\(\int \frac {32-48 x-8 x^2+e^5 (8-12 x-2 x^2)+(16+e^5 (4-x)-4 x) \log (x)+(-8 x^2-2 e^5 x^2) \log (x^2)}{576+432 x+108 x^2+9 x^3+(384+288 x+72 x^2+6 x^3) \log (x)+(64+48 x+12 x^2+x^3) \log ^2(x)+(384 x+288 x^2+72 x^3+6 x^4+(128 x+96 x^2+24 x^3+2 x^4) \log (x)) \log (x^2)+(64 x^2+48 x^3+12 x^4+x^5) \log ^2(x^2)} \, dx\) [692]

Optimal result

Integrand size = 186, antiderivative size = 26 \[ \int \frac {32-48 x-8 x^2+e^5 \left (8-12 x-2 x^2\right )+\left (16+e^5 (4-x)-4 x\right ) \log (x)+\left (-8 x^2-2 e^5 x^2\right ) \log \left (x^2\right )}{576+432 x+108 x^2+9 x^3+\left (384+288 x+72 x^2+6 x^3\right ) \log (x)+\left (64+48 x+12 x^2+x^3\right ) \log ^2(x)+\left (384 x+288 x^2+72 x^3+6 x^4+\left (128 x+96 x^2+24 x^3+2 x^4\right ) \log (x)\right ) \log \left (x^2\right )+\left (64 x^2+48 x^3+12 x^4+x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {\left (4+e^5\right ) x}{(-4-x)^2 \left (3+\log (x)+x \log \left (x^2\right )\right )} \] Output:

x/(x*ln(x^2)+ln(x)+3)*(4+exp(5))/(-4-x)^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {32-48 x-8 x^2+e^5 \left (8-12 x-2 x^2\right )+\left (16+e^5 (4-x)-4 x\right ) \log (x)+\left (-8 x^2-2 e^5 x^2\right ) \log \left (x^2\right )}{576+432 x+108 x^2+9 x^3+\left (384+288 x+72 x^2+6 x^3\right ) \log (x)+\left (64+48 x+12 x^2+x^3\right ) \log ^2(x)+\left (384 x+288 x^2+72 x^3+6 x^4+\left (128 x+96 x^2+24 x^3+2 x^4\right ) \log (x)\right ) \log \left (x^2\right )+\left (64 x^2+48 x^3+12 x^4+x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {\left (4+e^5\right ) x}{(4+x)^2 \left (3+\log (x)+x \log \left (x^2\right )\right )} \] Input:

Integrate[(32 - 48*x - 8*x^2 + E^5*(8 - 12*x - 2*x^2) + (16 + E^5*(4 - x) 
- 4*x)*Log[x] + (-8*x^2 - 2*E^5*x^2)*Log[x^2])/(576 + 432*x + 108*x^2 + 9* 
x^3 + (384 + 288*x + 72*x^2 + 6*x^3)*Log[x] + (64 + 48*x + 12*x^2 + x^3)*L 
og[x]^2 + (384*x + 288*x^2 + 72*x^3 + 6*x^4 + (128*x + 96*x^2 + 24*x^3 + 2 
*x^4)*Log[x])*Log[x^2] + (64*x^2 + 48*x^3 + 12*x^4 + x^5)*Log[x^2]^2),x]
 

Output:

((4 + E^5)*x)/((4 + x)^2*(3 + Log[x] + x*Log[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.

\(\displaystyle \int \frac {-8 x^2+e^5 \left (-2 x^2-12 x+8\right )+\left (-2 e^5 x^2-8 x^2\right ) \log \left (x^2\right )-48 x+\left (e^5 (4-x)-4 x+16\right ) \log (x)+32}{9 x^3+108 x^2+\left (x^3+12 x^2+48 x+64\right ) \log ^2(x)+\left (6 x^3+72 x^2+288 x+384\right ) \log (x)+\left (6 x^4+72 x^3+288 x^2+\left (2 x^4+24 x^3+96 x^2+128 x\right ) \log (x)+384 x\right ) \log \left (x^2\right )+\left (x^5+12 x^4+48 x^3+64 x^2\right ) \log ^2\left (x^2\right )+432 x+576} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (4+e^5\right ) \left (-2 \left (x^2+x^2 \log \left (x^2\right )+6 x-4\right )-((x-4) \log (x))\right )}{(x+4)^3 \left (x \log \left (x^2\right )+\log (x)+3\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \left (4+e^5\right ) \int \frac {(4-x) \log (x)+2 \left (-\log \left (x^2\right ) x^2-x^2-6 x+4\right )}{(x+4)^3 \left (\log (x)+x \log \left (x^2\right )+3\right )^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \left (4+e^5\right ) \left (10 \int \frac {1}{(x+4)^2 \left (\log (x)+x \log \left (x^2\right )+3\right )^2}dx-2 \int \frac {1}{(x+4) \left (\log (x)+x \log \left (x^2\right )+3\right )^2}dx+\int \frac {\log (x)}{(x+4)^2 \left (\log (x)+x \log \left (x^2\right )+3\right )^2}dx+8 \int \frac {1}{(x+4)^3 \left (\log (x)+x \log \left (x^2\right )+3\right )}dx-2 \int \frac {1}{(x+4)^2 \left (\log (x)+x \log \left (x^2\right )+3\right )}dx\right )\)

Input:

Int[(32 - 48*x - 8*x^2 + E^5*(8 - 12*x - 2*x^2) + (16 + E^5*(4 - x) - 4*x) 
*Log[x] + (-8*x^2 - 2*E^5*x^2)*Log[x^2])/(576 + 432*x + 108*x^2 + 9*x^3 + 
(384 + 288*x + 72*x^2 + 6*x^3)*Log[x] + (64 + 48*x + 12*x^2 + x^3)*Log[x]^ 
2 + (384*x + 288*x^2 + 72*x^3 + 6*x^4 + (128*x + 96*x^2 + 24*x^3 + 2*x^4)* 
Log[x])*Log[x^2] + (64*x^2 + 48*x^3 + 12*x^4 + x^5)*Log[x^2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.50 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42

method	result	size
default	\(-\frac {\left (-{\mathrm e}^{5}-4\right ) x}{\left (4+x \right )^{2} \left (2 x \ln \left (x \right )+x \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )+3+\ln \left (x \right )\right )}\)	\(37\)
parallelrisch	\(\frac {4 x +x \,{\mathrm e}^{5}}{x^{3} \ln \left (x^{2}\right )+x^{2} \ln \left (x \right )+8 x^{2} \ln \left (x^{2}\right )+3 x^{2}+8 x \ln \left (x \right )+16 x \ln \left (x^{2}\right )+24 x +16 \ln \left (x \right )+48}\)	\(61\)
risch	\(\frac {2 i x \left (4+{\mathrm e}^{5}\right )}{\left (x^{2}+8 x +16\right ) \left (\pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi x \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i x \ln \left (x \right )+2 i \ln \left (x \right )+6 i\right )}\)	\(82\)

Input:

int(((-2*x^2*exp(5)-8*x^2)*ln(x^2)+((-x+4)*exp(5)-4*x+16)*ln(x)+(-2*x^2-12 
*x+8)*exp(5)-8*x^2-48*x+32)/((x^5+12*x^4+48*x^3+64*x^2)*ln(x^2)^2+((2*x^4+ 
24*x^3+96*x^2+128*x)*ln(x)+6*x^4+72*x^3+288*x^2+384*x)*ln(x^2)+(x^3+12*x^2 
+48*x+64)*ln(x)^2+(6*x^3+72*x^2+288*x+384)*ln(x)+9*x^3+108*x^2+432*x+576), 
x,method=_RETURNVERBOSE)
 

Output:

-(-exp(5)-4)*x/(4+x)^2/(2*x*ln(x)+x*(ln(x^2)-2*ln(x))+3+ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {32-48 x-8 x^2+e^5 \left (8-12 x-2 x^2\right )+\left (16+e^5 (4-x)-4 x\right ) \log (x)+\left (-8 x^2-2 e^5 x^2\right ) \log \left (x^2\right )}{576+432 x+108 x^2+9 x^3+\left (384+288 x+72 x^2+6 x^3\right ) \log (x)+\left (64+48 x+12 x^2+x^3\right ) \log ^2(x)+\left (384 x+288 x^2+72 x^3+6 x^4+\left (128 x+96 x^2+24 x^3+2 x^4\right ) \log (x)\right ) \log \left (x^2\right )+\left (64 x^2+48 x^3+12 x^4+x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {x e^{5} + 4 \, x}{3 \, x^{2} + {\left (2 \, x^{3} + 17 \, x^{2} + 40 \, x + 16\right )} \log \left (x\right ) + 24 \, x + 48} \] Input:

integrate(((-2*x^2*exp(5)-8*x^2)*log(x^2)+((-x+4)*exp(5)-4*x+16)*log(x)+(- 
2*x^2-12*x+8)*exp(5)-8*x^2-48*x+32)/((x^5+12*x^4+48*x^3+64*x^2)*log(x^2)^2 
+((2*x^4+24*x^3+96*x^2+128*x)*log(x)+6*x^4+72*x^3+288*x^2+384*x)*log(x^2)+ 
(x^3+12*x^2+48*x+64)*log(x)^2+(6*x^3+72*x^2+288*x+384)*log(x)+9*x^3+108*x^ 
2+432*x+576),x, algorithm="fricas")
 

Output:

(x*e^5 + 4*x)/(3*x^2 + (2*x^3 + 17*x^2 + 40*x + 16)*log(x) + 24*x + 48)
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {32-48 x-8 x^2+e^5 \left (8-12 x-2 x^2\right )+\left (16+e^5 (4-x)-4 x\right ) \log (x)+\left (-8 x^2-2 e^5 x^2\right ) \log \left (x^2\right )}{576+432 x+108 x^2+9 x^3+\left (384+288 x+72 x^2+6 x^3\right ) \log (x)+\left (64+48 x+12 x^2+x^3\right ) \log ^2(x)+\left (384 x+288 x^2+72 x^3+6 x^4+\left (128 x+96 x^2+24 x^3+2 x^4\right ) \log (x)\right ) \log \left (x^2\right )+\left (64 x^2+48 x^3+12 x^4+x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {4 x + x e^{5}}{3 x^{2} + 24 x + \left (2 x^{3} + 17 x^{2} + 40 x + 16\right ) \log {\left (x \right )} + 48} \] Input:

integrate(((-2*x**2*exp(5)-8*x**2)*ln(x**2)+((-x+4)*exp(5)-4*x+16)*ln(x)+( 
-2*x**2-12*x+8)*exp(5)-8*x**2-48*x+32)/((x**5+12*x**4+48*x**3+64*x**2)*ln( 
x**2)**2+((2*x**4+24*x**3+96*x**2+128*x)*ln(x)+6*x**4+72*x**3+288*x**2+384 
*x)*ln(x**2)+(x**3+12*x**2+48*x+64)*ln(x)**2+(6*x**3+72*x**2+288*x+384)*ln 
(x)+9*x**3+108*x**2+432*x+576),x)
 

Output:

(4*x + x*exp(5))/(3*x**2 + 24*x + (2*x**3 + 17*x**2 + 40*x + 16)*log(x) + 
48)
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {32-48 x-8 x^2+e^5 \left (8-12 x-2 x^2\right )+\left (16+e^5 (4-x)-4 x\right ) \log (x)+\left (-8 x^2-2 e^5 x^2\right ) \log \left (x^2\right )}{576+432 x+108 x^2+9 x^3+\left (384+288 x+72 x^2+6 x^3\right ) \log (x)+\left (64+48 x+12 x^2+x^3\right ) \log ^2(x)+\left (384 x+288 x^2+72 x^3+6 x^4+\left (128 x+96 x^2+24 x^3+2 x^4\right ) \log (x)\right ) \log \left (x^2\right )+\left (64 x^2+48 x^3+12 x^4+x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {x {\left (e^{5} + 4\right )}}{3 \, x^{2} + {\left (2 \, x^{3} + 17 \, x^{2} + 40 \, x + 16\right )} \log \left (x\right ) + 24 \, x + 48} \] Input:

integrate(((-2*x^2*exp(5)-8*x^2)*log(x^2)+((-x+4)*exp(5)-4*x+16)*log(x)+(- 
2*x^2-12*x+8)*exp(5)-8*x^2-48*x+32)/((x^5+12*x^4+48*x^3+64*x^2)*log(x^2)^2 
+((2*x^4+24*x^3+96*x^2+128*x)*log(x)+6*x^4+72*x^3+288*x^2+384*x)*log(x^2)+ 
(x^3+12*x^2+48*x+64)*log(x)^2+(6*x^3+72*x^2+288*x+384)*log(x)+9*x^3+108*x^ 
2+432*x+576),x, algorithm="maxima")
 

Output:

x*(e^5 + 4)/(3*x^2 + (2*x^3 + 17*x^2 + 40*x + 16)*log(x) + 24*x + 48)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {32-48 x-8 x^2+e^5 \left (8-12 x-2 x^2\right )+\left (16+e^5 (4-x)-4 x\right ) \log (x)+\left (-8 x^2-2 e^5 x^2\right ) \log \left (x^2\right )}{576+432 x+108 x^2+9 x^3+\left (384+288 x+72 x^2+6 x^3\right ) \log (x)+\left (64+48 x+12 x^2+x^3\right ) \log ^2(x)+\left (384 x+288 x^2+72 x^3+6 x^4+\left (128 x+96 x^2+24 x^3+2 x^4\right ) \log (x)\right ) \log \left (x^2\right )+\left (64 x^2+48 x^3+12 x^4+x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {x e^{5} + 4 \, x}{2 \, x^{3} \log \left (x\right ) + 17 \, x^{2} \log \left (x\right ) + 3 \, x^{2} + 40 \, x \log \left (x\right ) + 24 \, x + 16 \, \log \left (x\right ) + 48} \] Input:

integrate(((-2*x^2*exp(5)-8*x^2)*log(x^2)+((-x+4)*exp(5)-4*x+16)*log(x)+(- 
2*x^2-12*x+8)*exp(5)-8*x^2-48*x+32)/((x^5+12*x^4+48*x^3+64*x^2)*log(x^2)^2 
+((2*x^4+24*x^3+96*x^2+128*x)*log(x)+6*x^4+72*x^3+288*x^2+384*x)*log(x^2)+ 
(x^3+12*x^2+48*x+64)*log(x)^2+(6*x^3+72*x^2+288*x+384)*log(x)+9*x^3+108*x^ 
2+432*x+576),x, algorithm="giac")
 

Output:

(x*e^5 + 4*x)/(2*x^3*log(x) + 17*x^2*log(x) + 3*x^2 + 40*x*log(x) + 24*x + 
 16*log(x) + 48)
 

Mupad [B] (verification not implemented)

Time = 3.18 (sec) , antiderivative size = 147, normalized size of antiderivative = 5.65 \[ \int \frac {32-48 x-8 x^2+e^5 \left (8-12 x-2 x^2\right )+\left (16+e^5 (4-x)-4 x\right ) \log (x)+\left (-8 x^2-2 e^5 x^2\right ) \log \left (x^2\right )}{576+432 x+108 x^2+9 x^3+\left (384+288 x+72 x^2+6 x^3\right ) \log (x)+\left (64+48 x+12 x^2+x^3\right ) \log ^2(x)+\left (384 x+288 x^2+72 x^3+6 x^4+\left (128 x+96 x^2+24 x^3+2 x^4\right ) \log (x)\right ) \log \left (x^2\right )+\left (64 x^2+48 x^3+12 x^4+x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {\ln \left (x^2\right )\,\left (\left ({\mathrm {e}}^5+4\right )\,x^4+\left (4\,{\mathrm {e}}^5+16\right )\,x^3\right )-\ln \left (x\right )\,\left (\left (2\,{\mathrm {e}}^5+8\right )\,x^4+\left (8\,{\mathrm {e}}^5+32\right )\,x^3\right )+x^2\,\left (4\,{\mathrm {e}}^5+16\right )+x^5\,\left (4\,{\mathrm {e}}^5+16\right )-x^3\,\left (7\,{\mathrm {e}}^5+28\right )+x^4\,\left (14\,{\mathrm {e}}^5+56\right )}{{\left (x+4\right )}^3\,\left (\ln \left (x\right )\,\left (2\,x+1\right )+x\,\left (\ln \left (x^2\right )-2\,\ln \left (x\right )\right )+3\right )\,\left (x+x^2\,\left (\ln \left (x^2\right )-2\,\ln \left (x\right )\right )-2\,x^2+4\,x^3\right )} \] Input:

int(-(48*x + log(x)*(4*x + exp(5)*(x - 4) - 16) + exp(5)*(12*x + 2*x^2 - 8 
) + log(x^2)*(2*x^2*exp(5) + 8*x^2) + 8*x^2 - 32)/(432*x + log(x^2)*(384*x 
 + log(x)*(128*x + 96*x^2 + 24*x^3 + 2*x^4) + 288*x^2 + 72*x^3 + 6*x^4) + 
108*x^2 + 9*x^3 + log(x^2)^2*(64*x^2 + 48*x^3 + 12*x^4 + x^5) + log(x)*(28 
8*x + 72*x^2 + 6*x^3 + 384) + log(x)^2*(48*x + 12*x^2 + x^3 + 64) + 576),x 
)
 

Output:

(log(x^2)*(x^3*(4*exp(5) + 16) + x^4*(exp(5) + 4)) - log(x)*(x^4*(2*exp(5) 
 + 8) + x^3*(8*exp(5) + 32)) + x^2*(4*exp(5) + 16) + x^5*(4*exp(5) + 16) - 
 x^3*(7*exp(5) + 28) + x^4*(14*exp(5) + 56))/((x + 4)^3*(log(x)*(2*x + 1) 
+ x*(log(x^2) - 2*log(x)) + 3)*(x + x^2*(log(x^2) - 2*log(x)) - 2*x^2 + 4* 
x^3))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.23 \[ \int \frac {32-48 x-8 x^2+e^5 \left (8-12 x-2 x^2\right )+\left (16+e^5 (4-x)-4 x\right ) \log (x)+\left (-8 x^2-2 e^5 x^2\right ) \log \left (x^2\right )}{576+432 x+108 x^2+9 x^3+\left (384+288 x+72 x^2+6 x^3\right ) \log (x)+\left (64+48 x+12 x^2+x^3\right ) \log ^2(x)+\left (384 x+288 x^2+72 x^3+6 x^4+\left (128 x+96 x^2+24 x^3+2 x^4\right ) \log (x)\right ) \log \left (x^2\right )+\left (64 x^2+48 x^3+12 x^4+x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {x \left (e^{5}+4\right )}{\mathrm {log}\left (x^{2}\right ) x^{3}+8 \,\mathrm {log}\left (x^{2}\right ) x^{2}+16 \,\mathrm {log}\left (x^{2}\right ) x +\mathrm {log}\left (x \right ) x^{2}+8 \,\mathrm {log}\left (x \right ) x +16 \,\mathrm {log}\left (x \right )+3 x^{2}+24 x +48} \] Input:

int(((-2*x^2*exp(5)-8*x^2)*log(x^2)+((-x+4)*exp(5)-4*x+16)*log(x)+(-2*x^2- 
12*x+8)*exp(5)-8*x^2-48*x+32)/((x^5+12*x^4+48*x^3+64*x^2)*log(x^2)^2+((2*x 
^4+24*x^3+96*x^2+128*x)*log(x)+6*x^4+72*x^3+288*x^2+384*x)*log(x^2)+(x^3+1 
2*x^2+48*x+64)*log(x)^2+(6*x^3+72*x^2+288*x+384)*log(x)+9*x^3+108*x^2+432* 
x+576),x)
 

Output:

(x*(e**5 + 4))/(log(x**2)*x**3 + 8*log(x**2)*x**2 + 16*log(x**2)*x + log(x 
)*x**2 + 8*log(x)*x + 16*log(x) + 3*x**2 + 24*x + 48)