\(\int \frac {2-33 x+16 x^2-2 x^3+(16 x-8 x^2+x^3) \log (3 e^{\frac {-1+(-8 x+2 x^2) \log (x)}{-8 x+2 x^2}} x)}{(16 x-8 x^2+x^3) \log ^2(3 e^{\frac {-1+(-8 x+2 x^2) \log (x)}{-8 x+2 x^2}} x)} \, dx\) [2086]

Optimal result

Integrand size = 109, antiderivative size = 26 \[ \int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )} \, dx=\frac {x}{\log \left (3 e^{\frac {1}{2 (4-x) x}} x^2\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )} \, dx=\frac {x}{\log \left (3 e^{\frac {1}{8 x-2 x^2}} x^2\right )} \] Input:

Integrate[(2 - 33*x + 16*x^2 - 2*x^3 + (16*x - 8*x^2 + x^3)*Log[3*E^((-1 + 
 (-8*x + 2*x^2)*Log[x])/(-8*x + 2*x^2))*x])/((16*x - 8*x^2 + x^3)*Log[3*E^ 
((-1 + (-8*x + 2*x^2)*Log[x])/(-8*x + 2*x^2))*x]^2),x]
 

Output:

x/Log[3*E^(8*x - 2*x^2)^(-1)*x^2]
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2026, 7277, 27, 7238}

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.

\(\displaystyle \int \frac {-2 x^3+16 x^2+\left (x^3-8 x^2+16 x\right ) \log \left (3 x e^{\frac {\left (2 x^2-8 x\right ) \log (x)-1}{2 x^2-8 x}}\right )-33 x+2}{\left (x^3-8 x^2+16 x\right ) \log ^2\left (3 x e^{\frac {\left (2 x^2-8 x\right ) \log (x)-1}{2 x^2-8 x}}\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {-2 x^3+16 x^2+\left (x^3-8 x^2+16 x\right ) \log \left (3 x e^{\frac {\left (2 x^2-8 x\right ) \log (x)-1}{2 x^2-8 x}}\right )-33 x+2}{x \left (x^2-8 x+16\right ) \log ^2\left (3 x e^{\frac {\left (2 x^2-8 x\right ) \log (x)-1}{2 x^2-8 x}}\right )}dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 4 \int \frac {-2 x^3+16 x^2-33 x+\left (x^3-8 x^2+16 x\right ) \log \left (3 e^{\frac {1}{2 \left (4 x-x^2\right )}} x^2\right )+2}{4 (4-x)^2 x \log ^2\left (3 e^{\frac {1}{2 \left (4 x-x^2\right )}} x^2\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {-2 x^3+16 x^2+\left (x^3-8 x^2+16 x\right ) \log \left (3 e^{\frac {1}{2 \left (4 x-x^2\right )}} x^2\right )-33 x+2}{(4-x)^2 x \log ^2\left (3 e^{\frac {1}{2 \left (4 x-x^2\right )}} x^2\right )}dx\)

\(\Big \downarrow \) 7238

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

Input:

Int[(2 - 33*x + 16*x^2 - 2*x^3 + (16*x - 8*x^2 + x^3)*Log[3*E^((-1 + (-8*x 
 + 2*x^2)*Log[x])/(-8*x + 2*x^2))*x])/((16*x - 8*x^2 + x^3)*Log[3*E^((-1 + 
 (-8*x + 2*x^2)*Log[x])/(-8*x + 2*x^2))*x]^2),x]
 

Output:

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

Defintions of rubi rules used

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 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 7238

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y* 
z, u*z^(n - m), x]}, Simp[q*y^(m + 1)*(z^(m + 1)/(m + 1)), x] /;  !FalseQ[q 
]] /; FreeQ[{m, n}, x] && NeQ[m, -1]
 

rule 7277

Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> 
 Simp[1/(4^p*c^p)   Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} 
, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] &&  !AlgebraicFu 
nctionQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(57\) vs. \(2(24)=48\).

Time = 3.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.23

method	result	size
parallelrisch	\(\frac {98304 x^{5}-786432 x^{4}+1572864 x^{3}}{98304 x^{2} \ln \left (3 x \,{\mathrm e}^{\frac {\left (2 x^{2}-8 x \right ) \ln \left (x \right )-1}{2 \left (x -4\right ) x}}\right ) \left (x -4\right )^{2}}\)	\(58\)
risch	\(\frac {2 i x}{-\pi \,\operatorname {csgn}\left (i x^{\frac {x}{x -4}} x^{-\frac {4}{x -4}} {\mathrm e}^{-\frac {1}{2 \left (x -4\right ) x}}\right ) \operatorname {csgn}\left (i x \,x^{\frac {x}{x -4}} x^{-\frac {4}{x -4}} {\mathrm e}^{-\frac {1}{2 \left (x -4\right ) x}}\right )^{2}+\pi \,\operatorname {csgn}\left (i x^{\frac {x}{x -4}} x^{-\frac {4}{x -4}} {\mathrm e}^{-\frac {1}{2 \left (x -4\right ) x}}\right ) \operatorname {csgn}\left (i x \,x^{\frac {x}{x -4}} x^{-\frac {4}{x -4}} {\mathrm e}^{-\frac {1}{2 \left (x -4\right ) x}}\right ) \operatorname {csgn}\left (i x \right )+\pi \operatorname {csgn}\left (i x \,x^{\frac {x}{x -4}} x^{-\frac {4}{x -4}} {\mathrm e}^{-\frac {1}{2 \left (x -4\right ) x}}\right )^{3}-\pi \operatorname {csgn}\left (i x \,x^{\frac {x}{x -4}} x^{-\frac {4}{x -4}} {\mathrm e}^{-\frac {1}{2 \left (x -4\right ) x}}\right )^{2} \operatorname {csgn}\left (i x \right )+2 i \ln \left (3\right )+2 i \ln \left (x \right )+2 i \ln \left (x^{\frac {x}{x -4}} x^{-\frac {4}{x -4}} {\mathrm e}^{-\frac {1}{2 \left (x -4\right ) x}}\right )}\)	\(280\)

Input:

int(((x^3-8*x^2+16*x)*ln(3*x*exp(((2*x^2-8*x)*ln(x)-1)/(2*x^2-8*x)))-2*x^3 
+16*x^2-33*x+2)/(x^3-8*x^2+16*x)/ln(3*x*exp(((2*x^2-8*x)*ln(x)-1)/(2*x^2-8 
*x)))^2,x,method=_RETURNVERBOSE)
 

Output:

1/98304*(98304*x^5-786432*x^4+1572864*x^3)/x^2/ln(3*x*exp(1/2*((2*x^2-8*x) 
*ln(x)-1)/(x-4)/x))/(x-4)^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )} \, dx=\frac {2 \, {\left (x^{3} - 4 \, x^{2}\right )}}{2 \, {\left (x^{2} - 4 \, x\right )} \log \left (3\right ) + 4 \, {\left (x^{2} - 4 \, x\right )} \log \left (x\right ) - 1} \] Input:

integrate(((x^3-8*x^2+16*x)*log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x) 
))-2*x^3+16*x^2-33*x+2)/(x^3-8*x^2+16*x)/log(3*x*exp(((2*x^2-8*x)*log(x)-1 
)/(2*x^2-8*x)))^2,x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).

Time = 0.42 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )} \, dx=\frac {2 x^{3} - 8 x^{2}}{2 x^{2} \log {\left (3 \right )} - 8 x \log {\left (3 \right )} + \left (4 x^{2} - 16 x\right ) \log {\left (x \right )} - 1} \] Input:

integrate(((x**3-8*x**2+16*x)*ln(3*x*exp(((2*x**2-8*x)*ln(x)-1)/(2*x**2-8* 
x)))-2*x**3+16*x**2-33*x+2)/(x**3-8*x**2+16*x)/ln(3*x*exp(((2*x**2-8*x)*ln 
(x)-1)/(2*x**2-8*x)))**2,x)
 

Output:

(2*x**3 - 8*x**2)/(2*x**2*log(3) - 8*x*log(3) + (4*x**2 - 16*x)*log(x) - 1 
)
 

Maxima [F]

\[ \int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )} \, dx=\int { -\frac {2 \, x^{3} - 16 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \left (3 \, x e^{\left (\frac {2 \, {\left (x^{2} - 4 \, x\right )} \log \left (x\right ) - 1}{2 \, {\left (x^{2} - 4 \, x\right )}}\right )}\right ) + 33 \, x - 2}{{\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \left (3 \, x e^{\left (\frac {2 \, {\left (x^{2} - 4 \, x\right )} \log \left (x\right ) - 1}{2 \, {\left (x^{2} - 4 \, x\right )}}\right )}\right )^{2}} \,d x } \] Input:

integrate(((x^3-8*x^2+16*x)*log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x) 
))-2*x^3+16*x^2-33*x+2)/(x^3-8*x^2+16*x)/log(3*x*exp(((2*x^2-8*x)*log(x)-1 
)/(2*x^2-8*x)))^2,x, algorithm="maxima")
 

Output:

-2*(2*x^6 - 24*x^5 - 2*(x^6 - 12*x^5 + 48*x^4 - 64*x^3)*x^3 + 96*x^4 + 16* 
(x^6 - 12*x^5 + 48*x^4 - 64*x^3)*x^2 - 128*x^3 - 33*(x^6 - 12*x^5 + 48*x^4 
 - 64*x^3)*x)/((4*x^5*log(3) - 48*x^4*log(3) + 2*x^3*(97*log(3) - 1) - 4*x 
^2*(67*log(3) - 4) + x*(16*log(3) - 33) + 2)*x^3 - 8*(4*x^5*log(3) - 48*x^ 
4*log(3) + 2*x^3*(97*log(3) - 1) - 4*x^2*(67*log(3) - 4) + x*(16*log(3) - 
33) + 2)*x^2 + 16*(4*x^5*log(3) - 48*x^4*log(3) + 2*x^3*(97*log(3) - 1) - 
4*x^2*(67*log(3) - 4) + x*(16*log(3) - 33) + 2)*x + 4*((2*x^5 - 24*x^4 + 9 
7*x^3 - 134*x^2 + 8*x)*x^3 - 8*(2*x^5 - 24*x^4 + 97*x^3 - 134*x^2 + 8*x)*x 
^2 + 16*(2*x^5 - 24*x^4 + 97*x^3 - 134*x^2 + 8*x)*x)*log(x)) - integrate(2 
*(16*x^6 + (2*x^6 - 16*x^5 + 31*x^4 + 8*x^3 - 20*x^2 + 16*x)*x^4 - 176*x^5 
 - 12*(2*x^6 - 16*x^5 + 31*x^4 + 8*x^3 - 20*x^2 + 16*x)*x^3 + 704*x^4 - (2 
*x^8 - 40*x^7 + 223*x^6 - 504*x^5 + 1056*x^4 - 2432*x^3 + 960*x^2 - 768*x) 
*x^2 - 1280*x^3 - 4*(35*x^6 - 289*x^5 + 628*x^4 - 112*x^3 - 128*x^2 + 256* 
x)*x + 1024*x^2)/((8*x^8*log(3) - 160*x^7*log(3) + 4*x^6*(322*log(3) - 1) 
- 16*x^5*(327*log(3) - 4) + 2*x^4*(5409*log(3) - 194) - 8*x^3*(1186*log(3) 
 - 133) + x^2*(1064*log(3) - 1153) - 4*x*(8*log(3) - 33) - 4)*x^4 - 12*(8* 
x^8*log(3) - 160*x^7*log(3) + 4*x^6*(322*log(3) - 1) - 16*x^5*(327*log(3) 
- 4) + 2*x^4*(5409*log(3) - 194) - 8*x^3*(1186*log(3) - 133) + x^2*(1064*l 
og(3) - 1153) - 4*x*(8*log(3) - 33) - 4)*x^3 + 48*(8*x^8*log(3) - 160*x^7* 
log(3) + 4*x^6*(322*log(3) - 1) - 16*x^5*(327*log(3) - 4) + 2*x^4*(5409...
 

Giac [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )} \, dx=\frac {2 \, {\left (x^{3} - 4 \, x^{2}\right )}}{2 \, x^{2} \log \left (3\right ) + 4 \, x^{2} \log \left (x\right ) - 8 \, x \log \left (3\right ) - 16 \, x \log \left (x\right ) - 1} \] Input:

integrate(((x^3-8*x^2+16*x)*log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x) 
))-2*x^3+16*x^2-33*x+2)/(x^3-8*x^2+16*x)/log(3*x*exp(((2*x^2-8*x)*log(x)-1 
)/(2*x^2-8*x)))^2,x, algorithm="giac")
 

Output:

2*(x^3 - 4*x^2)/(2*x^2*log(3) + 4*x^2*log(x) - 8*x*log(3) - 16*x*log(x) - 
1)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )} \, dx=\int \frac {\ln \left (3\,x\,{\mathrm {e}}^{\frac {\ln \left (x\right )\,\left (8\,x-2\,x^2\right )+1}{8\,x-2\,x^2}}\right )\,\left (x^3-8\,x^2+16\,x\right )-33\,x+16\,x^2-2\,x^3+2}{{\ln \left (3\,x\,{\mathrm {e}}^{\frac {\ln \left (x\right )\,\left (8\,x-2\,x^2\right )+1}{8\,x-2\,x^2}}\right )}^2\,\left (x^3-8\,x^2+16\,x\right )} \,d x \] Input:

int((log(3*x*exp((log(x)*(8*x - 2*x^2) + 1)/(8*x - 2*x^2)))*(16*x - 8*x^2 
+ x^3) - 33*x + 16*x^2 - 2*x^3 + 2)/(log(3*x*exp((log(x)*(8*x - 2*x^2) + 1 
)/(8*x - 2*x^2)))^2*(16*x - 8*x^2 + x^3)),x)
 

Output:

int((log(3*x*exp((log(x)*(8*x - 2*x^2) + 1)/(8*x - 2*x^2)))*(16*x - 8*x^2 
+ x^3) - 33*x + 16*x^2 - 2*x^3 + 2)/(log(3*x*exp((log(x)*(8*x - 2*x^2) + 1 
)/(8*x - 2*x^2)))^2*(16*x - 8*x^2 + x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {2-33 x+16 x^2-2 x^3+\left (16 x-8 x^2+x^3\right ) \log \left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (3 e^{\frac {-1+\left (-8 x+2 x^2\right ) \log (x)}{-8 x+2 x^2}} x\right )} \, dx=\frac {x}{\mathrm {log}\left (\frac {3 x^{2}}{e^{\frac {1}{2 x^{2}-8 x}}}\right )} \] Input:

int(((x^3-8*x^2+16*x)*log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x^2-8*x)))-2*x 
^3+16*x^2-33*x+2)/(x^3-8*x^2+16*x)/log(3*x*exp(((2*x^2-8*x)*log(x)-1)/(2*x 
^2-8*x)))^2,x)
 

Output:

x/log((3*x**2)/e**(1/(2*x**2 - 8*x)))