\(\int \frac {-9 x+6 x^2-x^3+(6 x-2 x^2) \log (\frac {4}{x^3})-x \log ^2(\frac {4}{x^3})+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log (\frac {4}{x^3})}} (e^{9+12 x+4 x^2} (3-37 x-12 x^2+8 x^3)+e^{9+12 x+4 x^2} (12 x+8 x^2) \log (\frac {4}{x^3}))}{9 x-6 x^2+x^3+(-6 x+2 x^2) \log (\frac {4}{x^3})+x \log ^2(\frac {4}{x^3})} \, dx\) [2181]

Optimal result

Integrand size = 166, antiderivative size = 27 \[ \int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx=e^{\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}}-x \] Output:

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx=e^{\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}}-x \] Input:

Integrate[(-9*x + 6*x^2 - x^3 + (6*x - 2*x^2)*Log[4/x^3] - x*Log[4/x^3]^2 
+ E^(E^(9 + 12*x + 4*x^2)/(-3 + x + Log[4/x^3]))*(E^(9 + 12*x + 4*x^2)*(3 
- 37*x - 12*x^2 + 8*x^3) + E^(9 + 12*x + 4*x^2)*(12*x + 8*x^2)*Log[4/x^3]) 
)/(9*x - 6*x^2 + x^3 + (-6*x + 2*x^2)*Log[4/x^3] + x*Log[4/x^3]^2),x]
 

Output:

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

\(\displaystyle \int \frac {-x^3-x \log ^2\left (\frac {4}{x^3}\right )+6 x^2+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )+e^{\frac {e^{4 x^2+12 x+9}}{\log \left (\frac {4}{x^3}\right )+x-3}} \left (e^{4 x^2+12 x+9} \left (8 x^3-12 x^2-37 x+3\right )+e^{4 x^2+12 x+9} \left (8 x^2+12 x\right ) \log \left (\frac {4}{x^3}\right )\right )-9 x}{x^3+x \log ^2\left (\frac {4}{x^3}\right )-6 x^2+\left (2 x^2-6 x\right ) \log \left (\frac {4}{x^3}\right )+9 x} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-x^3-x \log ^2\left (\frac {4}{x^3}\right )+6 x^2+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )+e^{\frac {e^{4 x^2+12 x+9}}{\log \left (\frac {4}{x^3}\right )+x-3}} \left (e^{4 x^2+12 x+9} \left (8 x^3-12 x^2-37 x+3\right )+e^{4 x^2+12 x+9} \left (8 x^2+12 x\right ) \log \left (\frac {4}{x^3}\right )\right )-9 x}{x \left (-\log \left (\frac {4}{x^3}\right )-x+3\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\left (8 x^3+12 x \log \left (\frac {4}{x^3}\right )-12 x^2+8 x^2 \log \left (\frac {4}{x^3}\right )-37 x+3\right ) \exp \left (\frac {e^{(2 x+3)^2}}{\log \left (\frac {4}{x^3}\right )+x-3}+4 x^2+12 x+9\right )}{x \left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}-\frac {\log ^2\left (\frac {4}{x^3}\right )}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}+\frac {6 x}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}-\frac {2 (x-3) \log \left (\frac {4}{x^3}\right )}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}-\frac {9}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}-\frac {x^2}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (\frac {\left (8 x^3+12 x \log \left (\frac {4}{x^3}\right )-12 x^2+8 x^2 \log \left (\frac {4}{x^3}\right )-37 x+3\right ) \exp \left (\frac {e^{(2 x+3)^2}}{\log \left (\frac {4}{x^3}\right )+x-3}+4 x^2+12 x+9\right )}{x \left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}-\frac {\log ^2\left (\frac {4}{x^3}\right )}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}+\frac {6 x}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}-\frac {2 (x-3) \log \left (\frac {4}{x^3}\right )}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}-\frac {9}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}-\frac {x^2}{\left (\log \left (\frac {4}{x^3}\right )+x-3\right )^2}\right )dx\)

Input:

Int[(-9*x + 6*x^2 - x^3 + (6*x - 2*x^2)*Log[4/x^3] - x*Log[4/x^3]^2 + E^(E 
^(9 + 12*x + 4*x^2)/(-3 + x + Log[4/x^3]))*(E^(9 + 12*x + 4*x^2)*(3 - 37*x 
 - 12*x^2 + 8*x^3) + E^(9 + 12*x + 4*x^2)*(12*x + 8*x^2)*Log[4/x^3]))/(9*x 
 - 6*x^2 + x^3 + (-6*x + 2*x^2)*Log[4/x^3] + x*Log[4/x^3]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 12.53 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(-6-x +{\mathrm e}^{\frac {{\mathrm e}^{4 x^{2}+12 x +9}}{\ln \left (\frac {4}{x^{3}}\right )+x -3}}\)	\(30\)
risch	\(-x +{\mathrm e}^{\frac {2 \,{\mathrm e}^{\left (3+2 x \right )^{2}}}{i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}+4 \ln \left (2\right )-6 \ln \left (x \right )+2 x -6}}\)	\(154\)

Input:

int((((8*x^2+12*x)*exp(4*x^2+12*x+9)*ln(4/x^3)+(8*x^3-12*x^2-37*x+3)*exp(4 
*x^2+12*x+9))*exp(exp(4*x^2+12*x+9)/(ln(4/x^3)+x-3))-x*ln(4/x^3)^2+(-2*x^2 
+6*x)*ln(4/x^3)-x^3+6*x^2-9*x)/(x*ln(4/x^3)^2+(2*x^2-6*x)*ln(4/x^3)+x^3-6* 
x^2+9*x),x,method=_RETURNVERBOSE)
 

Output:

-6-x+exp(exp(4*x^2+12*x+9)/(ln(4/x^3)+x-3))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx=-x + e^{\left (\frac {e^{\left (4 \, x^{2} + 12 \, x + 9\right )}}{x + \log \left (\frac {4}{x^{3}}\right ) - 3}\right )} \] Input:

integrate((((8*x^2+12*x)*exp(4*x^2+12*x+9)*log(4/x^3)+(8*x^3-12*x^2-37*x+3 
)*exp(4*x^2+12*x+9))*exp(exp(4*x^2+12*x+9)/(log(4/x^3)+x-3))-x*log(4/x^3)^ 
2+(-2*x^2+6*x)*log(4/x^3)-x^3+6*x^2-9*x)/(x*log(4/x^3)^2+(2*x^2-6*x)*log(4 
/x^3)+x^3-6*x^2+9*x),x, algorithm="fricas")
 

Output:

-x + e^(e^(4*x^2 + 12*x + 9)/(x + log(4/x^3) - 3))
 

Sympy [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx=- x + e^{\frac {e^{4 x^{2} + 12 x + 9}}{x + \log {\left (\frac {4}{x^{3}} \right )} - 3}} \] Input:

integrate((((8*x**2+12*x)*exp(4*x**2+12*x+9)*ln(4/x**3)+(8*x**3-12*x**2-37 
*x+3)*exp(4*x**2+12*x+9))*exp(exp(4*x**2+12*x+9)/(ln(4/x**3)+x-3))-x*ln(4/ 
x**3)**2+(-2*x**2+6*x)*ln(4/x**3)-x**3+6*x**2-9*x)/(x*ln(4/x**3)**2+(2*x** 
2-6*x)*ln(4/x**3)+x**3-6*x**2+9*x),x)
 

Output:

-x + exp(exp(4*x**2 + 12*x + 9)/(x + log(4/x**3) - 3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).

Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.19 \[ \int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx=-{\left (x e^{\left (-\frac {e^{\left (4 \, x^{2} + 12 \, x + 9\right )}}{x + 2 \, \log \left (2\right ) - 3 \, \log \left (x\right ) - 3}\right )} - 1\right )} e^{\left (\frac {e^{\left (4 \, x^{2} + 12 \, x + 9\right )}}{x + 2 \, \log \left (2\right ) - 3 \, \log \left (x\right ) - 3}\right )} \] Input:

integrate((((8*x^2+12*x)*exp(4*x^2+12*x+9)*log(4/x^3)+(8*x^3-12*x^2-37*x+3 
)*exp(4*x^2+12*x+9))*exp(exp(4*x^2+12*x+9)/(log(4/x^3)+x-3))-x*log(4/x^3)^ 
2+(-2*x^2+6*x)*log(4/x^3)-x^3+6*x^2-9*x)/(x*log(4/x^3)^2+(2*x^2-6*x)*log(4 
/x^3)+x^3-6*x^2+9*x),x, algorithm="maxima")
 

Output:

-(x*e^(-e^(4*x^2 + 12*x + 9)/(x + 2*log(2) - 3*log(x) - 3)) - 1)*e^(e^(4*x 
^2 + 12*x + 9)/(x + 2*log(2) - 3*log(x) - 3))
 

Giac [F]

\[ \int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx=\int { -\frac {x^{3} + x \log \left (\frac {4}{x^{3}}\right )^{2} - 6 \, x^{2} - {\left (4 \, {\left (2 \, x^{2} + 3 \, x\right )} e^{\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (\frac {4}{x^{3}}\right ) + {\left (8 \, x^{3} - 12 \, x^{2} - 37 \, x + 3\right )} e^{\left (4 \, x^{2} + 12 \, x + 9\right )}\right )} e^{\left (\frac {e^{\left (4 \, x^{2} + 12 \, x + 9\right )}}{x + \log \left (\frac {4}{x^{3}}\right ) - 3}\right )} + 2 \, {\left (x^{2} - 3 \, x\right )} \log \left (\frac {4}{x^{3}}\right ) + 9 \, x}{x^{3} + x \log \left (\frac {4}{x^{3}}\right )^{2} - 6 \, x^{2} + 2 \, {\left (x^{2} - 3 \, x\right )} \log \left (\frac {4}{x^{3}}\right ) + 9 \, x} \,d x } \] Input:

integrate((((8*x^2+12*x)*exp(4*x^2+12*x+9)*log(4/x^3)+(8*x^3-12*x^2-37*x+3 
)*exp(4*x^2+12*x+9))*exp(exp(4*x^2+12*x+9)/(log(4/x^3)+x-3))-x*log(4/x^3)^ 
2+(-2*x^2+6*x)*log(4/x^3)-x^3+6*x^2-9*x)/(x*log(4/x^3)^2+(2*x^2-6*x)*log(4 
/x^3)+x^3-6*x^2+9*x),x, algorithm="giac")
 

Output:

integrate(-(x^3 + x*log(4/x^3)^2 - 6*x^2 - (4*(2*x^2 + 3*x)*e^(4*x^2 + 12* 
x + 9)*log(4/x^3) + (8*x^3 - 12*x^2 - 37*x + 3)*e^(4*x^2 + 12*x + 9))*e^(e 
^(4*x^2 + 12*x + 9)/(x + log(4/x^3) - 3)) + 2*(x^2 - 3*x)*log(4/x^3) + 9*x 
)/(x^3 + x*log(4/x^3)^2 - 6*x^2 + 2*(x^2 - 3*x)*log(4/x^3) + 9*x), x)
 

Mupad [B] (verification not implemented)

Time = 3.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{12\,x}\,{\mathrm {e}}^9\,{\mathrm {e}}^{4\,x^2}}{x+\ln \left (\frac {4}{x^3}\right )-3}}-x \] Input:

int(-(9*x + exp(exp(12*x + 4*x^2 + 9)/(x + log(4/x^3) - 3))*(exp(12*x + 4* 
x^2 + 9)*(37*x + 12*x^2 - 8*x^3 - 3) - exp(12*x + 4*x^2 + 9)*log(4/x^3)*(1 
2*x + 8*x^2)) - log(4/x^3)*(6*x - 2*x^2) - 6*x^2 + x^3 + x*log(4/x^3)^2)/( 
9*x - log(4/x^3)*(6*x - 2*x^2) - 6*x^2 + x^3 + x*log(4/x^3)^2),x)
 

Output:

exp((exp(12*x)*exp(9)*exp(4*x^2))/(x + log(4/x^3) - 3)) - x
 

Reduce [F]

\[ \int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx=\int \frac {\left (\left (8 x^{2}+12 x \right ) {\mathrm e}^{4 x^{2}+12 x +9} \mathrm {log}\left (\frac {4}{x^{3}}\right )+\left (8 x^{3}-12 x^{2}-37 x +3\right ) {\mathrm e}^{4 x^{2}+12 x +9}\right ) {\mathrm e}^{\frac {{\mathrm e}^{4 x^{2}+12 x +9}}{\mathrm {log}\left (\frac {4}{x^{3}}\right )+x -3}}-x \mathrm {log}\left (\frac {4}{x^{3}}\right )^{2}+\left (-2 x^{2}+6 x \right ) \mathrm {log}\left (\frac {4}{x^{3}}\right )-x^{3}+6 x^{2}-9 x}{x \mathrm {log}\left (\frac {4}{x^{3}}\right )^{2}+\left (2 x^{2}-6 x \right ) \mathrm {log}\left (\frac {4}{x^{3}}\right )+x^{3}-6 x^{2}+9 x}d x \] Input:

int((((8*x^2+12*x)*exp(4*x^2+12*x+9)*log(4/x^3)+(8*x^3-12*x^2-37*x+3)*exp( 
4*x^2+12*x+9))*exp(exp(4*x^2+12*x+9)/(log(4/x^3)+x-3))-x*log(4/x^3)^2+(-2* 
x^2+6*x)*log(4/x^3)-x^3+6*x^2-9*x)/(x*log(4/x^3)^2+(2*x^2-6*x)*log(4/x^3)+ 
x^3-6*x^2+9*x),x)
 

Output:

int((((8*x^2+12*x)*exp(4*x^2+12*x+9)*log(4/x^3)+(8*x^3-12*x^2-37*x+3)*exp( 
4*x^2+12*x+9))*exp(exp(4*x^2+12*x+9)/(log(4/x^3)+x-3))-x*log(4/x^3)^2+(-2* 
x^2+6*x)*log(4/x^3)-x^3+6*x^2-9*x)/(x*log(4/x^3)^2+(2*x^2-6*x)*log(4/x^3)+ 
x^3-6*x^2+9*x),x)