\(\int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+(384-40 x+5 x^2+(96-10 x) \log (4)) \log (x)+(80+20 \log (4)) \log ^2(x)}{(256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)) \log (5)+(256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)) \log (5) \log (x)+(64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)) \log (5) \log ^2(x)} \, dx\) [821]

Optimal result

Integrand size = 141, antiderivative size = 34 \[ \int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+\left (384-40 x+5 x^2+(96-10 x) \log (4)\right ) \log (x)+(80+20 \log (4)) \log ^2(x)}{\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5)+\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5) \log (x)+\left (64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)\right ) \log (5) \log ^2(x)} \, dx=\frac {5+\frac {4-\frac {5 x}{4}}{2+\log (x)}}{-\log (5)+\frac {(4+\log (4)) \log (5)}{x}} \] Output:

(5+(4-5/4*x)/(ln(x)+2))/((4+2*ln(2))*ln(5)/x-ln(5))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 5.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+\left (384-40 x+5 x^2+(96-10 x) \log (4)\right ) \log (x)+(80+20 \log (4)) \log ^2(x)}{\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5)+\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5) \log (x)+\left (64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)\right ) \log (5) \log ^2(x)} \, dx=\frac {-20 (4+\log (4))+\frac {x (-16+5 x)}{2+\log (x)}}{4 (-4+x-\log (4)) \log (5)} \] Input:

Integrate[(384 - 44*x + 5*x^2 + (96 - 15*x)*Log[4] + (384 - 40*x + 5*x^2 + 
 (96 - 10*x)*Log[4])*Log[x] + (80 + 20*Log[4])*Log[x]^2)/((256 - 128*x + 1 
6*x^2 + (128 - 32*x)*Log[4] + 16*Log[4]^2)*Log[5] + (256 - 128*x + 16*x^2 
+ (128 - 32*x)*Log[4] + 16*Log[4]^2)*Log[5]*Log[x] + (64 - 32*x + 4*x^2 + 
(32 - 8*x)*Log[4] + 4*Log[4]^2)*Log[5]*Log[x]^2),x]
 

Output:

(-20*(4 + Log[4]) + (x*(-16 + 5*x))/(2 + Log[x]))/(4*(-4 + x - Log[4])*Log 
[5])
 

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.

\(\displaystyle \int \frac {5 x^2+\left (5 x^2-40 x+(96-10 x) \log (4)+384\right ) \log (x)-44 x+(80+20 \log (4)) \log ^2(x)+(96-15 x) \log (4)+384}{\log (5) \left (4 x^2-32 x+(32-8 x) \log (4)+64+4 \log ^2(4)\right ) \log ^2(x)+\log (5) \left (16 x^2-128 x+(128-32 x) \log (4)+256+16 \log ^2(4)\right ) \log (x)+\log (5) \left (16 x^2-128 x+(128-32 x) \log (4)+256+16 \log ^2(4)\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {5 x^2+\left (5 x^2-10 x (4+\log (4))+96 (4+\log (4))\right ) \log (x)+20 (4+\log (4)) \log ^2(x)-x (44+15 \log (4))+96 (4+\log (4))}{4 \log (5) (-x+4+\log (4))^2 (\log (x)+2)^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {5 x^2-(44+15 \log (4)) x+20 (4+\log (4)) \log ^2(x)+\left (5 x^2-10 (4+\log (4)) x+96 (4+\log (4))\right ) \log (x)+96 (4+\log (4))}{(-x+\log (4)+4)^2 (\log (x)+2)^2}dx}{4 \log (5)}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {\int \left (\frac {16-5 x}{(x-\log (4)-4) (\log (x)+2)^2}+\frac {5 x^2-10 (4+\log (4)) x+16 (4+\log (4))}{(-x+\log (4)+4)^2 (\log (x)+2)}+\frac {20 (4+\log (4))}{(-x+\log (4)+4)^2}\right )dx}{4 \log (5)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\int \frac {5 x^2-10 (4+\log (4)) x+16 (4+\log (4))}{(-x+\log (4)+4)^2 (\log (x)+2)}dx+\int \frac {16-5 x}{(x-\log (4)-4) (\log (x)+2)^2}dx+\frac {20 (4+\log (4))}{-x+4+\log (4)}}{4 \log (5)}\)

Input:

Int[(384 - 44*x + 5*x^2 + (96 - 15*x)*Log[4] + (384 - 40*x + 5*x^2 + (96 - 
 10*x)*Log[4])*Log[x] + (80 + 20*Log[4])*Log[x]^2)/((256 - 128*x + 16*x^2 
+ (128 - 32*x)*Log[4] + 16*Log[4]^2)*Log[5] + (256 - 128*x + 16*x^2 + (128 
 - 32*x)*Log[4] + 16*Log[4]^2)*Log[5]*Log[x] + (64 - 32*x + 4*x^2 + (32 - 
8*x)*Log[4] + 4*Log[4]^2)*Log[5]*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {16 x +\left (40 \ln \left (2\right )+80\right ) \ln \left (x \right )-5 x^{2}+80 \ln \left (2\right )+160}{4 \ln \left (5\right ) \left (\ln \left (x \right )+2\right ) \left (2 \ln \left (2\right )-x +4\right )}\)	\(47\)
parallelrisch	\(\frac {160+40 \ln \left (2\right ) \ln \left (x \right )-5 x^{2}+80 \ln \left (2\right )+16 x +80 \ln \left (x \right )}{4 \ln \left (5\right ) \left (\ln \left (x \right )+2\right ) \left (2 \ln \left (2\right )-x +4\right )}\)	\(48\)
norman	\(\frac {\frac {4 x}{\ln \left (5\right )}+\frac {10 \left (\ln \left (2\right )+2\right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {5 x^{2}}{4 \ln \left (5\right )}+\frac {20 \ln \left (2\right )+40}{\ln \left (5\right )}}{\left (\ln \left (x \right )+2\right ) \left (2 \ln \left (2\right )-x +4\right )}\)	\(58\)
risch	\(\frac {10 \ln \left (2\right )}{\left (2 \ln \left (2\right )-x +4\right ) \ln \left (5\right )}+\frac {20}{\left (2 \ln \left (2\right )-x +4\right ) \ln \left (5\right )}-\frac {x \left (-16+5 x \right )}{4 \ln \left (5\right ) \left (2 \ln \left (2\right )-x +4\right ) \left (\ln \left (x \right )+2\right )}\)	\(67\)

Input:

int(((40*ln(2)+80)*ln(x)^2+(2*(-10*x+96)*ln(2)+5*x^2-40*x+384)*ln(x)+2*(-1 
5*x+96)*ln(2)+5*x^2-44*x+384)/((16*ln(2)^2+2*(-8*x+32)*ln(2)+4*x^2-32*x+64 
)*ln(5)*ln(x)^2+(64*ln(2)^2+2*(-32*x+128)*ln(2)+16*x^2-128*x+256)*ln(5)*ln 
(x)+(64*ln(2)^2+2*(-32*x+128)*ln(2)+16*x^2-128*x+256)*ln(5)),x,method=_RET 
URNVERBOSE)
 

Output:

1/4/ln(5)*(16*x+(40*ln(2)+80)*ln(x)-5*x^2+80*ln(2)+160)/(ln(x)+2)/(2*ln(2) 
-x+4)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.47 \[ \int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+\left (384-40 x+5 x^2+(96-10 x) \log (4)\right ) \log (x)+(80+20 \log (4)) \log ^2(x)}{\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5)+\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5) \log (x)+\left (64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)\right ) \log (5) \log ^2(x)} \, dx=\frac {5 \, x^{2} - 40 \, {\left (\log \left (2\right ) + 2\right )} \log \left (x\right ) - 16 \, x - 80 \, \log \left (2\right ) - 160}{4 \, {\left ({\left (x - 2 \, \log \left (2\right ) - 4\right )} \log \left (5\right ) \log \left (x\right ) + 2 \, {\left (x - 2 \, \log \left (2\right ) - 4\right )} \log \left (5\right )\right )}} \] Input:

integrate(((40*log(2)+80)*log(x)^2+(2*(-10*x+96)*log(2)+5*x^2-40*x+384)*lo 
g(x)+2*(-15*x+96)*log(2)+5*x^2-44*x+384)/((16*log(2)^2+2*(-8*x+32)*log(2)+ 
4*x^2-32*x+64)*log(5)*log(x)^2+(64*log(2)^2+2*(-32*x+128)*log(2)+16*x^2-12 
8*x+256)*log(5)*log(x)+(64*log(2)^2+2*(-32*x+128)*log(2)+16*x^2-128*x+256) 
*log(5)),x, algorithm="fricas")
 

Output:

1/4*(5*x^2 - 40*(log(2) + 2)*log(x) - 16*x - 80*log(2) - 160)/((x - 2*log( 
2) - 4)*log(5)*log(x) + 2*(x - 2*log(2) - 4)*log(5))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (27) = 54\).

Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.24 \[ \int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+\left (384-40 x+5 x^2+(96-10 x) \log (4)\right ) \log (x)+(80+20 \log (4)) \log ^2(x)}{\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5)+\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5) \log (x)+\left (64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)\right ) \log (5) \log ^2(x)} \, dx=\frac {5 x^{2} - 16 x}{8 x \log {\left (5 \right )} + \left (4 x \log {\left (5 \right )} - 16 \log {\left (5 \right )} - 8 \log {\left (2 \right )} \log {\left (5 \right )}\right ) \log {\left (x \right )} - 32 \log {\left (5 \right )} - 16 \log {\left (2 \right )} \log {\left (5 \right )}} - \frac {10 \log {\left (2 \right )} + 20}{x \log {\left (5 \right )} - 4 \log {\left (5 \right )} - 2 \log {\left (2 \right )} \log {\left (5 \right )}} \] Input:

integrate(((40*ln(2)+80)*ln(x)**2+(2*(-10*x+96)*ln(2)+5*x**2-40*x+384)*ln( 
x)+2*(-15*x+96)*ln(2)+5*x**2-44*x+384)/((16*ln(2)**2+2*(-8*x+32)*ln(2)+4*x 
**2-32*x+64)*ln(5)*ln(x)**2+(64*ln(2)**2+2*(-32*x+128)*ln(2)+16*x**2-128*x 
+256)*ln(5)*ln(x)+(64*ln(2)**2+2*(-32*x+128)*ln(2)+16*x**2-128*x+256)*ln(5 
)),x)
 

Output:

(5*x**2 - 16*x)/(8*x*log(5) + (4*x*log(5) - 16*log(5) - 8*log(2)*log(5))*l 
og(x) - 32*log(5) - 16*log(2)*log(5)) - (10*log(2) + 20)/(x*log(5) - 4*log 
(5) - 2*log(2)*log(5))
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.76 \[ \int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+\left (384-40 x+5 x^2+(96-10 x) \log (4)\right ) \log (x)+(80+20 \log (4)) \log ^2(x)}{\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5)+\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5) \log (x)+\left (64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)\right ) \log (5) \log ^2(x)} \, dx=\frac {5 \, x^{2} - 40 \, {\left (\log \left (2\right ) + 2\right )} \log \left (x\right ) - 16 \, x - 80 \, \log \left (2\right ) - 160}{4 \, {\left (2 \, x \log \left (5\right ) - 4 \, \log \left (5\right ) \log \left (2\right ) + {\left (x \log \left (5\right ) - 2 \, \log \left (5\right ) \log \left (2\right ) - 4 \, \log \left (5\right )\right )} \log \left (x\right ) - 8 \, \log \left (5\right )\right )}} \] Input:

integrate(((40*log(2)+80)*log(x)^2+(2*(-10*x+96)*log(2)+5*x^2-40*x+384)*lo 
g(x)+2*(-15*x+96)*log(2)+5*x^2-44*x+384)/((16*log(2)^2+2*(-8*x+32)*log(2)+ 
4*x^2-32*x+64)*log(5)*log(x)^2+(64*log(2)^2+2*(-32*x+128)*log(2)+16*x^2-12 
8*x+256)*log(5)*log(x)+(64*log(2)^2+2*(-32*x+128)*log(2)+16*x^2-128*x+256) 
*log(5)),x, algorithm="maxima")
 

Output:

1/4*(5*x^2 - 40*(log(2) + 2)*log(x) - 16*x - 80*log(2) - 160)/(2*x*log(5) 
- 4*log(5)*log(2) + (x*log(5) - 2*log(5)*log(2) - 4*log(5))*log(x) - 8*log 
(5))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (34) = 68\).

Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.15 \[ \int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+\left (384-40 x+5 x^2+(96-10 x) \log (4)\right ) \log (x)+(80+20 \log (4)) \log ^2(x)}{\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5)+\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5) \log (x)+\left (64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)\right ) \log (5) \log ^2(x)} \, dx=\frac {5 \, x^{2} - 16 \, x}{4 \, {\left (x \log \left (5\right ) \log \left (x\right ) - 2 \, \log \left (5\right ) \log \left (2\right ) \log \left (x\right ) + 2 \, x \log \left (5\right ) - 4 \, \log \left (5\right ) \log \left (2\right ) - 4 \, \log \left (5\right ) \log \left (x\right ) - 8 \, \log \left (5\right )\right )}} - \frac {10 \, {\left (\log \left (2\right ) + 2\right )}}{x \log \left (5\right ) - 2 \, \log \left (5\right ) \log \left (2\right ) - 4 \, \log \left (5\right )} \] Input:

integrate(((40*log(2)+80)*log(x)^2+(2*(-10*x+96)*log(2)+5*x^2-40*x+384)*lo 
g(x)+2*(-15*x+96)*log(2)+5*x^2-44*x+384)/((16*log(2)^2+2*(-8*x+32)*log(2)+ 
4*x^2-32*x+64)*log(5)*log(x)^2+(64*log(2)^2+2*(-32*x+128)*log(2)+16*x^2-12 
8*x+256)*log(5)*log(x)+(64*log(2)^2+2*(-32*x+128)*log(2)+16*x^2-128*x+256) 
*log(5)),x, algorithm="giac")
 

Output:

1/4*(5*x^2 - 16*x)/(x*log(5)*log(x) - 2*log(5)*log(2)*log(x) + 2*x*log(5) 
- 4*log(5)*log(2) - 4*log(5)*log(x) - 8*log(5)) - 10*(log(2) + 2)/(x*log(5 
) - 2*log(5)*log(2) - 4*log(5))
 

Mupad [B] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 1650, normalized size of antiderivative = 48.53 \[ \int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+\left (384-40 x+5 x^2+(96-10 x) \log (4)\right ) \log (x)+(80+20 \log (4)) \log ^2(x)}{\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5)+\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5) \log (x)+\left (64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)\right ) \log (5) \log ^2(x)} \, dx=\text {Too large to display} \] Input:

int(-(44*x + 2*log(2)*(15*x - 96) + log(x)*(40*x + 2*log(2)*(10*x - 96) - 
5*x^2 - 384) - log(x)^2*(40*log(2) + 80) - 5*x^2 - 384)/(log(5)*(64*log(2) 
^2 - 2*log(2)*(32*x - 128) - 128*x + 16*x^2 + 256) + log(5)*log(x)*(64*log 
(2)^2 - 2*log(2)*(32*x - 128) - 128*x + 16*x^2 + 256) + log(5)*log(x)^2*(1 
6*log(2)^2 - 2*log(2)*(8*x - 32) - 32*x + 4*x^2 + 64)),x)
 

Output:

(5*x)/(4*log(5)) - (16*x)/(32*log(5) + 32*log(2)*log(5) - 16*x*log(5) + 8* 
log(2)^2*log(5) + 2*x^2*log(5) + 16*log(5)*log(x) + 16*log(2)*log(5)*log(x 
) - 8*x*log(5)*log(x) + 4*log(2)^2*log(5)*log(x) + x^2*log(5)*log(x) - 8*x 
*log(2)*log(5) - 4*x*log(2)*log(5)*log(x)) + (11*x^2)/(32*log(5) + 32*log( 
2)*log(5) - 16*x*log(5) + 8*log(2)^2*log(5) + 2*x^2*log(5) + 16*log(5)*log 
(x) + 16*log(2)*log(5)*log(x) - 8*x*log(5)*log(x) + 4*log(2)^2*log(5)*log( 
x) + x^2*log(5)*log(x) - 8*x*log(2)*log(5) - 4*x*log(2)*log(5)*log(x)) - ( 
5*x^3)/(4*(32*log(5) + 32*log(2)*log(5) - 16*x*log(5) + 8*log(2)^2*log(5) 
+ 2*x^2*log(5) + 16*log(5)*log(x) + 16*log(2)*log(5)*log(x) - 8*x*log(5)*l 
og(x) + 4*log(2)^2*log(5)*log(x) + x^2*log(5)*log(x) - 8*x*log(2)*log(5) - 
 4*x*log(2)*log(5)*log(x))) + symsum(log(576*x - 1984*log(2) - 384*root(22 
020096*log(2) + 287047680*log(2)^4 + 236322816*log(2)^5 + 214958080*log(2) 
^3 + 121765888*log(2)^6 + 94371840*log(2)^2 + 38338560*log(2)^7 + 6758400* 
log(2)^8 + 512000*log(2)^9 + 2097152, z, k)*log(5) + 1056*x*log(2) + 724*x 
*log(2)^2 + 220*x*log(2)^3 + 25*x*log(2)^4 - 272*log(2)^2 + 648*log(2)^3 + 
 340*log(2)^4 + 50*log(2)^5 - 1536*root(22020096*log(2) + 287047680*log(2) 
^4 + 236322816*log(2)^5 + 214958080*log(2)^3 + 121765888*log(2)^6 + 943718 
40*log(2)^2 + 38338560*log(2)^7 + 6758400*log(2)^8 + 512000*log(2)^9 + 209 
7152, z, k)*log(2)*log(5) + 96*root(22020096*log(2) + 287047680*log(2)^4 + 
 236322816*log(2)^5 + 214958080*log(2)^3 + 121765888*log(2)^6 + 9437184...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {384-44 x+5 x^2+(96-15 x) \log (4)+\left (384-40 x+5 x^2+(96-10 x) \log (4)\right ) \log (x)+(80+20 \log (4)) \log ^2(x)}{\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5)+\left (256-128 x+16 x^2+(128-32 x) \log (4)+16 \log ^2(4)\right ) \log (5) \log (x)+\left (64-32 x+4 x^2+(32-8 x) \log (4)+4 \log ^2(4)\right ) \log (5) \log ^2(x)} \, dx=\frac {x \left (20 \,\mathrm {log}\left (x \right )-5 x +56\right )}{4 \,\mathrm {log}\left (5\right ) \left (2 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )-\mathrm {log}\left (x \right ) x +4 \,\mathrm {log}\left (x \right )+4 \,\mathrm {log}\left (2\right )-2 x +8\right )} \] Input:

int(((40*log(2)+80)*log(x)^2+(2*(-10*x+96)*log(2)+5*x^2-40*x+384)*log(x)+2 
*(-15*x+96)*log(2)+5*x^2-44*x+384)/((16*log(2)^2+2*(-8*x+32)*log(2)+4*x^2- 
32*x+64)*log(5)*log(x)^2+(64*log(2)^2+2*(-32*x+128)*log(2)+16*x^2-128*x+25 
6)*log(5)*log(x)+(64*log(2)^2+2*(-32*x+128)*log(2)+16*x^2-128*x+256)*log(5 
)),x)
 

Output:

(x*(20*log(x) - 5*x + 56))/(4*log(5)*(2*log(x)*log(2) - log(x)*x + 4*log(x 
) + 4*log(2) - 2*x + 8))