\(\int \frac {(-20 x+e^{3+2 x} (16 x+16 x^2+4 x^3)) \log (x) \log (\frac {-5 x+e^{3+2 x} (2+x)}{2+x})+(20 x+10 x^2+e^{3+2 x} (-8-8 x-2 x^2)+(-10 x-5 x^2+e^{3+2 x} (4+4 x+x^2)) \log (x)) \log ^2(\frac {-5 x+e^{3+2 x} (2+x)}{2+x})}{(-10 x-5 x^2+e^{3+2 x} (4+4 x+x^2)) \log ^3(x)} \, dx\) [1164]

Optimal result

Integrand size = 164, antiderivative size = 25 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \] Output:

x*ln(exp(3+2*x)-5*x/(2+x))^2/ln(x)^2
 

Mathematica [F]

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

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

Output:

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

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {20 x \left (x^2+2 x-1\right ) \log \left (e^{2 x+3}-\frac {5 x}{x+2}\right )}{(x+2) \left (e^{2 x+3} x-5 x+2 e^{2 x+3}\right ) \log ^2(x)}+\frac {\left (4 x \log (x)+\log \left (e^{2 x+3}-\frac {5 x}{x+2}\right ) \log (x)-2 \log \left (e^{2 x+3}-\frac {5 x}{x+2}\right )\right ) \log \left (e^{2 x+3}-\frac {5 x}{x+2}\right )}{\log ^3(x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 20 \int \frac {x^2 \log \left (e^{2 x+3}-\frac {5 x}{x+2}\right )}{\left (e^{2 x+3} x-5 x+2 e^{2 x+3}\right ) \log ^2(x)}dx+4 \int \frac {x \log \left (e^{2 x+3}-\frac {5 x}{x+2}\right )}{\log ^2(x)}dx-20 \int \frac {\log \left (e^{2 x+3}-\frac {5 x}{x+2}\right )}{\left (e^{2 x+3} x-5 x+2 e^{2 x+3}\right ) \log ^2(x)}dx+40 \int \frac {\log \left (e^{2 x+3}-\frac {5 x}{x+2}\right )}{(x+2) \left (e^{2 x+3} x-5 x+2 e^{2 x+3}\right ) \log ^2(x)}dx+\int \frac {\log ^2\left (e^{2 x+3}-\frac {5 x}{x+2}\right )}{\log ^2(x)}dx-2 \int \frac {\log ^2\left (e^{2 x+3}-\frac {5 x}{x+2}\right )}{\log ^3(x)}dx\)

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 41.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(\frac {\ln \left (\frac {\left (2+x \right ) {\mathrm e}^{3+2 x}-5 x}{2+x}\right )^{2} x}{\ln \left (x \right )^{2}}\)	\(30\)
risch	\(\text {Expression too large to display}\)	\(1011\)

Input:

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

Output:

ln(((2+x)*exp(3+2*x)-5*x)/(2+x))^2*x/ln(x)^2
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x \log \left (\frac {{\left (x + 2\right )} e^{\left (2 \, x + 3\right )} - 5 \, x}{x + 2}\right )^{2}}{\log \left (x\right )^{2}} \] Input:

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

Output:

x*log(((x + 2)*e^(2*x + 3) - 5*x)/(x + 2))^2/log(x)^2
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x \log \left ({\left (x e^{3} + 2 \, e^{3}\right )} e^{\left (2 \, x\right )} - 5 \, x\right )^{2} - 2 \, x \log \left ({\left (x e^{3} + 2 \, e^{3}\right )} e^{\left (2 \, x\right )} - 5 \, x\right ) \log \left (x + 2\right ) + x \log \left (x + 2\right )^{2}}{\log \left (x\right )^{2}} \] Input:

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

Output:

(x*log((x*e^3 + 2*e^3)*e^(2*x) - 5*x)^2 - 2*x*log((x*e^3 + 2*e^3)*e^(2*x) 
- 5*x)*log(x + 2) + x*log(x + 2)^2)/log(x)^2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (24) = 48\).

Time = 0.63 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.68 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x \log \left (x e^{\left (2 \, x + 3\right )} - 5 \, x + 2 \, e^{\left (2 \, x + 3\right )}\right )^{2} - 2 \, x \log \left (x e^{\left (2 \, x + 3\right )} - 5 \, x + 2 \, e^{\left (2 \, x + 3\right )}\right ) \log \left (x + 2\right ) + x \log \left (x + 2\right )^{2}}{\log \left (x\right )^{2}} \] Input:

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

Output:

(x*log(x*e^(2*x + 3) - 5*x + 2*e^(2*x + 3))^2 - 2*x*log(x*e^(2*x + 3) - 5* 
x + 2*e^(2*x + 3))*log(x + 2) + x*log(x + 2)^2)/log(x)^2
 

Mupad [B] (verification not implemented)

Time = 2.88 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x\,{\ln \left (-\frac {5\,x-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3\,\left (x+2\right )}{x+2}\right )}^2}{{\ln \left (x\right )}^2} \] Input:

int(-(log(-(5*x - exp(2*x + 3)*(x + 2))/(x + 2))^2*(20*x - exp(2*x + 3)*(8 
*x + 2*x^2 + 8) - log(x)*(10*x - exp(2*x + 3)*(4*x + x^2 + 4) + 5*x^2) + 1 
0*x^2) - log(-(5*x - exp(2*x + 3)*(x + 2))/(x + 2))*log(x)*(20*x - exp(2*x 
 + 3)*(16*x + 16*x^2 + 4*x^3)))/(log(x)^3*(10*x - exp(2*x + 3)*(4*x + x^2 
+ 4) + 5*x^2)),x)
 

Output:

(x*log(-(5*x - exp(2*x)*exp(3)*(x + 2))/(x + 2))^2)/log(x)^2
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {\mathrm {log}\left (\frac {e^{2 x} e^{3} x +2 e^{2 x} e^{3}-5 x}{x +2}\right )^{2} x}{\mathrm {log}\left (x \right )^{2}} \] Input:

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

Output:

(log((e**(2*x)*e**3*x + 2*e**(2*x)*e**3 - 5*x)/(x + 2))**2*x)/log(x)**2