\(\int \frac {(-3-3 \log (x)) \log (\frac {3+16 x \log (-2+e^5) \log (x)}{4 x \log (-2+e^5) \log (x)})}{6 x \log (x)+32 x^2 \log (-2+e^5) \log ^2(x)} \, dx\) [197]

Optimal result

Integrand size = 63, antiderivative size = 28 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \] Output:

1/4*ln(4+3/4/x/ln(x)/ln(exp(5)-2))^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \] Input:

Integrate[((-3 - 3*Log[x])*Log[(3 + 16*x*Log[-2 + E^5]*Log[x])/(4*x*Log[-2 
 + E^5]*Log[x])])/(6*x*Log[x] + 32*x^2*Log[-2 + E^5]*Log[x]^2),x]
 

Output:

Log[4 + 3/(4*x*Log[-2 + E^5]*Log[x])]^2/4
 

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.

\(\displaystyle \int \frac {(-3 \log (x)-3) \log \left (\frac {16 x \log \left (e^5-2\right ) \log (x)+3}{4 x \log \left (e^5-2\right ) \log (x)}\right )}{32 x^2 \log \left (e^5-2\right ) \log ^2(x)+6 x \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {3 (-\log (x)-1) \log \left (\frac {16 x \log \left (e^5-2\right ) \log (x)+3}{4 x \log \left (e^5-2\right ) \log (x)}\right )}{2 x \log (x) \left (16 x \log \left (e^5-2\right ) \log (x)+3\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{2} \int -\frac {(\log (x)+1) \log \left (\frac {16 x \log \left (-2+e^5\right ) \log (x)+3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {3}{2} \int \frac {(\log (x)+1) \log \left (\frac {16 x \log \left (-2+e^5\right ) \log (x)+3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {3}{2} \int \left (\frac {\log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}+\frac {\log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3}{2} \left (\int \frac {\log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}dx+\int \frac {\log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}dx\right )\)

Input:

Int[((-3 - 3*Log[x])*Log[(3 + 16*x*Log[-2 + E^5]*Log[x])/(4*x*Log[-2 + E^5 
]*Log[x])])/(6*x*Log[x] + 32*x^2*Log[-2 + E^5]*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(88\) vs. \(2(23)=46\).

Time = 2.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.18

method	result	size
default	\(\frac {3 \ln \left (\ln \left ({\mathrm e}^{5}-2\right )\right ) \left (\frac {\ln \left (x \right )}{3}+\frac {\ln \left (\ln \left (x \right )\right )}{3}-\frac {\ln \left (16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3\right )}{3}\right )}{2}+3 \ln \left (2\right ) \left (\frac {\ln \left (x \right )}{3}+\frac {\ln \left (\ln \left (x \right )\right )}{3}-\frac {\ln \left (16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3\right )}{3}\right )+\frac {{\ln \left (\frac {16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3}{x \ln \left (x \right )}\right )}^{2}}{4}\)	\(89\)

Input:

int((-3*ln(x)-3)*ln(1/4*(16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x)/ln(exp(5)-2))/ 
(32*x^2*ln(x)^2*ln(exp(5)-2)+6*x*ln(x)),x,method=_RETURNVERBOSE)
 

Output:

3/2*ln(ln(exp(5)-2))*(1/3*ln(x)+1/3*ln(ln(x))-1/3*ln(16*x*ln(x)*ln(exp(5)- 
2)+3))+3*ln(2)*(1/3*ln(x)+1/3*ln(ln(x))-1/3*ln(16*x*ln(x)*ln(exp(5)-2)+3)) 
+1/4*ln((16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x))^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \, \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )^{2} \] Input:

integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log 
(exp(5)-2))/(32*x^2*log(x)^2*log(exp(5)-2)+6*x*log(x)),x, algorithm="frica 
s")
 

Output:

1/4*log(1/4*(16*x*log(x)*log(e^5 - 2) + 3)/(x*log(x)*log(e^5 - 2)))^2
 

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {\log {\left (\frac {4 x \log {\left (x \right )} \log {\left (-2 + e^{5} \right )} + \frac {3}{4}}{x \log {\left (x \right )} \log {\left (-2 + e^{5} \right )}} \right )}^{2}}{4} \] Input:

integrate((-3*ln(x)-3)*ln(1/4*(16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x)/ln(exp(5 
)-2))/(32*x**2*ln(x)**2*ln(exp(5)-2)+6*x*ln(x)),x)
 

Output:

log((4*x*log(x)*log(-2 + exp(5)) + 3/4)/(x*log(x)*log(-2 + exp(5))))**2/4
 

Maxima [F]

\[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\int { -\frac {3 \, {\left (\log \left (x\right ) + 1\right )} \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )}{2 \, {\left (16 \, x^{2} \log \left (x\right )^{2} \log \left (e^{5} - 2\right ) + 3 \, x \log \left (x\right )\right )}} \,d x } \] Input:

integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log 
(exp(5)-2))/(32*x^2*log(x)^2*log(exp(5)-2)+6*x*log(x)),x, algorithm="maxim 
a")
 

Output:

-3/2*integrate((log(x) + 1)*log(1/4*(16*x*log(x)*log(e^5 - 2) + 3)/(x*log( 
x)*log(e^5 - 2)))/(16*x^2*log(x)^2*log(e^5 - 2) + 3*x*log(x)), x)
 

Giac [F]

\[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\int { -\frac {3 \, {\left (\log \left (x\right ) + 1\right )} \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )}{2 \, {\left (16 \, x^{2} \log \left (x\right )^{2} \log \left (e^{5} - 2\right ) + 3 \, x \log \left (x\right )\right )}} \,d x } \] Input:

integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log 
(exp(5)-2))/(32*x^2*log(x)^2*log(exp(5)-2)+6*x*log(x)),x, algorithm="giac" 
)
 

Output:

integrate(-3/2*(log(x) + 1)*log(1/4*(16*x*log(x)*log(e^5 - 2) + 3)/(x*log( 
x)*log(e^5 - 2)))/(16*x^2*log(x)^2*log(e^5 - 2) + 3*x*log(x)), x)
 

Mupad [B] (verification not implemented)

Time = 4.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {{\ln \left (\frac {4\,x\,\ln \left ({\mathrm {e}}^5-2\right )\,\ln \left (x\right )+\frac {3}{4}}{x\,\ln \left ({\mathrm {e}}^5-2\right )\,\ln \left (x\right )}\right )}^2}{4} \] Input:

int(-(log((4*x*log(exp(5) - 2)*log(x) + 3/4)/(x*log(exp(5) - 2)*log(x)))*( 
3*log(x) + 3))/(6*x*log(x) + 32*x^2*log(exp(5) - 2)*log(x)^2),x)
 

Output:

log((4*x*log(exp(5) - 2)*log(x) + 3/4)/(x*log(exp(5) - 2)*log(x)))^2/4
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {{\mathrm {log}\left (\frac {16 \,\mathrm {log}\left (e^{5}-2\right ) \mathrm {log}\left (x \right ) x +3}{4 \,\mathrm {log}\left (e^{5}-2\right ) \mathrm {log}\left (x \right ) x}\right )}^{2}}{4} \] Input:

int((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log(exp(5 
)-2))/(32*x^2*log(x)^2*log(exp(5)-2)+6*x*log(x)),x)
 

Output:

log((16*log(e**5 - 2)*log(x)*x + 3)/(4*log(e**5 - 2)*log(x)*x))**2/4