\(\int \frac {-24+e^x (6-3 x+3 x^3)+(12-12 x^2+e^x (-3+3 x^2)) \log (\frac {x^2}{4-4 x^2+e^x (-1+x^2)})+(-12+12 x^2+e^x (3-3 x^2)) \log ^2(\frac {x^2}{4-4 x^2+e^x (-1+x^2)})}{(4-4 x^2+e^x (-1+x^2)) \log ^2(\frac {x^2}{4-4 x^2+e^x (-1+x^2)})} \, dx\) [2883]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 146, antiderivative size = 27 \[ \int \frac {-24+e^x \left (6-3 x+3 x^3\right )+\left (12-12 x^2+e^x \left (-3+3 x^2\right )\right ) \log \left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )+\left (-12+12 x^2+e^x \left (3-3 x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )}{\left (4-4 x^2+e^x \left (-1+x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )} \, dx=x \left (-3+\frac {3}{\log \left (\frac {x}{\left (-4+e^x\right ) \left (-\frac {1}{x}+x\right )}\right )}\right ) \] Output:

x*(3/ln(x/(exp(x)-4)/(x-1/x))-3)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-24+e^x \left (6-3 x+3 x^3\right )+\left (12-12 x^2+e^x \left (-3+3 x^2\right )\right ) \log \left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )+\left (-12+12 x^2+e^x \left (3-3 x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )}{\left (4-4 x^2+e^x \left (-1+x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )} \, dx=3 \left (-x+\frac {x}{\log \left (\frac {x^2}{\left (-4+e^x\right ) \left (-1+x^2\right )}\right )}\right ) \] Input:

Integrate[(-24 + E^x*(6 - 3*x + 3*x^3) + (12 - 12*x^2 + E^x*(-3 + 3*x^2))* 
Log[x^2/(4 - 4*x^2 + E^x*(-1 + x^2))] + (-12 + 12*x^2 + E^x*(3 - 3*x^2))*L 
og[x^2/(4 - 4*x^2 + E^x*(-1 + x^2))]^2)/((4 - 4*x^2 + E^x*(-1 + x^2))*Log[ 
x^2/(4 - 4*x^2 + E^x*(-1 + x^2))]^2),x]
 

Output:

3*(-x + x/Log[x^2/((-4 + E^x)*(-1 + 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.

\(\displaystyle \int \frac {e^x \left (3 x^3-3 x+6\right )+\left (12 x^2+e^x \left (3-3 x^2\right )-12\right ) \log ^2\left (\frac {x^2}{-4 x^2+e^x \left (x^2-1\right )+4}\right )+\left (-12 x^2+e^x \left (3 x^2-3\right )+12\right ) \log \left (\frac {x^2}{-4 x^2+e^x \left (x^2-1\right )+4}\right )-24}{\left (-4 x^2+e^x \left (x^2-1\right )+4\right ) \log ^2\left (\frac {x^2}{-4 x^2+e^x \left (x^2-1\right )+4}\right )} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2009

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

Input:

Int[(-24 + E^x*(6 - 3*x + 3*x^3) + (12 - 12*x^2 + E^x*(-3 + 3*x^2))*Log[x^ 
2/(4 - 4*x^2 + E^x*(-1 + x^2))] + (-12 + 12*x^2 + E^x*(3 - 3*x^2))*Log[x^2 
/(4 - 4*x^2 + E^x*(-1 + x^2))]^2)/((4 - 4*x^2 + E^x*(-1 + x^2))*Log[x^2/(4 
 - 4*x^2 + E^x*(-1 + x^2))]^2),x]
 

Output:

$Aborted
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(59\) vs. \(2(26)=52\).

Time = 1.90 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22

method result size
parallelrisch \(-\frac {48 \ln \left (\frac {x^{2}}{{\mathrm e}^{x} x^{2}-4 x^{2}-{\mathrm e}^{x}+4}\right ) x -48 x}{16 \ln \left (\frac {x^{2}}{{\mathrm e}^{x} x^{2}-4 x^{2}-{\mathrm e}^{x}+4}\right )}\) \(60\)
risch \(-3 x +\frac {6 i x}{\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}-4}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}-1}\right ) \operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x}-4\right ) \left (x^{2}-1\right )}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}-4}\right ) {\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x}-4\right ) \left (x^{2}-1\right )}\right )}^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{x^{2}-1}\right ) {\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x}-4\right ) \left (x^{2}-1\right )}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x}-4\right ) \left (x^{2}-1\right )}\right )}^{3}-\pi \,\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x}-4\right ) \left (x^{2}-1\right )}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\left (x^{2}-1\right ) \left ({\mathrm e}^{x}-4\right )}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x}-4\right ) \left (x^{2}-1\right )}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\left (x^{2}-1\right ) \left ({\mathrm e}^{x}-4\right )}\right ) \operatorname {csgn}\left (i x^{2}\right )+\pi \operatorname {csgn}\left (\frac {i x^{2}}{\left (x^{2}-1\right ) \left ({\mathrm e}^{x}-4\right )}\right )^{3}-\pi \operatorname {csgn}\left (\frac {i x^{2}}{\left (x^{2}-1\right ) \left ({\mathrm e}^{x}-4\right )}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )-2 i \ln \left (x^{2}-1\right )-2 i \ln \left ({\mathrm e}^{x}-4\right )}\) \(348\)

Input:

int((((-3*x^2+3)*exp(x)+12*x^2-12)*ln(x^2/((x^2-1)*exp(x)-4*x^2+4))^2+((3* 
x^2-3)*exp(x)-12*x^2+12)*ln(x^2/((x^2-1)*exp(x)-4*x^2+4))+(3*x^3-3*x+6)*ex 
p(x)-24)/((x^2-1)*exp(x)-4*x^2+4)/ln(x^2/((x^2-1)*exp(x)-4*x^2+4))^2,x,met 
hod=_RETURNVERBOSE)
 

Output:

-1/16*(48*ln(x^2/(exp(x)*x^2-4*x^2-exp(x)+4))*x-48*x)/ln(x^2/(exp(x)*x^2-4 
*x^2-exp(x)+4))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (25) = 50\).

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {-24+e^x \left (6-3 x+3 x^3\right )+\left (12-12 x^2+e^x \left (-3+3 x^2\right )\right ) \log \left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )+\left (-12+12 x^2+e^x \left (3-3 x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )}{\left (4-4 x^2+e^x \left (-1+x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )} \, dx=-\frac {3 \, {\left (x \log \left (-\frac {x^{2}}{4 \, x^{2} - {\left (x^{2} - 1\right )} e^{x} - 4}\right ) - x\right )}}{\log \left (-\frac {x^{2}}{4 \, x^{2} - {\left (x^{2} - 1\right )} e^{x} - 4}\right )} \] Input:

integrate((((-3*x^2+3)*exp(x)+12*x^2-12)*log(x^2/((x^2-1)*exp(x)-4*x^2+4)) 
^2+((3*x^2-3)*exp(x)-12*x^2+12)*log(x^2/((x^2-1)*exp(x)-4*x^2+4))+(3*x^3-3 
*x+6)*exp(x)-24)/((x^2-1)*exp(x)-4*x^2+4)/log(x^2/((x^2-1)*exp(x)-4*x^2+4) 
)^2,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-24+e^x \left (6-3 x+3 x^3\right )+\left (12-12 x^2+e^x \left (-3+3 x^2\right )\right ) \log \left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )+\left (-12+12 x^2+e^x \left (3-3 x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )}{\left (4-4 x^2+e^x \left (-1+x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )} \, dx=- 3 x + \frac {3 x}{\log {\left (\frac {x^{2}}{- 4 x^{2} + \left (x^{2} - 1\right ) e^{x} + 4} \right )}} \] Input:

integrate((((-3*x**2+3)*exp(x)+12*x**2-12)*ln(x**2/((x**2-1)*exp(x)-4*x**2 
+4))**2+((3*x**2-3)*exp(x)-12*x**2+12)*ln(x**2/((x**2-1)*exp(x)-4*x**2+4)) 
+(3*x**3-3*x+6)*exp(x)-24)/((x**2-1)*exp(x)-4*x**2+4)/ln(x**2/((x**2-1)*ex 
p(x)-4*x**2+4))**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {-24+e^x \left (6-3 x+3 x^3\right )+\left (12-12 x^2+e^x \left (-3+3 x^2\right )\right ) \log \left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )+\left (-12+12 x^2+e^x \left (3-3 x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )}{\left (4-4 x^2+e^x \left (-1+x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )} \, dx=-\frac {3 \, {\left (x \log \left (x + 1\right ) + x \log \left (x - 1\right ) - 2 \, x \log \left (x\right ) + x \log \left (e^{x} - 4\right ) + x\right )}}{\log \left (x + 1\right ) + \log \left (x - 1\right ) - 2 \, \log \left (x\right ) + \log \left (e^{x} - 4\right )} \] Input:

integrate((((-3*x^2+3)*exp(x)+12*x^2-12)*log(x^2/((x^2-1)*exp(x)-4*x^2+4)) 
^2+((3*x^2-3)*exp(x)-12*x^2+12)*log(x^2/((x^2-1)*exp(x)-4*x^2+4))+(3*x^3-3 
*x+6)*exp(x)-24)/((x^2-1)*exp(x)-4*x^2+4)/log(x^2/((x^2-1)*exp(x)-4*x^2+4) 
)^2,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (25) = 50\).

Time = 0.71 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {-24+e^x \left (6-3 x+3 x^3\right )+\left (12-12 x^2+e^x \left (-3+3 x^2\right )\right ) \log \left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )+\left (-12+12 x^2+e^x \left (3-3 x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )}{\left (4-4 x^2+e^x \left (-1+x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )} \, dx=-\frac {3 \, {\left (x \log \left (\frac {x^{2}}{x^{2} e^{x} - 4 \, x^{2} - e^{x} + 4}\right ) - x\right )}}{\log \left (\frac {x^{2}}{x^{2} e^{x} - 4 \, x^{2} - e^{x} + 4}\right )} \] Input:

integrate((((-3*x^2+3)*exp(x)+12*x^2-12)*log(x^2/((x^2-1)*exp(x)-4*x^2+4)) 
^2+((3*x^2-3)*exp(x)-12*x^2+12)*log(x^2/((x^2-1)*exp(x)-4*x^2+4))+(3*x^3-3 
*x+6)*exp(x)-24)/((x^2-1)*exp(x)-4*x^2+4)/log(x^2/((x^2-1)*exp(x)-4*x^2+4) 
)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-24+e^x \left (6-3 x+3 x^3\right )+\left (12-12 x^2+e^x \left (-3+3 x^2\right )\right ) \log \left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )+\left (-12+12 x^2+e^x \left (3-3 x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )}{\left (4-4 x^2+e^x \left (-1+x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )} \, dx=\int -\frac {\left ({\mathrm {e}}^x\,\left (3\,x^2-3\right )-12\,x^2+12\right )\,{\ln \left (\frac {x^2}{{\mathrm {e}}^x\,\left (x^2-1\right )-4\,x^2+4}\right )}^2+\left (12\,x^2-{\mathrm {e}}^x\,\left (3\,x^2-3\right )-12\right )\,\ln \left (\frac {x^2}{{\mathrm {e}}^x\,\left (x^2-1\right )-4\,x^2+4}\right )-{\mathrm {e}}^x\,\left (3\,x^3-3\,x+6\right )+24}{{\ln \left (\frac {x^2}{{\mathrm {e}}^x\,\left (x^2-1\right )-4\,x^2+4}\right )}^2\,\left ({\mathrm {e}}^x\,\left (x^2-1\right )-4\,x^2+4\right )} \,d x \] Input:

int(-(log(x^2/(exp(x)*(x^2 - 1) - 4*x^2 + 4))^2*(exp(x)*(3*x^2 - 3) - 12*x 
^2 + 12) - exp(x)*(3*x^3 - 3*x + 6) - log(x^2/(exp(x)*(x^2 - 1) - 4*x^2 + 
4))*(exp(x)*(3*x^2 - 3) - 12*x^2 + 12) + 24)/(log(x^2/(exp(x)*(x^2 - 1) - 
4*x^2 + 4))^2*(exp(x)*(x^2 - 1) - 4*x^2 + 4)),x)
 

Output:

int(-(log(x^2/(exp(x)*(x^2 - 1) - 4*x^2 + 4))^2*(exp(x)*(3*x^2 - 3) - 12*x 
^2 + 12) - exp(x)*(3*x^3 - 3*x + 6) - log(x^2/(exp(x)*(x^2 - 1) - 4*x^2 + 
4))*(exp(x)*(3*x^2 - 3) - 12*x^2 + 12) + 24)/(log(x^2/(exp(x)*(x^2 - 1) - 
4*x^2 + 4))^2*(exp(x)*(x^2 - 1) - 4*x^2 + 4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.26 \[ \int \frac {-24+e^x \left (6-3 x+3 x^3\right )+\left (12-12 x^2+e^x \left (-3+3 x^2\right )\right ) \log \left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )+\left (-12+12 x^2+e^x \left (3-3 x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )}{\left (4-4 x^2+e^x \left (-1+x^2\right )\right ) \log ^2\left (\frac {x^2}{4-4 x^2+e^x \left (-1+x^2\right )}\right )} \, dx=\frac {3 x \left (-\mathrm {log}\left (\frac {x^{2}}{e^{x} x^{2}-e^{x}-4 x^{2}+4}\right )+1\right )}{\mathrm {log}\left (\frac {x^{2}}{e^{x} x^{2}-e^{x}-4 x^{2}+4}\right )} \] Input:

int((((-3*x^2+3)*exp(x)+12*x^2-12)*log(x^2/((x^2-1)*exp(x)-4*x^2+4))^2+((3 
*x^2-3)*exp(x)-12*x^2+12)*log(x^2/((x^2-1)*exp(x)-4*x^2+4))+(3*x^3-3*x+6)* 
exp(x)-24)/((x^2-1)*exp(x)-4*x^2+4)/log(x^2/((x^2-1)*exp(x)-4*x^2+4))^2,x)
 

Output:

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