\(\int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} (-48 x+18 x^2))}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} (-48 x+16 x^2)+e^{6 x} (216 x-144 x^2+24 x^3)+e^{3 x} (-432 x+432 x^2-144 x^3+16 x^4)+(-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} (48 x-16 x^2)) \log (x)+4 x \log ^2(x)} \, dx\) [416]

Optimal result

Integrand size = 209, antiderivative size = 27 \[ \int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} \left (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} \left (-48 x+18 x^2\right )\right )}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} \left (-48 x+16 x^2\right )+e^{6 x} \left (216 x-144 x^2+24 x^3\right )+e^{3 x} \left (-432 x+432 x^2-144 x^3+16 x^4\right )+\left (-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} \left (48 x-16 x^2\right )\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3}{4} e^{\frac {1}{-\left (3-e^{3 x}-x\right )^2+\log (x)}} \] Output:

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

Mathematica [A] (verified)

Time = 2.93 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} \left (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} \left (-48 x+18 x^2\right )\right )}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} \left (-48 x+16 x^2\right )+e^{6 x} \left (216 x-144 x^2+24 x^3\right )+e^{3 x} \left (-432 x+432 x^2-144 x^3+16 x^4\right )+\left (-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} \left (48 x-16 x^2\right )\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3}{4} e^{-\frac {1}{\left (-3+e^{3 x}+x\right )^2-\log (x)}} \] Input:

Integrate[(E^(-9 - E^(6*x) + E^(3*x)*(6 - 2*x) + 6*x - x^2 + Log[x])^(-1)* 
(-3 - 18*x + 18*E^(6*x)*x + 6*x^2 + E^(3*x)*(-48*x + 18*x^2)))/(324*x + 4* 
E^(12*x)*x - 432*x^2 + 216*x^3 - 48*x^4 + 4*x^5 + E^(9*x)*(-48*x + 16*x^2) 
 + E^(6*x)*(216*x - 144*x^2 + 24*x^3) + E^(3*x)*(-432*x + 432*x^2 - 144*x^ 
3 + 16*x^4) + (-72*x - 8*E^(6*x)*x + 48*x^2 - 8*x^3 + E^(3*x)*(48*x - 16*x 
^2))*Log[x] + 4*x*Log[x]^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 3.99 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {7239, 27, 25, 7257}

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.

\(\displaystyle \int \frac {\left (6 x^2+e^{3 x} \left (18 x^2-48 x\right )+18 e^{6 x} x-18 x-3\right ) \exp \left (\frac {1}{-x^2+6 x-e^{6 x}+e^{3 x} (6-2 x)+\log (x)-9}\right )}{4 x^5-48 x^4+216 x^3-432 x^2+e^{9 x} \left (16 x^2-48 x\right )+e^{6 x} \left (24 x^3-144 x^2+216 x\right )+\left (-8 x^3+48 x^2+e^{3 x} \left (48 x-16 x^2\right )-8 e^{6 x} x-72 x\right ) \log (x)+e^{3 x} \left (16 x^4-144 x^3+432 x^2-432 x\right )+4 e^{12 x} x+324 x+4 x \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {3 \left (\left (6 e^{3 x}+2\right ) x^2+2 \left (-8 e^{3 x}+3 e^{6 x}-3\right ) x-1\right ) e^{-\frac {1}{\left (x+e^{3 x}-3\right )^2-\log (x)}}}{4 x \left (\left (x+e^{3 x}-3\right )^2-\log (x)\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{4} \int -\frac {e^{-\frac {1}{\left (x+e^{3 x}-3\right )^2-\log (x)}} \left (-2 \left (1+3 e^{3 x}\right ) x^2+2 \left (3+8 e^{3 x}-3 e^{6 x}\right ) x+1\right )}{x \left (\left (x+e^{3 x}-3\right )^2-\log (x)\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {3}{4} \int \frac {e^{-\frac {1}{\left (x+e^{3 x}-3\right )^2-\log (x)}} \left (-2 \left (1+3 e^{3 x}\right ) x^2+2 \left (3+8 e^{3 x}-3 e^{6 x}\right ) x+1\right )}{x \left (\left (x+e^{3 x}-3\right )^2-\log (x)\right )^2}dx\)

\(\Big \downarrow \) 7257

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

Input:

Int[(E^(-9 - E^(6*x) + E^(3*x)*(6 - 2*x) + 6*x - x^2 + Log[x])^(-1)*(-3 - 
18*x + 18*E^(6*x)*x + 6*x^2 + E^(3*x)*(-48*x + 18*x^2)))/(324*x + 4*E^(12* 
x)*x - 432*x^2 + 216*x^3 - 48*x^4 + 4*x^5 + E^(9*x)*(-48*x + 16*x^2) + E^( 
6*x)*(216*x - 144*x^2 + 24*x^3) + E^(3*x)*(-432*x + 432*x^2 - 144*x^3 + 16 
*x^4) + (-72*x - 8*E^(6*x)*x + 48*x^2 - 8*x^3 + E^(3*x)*(48*x - 16*x^2))*L 
og[x] + 4*x*Log[x]^2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7257

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim 
p[q*(F^v/Log[F]), x] /;  !FalseQ[q]] /; FreeQ[F, x]
 

Maple [A] (verified)

Time = 47.41 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{\frac {1}{-2 x \,{\mathrm e}^{3 x}-x^{2}-{\mathrm e}^{6 x}+\ln \left (x \right )+6 \,{\mathrm e}^{3 x}+6 x -9}}}{4}\)	\(37\)
parallelrisch	\(\frac {3 \,{\mathrm e}^{\frac {1}{-2 x \,{\mathrm e}^{3 x}-x^{2}-{\mathrm e}^{6 x}+\ln \left (x \right )+6 \,{\mathrm e}^{3 x}+6 x -9}}}{4}\)	\(39\)

Input:

int((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(ln(x)-exp 
(3*x)^2+(6-2*x)*exp(3*x)-x^2+6*x-9))/(4*x*ln(x)^2+(-8*x*exp(3*x)^2+(-16*x^ 
2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*ln(x)+4*x*exp(3*x)^4+(16*x^2-48*x)*exp 
(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2-432*x)*e 
xp(3*x)+4*x^5-48*x^4+216*x^3-432*x^2+324*x),x,method=_RETURNVERBOSE)
 

Output:

3/4*exp(1/(-2*x*exp(3*x)-x^2-exp(6*x)+ln(x)+6*exp(3*x)+6*x-9))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} \left (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} \left (-48 x+18 x^2\right )\right )}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} \left (-48 x+16 x^2\right )+e^{6 x} \left (216 x-144 x^2+24 x^3\right )+e^{3 x} \left (-432 x+432 x^2-144 x^3+16 x^4\right )+\left (-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} \left (48 x-16 x^2\right )\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3}{4} \, e^{\left (-\frac {1}{x^{2} + 2 \, {\left (x - 3\right )} e^{\left (3 \, x\right )} - 6 \, x + e^{\left (6 \, x\right )} - \log \left (x\right ) + 9}\right )} \] Input:

integrate((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(log 
(x)-exp(3*x)^2+(6-2*x)*exp(3*x)-x^2+6*x-9))/(4*x*log(x)^2+(-8*x*exp(3*x)^2 
+(-16*x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*log(x)+4*x*exp(3*x)^4+(16*x^2- 
48*x)*exp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2 
-432*x)*exp(3*x)+4*x^5-48*x^4+216*x^3-432*x^2+324*x),x, algorithm="fricas" 
)
 

Output:

3/4*e^(-1/(x^2 + 2*(x - 3)*e^(3*x) - 6*x + e^(6*x) - log(x) + 9))
 

Sympy [A] (verification not implemented)

Time = 8.75 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} \left (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} \left (-48 x+18 x^2\right )\right )}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} \left (-48 x+16 x^2\right )+e^{6 x} \left (216 x-144 x^2+24 x^3\right )+e^{3 x} \left (-432 x+432 x^2-144 x^3+16 x^4\right )+\left (-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} \left (48 x-16 x^2\right )\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3 e^{\frac {1}{- x^{2} + 6 x + \left (6 - 2 x\right ) e^{3 x} - e^{6 x} + \log {\left (x \right )} - 9}}}{4} \] Input:

integrate((18*x*exp(3*x)**2+(18*x**2-48*x)*exp(3*x)+6*x**2-18*x-3)*exp(1/( 
ln(x)-exp(3*x)**2+(6-2*x)*exp(3*x)-x**2+6*x-9))/(4*x*ln(x)**2+(-8*x*exp(3* 
x)**2+(-16*x**2+48*x)*exp(3*x)-8*x**3+48*x**2-72*x)*ln(x)+4*x*exp(3*x)**4+ 
(16*x**2-48*x)*exp(3*x)**3+(24*x**3-144*x**2+216*x)*exp(3*x)**2+(16*x**4-1 
44*x**3+432*x**2-432*x)*exp(3*x)+4*x**5-48*x**4+216*x**3-432*x**2+324*x),x 
)
 

Output:

3*exp(1/(-x**2 + 6*x + (6 - 2*x)*exp(3*x) - exp(6*x) + log(x) - 9))/4
 

Maxima [A] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} \left (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} \left (-48 x+18 x^2\right )\right )}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} \left (-48 x+16 x^2\right )+e^{6 x} \left (216 x-144 x^2+24 x^3\right )+e^{3 x} \left (-432 x+432 x^2-144 x^3+16 x^4\right )+\left (-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} \left (48 x-16 x^2\right )\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3}{4} \, e^{\left (-\frac {1}{x^{2} + 2 \, {\left (x - 3\right )} e^{\left (3 \, x\right )} - 6 \, x + e^{\left (6 \, x\right )} - \log \left (x\right ) + 9}\right )} \] Input:

integrate((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(log 
(x)-exp(3*x)^2+(6-2*x)*exp(3*x)-x^2+6*x-9))/(4*x*log(x)^2+(-8*x*exp(3*x)^2 
+(-16*x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*log(x)+4*x*exp(3*x)^4+(16*x^2- 
48*x)*exp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2 
-432*x)*exp(3*x)+4*x^5-48*x^4+216*x^3-432*x^2+324*x),x, algorithm="maxima" 
)
 

Output:

3/4*e^(-1/(x^2 + 2*(x - 3)*e^(3*x) - 6*x + e^(6*x) - log(x) + 9))
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} \left (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} \left (-48 x+18 x^2\right )\right )}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} \left (-48 x+16 x^2\right )+e^{6 x} \left (216 x-144 x^2+24 x^3\right )+e^{3 x} \left (-432 x+432 x^2-144 x^3+16 x^4\right )+\left (-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} \left (48 x-16 x^2\right )\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3}{4} \, e^{\left (-\frac {1}{x^{2} + 2 \, x e^{\left (3 \, x\right )} - 6 \, x + e^{\left (6 \, x\right )} - 6 \, e^{\left (3 \, x\right )} - \log \left (x\right ) + 9}\right )} \] Input:

integrate((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(log 
(x)-exp(3*x)^2+(6-2*x)*exp(3*x)-x^2+6*x-9))/(4*x*log(x)^2+(-8*x*exp(3*x)^2 
+(-16*x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*log(x)+4*x*exp(3*x)^4+(16*x^2- 
48*x)*exp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2 
-432*x)*exp(3*x)+4*x^5-48*x^4+216*x^3-432*x^2+324*x),x, algorithm="giac")
 

Output:

3/4*e^(-1/(x^2 + 2*x*e^(3*x) - 6*x + e^(6*x) - 6*e^(3*x) - log(x) + 9))
 

Mupad [B] (verification not implemented)

Time = 3.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} \left (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} \left (-48 x+18 x^2\right )\right )}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} \left (-48 x+16 x^2\right )+e^{6 x} \left (216 x-144 x^2+24 x^3\right )+e^{3 x} \left (-432 x+432 x^2-144 x^3+16 x^4\right )+\left (-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} \left (48 x-16 x^2\right )\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3\,{\mathrm {e}}^{-\frac {1}{{\mathrm {e}}^{6\,x}-6\,{\mathrm {e}}^{3\,x}-6\,x-\ln \left (x\right )+2\,x\,{\mathrm {e}}^{3\,x}+x^2+9}}}{4} \] Input:

int(-(exp(-1/(exp(6*x) - 6*x - log(x) + exp(3*x)*(2*x - 6) + x^2 + 9))*(18 
*x + exp(3*x)*(48*x - 18*x^2) - 18*x*exp(6*x) - 6*x^2 + 3))/(324*x - log(x 
)*(72*x - exp(3*x)*(48*x - 16*x^2) + 8*x*exp(6*x) - 48*x^2 + 8*x^3) - exp( 
9*x)*(48*x - 16*x^2) + 4*x*exp(12*x) + 4*x*log(x)^2 + exp(6*x)*(216*x - 14 
4*x^2 + 24*x^3) - exp(3*x)*(432*x - 432*x^2 + 144*x^3 - 16*x^4) - 432*x^2 
+ 216*x^3 - 48*x^4 + 4*x^5),x)
 

Output:

(3*exp(-1/(exp(6*x) - 6*exp(3*x) - 6*x - log(x) + 2*x*exp(3*x) + x^2 + 9)) 
)/4
 

Reduce [F]

\[ \int \frac {e^{\frac {1}{-9-e^{6 x}+e^{3 x} (6-2 x)+6 x-x^2+\log (x)}} \left (-3-18 x+18 e^{6 x} x+6 x^2+e^{3 x} \left (-48 x+18 x^2\right )\right )}{324 x+4 e^{12 x} x-432 x^2+216 x^3-48 x^4+4 x^5+e^{9 x} \left (-48 x+16 x^2\right )+e^{6 x} \left (216 x-144 x^2+24 x^3\right )+e^{3 x} \left (-432 x+432 x^2-144 x^3+16 x^4\right )+\left (-72 x-8 e^{6 x} x+48 x^2-8 x^3+e^{3 x} \left (48 x-16 x^2\right )\right ) \log (x)+4 x \log ^2(x)} \, dx=\int \frac {\left (18 x \left ({\mathrm e}^{3 x}\right )^{2}+\left (18 x^{2}-48 x \right ) {\mathrm e}^{3 x}+6 x^{2}-18 x -3\right ) {\mathrm e}^{\frac {1}{\mathrm {log}\left (x \right )-\left ({\mathrm e}^{3 x}\right )^{2}+\left (6-2 x \right ) {\mathrm e}^{3 x}-x^{2}+6 x -9}}}{4 \mathrm {log}\left (x \right )^{2} x +\left (-8 x \left ({\mathrm e}^{3 x}\right )^{2}+\left (-16 x^{2}+48 x \right ) {\mathrm e}^{3 x}-8 x^{3}+48 x^{2}-72 x \right ) \mathrm {log}\left (x \right )+4 x \left ({\mathrm e}^{3 x}\right )^{4}+\left (16 x^{2}-48 x \right ) \left ({\mathrm e}^{3 x}\right )^{3}+\left (24 x^{3}-144 x^{2}+216 x \right ) \left ({\mathrm e}^{3 x}\right )^{2}+\left (16 x^{4}-144 x^{3}+432 x^{2}-432 x \right ) {\mathrm e}^{3 x}+4 x^{5}-48 x^{4}+216 x^{3}-432 x^{2}+324 x}d x \] Input:

int((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(log(x)-ex 
p(3*x)^2+(6-2*x)*exp(3*x)-x^2+6*x-9))/(4*x*log(x)^2+(-8*x*exp(3*x)^2+(-16* 
x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*log(x)+4*x*exp(3*x)^4+(16*x^2-48*x)* 
exp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2-432*x 
)*exp(3*x)+4*x^5-48*x^4+216*x^3-432*x^2+324*x),x)
 

Output:

int((18*x*exp(3*x)^2+(18*x^2-48*x)*exp(3*x)+6*x^2-18*x-3)*exp(1/(log(x)-ex 
p(3*x)^2+(6-2*x)*exp(3*x)-x^2+6*x-9))/(4*x*log(x)^2+(-8*x*exp(3*x)^2+(-16* 
x^2+48*x)*exp(3*x)-8*x^3+48*x^2-72*x)*log(x)+4*x*exp(3*x)^4+(16*x^2-48*x)* 
exp(3*x)^3+(24*x^3-144*x^2+216*x)*exp(3*x)^2+(16*x^4-144*x^3+432*x^2-432*x 
)*exp(3*x)+4*x^5-48*x^4+216*x^3-432*x^2+324*x),x)