\(\int \frac {4+11 x+5 x^2+x^3+(20+55 x+25 x^2+5 x^3) \log (3)+(-4-7 x-2 x^2+(-20-35 x-10 x^2) \log (3)) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+(8 x-2 x^2-2 x^3) \log (x)+(-8 x-2 x^2) \log ^2(x)+2 x \log ^3(x)} \, dx\) [1076]

Optimal result

Integrand size = 130, antiderivative size = 30 \[ \int \frac {4+11 x+5 x^2+x^3+\left (20+55 x+25 x^2+5 x^3\right ) \log (3)+\left (-4-7 x-2 x^2+\left (-20-35 x-10 x^2\right ) \log (3)\right ) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+\left (8 x-2 x^2-2 x^3\right ) \log (x)+\left (-8 x-2 x^2\right ) \log ^2(x)+2 x \log ^3(x)} \, dx=\frac {(x+5 x \log (3)) \log \left (x+\frac {x}{2+x-\log (x)}+\log (x)\right )}{2 x} \] Output:

1/2*(5*x*ln(3)+x)*ln(ln(x)+x/(2-ln(x)+x)+x)/x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {4+11 x+5 x^2+x^3+\left (20+55 x+25 x^2+5 x^3\right ) \log (3)+\left (-4-7 x-2 x^2+\left (-20-35 x-10 x^2\right ) \log (3)\right ) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+\left (8 x-2 x^2-2 x^3\right ) \log (x)+\left (-8 x-2 x^2\right ) \log ^2(x)+2 x \log ^3(x)} \, dx=\frac {1}{2} (1+\log (243)) \left (-\log (2+x-\log (x))+\log \left (3 x+x^2+2 \log (x)-\log ^2(x)\right )\right ) \] Input:

Integrate[(4 + 11*x + 5*x^2 + x^3 + (20 + 55*x + 25*x^2 + 5*x^3)*Log[3] + 
(-4 - 7*x - 2*x^2 + (-20 - 35*x - 10*x^2)*Log[3])*Log[x] + (1 + x + (5 + 5 
*x)*Log[3])*Log[x]^2)/(12*x^2 + 10*x^3 + 2*x^4 + (8*x - 2*x^2 - 2*x^3)*Log 
[x] + (-8*x - 2*x^2)*Log[x]^2 + 2*x*Log[x]^3),x]
 

Output:

((1 + Log[243])*(-Log[2 + x - Log[x]] + Log[3*x + x^2 + 2*Log[x] - Log[x]^ 
2]))/2
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {7292, 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.

\(\displaystyle \int \frac {x^3+5 x^2+\left (-2 x^2+\left (-10 x^2-35 x-20\right ) \log (3)-7 x-4\right ) \log (x)+\left (5 x^3+25 x^2+55 x+20\right ) \log (3)+11 x+(x+(5 x+5) \log (3)+1) \log ^2(x)+4}{2 x^4+10 x^3+12 x^2+\left (-2 x^2-8 x\right ) \log ^2(x)+\left (-2 x^3-2 x^2+8 x\right ) \log (x)+2 x \log ^3(x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {(1+\log (243)) \left (x^3+5 x^2-2 x^2 \log (x)+11 x+x \log ^2(x)+\log ^2(x)-7 x \log (x)-4 \log (x)+4\right )}{2 x \left (x^3+5 x^2-x^2 \log (x)+6 x+\log ^3(x)-x \log ^2(x)-4 \log ^2(x)-x \log (x)+4 \log (x)\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} (1+\log (243)) \int \frac {x^3-2 \log (x) x^2+5 x^2+\log ^2(x) x-7 \log (x) x+11 x+\log ^2(x)-4 \log (x)+4}{x \left (x^3-\log (x) x^2+5 x^2-\log ^2(x) x-\log (x) x+6 x+\log ^3(x)-4 \log ^2(x)+4 \log (x)\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{2} (1+\log (243)) \int \left (\frac {1-x}{x (x-\log (x)+2)}+\frac {2 x^2+3 x-2 \log (x)+2}{x \left (x^2+3 x-\log ^2(x)+2 \log (x)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} (1+\log (243)) \left (\log \left (x^2+3 x-\log ^2(x)+2 \log (x)\right )-\log (x-\log (x)+2)\right )\)

Input:

Int[(4 + 11*x + 5*x^2 + x^3 + (20 + 55*x + 25*x^2 + 5*x^3)*Log[3] + (-4 - 
7*x - 2*x^2 + (-20 - 35*x - 10*x^2)*Log[3])*Log[x] + (1 + x + (5 + 5*x)*Lo 
g[3])*Log[x]^2)/(12*x^2 + 10*x^3 + 2*x^4 + (8*x - 2*x^2 - 2*x^3)*Log[x] + 
(-8*x - 2*x^2)*Log[x]^2 + 2*x*Log[x]^3),x]
 

Output:

((1 + Log[243])*(-Log[2 + x - Log[x]] + Log[3*x + x^2 + 2*Log[x] - Log[x]^ 
2]))/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 2009

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

rule 7292

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

rule 7293

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

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40

method	result	size
norman	\(\left (-\frac {5 \ln \left (3\right )}{2}-\frac {1}{2}\right ) \ln \left (2-\ln \left (x \right )+x \right )+\left (\frac {5 \ln \left (3\right )}{2}+\frac {1}{2}\right ) \ln \left (x^{2}-\ln \left (x \right )^{2}+3 x +2 \ln \left (x \right )\right )\)	\(42\)
default	\(-\frac {\ln \left (-x -2+\ln \left (x \right )\right )}{2}+\frac {\ln \left (\ln \left (x \right )^{2}-x^{2}-2 \ln \left (x \right )-3 x \right )}{2}+\frac {5 \ln \left (3\right ) \left (-\ln \left (-x -2+\ln \left (x \right )\right )+\ln \left (\ln \left (x \right )^{2}-x^{2}-2 \ln \left (x \right )-3 x \right )\right )}{2}\)	\(65\)
risch	\(-\frac {5 \ln \left (-x -2+\ln \left (x \right )\right ) \ln \left (3\right )}{2}-\frac {\ln \left (-x -2+\ln \left (x \right )\right )}{2}+\frac {5 \ln \left (\ln \left (x \right )^{2}-x^{2}-2 \ln \left (x \right )-3 x \right ) \ln \left (3\right )}{2}+\frac {\ln \left (\ln \left (x \right )^{2}-x^{2}-2 \ln \left (x \right )-3 x \right )}{2}\)	\(66\)
parallelrisch	\(-\frac {5 \ln \left (2-\ln \left (x \right )+x \right ) \ln \left (3\right )}{2}-\frac {\ln \left (2-\ln \left (x \right )+x \right )}{2}+\frac {5 \ln \left (x^{2}-\ln \left (x \right )^{2}+3 x +2 \ln \left (x \right )\right ) \ln \left (3\right )}{2}+\frac {\ln \left (x^{2}-\ln \left (x \right )^{2}+3 x +2 \ln \left (x \right )\right )}{2}\)	\(66\)

Input:

int((((5*x+5)*ln(3)+x+1)*ln(x)^2+((-10*x^2-35*x-20)*ln(3)-2*x^2-7*x-4)*ln( 
x)+(5*x^3+25*x^2+55*x+20)*ln(3)+x^3+5*x^2+11*x+4)/(2*x*ln(x)^3+(-2*x^2-8*x 
)*ln(x)^2+(-2*x^3-2*x^2+8*x)*ln(x)+2*x^4+10*x^3+12*x^2),x,method=_RETURNVE 
RBOSE)
 

Output:

(-5/2*ln(3)-1/2)*ln(2-ln(x)+x)+(5/2*ln(3)+1/2)*ln(x^2-ln(x)^2+3*x+2*ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {4+11 x+5 x^2+x^3+\left (20+55 x+25 x^2+5 x^3\right ) \log (3)+\left (-4-7 x-2 x^2+\left (-20-35 x-10 x^2\right ) \log (3)\right ) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+\left (8 x-2 x^2-2 x^3\right ) \log (x)+\left (-8 x-2 x^2\right ) \log ^2(x)+2 x \log ^3(x)} \, dx=\frac {1}{2} \, {\left (5 \, \log \left (3\right ) + 1\right )} \log \left (-x^{2} + \log \left (x\right )^{2} - 3 \, x - 2 \, \log \left (x\right )\right ) - \frac {1}{2} \, {\left (5 \, \log \left (3\right ) + 1\right )} \log \left (-x + \log \left (x\right ) - 2\right ) \] Input:

integrate((((5*x+5)*log(3)+x+1)*log(x)^2+((-10*x^2-35*x-20)*log(3)-2*x^2-7 
*x-4)*log(x)+(5*x^3+25*x^2+55*x+20)*log(3)+x^3+5*x^2+11*x+4)/(2*x*log(x)^3 
+(-2*x^2-8*x)*log(x)^2+(-2*x^3-2*x^2+8*x)*log(x)+2*x^4+10*x^3+12*x^2),x, a 
lgorithm="fricas")
 

Output:

1/2*(5*log(3) + 1)*log(-x^2 + log(x)^2 - 3*x - 2*log(x)) - 1/2*(5*log(3) + 
 1)*log(-x + log(x) - 2)
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {4+11 x+5 x^2+x^3+\left (20+55 x+25 x^2+5 x^3\right ) \log (3)+\left (-4-7 x-2 x^2+\left (-20-35 x-10 x^2\right ) \log (3)\right ) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+\left (8 x-2 x^2-2 x^3\right ) \log (x)+\left (-8 x-2 x^2\right ) \log ^2(x)+2 x \log ^3(x)} \, dx=- \frac {\left (1 + 5 \log {\left (3 \right )}\right ) \log {\left (- x + \log {\left (x \right )} - 2 \right )}}{2} + \frac {\left (1 + 5 \log {\left (3 \right )}\right ) \log {\left (- x^{2} - 3 x + \log {\left (x \right )}^{2} - 2 \log {\left (x \right )} \right )}}{2} \] Input:

integrate((((5*x+5)*ln(3)+x+1)*ln(x)**2+((-10*x**2-35*x-20)*ln(3)-2*x**2-7 
*x-4)*ln(x)+(5*x**3+25*x**2+55*x+20)*ln(3)+x**3+5*x**2+11*x+4)/(2*x*ln(x)* 
*3+(-2*x**2-8*x)*ln(x)**2+(-2*x**3-2*x**2+8*x)*ln(x)+2*x**4+10*x**3+12*x** 
2),x)
 

Output:

-(1 + 5*log(3))*log(-x + log(x) - 2)/2 + (1 + 5*log(3))*log(-x**2 - 3*x + 
log(x)**2 - 2*log(x))/2
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {4+11 x+5 x^2+x^3+\left (20+55 x+25 x^2+5 x^3\right ) \log (3)+\left (-4-7 x-2 x^2+\left (-20-35 x-10 x^2\right ) \log (3)\right ) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+\left (8 x-2 x^2-2 x^3\right ) \log (x)+\left (-8 x-2 x^2\right ) \log ^2(x)+2 x \log ^3(x)} \, dx=\frac {1}{2} \, {\left (5 \, \log \left (3\right ) + 1\right )} \log \left (-x^{2} + \log \left (x\right )^{2} - 3 \, x - 2 \, \log \left (x\right )\right ) - \frac {1}{2} \, {\left (5 \, \log \left (3\right ) + 1\right )} \log \left (-x + \log \left (x\right ) - 2\right ) \] Input:

integrate((((5*x+5)*log(3)+x+1)*log(x)^2+((-10*x^2-35*x-20)*log(3)-2*x^2-7 
*x-4)*log(x)+(5*x^3+25*x^2+55*x+20)*log(3)+x^3+5*x^2+11*x+4)/(2*x*log(x)^3 
+(-2*x^2-8*x)*log(x)^2+(-2*x^3-2*x^2+8*x)*log(x)+2*x^4+10*x^3+12*x^2),x, a 
lgorithm="maxima")
 

Output:

1/2*(5*log(3) + 1)*log(-x^2 + log(x)^2 - 3*x - 2*log(x)) - 1/2*(5*log(3) + 
 1)*log(-x + log(x) - 2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {4+11 x+5 x^2+x^3+\left (20+55 x+25 x^2+5 x^3\right ) \log (3)+\left (-4-7 x-2 x^2+\left (-20-35 x-10 x^2\right ) \log (3)\right ) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+\left (8 x-2 x^2-2 x^3\right ) \log (x)+\left (-8 x-2 x^2\right ) \log ^2(x)+2 x \log ^3(x)} \, dx=\frac {1}{2} \, {\left (5 \, \log \left (3\right ) + 1\right )} \log \left (x^{2} - \log \left (x\right )^{2} + 3 \, x + 2 \, \log \left (x\right )\right ) - \frac {1}{2} \, {\left (5 \, \log \left (3\right ) + 1\right )} \log \left (x - \log \left (x\right ) + 2\right ) \] Input:

integrate((((5*x+5)*log(3)+x+1)*log(x)^2+((-10*x^2-35*x-20)*log(3)-2*x^2-7 
*x-4)*log(x)+(5*x^3+25*x^2+55*x+20)*log(3)+x^3+5*x^2+11*x+4)/(2*x*log(x)^3 
+(-2*x^2-8*x)*log(x)^2+(-2*x^3-2*x^2+8*x)*log(x)+2*x^4+10*x^3+12*x^2),x, a 
lgorithm="giac")
 

Output:

1/2*(5*log(3) + 1)*log(x^2 - log(x)^2 + 3*x + 2*log(x)) - 1/2*(5*log(3) + 
1)*log(x - log(x) + 2)
 

Mupad [B] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {4+11 x+5 x^2+x^3+\left (20+55 x+25 x^2+5 x^3\right ) \log (3)+\left (-4-7 x-2 x^2+\left (-20-35 x-10 x^2\right ) \log (3)\right ) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+\left (8 x-2 x^2-2 x^3\right ) \log (x)+\left (-8 x-2 x^2\right ) \log ^2(x)+2 x \log ^3(x)} \, dx=\left (\frac {\ln \left (243\right )}{2}+\frac {1}{2}\right )\,\left (\ln \left (x^2+3\,x-{\ln \left (x\right )}^2+2\,\ln \left (x\right )\right )-\ln \left (x-\ln \left (x\right )+2\right )\right ) \] Input:

int((11*x + log(3)*(55*x + 25*x^2 + 5*x^3 + 20) + log(x)^2*(x + log(3)*(5* 
x + 5) + 1) + 5*x^2 + x^3 - log(x)*(7*x + log(3)*(35*x + 10*x^2 + 20) + 2* 
x^2 + 4) + 4)/(2*x*log(x)^3 - log(x)^2*(8*x + 2*x^2) + 12*x^2 + 10*x^3 + 2 
*x^4 - log(x)*(2*x^2 - 8*x + 2*x^3)),x)
 

Output:

(log(243)/2 + 1/2)*(log(3*x + 2*log(x) - log(x)^2 + x^2) - log(x - log(x) 
+ 2))
 

Reduce [F]

\[ \int \frac {4+11 x+5 x^2+x^3+\left (20+55 x+25 x^2+5 x^3\right ) \log (3)+\left (-4-7 x-2 x^2+\left (-20-35 x-10 x^2\right ) \log (3)\right ) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+\left (8 x-2 x^2-2 x^3\right ) \log (x)+\left (-8 x-2 x^2\right ) \log ^2(x)+2 x \log ^3(x)} \, dx =\text {Too large to display} \] Input:

int((((5*x+5)*log(3)+x+1)*log(x)^2+((-10*x^2-35*x-20)*log(3)-2*x^2-7*x-4)* 
log(x)+(5*x^3+25*x^2+55*x+20)*log(3)+x^3+5*x^2+11*x+4)/(2*x*log(x)^3+(-2*x 
^2-8*x)*log(x)^2+(-2*x^3-2*x^2+8*x)*log(x)+2*x^4+10*x^3+12*x^2),x)
 

Output:

(5*int(log(x)**2/(log(x)**3 - log(x)**2*x - 4*log(x)**2 - log(x)*x**2 - lo 
g(x)*x + 4*log(x) + x**3 + 5*x**2 + 6*x),x)*log(3) + int(log(x)**2/(log(x) 
**3 - log(x)**2*x - 4*log(x)**2 - log(x)*x**2 - log(x)*x + 4*log(x) + x**3 
 + 5*x**2 + 6*x),x) + 5*int(log(x)**2/(log(x)**3*x - log(x)**2*x**2 - 4*lo 
g(x)**2*x - log(x)*x**3 - log(x)*x**2 + 4*log(x)*x + x**4 + 5*x**3 + 6*x** 
2),x)*log(3) + int(log(x)**2/(log(x)**3*x - log(x)**2*x**2 - 4*log(x)**2*x 
 - log(x)*x**3 - log(x)*x**2 + 4*log(x)*x + x**4 + 5*x**3 + 6*x**2),x) + 5 
*int(x**2/(log(x)**3 - log(x)**2*x - 4*log(x)**2 - log(x)*x**2 - log(x)*x 
+ 4*log(x) + x**3 + 5*x**2 + 6*x),x)*log(3) + int(x**2/(log(x)**3 - log(x) 
**2*x - 4*log(x)**2 - log(x)*x**2 - log(x)*x + 4*log(x) + x**3 + 5*x**2 + 
6*x),x) - 35*int(log(x)/(log(x)**3 - log(x)**2*x - 4*log(x)**2 - log(x)*x* 
*2 - log(x)*x + 4*log(x) + x**3 + 5*x**2 + 6*x),x)*log(3) - 7*int(log(x)/( 
log(x)**3 - log(x)**2*x - 4*log(x)**2 - log(x)*x**2 - log(x)*x + 4*log(x) 
+ x**3 + 5*x**2 + 6*x),x) - 20*int(log(x)/(log(x)**3*x - log(x)**2*x**2 - 
4*log(x)**2*x - log(x)*x**3 - log(x)*x**2 + 4*log(x)*x + x**4 + 5*x**3 + 6 
*x**2),x)*log(3) - 4*int(log(x)/(log(x)**3*x - log(x)**2*x**2 - 4*log(x)** 
2*x - log(x)*x**3 - log(x)*x**2 + 4*log(x)*x + x**4 + 5*x**3 + 6*x**2),x) 
- 10*int((log(x)*x)/(log(x)**3 - log(x)**2*x - 4*log(x)**2 - log(x)*x**2 - 
 log(x)*x + 4*log(x) + x**3 + 5*x**2 + 6*x),x)*log(3) - 2*int((log(x)*x)/( 
log(x)**3 - log(x)**2*x - 4*log(x)**2 - log(x)*x**2 - log(x)*x + 4*log(...