ) \log (2 x) \log ^2\left (x^2\right ) \] Output:

ln(2*x)*ln(x^2)^2*(10-exp(x)*x^2/(2+x))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=-\frac {\left (-20-10 x+e^x x^2\right ) \log (2 x) \log ^2\left (x^2\right )}{2+x} \] Input:

Integrate[((160 + 160*x + 40*x^2 + E^x*(-8*x^2 - 4*x^3))*Log[2*x]*Log[x^2] 
 + (40 + 40*x + 10*x^2 + E^x*(-2*x^2 - x^3) + E^x*(-4*x^2 - 3*x^3 - x^4)*L 
og[2*x])*Log[x^2]^2)/(4*x + 4*x^2 + x^3),x]
 

Output:

-(((-20 - 10*x + E^x*x^2)*Log[2*x]*Log[x^2]^2)/(2 + 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.

\(\displaystyle \int \frac {\left (40 x^2+e^x \left (-4 x^3-8 x^2\right )+160 x+160\right ) \log (2 x) \log \left (x^2\right )+\left (10 x^2+e^x \left (-x^3-2 x^2\right )+e^x \left (-x^4-3 x^3-4 x^2\right ) \log (2 x)+40 x+40\right ) \log ^2\left (x^2\right )}{x^3+4 x^2+4 x} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (40 x^2+e^x \left (-4 x^3-8 x^2\right )+160 x+160\right ) \log (2 x) \log \left (x^2\right )+\left (10 x^2+e^x \left (-x^3-2 x^2\right )+e^x \left (-x^4-3 x^3-4 x^2\right ) \log (2 x)+40 x+40\right ) \log ^2\left (x^2\right )}{x \left (x^2+4 x+4\right )}dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {\left (40 x^2+e^x \left (-4 x^3-8 x^2\right )+160 x+160\right ) \log (2 x) \log \left (x^2\right )+\left (10 x^2+e^x \left (-x^3-2 x^2\right )+e^x \left (-x^4-3 x^3-4 x^2\right ) \log (2 x)+40 x+40\right ) \log ^2\left (x^2\right )}{x (x+2)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {10 x \log ^2\left (x^2\right )}{(x+2)^2}+\frac {40 \log ^2\left (x^2\right )}{x (x+2)^2}+\frac {40 \log ^2\left (x^2\right )}{(x+2)^2}+\frac {40 x \log (2 x) \log \left (x^2\right )}{(x+2)^2}+\frac {160 \log (2 x) \log \left (x^2\right )}{x (x+2)^2}+\frac {160 \log (2 x) \log \left (x^2\right )}{(x+2)^2}-\frac {e^x x \left (x^2 \log (2 x) \log \left (x^2\right )+3 x \log (2 x) \log \left (x^2\right )+x \log \left (x^2\right )+4 \log (2 x) \log \left (x^2\right )+2 \log \left (x^2\right )+4 x \log (2 x)+8 \log (2 x)\right ) \log \left (x^2\right )}{(x+2)^2}\right )dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 7299

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

Input:

Int[((160 + 160*x + 40*x^2 + E^x*(-8*x^2 - 4*x^3))*Log[2*x]*Log[x^2] + (40 
 + 40*x + 10*x^2 + E^x*(-2*x^2 - x^3) + E^x*(-4*x^2 - 3*x^3 - x^4)*Log[2*x 
])*Log[x^2]^2)/(4*x + 4*x^2 + x^3),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.14 (sec) , antiderivative size = 806, normalized size of antiderivative = 31.00

\[\text {Expression too large to display}\]

Input:

int((((-x^4-3*x^3-4*x^2)*exp(x)*ln(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+40*x+40 
)*ln(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*ln(2*x)*ln(x^2))/(x^3 
+4*x^2+4*x),x)
 

Output:

-4*(exp(x)*x^2-10*x-20)/(2+x)*ln(x)^3-2*(-20*I*Pi*x*csgn(I*x)*csgn(I*x^2)^ 
2-I*Pi*x^2*csgn(I*x^2)^3*exp(x)-40*I*Pi*csgn(I*x)*csgn(I*x^2)^2+10*I*Pi*x* 
csgn(I*x^2)^3+20*I*Pi*csgn(I*x^2)^3+20*I*Pi*csgn(I*x)^2*csgn(I*x^2)+2*I*Pi 
*x^2*csgn(I*x)*csgn(I*x^2)^2*exp(x)+10*I*Pi*x*csgn(I*x)^2*csgn(I*x^2)-I*Pi 
*x^2*csgn(I*x)^2*csgn(I*x^2)*exp(x)+2*x^2*ln(2)*exp(x)-20*x*ln(2)-40*ln(2) 
)/(2+x)*ln(x)^2+1/4*Pi*csgn(I*x^2)*(csgn(I*x^2)*Pi*csgn(I*x)^4-4*csgn(I*x^ 
2)^2*Pi*csgn(I*x)^3+6*csgn(I*x^2)^3*Pi*csgn(I*x)^2-4*csgn(I*x^2)^4*Pi*csgn 
(I*x)+csgn(I*x^2)^5*Pi+8*I*ln(2)*csgn(I*x^2)^2+8*I*ln(2)*csgn(I*x)^2-16*I* 
ln(2)*csgn(I*x)*csgn(I*x^2))*x^2/(2+x)*exp(x)*ln(x)+1/8*I*Pi*csgn(I*x^2)*( 
-160*ln(x)*ln(2)*csgn(I*x)^2*x-160*ln(x)*ln(2)*csgn(I*x^2)^2*x+640*ln(x)*l 
n(2)*csgn(I*x)*csgn(I*x^2)+20*I*ln(x)*Pi*x*csgn(I*x^2)^5+320*ln(x)*ln(2)*c 
sgn(I*x)*csgn(I*x^2)*x-80*I*ln(x)*Pi*x*csgn(I*x)*csgn(I*x^2)^4+40*I*ln(x)* 
Pi*csgn(I*x)^4*csgn(I*x^2)+240*I*ln(x)*Pi*csgn(I*x)^2*csgn(I*x^2)^3-160*I* 
ln(x)*Pi*csgn(I*x)^3*csgn(I*x^2)^2-320*ln(x)*ln(2)*csgn(I*x)^2-320*ln(x)*l 
n(2)*csgn(I*x^2)^2-2*I*Pi*ln(2)*x^2*csgn(I*x^2)^5*exp(x)-12*I*Pi*ln(2)*x^2 
*csgn(I*x)^2*csgn(I*x^2)^3*exp(x)+120*I*ln(x)*Pi*x*csgn(I*x)^2*csgn(I*x^2) 
^3+20*I*ln(x)*Pi*x*csgn(I*x)^4*csgn(I*x^2)+40*I*ln(x)*Pi*csgn(I*x^2)^5-2*I 
*Pi*ln(2)*x^2*csgn(I*x)^4*csgn(I*x^2)*exp(x)-160*I*ln(x)*Pi*csgn(I*x)*csgn 
(I*x^2)^4+8*I*Pi*ln(2)*x^2*csgn(I*x)*csgn(I*x^2)^4*exp(x)+8*I*Pi*ln(2)*x^2 
*csgn(I*x)^3*csgn(I*x^2)^2*exp(x)-80*I*ln(x)*Pi*x*csgn(I*x)^3*csgn(I*x^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (25) = 50\).

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=-\frac {4 \, {\left ({\left (x^{2} e^{x} - 10 \, x - 20\right )} \log \left (2 \, x\right )^{3} - 2 \, {\left (x^{2} e^{x} \log \left (2\right ) - 10 \, {\left (x + 2\right )} \log \left (2\right )\right )} \log \left (2 \, x\right )^{2} + {\left (x^{2} e^{x} \log \left (2\right )^{2} - 10 \, {\left (x + 2\right )} \log \left (2\right )^{2}\right )} \log \left (2 \, x\right )\right )}}{x + 2} \] Input:

integrate((((-x^4-3*x^3-4*x^2)*exp(x)*log(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+ 
40*x+40)*log(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*log(2*x)*log( 
x^2))/(x^3+4*x^2+4*x),x, algorithm="fricas")
 

Output:

-4*((x^2*e^x - 10*x - 20)*log(2*x)^3 - 2*(x^2*e^x*log(2) - 10*(x + 2)*log( 
2))*log(2*x)^2 + (x^2*e^x*log(2)^2 - 10*(x + 2)*log(2)^2)*log(2*x))/(x + 2 
)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (22) = 44\).

Time = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.00 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=40 \log {\left (2 \right )}^{2} \log {\left (x \right )} + 40 \log {\left (2 x \right )}^{3} - 80 \log {\left (2 \right )} \log {\left (2 x \right )}^{2} + \frac {\left (- 4 x^{2} \log {\left (2 x \right )}^{3} + 8 x^{2} \log {\left (2 \right )} \log {\left (2 x \right )}^{2} - 4 x^{2} \log {\left (2 \right )}^{2} \log {\left (2 x \right )}\right ) e^{x}}{x + 2} \] Input:

integrate((((-x**4-3*x**3-4*x**2)*exp(x)*ln(2*x)+(-x**3-2*x**2)*exp(x)+10* 
x**2+40*x+40)*ln(x**2)**2+((-4*x**3-8*x**2)*exp(x)+40*x**2+160*x+160)*ln(2 
*x)*ln(x**2))/(x**3+4*x**2+4*x),x)
 

Output:

40*log(2)**2*log(x) + 40*log(2*x)**3 - 80*log(2)*log(2*x)**2 + (-4*x**2*lo 
g(2*x)**3 + 8*x**2*log(2)*log(2*x)**2 - 4*x**2*log(2)**2*log(2*x))*exp(x)/ 
(x + 2)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (25) = 50\).

Time = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=\frac {4 \, {\left (10 \, {\left (x + 2\right )} \log \left (x\right )^{3} + 10 \, {\left (x \log \left (2\right ) + 2 \, \log \left (2\right )\right )} \log \left (x\right )^{2} - {\left (x^{2} \log \left (2\right ) \log \left (x\right )^{2} + x^{2} \log \left (x\right )^{3}\right )} e^{x}\right )}}{x + 2} \] Input:

integrate((((-x^4-3*x^3-4*x^2)*exp(x)*log(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+ 
40*x+40)*log(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*log(2*x)*log( 
x^2))/(x^3+4*x^2+4*x),x, algorithm="maxima")
 

Output:

4*(10*(x + 2)*log(x)^3 + 10*(x*log(2) + 2*log(2))*log(x)^2 - (x^2*log(2)*l 
og(x)^2 + x^2*log(x)^3)*e^x)/(x + 2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (25) = 50\).

Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=-\frac {4 \, {\left (x^{2} e^{x} \log \left (2\right ) \log \left (x\right )^{2} + x^{2} e^{x} \log \left (x\right )^{3} - 10 \, x \log \left (2\right ) \log \left (x\right )^{2} - 10 \, x \log \left (x\right )^{3} - 20 \, \log \left (2\right ) \log \left (x\right )^{2} - 20 \, \log \left (x\right )^{3}\right )}}{x + 2} \] Input:

integrate((((-x^4-3*x^3-4*x^2)*exp(x)*log(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+ 
40*x+40)*log(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*log(2*x)*log( 
x^2))/(x^3+4*x^2+4*x),x, algorithm="giac")
 

Output:

-4*(x^2*e^x*log(2)*log(x)^2 + x^2*e^x*log(x)^3 - 10*x*log(2)*log(x)^2 - 10 
*x*log(x)^3 - 20*log(2)*log(x)^2 - 20*log(x)^3)/(x + 2)
 

Mupad [B] (verification not implemented)

Time = 1.61 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.35 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=40\,\ln \left (x^2\right )\,{\ln \left (2\right )}^2-80\,{\ln \left (2\right )}^2\,\ln \left (x\right )-40\,\ln \left (2\right )\,{\ln \left (x\right )}^2+10\,{\ln \left (x^2\right )}^2\,\ln \left (x\right )+40\,\ln \left (x^2\right )\,\ln \left (2\right )\,\ln \left (x\right )-\frac {x^2\,{\ln \left (x^2\right )}^2\,{\mathrm {e}}^x\,\ln \left (2\right )}{x+2}-\frac {x^2\,{\ln \left (x^2\right )}^2\,{\mathrm {e}}^x\,\ln \left (x\right )}{x+2} \] Input:

int((log(x^2)^2*(40*x - exp(x)*(2*x^2 + x^3) + 10*x^2 - log(2*x)*exp(x)*(4 
*x^2 + 3*x^3 + x^4) + 40) + log(2*x)*log(x^2)*(160*x - exp(x)*(8*x^2 + 4*x 
^3) + 40*x^2 + 160))/(4*x + 4*x^2 + x^3),x)
 

Output:

40*log(x^2)*log(2)^2 - 80*log(2)^2*log(x) - 40*log(2)*log(x)^2 + 10*log(x^ 
2)^2*log(x) + 40*log(x^2)*log(2)*log(x) - (x^2*log(x^2)^2*exp(x)*log(2))/( 
x + 2) - (x^2*log(x^2)^2*exp(x)*log(x))/(x + 2)
 

Reduce [F]

\[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=\frac {-3 e^{x} \mathrm {log}\left (x^{2}\right )^{2} \mathrm {log}\left (2 x \right ) x^{2}+120 \left (\int \frac {\mathrm {log}\left (x^{2}\right ) \mathrm {log}\left (2 x \right )}{x}d x \right ) x +240 \left (\int \frac {\mathrm {log}\left (x^{2}\right ) \mathrm {log}\left (2 x \right )}{x}d x \right )+5 \mathrm {log}\left (x^{2}\right )^{3} x +10 \mathrm {log}\left (x^{2}\right )^{3}}{3 x +6} \] Input:

int((((-x^4-3*x^3-4*x^2)*exp(x)*log(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+40*x+4 
0)*log(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*log(2*x)*log(x^2))/ 
(x^3+4*x^2+4*x),x)
 

Output:

( - 3*e**x*log(x**2)**2*log(2*x)*x**2 + 120*int((log(x**2)*log(2*x))/x,x)* 
x + 240*int((log(x**2)*log(2*x))/x,x) + 5*log(x**2)**3*x + 10*log(x**2)**3 
)/(3*(x + 2))