\(\int \frac {-10-x+x \log (x)+(-1-2 x-x^2+(2 x+2 x^2) \log (x)) \log (\frac {\log (5)}{25})}{2 x \log ^2(x)} \, dx\) [439]

Optimal result

Integrand size = 50, antiderivative size = 28 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {5+\frac {1}{2} \left (x+\log (x)+(1+x)^2 \log \left (\frac {\log (5)}{25}\right )\right )}{\log (x)} \] Output:

(5+1/2*x+1/2*ln(1/25*ln(5))*(1+x)^2+1/2*ln(x))/ln(x)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {10+x+\log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )+x^2 \log \left (\frac {\log (5)}{25}\right )}{2 \log (x)} \] Input:

Integrate[(-10 - x + x*Log[x] + (-1 - 2*x - x^2 + (2*x + 2*x^2)*Log[x])*Lo 
g[Log[5]/25])/(2*x*Log[x]^2),x]
 

Output:

(10 + x + Log[Log[5]/25] + 2*x*Log[Log[5]/25] + x^2*Log[Log[5]/25])/(2*Log 
[x])
 

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.

\(\displaystyle \int \frac {\log \left (\frac {\log (5)}{25}\right ) \left (-x^2+\left (2 x^2+2 x\right ) \log (x)-2 x-1\right )-x+x \log (x)-10}{2 x \log ^2(x)} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int -\frac {-\log (x) x+x+\left (x^2+2 x-2 \left (x^2+x\right ) \log (x)+1\right ) \log \left (\frac {\log (5)}{25}\right )+10}{x \log ^2(x)}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{2} \int \frac {-\log (x) x+x+\left (x^2+2 x-2 \left (x^2+x\right ) \log (x)+1\right ) \log \left (\frac {\log (5)}{25}\right )+10}{x \log ^2(x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{2} \int \left (\frac {\log \left (\frac {\log (5)}{25}\right ) x^2+\left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) x+\log \left (\frac {\log (5)}{25}\right )+10}{x \log ^2(x)}+\frac {-2 \log \left (\frac {\log (5)}{25}\right ) x-2 \log (\log (5))+\log (625)-1}{\log (x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\int \frac {\log \left (\frac {\log (5)}{25}\right ) x^2+\left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) x+\log \left (\frac {\log (5)}{25}\right )+10}{x \log ^2(x)}dx+2 \log \left (\frac {\log (5)}{25}\right ) \operatorname {ExpIntegralEi}(2 \log (x))+(1-\log (625)+2 \log (\log (5))) \operatorname {LogIntegral}(x)\right )\)

Input:

Int[(-10 - x + x*Log[x] + (-1 - 2*x - x^2 + (2*x + 2*x^2)*Log[x])*Log[Log[ 
5]/25])/(2*x*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(\frac {\ln \left (\frac {\ln \left (5\right )}{25}\right ) x^{2}+2 \ln \left (\frac {\ln \left (5\right )}{25}\right ) x +10+\ln \left (\frac {\ln \left (5\right )}{25}\right )+x}{2 \ln \left (x \right )}\)	\(32\)
norman	\(\frac {\left (-\ln \left (5\right )+\frac {\ln \left (\ln \left (5\right )\right )}{2}\right ) x^{2}+\left (-2 \ln \left (5\right )+\ln \left (\ln \left (5\right )\right )+\frac {1}{2}\right ) x +5-\ln \left (5\right )+\frac {\ln \left (\ln \left (5\right )\right )}{2}}{\ln \left (x \right )}\)	\(42\)
risch	\(-\frac {2 x^{2} \ln \left (5\right )-x^{2} \ln \left (\ln \left (5\right )\right )+4 x \ln \left (5\right )-2 x \ln \left (\ln \left (5\right )\right )+2 \ln \left (5\right )-\ln \left (\ln \left (5\right )\right )-x -10}{2 \ln \left (x \right )}\)	\(47\)
default	\(2 \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )\right )+2 \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )+\ln \left (5\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {\ln \left (\ln \left (5\right )\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )\right )\right )}{2}+2 \ln \left (5\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )\right )-\ln \left (\ln \left (5\right )\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )\right )-\frac {\ln \left (5\right )}{\ln \left (x \right )}+\frac {\ln \left (\ln \left (5\right )\right )}{2 \ln \left (x \right )}+\frac {x}{2 \ln \left (x \right )}+\frac {5}{\ln \left (x \right )}\)	\(159\)
parts	\(2 \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )\right )+2 \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )+\ln \left (5\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {\ln \left (\ln \left (5\right )\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )\right )\right )}{2}+2 \ln \left (5\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )\right )-\ln \left (\ln \left (5\right )\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )\right )-\frac {\ln \left (5\right )}{\ln \left (x \right )}+\frac {\ln \left (\ln \left (5\right )\right )}{2 \ln \left (x \right )}+\frac {x}{2 \ln \left (x \right )}+\frac {5}{\ln \left (x \right )}\)	\(159\)

Input:

int(1/2*(((2*x^2+2*x)*ln(x)-x^2-2*x-1)*ln(1/25*ln(5))+x*ln(x)-x-10)/x/ln(x 
)^2,x,method=_RETURNVERBOSE)
 

Output:

1/2*(ln(1/25*ln(5))*x^2+2*ln(1/25*ln(5))*x+10+ln(1/25*ln(5))+x)/ln(x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + x + 10}{2 \, \log \left (x\right )} \] Input:

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x- 
10)/x/log(x)^2,x, algorithm="fricas")
 

Output:

1/2*((x^2 + 2*x + 1)*log(1/25*log(5)) + x + 10)/log(x)
 

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {- 2 x^{2} \log {\left (5 \right )} + x^{2} \log {\left (\log {\left (5 \right )} \right )} - 4 x \log {\left (5 \right )} + 2 x \log {\left (\log {\left (5 \right )} \right )} + x - 2 \log {\left (5 \right )} + \log {\left (\log {\left (5 \right )} \right )} + 10}{2 \log {\left (x \right )}} \] Input:

integrate(1/2*(((2*x**2+2*x)*ln(x)-x**2-2*x-1)*ln(1/25*ln(5))+x*ln(x)-x-10 
)/x/ln(x)**2,x)
 

Output:

(-2*x**2*log(5) + x**2*log(log(5)) - 4*x*log(5) + 2*x*log(log(5)) + x - 2* 
log(5) + log(log(5)) + 10)/(2*log(x))
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.75 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx={\rm Ei}\left (2 \, \log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + {\rm Ei}\left (\log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) - \Gamma \left (-1, -\log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) - \Gamma \left (-1, -2 \, \log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + \frac {\log \left (\frac {1}{25} \, \log \left (5\right )\right )}{2 \, \log \left (x\right )} + \frac {5}{\log \left (x\right )} + \frac {1}{2} \, {\rm Ei}\left (\log \left (x\right )\right ) - \frac {1}{2} \, \Gamma \left (-1, -\log \left (x\right )\right ) \] Input:

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x- 
10)/x/log(x)^2,x, algorithm="maxima")
 

Output:

Ei(2*log(x))*log(1/25*log(5)) + Ei(log(x))*log(1/25*log(5)) - gamma(-1, -l 
og(x))*log(1/25*log(5)) - gamma(-1, -2*log(x))*log(1/25*log(5)) + 1/2*log( 
1/25*log(5))/log(x) + 5/log(x) + 1/2*Ei(log(x)) - 1/2*gamma(-1, -log(x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).

Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=-\frac {2 \, x^{2} \log \left (5\right ) - x^{2} \log \left (\log \left (5\right )\right ) + 4 \, x \log \left (5\right ) - 2 \, x \log \left (\log \left (5\right )\right ) - x + 2 \, \log \left (5\right ) - \log \left (\log \left (5\right )\right ) - 10}{2 \, \log \left (x\right )} \] Input:

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x- 
10)/x/log(x)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {\ln \left (\frac {\ln \left (5\right )}{25}\right )\,x^4+\left (2\,\ln \left (\frac {\ln \left (5\right )}{25}\right )+1\right )\,x^3+\left (\ln \left (\frac {\ln \left (5\right )}{25}\right )+10\right )\,x^2}{2\,x^2\,\ln \left (x\right )} \] Input:

int(-(x/2 + (log(log(5)/25)*(2*x - log(x)*(2*x + 2*x^2) + x^2 + 1))/2 - (x 
*log(x))/2 + 5)/(x*log(x)^2),x)
 

Output:

(x^3*(2*log(log(5)/25) + 1) + x^4*log(log(5)/25) + x^2*(log(log(5)/25) + 1 
0))/(2*x^2*log(x))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (5\right )}{25}\right ) x^{2}+2 \,\mathrm {log}\left (\frac {\mathrm {log}\left (5\right )}{25}\right ) x +\mathrm {log}\left (\frac {\mathrm {log}\left (5\right )}{25}\right )+x +10}{2 \,\mathrm {log}\left (x \right )} \] Input:

int(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x-10)/x/ 
log(x)^2,x)
 

Output:

(log(log(5)/25)*x**2 + 2*log(log(5)/25)*x + log(log(5)/25) + x + 10)/(2*lo 
g(x))