\(\int \frac {-36-6 x-3 x^2+e^x (-9+3 x-3 x^2)+(12+e^x (3-3 x)) \log (x^2)}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} (4-8 x+4 x^2)+e^x (32-56 x+16 x^2+8 x^3)+(-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x (-64+48 x+16 x^2)) \log (x^2)+(64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)) \log ^2(x^2)} \, dx\) [1309]

Optimal result

Integrand size = 176, antiderivative size = 22 \[ \int \frac {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx=\frac {3 x}{4 \left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )} \] Output:

3/4*x/(-1+x+ln(x^2))/(4+x+exp(x))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx=\frac {3 x}{4 \left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )} \] Input:

Integrate[(-36 - 6*x - 3*x^2 + E^x*(-9 + 3*x - 3*x^2) + (12 + E^x*(3 - 3*x 
))*Log[x^2])/(64 - 96*x + 4*x^2 + 24*x^3 + 4*x^4 + E^(2*x)*(4 - 8*x + 4*x^ 
2) + E^x*(32 - 56*x + 16*x^2 + 8*x^3) + (-128 + 64*x + 56*x^2 + 8*x^3 + E^ 
(2*x)*(-8 + 8*x) + E^x*(-64 + 48*x + 16*x^2))*Log[x^2] + (64 + 4*E^(2*x) + 
 32*x + 4*x^2 + E^x*(32 + 8*x))*Log[x^2]^2),x]
 

Output:

(3*x)/(4*(4 + E^x + x)*(-1 + 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.

\(\displaystyle \int \frac {-3 x^2+e^x \left (-3 x^2+3 x-9\right )+\left (e^x (3-3 x)+12\right ) \log \left (x^2\right )-6 x-36}{4 x^4+24 x^3+4 x^2+e^{2 x} \left (4 x^2-8 x+4\right )+\left (4 x^2+32 x+4 e^{2 x}+e^x (8 x+32)+64\right ) \log ^2\left (x^2\right )+e^x \left (8 x^3+16 x^2-56 x+32\right )+\left (8 x^3+56 x^2+e^x \left (16 x^2+48 x-64\right )+64 x+e^{2 x} (8 x-8)-128\right ) \log \left (x^2\right )-96 x+64} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {3 \left (-x^2-e^x \left (x^2-x+3\right )-\left (e^x (x-1)-4\right ) \log \left (x^2\right )-2 x-12\right )}{4 \left (x+e^x+4\right )^2 \left (-\log \left (x^2\right )-x+1\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{4} \int -\frac {x^2+2 x+e^x \left (x^2-x+3\right )-\left (e^x (1-x)+4\right ) \log \left (x^2\right )+12}{\left (x+e^x+4\right )^2 \left (-x-\log \left (x^2\right )+1\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {3}{4} \int \frac {x^2+2 x+e^x \left (x^2-x+3\right )-\left (e^x (1-x)+4\right ) \log \left (x^2\right )+12}{\left (x+e^x+4\right )^2 \left (-x-\log \left (x^2\right )+1\right )^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(-36 - 6*x - 3*x^2 + E^x*(-9 + 3*x - 3*x^2) + (12 + E^x*(3 - 3*x))*Log 
[x^2])/(64 - 96*x + 4*x^2 + 24*x^3 + 4*x^4 + E^(2*x)*(4 - 8*x + 4*x^2) + E 
^x*(32 - 56*x + 16*x^2 + 8*x^3) + (-128 + 64*x + 56*x^2 + 8*x^3 + E^(2*x)* 
(-8 + 8*x) + E^x*(-64 + 48*x + 16*x^2))*Log[x^2] + (64 + 4*E^(2*x) + 32*x 
+ 4*x^2 + E^x*(32 + 8*x))*Log[x^2]^2),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(40\) vs. \(2(19)=38\).

Time = 1.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86

method	result	size
parallelrisch	\(\frac {3 x}{4 \left ({\mathrm e}^{x} \ln \left (x^{2}\right )+{\mathrm e}^{x} x +x \ln \left (x^{2}\right )+x^{2}-{\mathrm e}^{x}+4 \ln \left (x^{2}\right )+3 x -4\right )}\)	\(41\)
risch	\(\frac {3 x}{2 \left (4+x +{\mathrm e}^{x}\right ) \left (-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+2 i \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right ) \pi -i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 x +4 \ln \left (x \right )-2\right )}\)	\(71\)

Input:

int((((-3*x+3)*exp(x)+12)*ln(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-36)/((4* 
exp(x)^2+(8*x+32)*exp(x)+4*x^2+32*x+64)*ln(x^2)^2+((8*x-8)*exp(x)^2+(16*x^ 
2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*ln(x^2)+(4*x^2-8*x+4)*exp(x)^2+(8 
*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x,method=_RETURNVE 
RBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx=\frac {3 \, x}{4 \, {\left (x^{2} + {\left (x - 1\right )} e^{x} + {\left (x + e^{x} + 4\right )} \log \left (x^{2}\right ) + 3 \, x - 4\right )}} \] Input:

integrate((((-3*x+3)*exp(x)+12)*log(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-3 
6)/((4*exp(x)^2+(8*x+32)*exp(x)+4*x^2+32*x+64)*log(x^2)^2+((8*x-8)*exp(x)^ 
2+(16*x^2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*log(x^2)+(4*x^2-8*x+4)*ex 
p(x)^2+(8*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x, algori 
thm="fricas")
 

Output:

3/4*x/(x^2 + (x - 1)*e^x + (x + e^x + 4)*log(x^2) + 3*x - 4)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).

Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx=\frac {3 x}{4 x^{2} + 4 x \log {\left (x^{2} \right )} + 12 x + \left (4 x + 4 \log {\left (x^{2} \right )} - 4\right ) e^{x} + 16 \log {\left (x^{2} \right )} - 16} \] Input:

integrate((((-3*x+3)*exp(x)+12)*ln(x**2)+(-3*x**2+3*x-9)*exp(x)-3*x**2-6*x 
-36)/((4*exp(x)**2+(8*x+32)*exp(x)+4*x**2+32*x+64)*ln(x**2)**2+((8*x-8)*ex 
p(x)**2+(16*x**2+48*x-64)*exp(x)+8*x**3+56*x**2+64*x-128)*ln(x**2)+(4*x**2 
-8*x+4)*exp(x)**2+(8*x**3+16*x**2-56*x+32)*exp(x)+4*x**4+24*x**3+4*x**2-96 
*x+64),x)
 

Output:

3*x/(4*x**2 + 4*x*log(x**2) + 12*x + (4*x + 4*log(x**2) - 4)*exp(x) + 16*l 
og(x**2) - 16)
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx=\frac {3 \, x}{4 \, {\left (x^{2} + {\left (x + 2 \, \log \left (x\right ) - 1\right )} e^{x} + 2 \, {\left (x + 4\right )} \log \left (x\right ) + 3 \, x - 4\right )}} \] Input:

integrate((((-3*x+3)*exp(x)+12)*log(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-3 
6)/((4*exp(x)^2+(8*x+32)*exp(x)+4*x^2+32*x+64)*log(x^2)^2+((8*x-8)*exp(x)^ 
2+(16*x^2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*log(x^2)+(4*x^2-8*x+4)*ex 
p(x)^2+(8*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x, algori 
thm="maxima")
 

Output:

3/4*x/(x^2 + (x + 2*log(x) - 1)*e^x + 2*(x + 4)*log(x) + 3*x - 4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (19) = 38\).

Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx=\frac {3 \, x}{4 \, {\left (x^{2} + x e^{x} + x \log \left (x^{2}\right ) + e^{x} \log \left (x^{2}\right ) + 3 \, x - e^{x} + 4 \, \log \left (x^{2}\right ) - 4\right )}} \] Input:

integrate((((-3*x+3)*exp(x)+12)*log(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-3 
6)/((4*exp(x)^2+(8*x+32)*exp(x)+4*x^2+32*x+64)*log(x^2)^2+((8*x-8)*exp(x)^ 
2+(16*x^2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*log(x^2)+(4*x^2-8*x+4)*ex 
p(x)^2+(8*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x, algori 
thm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx=\int -\frac {6\,x+\ln \left (x^2\right )\,\left ({\mathrm {e}}^x\,\left (3\,x-3\right )-12\right )+{\mathrm {e}}^x\,\left (3\,x^2-3\,x+9\right )+3\,x^2+36}{{\mathrm {e}}^{2\,x}\,\left (4\,x^2-8\,x+4\right )-96\,x+{\ln \left (x^2\right )}^2\,\left (32\,x+4\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (8\,x+32\right )+4\,x^2+64\right )+\ln \left (x^2\right )\,\left (64\,x+{\mathrm {e}}^x\,\left (16\,x^2+48\,x-64\right )+{\mathrm {e}}^{2\,x}\,\left (8\,x-8\right )+56\,x^2+8\,x^3-128\right )+4\,x^2+24\,x^3+4\,x^4+{\mathrm {e}}^x\,\left (8\,x^3+16\,x^2-56\,x+32\right )+64} \,d x \] Input:

int(-(6*x + log(x^2)*(exp(x)*(3*x - 3) - 12) + exp(x)*(3*x^2 - 3*x + 9) + 
3*x^2 + 36)/(exp(2*x)*(4*x^2 - 8*x + 4) - 96*x + log(x^2)^2*(32*x + 4*exp( 
2*x) + exp(x)*(8*x + 32) + 4*x^2 + 64) + log(x^2)*(64*x + exp(x)*(48*x + 1 
6*x^2 - 64) + exp(2*x)*(8*x - 8) + 56*x^2 + 8*x^3 - 128) + 4*x^2 + 24*x^3 
+ 4*x^4 + exp(x)*(16*x^2 - 56*x + 8*x^3 + 32) + 64),x)
 

Output:

int(-(6*x + log(x^2)*(exp(x)*(3*x - 3) - 12) + exp(x)*(3*x^2 - 3*x + 9) + 
3*x^2 + 36)/(exp(2*x)*(4*x^2 - 8*x + 4) - 96*x + log(x^2)^2*(32*x + 4*exp( 
2*x) + exp(x)*(8*x + 32) + 4*x^2 + 64) + log(x^2)*(64*x + exp(x)*(48*x + 1 
6*x^2 - 64) + exp(2*x)*(8*x - 8) + 56*x^2 + 8*x^3 - 128) + 4*x^2 + 24*x^3 
+ 4*x^4 + exp(x)*(16*x^2 - 56*x + 8*x^3 + 32) + 64), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.18 \[ \int \frac {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx=\frac {3 x}{4 e^{x} \mathrm {log}\left (x^{2}\right )+4 e^{x} x -4 e^{x}+4 \,\mathrm {log}\left (x^{2}\right ) x +16 \,\mathrm {log}\left (x^{2}\right )+4 x^{2}+12 x -16} \] Input:

int((((-3*x+3)*exp(x)+12)*log(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-36)/((4 
*exp(x)^2+(8*x+32)*exp(x)+4*x^2+32*x+64)*log(x^2)^2+((8*x-8)*exp(x)^2+(16* 
x^2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*log(x^2)+(4*x^2-8*x+4)*exp(x)^2 
+(8*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x)
 

Output:

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