\(\int \frac {x^3+(x+8 x^2-2 x^3) \log (3)+(-4+15 x-8 x^2+x^3) \log ^2(3)+(x \log (3)-x \log ^2(3)) \log (x)}{x^3+(8 x^2-2 x^3) \log (3)+(16 x-8 x^2+x^3) \log ^2(3)} \, dx\) [732]

Optimal result

Integrand size = 90, antiderivative size = 22 \[ \int \frac {x^3+\left (x+8 x^2-2 x^3\right ) \log (3)+\left (-4+15 x-8 x^2+x^3\right ) \log ^2(3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (8 x^2-2 x^3\right ) \log (3)+\left (16 x-8 x^2+x^3\right ) \log ^2(3)} \, dx=-6+x-\frac {2+\log (x)}{4-x+\frac {x}{\log (3)}} \] Output:

x-6-(ln(x)+2)/(4-x+x/ln(3))
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {x^3+\left (x+8 x^2-2 x^3\right ) \log (3)+\left (-4+15 x-8 x^2+x^3\right ) \log ^2(3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (8 x^2-2 x^3\right ) \log (3)+\left (16 x-8 x^2+x^3\right ) \log ^2(3)} \, dx=\frac {x^2 (-1+\log (3))-4 x \log (3)+\log (9)+\log (3) \log (x)}{x (-1+\log (3))-4 \log (3)} \] Input:

Integrate[(x^3 + (x + 8*x^2 - 2*x^3)*Log[3] + (-4 + 15*x - 8*x^2 + x^3)*Lo 
g[3]^2 + (x*Log[3] - x*Log[3]^2)*Log[x])/(x^3 + (8*x^2 - 2*x^3)*Log[3] + ( 
16*x - 8*x^2 + x^3)*Log[3]^2),x]
 

Output:

(x^2*(-1 + Log[3]) - 4*x*Log[3] + Log[9] + Log[3]*Log[x])/(x*(-1 + Log[3]) 
 - 4*Log[3])
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(154\) vs. \(2(22)=44\).

Time = 1.21 (sec) , antiderivative size = 154, normalized size of antiderivative = 7.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2026, 7277, 27, 7293, 2009}

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.

\(\displaystyle \int \frac {x^3+\left (x^3-8 x^2+15 x-4\right ) \log ^2(3)+\left (-2 x^3+8 x^2+x\right ) \log (3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (x^3-8 x^2+16 x\right ) \log ^2(3)+\left (8 x^2-2 x^3\right ) \log (3)} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {x^3+\left (x^3-8 x^2+15 x-4\right ) \log ^2(3)+\left (-2 x^3+8 x^2+x\right ) \log (3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x \left (x^2 (1-\log (3))^2+8 x (1-\log (3)) \log (3)+16 \log ^2(3)\right )}dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 4 (1-\log (3))^2 \int \frac {x^3+(1-\log (3)) \log (3) \log (x) x-\left (-x^3+8 x^2-15 x+4\right ) \log ^2(3)+\left (-2 x^3+8 x^2+x\right ) \log (3)}{4 x (1-\log (3))^2 ((1-\log (3)) x+4 \log (3))^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {x^3-\left (-x^3+8 x^2-15 x+4\right ) \log ^2(3)+\left (-2 x^3+8 x^2+x\right ) \log (3)+x (1-\log (3)) \log (3) \log (x)}{x (x (1-\log (3))+\log (81))^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^3 (1-\log (3))^2+8 x^2 (1-\log (3)) \log (3)+x \log (3) (1+15 \log (3))-4 \log ^2(3)}{x (x (1-\log (3))+\log (81))^2}+\frac {(1-\log (3)) \log (3) \log (x)}{(x (1-\log (3))+\log (81))^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle x-\frac {4 \log ^2(3) \log (x)}{\log ^2(81)}+\frac {4 \log ^2(3) \log (-(x (1-\log (3)))-\log (81))}{\log ^2(81)}+\frac {4 \log ^3(3)-\log ^3(81)-\log ^2(3) (4+15 \log (81))-\log (3) (1-8 \log (81)) \log (81)}{(1-\log (3)) \log (81) (x (1-\log (3))+\log (81))}+\frac {x (1-\log (3)) \log (3) \log (x)}{\log (81) (x (1-\log (3))+\log (81))}-\frac {\log (3) \log (-(x (1-\log (3)))-\log (81))}{\log (81)}\)

Input:

Int[(x^3 + (x + 8*x^2 - 2*x^3)*Log[3] + (-4 + 15*x - 8*x^2 + x^3)*Log[3]^2 
 + (x*Log[3] - x*Log[3]^2)*Log[x])/(x^3 + (8*x^2 - 2*x^3)*Log[3] + (16*x - 
 8*x^2 + x^3)*Log[3]^2),x]
 

Output:

x + (4*Log[3]^3 - Log[3]*(1 - 8*Log[81])*Log[81] - Log[81]^3 - Log[3]^2*(4 
 + 15*Log[81]))/((1 - Log[3])*Log[81]*(x*(1 - Log[3]) + Log[81])) - (4*Log 
[3]^2*Log[x])/Log[81]^2 + (x*(1 - Log[3])*Log[3]*Log[x])/(Log[81]*(x*(1 - 
Log[3]) + Log[81])) + (4*Log[3]^2*Log[-(x*(1 - Log[3])) - Log[81]])/Log[81 
]^2 - (Log[3]*Log[-(x*(1 - Log[3])) - Log[81]])/Log[81]
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 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]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73

method	result	size
norman	\(\frac {\ln \left (3\right ) \ln \left (x \right )+\left (\ln \left (3\right )-1\right ) x^{2}+\left (-\frac {7 \ln \left (3\right )}{2}-\frac {1}{2}\right ) x}{x \ln \left (3\right )-4 \ln \left (3\right )-x}\)	\(38\)
risch	\(\frac {\ln \left (3\right ) \ln \left (x \right )}{x \ln \left (3\right )-4 \ln \left (3\right )-x}+\frac {x^{2} \ln \left (3\right )-4 x \ln \left (3\right )-x^{2}+2 \ln \left (3\right )}{x \ln \left (3\right )-4 \ln \left (3\right )-x}\)	\(57\)
parallelrisch	\(\frac {4 x^{2} \ln \left (3\right )^{2}-14 x \ln \left (3\right )^{2}+4 \ln \left (3\right )^{2} \ln \left (x \right )-4 x^{2} \ln \left (3\right )-2 x \ln \left (3\right )}{4 \ln \left (3\right ) \left (x \ln \left (3\right )-4 \ln \left (3\right )-x \right )}\)	\(58\)
default	\(x +\frac {\left (\frac {\ln \left (3\right )}{4}-\frac {1}{4}\right ) \ln \left (x \ln \left (3\right )-4 \ln \left (3\right )-x \right )}{\ln \left (3\right )-1}+\frac {2 \ln \left (3\right )}{x \ln \left (3\right )-4 \ln \left (3\right )-x}-\frac {\ln \left (x \right )}{4}-\ln \left (3\right ) \left (\ln \left (3\right )-1\right ) \left (\frac {\ln \left (\left (\ln \left (3\right )-1\right ) x -4 \ln \left (3\right )\right )}{4 \ln \left (3\right ) \left (\ln \left (3\right )-1\right )}-\frac {\ln \left (x \right ) x}{4 \ln \left (3\right ) \left (x \ln \left (3\right )-4 \ln \left (3\right )-x \right )}\right )\)	\(107\)
parts	\(x +\frac {\left (\frac {\ln \left (3\right )}{4}-\frac {1}{4}\right ) \ln \left (x \ln \left (3\right )-4 \ln \left (3\right )-x \right )}{\ln \left (3\right )-1}+\frac {2 \ln \left (3\right )}{x \ln \left (3\right )-4 \ln \left (3\right )-x}-\frac {\ln \left (x \right )}{4}-\ln \left (3\right ) \left (\ln \left (3\right )-1\right ) \left (\frac {\ln \left (\left (\ln \left (3\right )-1\right ) x -4 \ln \left (3\right )\right )}{4 \ln \left (3\right ) \left (\ln \left (3\right )-1\right )}-\frac {\ln \left (x \right ) x}{4 \ln \left (3\right ) \left (x \ln \left (3\right )-4 \ln \left (3\right )-x \right )}\right )\)	\(107\)

Input:

int(((-x*ln(3)^2+x*ln(3))*ln(x)+(x^3-8*x^2+15*x-4)*ln(3)^2+(-2*x^3+8*x^2+x 
)*ln(3)+x^3)/((x^3-8*x^2+16*x)*ln(3)^2+(-2*x^3+8*x^2)*ln(3)+x^3),x,method= 
_RETURNVERBOSE)
 

Output:

(ln(3)*ln(x)+(ln(3)-1)*x^2+(-7/2*ln(3)-1/2)*x)/(x*ln(3)-4*ln(3)-x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {x^3+\left (x+8 x^2-2 x^3\right ) \log (3)+\left (-4+15 x-8 x^2+x^3\right ) \log ^2(3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (8 x^2-2 x^3\right ) \log (3)+\left (16 x-8 x^2+x^3\right ) \log ^2(3)} \, dx=-\frac {x^{2} - {\left (x^{2} - 4 \, x + 2\right )} \log \left (3\right ) - \log \left (3\right ) \log \left (x\right )}{{\left (x - 4\right )} \log \left (3\right ) - x} \] Input:

integrate(((-x*log(3)^2+x*log(3))*log(x)+(x^3-8*x^2+15*x-4)*log(3)^2+(-2*x 
^3+8*x^2+x)*log(3)+x^3)/((x^3-8*x^2+16*x)*log(3)^2+(-2*x^3+8*x^2)*log(3)+x 
^3),x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (15) = 30\).

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {x^3+\left (x+8 x^2-2 x^3\right ) \log (3)+\left (-4+15 x-8 x^2+x^3\right ) \log ^2(3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (8 x^2-2 x^3\right ) \log (3)+\left (16 x-8 x^2+x^3\right ) \log ^2(3)} \, dx=x + \frac {\log {\left (3 \right )} \log {\left (x \right )}}{- x + x \log {\left (3 \right )} - 4 \log {\left (3 \right )}} + \frac {2 \log {\left (3 \right )}}{x \left (-1 + \log {\left (3 \right )}\right ) - 4 \log {\left (3 \right )}} \] Input:

integrate(((-x*ln(3)**2+x*ln(3))*ln(x)+(x**3-8*x**2+15*x-4)*ln(3)**2+(-2*x 
**3+8*x**2+x)*ln(3)+x**3)/((x**3-8*x**2+16*x)*ln(3)**2+(-2*x**3+8*x**2)*ln 
(3)+x**3),x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 737 vs. \(2 (20) = 40\).

Time = 0.05 (sec) , antiderivative size = 737, normalized size of antiderivative = 33.50 \[ \int \frac {x^3+\left (x+8 x^2-2 x^3\right ) \log (3)+\left (-4+15 x-8 x^2+x^3\right ) \log ^2(3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (8 x^2-2 x^3\right ) \log (3)+\left (16 x-8 x^2+x^3\right ) \log ^2(3)} \, dx=\text {Too large to display} \] Input:

integrate(((-x*log(3)^2+x*log(3))*log(x)+(x^3-8*x^2+15*x-4)*log(3)^2+(-2*x 
^3+8*x^2+x)*log(3)+x^3)/((x^3-8*x^2+16*x)*log(3)^2+(-2*x^3+8*x^2)*log(3)+x 
^3),x, algorithm="maxima")
 

Output:

(16*log(3)^2/(4*log(3)^4 - 12*log(3)^3 - (log(3)^4 - 4*log(3)^3 + 6*log(3) 
^2 - 4*log(3) + 1)*x + 12*log(3)^2 - 4*log(3)) + 8*log(3)*log(x*(log(3) - 
1) - 4*log(3))/(log(3)^3 - 3*log(3)^2 + 3*log(3) - 1) + x/(log(3)^2 - 2*lo 
g(3) + 1))*log(3)^2 - 8*(4*log(3)/(4*log(3)^3 - (log(3)^3 - 3*log(3)^2 + 3 
*log(3) - 1)*x - 8*log(3)^2 + 4*log(3)) + log(x*(log(3) - 1) - 4*log(3))/( 
log(3)^2 - 2*log(3) + 1))*log(3)^2 - 1/4*(log(x*(log(3) - 1) - 4*log(3))/( 
log(3)^2 - log(3)) - log(x)/(log(3)^2 - log(3)))*log(3)^2 + 1/4*(4/((log(3 
)^2 - log(3))*x - 4*log(3)^2) + log(x*(log(3) - 1) - 4*log(3))/log(3)^2 - 
log(x)/log(3)^2)*log(3)^2 - 2*(16*log(3)^2/(4*log(3)^4 - 12*log(3)^3 - (lo 
g(3)^4 - 4*log(3)^3 + 6*log(3)^2 - 4*log(3) + 1)*x + 12*log(3)^2 - 4*log(3 
)) + 8*log(3)*log(x*(log(3) - 1) - 4*log(3))/(log(3)^3 - 3*log(3)^2 + 3*lo 
g(3) - 1) + x/(log(3)^2 - 2*log(3) + 1))*log(3) + 8*(4*log(3)/(4*log(3)^3 
- (log(3)^3 - 3*log(3)^2 + 3*log(3) - 1)*x - 8*log(3)^2 + 4*log(3)) + log( 
x*(log(3) - 1) - 4*log(3))/(log(3)^2 - 2*log(3) + 1))*log(3) + 1/4*(log(x* 
(log(3) - 1) - 4*log(3))/(log(3)^2 - log(3)) - log(x)/(log(3)^2 - log(3))) 
*log(3) + log(3)^2*log(x)/((log(3)^2 - 2*log(3) + 1)*x - 4*log(3)^2 + 4*lo 
g(3)) + 16*log(3)^2/(4*log(3)^4 - 12*log(3)^3 - (log(3)^4 - 4*log(3)^3 + 6 
*log(3)^2 - 4*log(3) + 1)*x + 12*log(3)^2 - 4*log(3)) - 15*log(3)^2/((log( 
3)^2 - 2*log(3) + 1)*x - 4*log(3)^2 + 4*log(3)) + 8*log(3)*log(x*(log(3) - 
 1) - 4*log(3))/(log(3)^3 - 3*log(3)^2 + 3*log(3) - 1) - log(3)*log(x)/...
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {x^3+\left (x+8 x^2-2 x^3\right ) \log (3)+\left (-4+15 x-8 x^2+x^3\right ) \log ^2(3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (8 x^2-2 x^3\right ) \log (3)+\left (16 x-8 x^2+x^3\right ) \log ^2(3)} \, dx=x + \frac {\log \left (3\right ) \log \left (x\right )}{x \log \left (3\right ) - x - 4 \, \log \left (3\right )} + \frac {2 \, \log \left (3\right )}{x \log \left (3\right ) - x - 4 \, \log \left (3\right )} \] Input:

integrate(((-x*log(3)^2+x*log(3))*log(x)+(x^3-8*x^2+15*x-4)*log(3)^2+(-2*x 
^3+8*x^2+x)*log(3)+x^3)/((x^3-8*x^2+16*x)*log(3)^2+(-2*x^3+8*x^2)*log(3)+x 
^3),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 12.65 (sec) , antiderivative size = 7411, normalized size of antiderivative = 336.86 \[ \int \frac {x^3+\left (x+8 x^2-2 x^3\right ) \log (3)+\left (-4+15 x-8 x^2+x^3\right ) \log ^2(3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (8 x^2-2 x^3\right ) \log (3)+\left (16 x-8 x^2+x^3\right ) \log ^2(3)} \, dx=\text {Too large to display} \] Input:

int((log(3)*(x + 8*x^2 - 2*x^3) + log(3)^2*(15*x - 8*x^2 + x^3 - 4) + log( 
x)*(x*log(3) - x*log(3)^2) + x^3)/(log(3)^2*(16*x - 8*x^2 + x^3) + log(3)* 
(8*x^2 - 2*x^3) + x^3),x)
 

Output:

log((((log(3)^2*(log(9) - 2*log(3))^(1/2))/2 - (13*log(3)^3*(log(9) - 2*lo 
g(3))^(1/2))/2 + (3*log(3)^4*(log(9) - 2*log(3))^(1/2))/8 - (log(3)^5*(log 
(9) - 2*log(3))^(1/2))/4 - log(9)^2*((log(3)*(log(9) - 2*log(3))^(1/2))/8 
- (log(9) - 2*log(3))^(1/2)/8) + (7*log(3)*(log(9) - 2*log(3))^(1/2))/4 - 
8*log(3)^3 + 8*log(3)^4 + log(9)*((7*log(3)^2*(log(9) - 2*log(3))^(1/2))/2 
 + (3*log(3)^3*(log(9) - 2*log(3))^(1/2))/8 - (13*log(3)*(log(9) - 2*log(3 
))^(1/2))/8 + 4*log(3)^2 - 4*log(3)^3 - (log(9) - 2*log(3))^(1/2)/4) + (lo 
g(9) - 2*log(3))^(1/2)/8)/(2*log(3) - log(9) - 4*log(3)*log(9) + 2*log(3)* 
log(9)^2 - 2*log(3)^2*log(9) - 4*log(3)^3*log(9) - log(3)^4*log(9) + 4*log 
(3)^3 + 2*log(3)^5 + 2*log(9)^2 - log(9)^3 + 2*log(3)^2*log(9)^2) + 1/8)*( 
(4*(104*log(3)^5*log(9) - 44*log(3)^4*log(9) - 16*log(3)^3*log(9) + 20*log 
(3)^6*log(9) + 8*log(3)^3 + 52*log(3)^4 + 152*log(3)^5 - 472*log(3)^6 + 14 
4*log(3)^7 - 12*log(3)^8 + 8*log(3)^3*log(9)^2 - 8*log(3)^4*log(9)^2))/(lo 
g(3)^2 - log(9) + 1) - (((log(3)^2*(log(9) - 2*log(3))^(1/2))/2 - (13*log( 
3)^3*(log(9) - 2*log(3))^(1/2))/2 + (3*log(3)^4*(log(9) - 2*log(3))^(1/2)) 
/8 - (log(3)^5*(log(9) - 2*log(3))^(1/2))/4 - log(9)^2*((log(3)*(log(9) - 
2*log(3))^(1/2))/8 - (log(9) - 2*log(3))^(1/2)/8) + (7*log(3)*(log(9) - 2* 
log(3))^(1/2))/4 - 8*log(3)^3 + 8*log(3)^4 + log(9)*((7*log(3)^2*(log(9) - 
 2*log(3))^(1/2))/2 + (3*log(3)^3*(log(9) - 2*log(3))^(1/2))/8 - (13*log(3 
)*(log(9) - 2*log(3))^(1/2))/8 + 4*log(3)^2 - 4*log(3)^3 - (log(9) - 2*...
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.95 \[ \int \frac {x^3+\left (x+8 x^2-2 x^3\right ) \log (3)+\left (-4+15 x-8 x^2+x^3\right ) \log ^2(3)+\left (x \log (3)-x \log ^2(3)\right ) \log (x)}{x^3+\left (8 x^2-2 x^3\right ) \log (3)+\left (16 x-8 x^2+x^3\right ) \log ^2(3)} \, dx=\frac {2 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (3\right )+2 \,\mathrm {log}\left (3\right ) x^{2}-7 \,\mathrm {log}\left (3\right ) x -2 x^{2}-x}{2 \,\mathrm {log}\left (3\right ) x -8 \,\mathrm {log}\left (3\right )-2 x} \] Input:

int(((-x*log(3)^2+x*log(3))*log(x)+(x^3-8*x^2+15*x-4)*log(3)^2+(-2*x^3+8*x 
^2+x)*log(3)+x^3)/((x^3-8*x^2+16*x)*log(3)^2+(-2*x^3+8*x^2)*log(3)+x^3),x)
 

Output:

(2*log(x)*log(3) + 2*log(3)*x**2 - 7*log(3)*x - 2*x**2 - x)/(2*(log(3)*x - 
 4*log(3) - x))