\(\int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} (-14-16 x+4 x^2+4 x^3-x^4)}}(x) (5 e^{\frac {1}{5} (-14-16 x+4 x^2+4 x^3-x^4)}+e^{\frac {1}{5} (-14-16 x+4 x^2+4 x^3-x^4)} (-16 x+8 x^2+12 x^3-4 x^4) \log (x) \log (\log (x)))}{5 x \log (x)} \, dx\) [2182]

Optimal result

Integrand size = 127, antiderivative size = 28 \[ \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx=3+x+\log ^{e^{\frac {1}{5} \left (2-\left (4+2 x-x^2\right )^2\right )}}(x) \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx=x+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \] Input:

Integrate[(5*x*Log[x] + Log[x]^E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*(5 
*E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5) + E^((-14 - 16*x + 4*x^2 + 4*x^3 
 - x^4)/5)*(-16*x + 8*x^2 + 12*x^3 - 4*x^4)*Log[x]*Log[Log[x]]))/(5*x*Log[ 
x]),x]
 

Output:

x + Log[x]^E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/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.

\(\displaystyle \int \frac {\left (5 e^{\frac {1}{5} \left (-x^4+4 x^3+4 x^2-16 x-14\right )}+e^{\frac {1}{5} \left (-x^4+4 x^3+4 x^2-16 x-14\right )} \left (-4 x^4+12 x^3+8 x^2-16 x\right ) \log (x) \log (\log (x))\right ) \log ^{e^{\frac {1}{5} \left (-x^4+4 x^3+4 x^2-16 x-14\right )}}(x)+5 x \log (x)}{5 x \log (x)} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {\left (5 e^{\frac {1}{5} \left (-x^4+4 x^3+4 x^2-16 x-14\right )}-4 e^{\frac {1}{5} \left (-x^4+4 x^3+4 x^2-16 x-14\right )} \left (x^4-3 x^3-2 x^2+4 x\right ) \log (x) \log (\log (x))\right ) \log ^{e^{\frac {1}{5} \left (-x^4+4 x^3+4 x^2-16 x-14\right )}}(x)+5 x \log (x)}{x \log (x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{5} \int \left (5-\frac {e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}} \log ^{-1+e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}}}(x) \left (4 \log (x) \log (\log (x)) x^4-12 \log (x) \log (\log (x)) x^3-8 \log (x) \log (\log (x)) x^2+16 \log (x) \log (\log (x)) x-5\right )}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} \left (5 \int \frac {e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}} \log ^{-1+e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}}}(x)}{x}dx-16 \int e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}} \log ^{e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}}}(x) \log (\log (x))dx+8 \int e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}} x \log ^{e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}}}(x) \log (\log (x))dx+12 \int e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}} x^2 \log ^{e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}}}(x) \log (\log (x))dx-4 \int e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}} x^3 \log ^{e^{-\frac {x^4}{5}+\frac {4 x^3}{5}+\frac {4 x^2}{5}-\frac {16 x}{5}-\frac {14}{5}}}(x) \log (\log (x))dx+5 x\right )\)

Input:

Int[(5*x*Log[x] + Log[x]^E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5)*(5*E^((- 
14 - 16*x + 4*x^2 + 4*x^3 - x^4)/5) + E^((-14 - 16*x + 4*x^2 + 4*x^3 - x^4 
)/5)*(-16*x + 8*x^2 + 12*x^3 - 4*x^4)*Log[x]*Log[Log[x]]))/(5*x*Log[x]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 6.59 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96

method	result	size
risch	\(x +\ln \left (x \right )^{{\mathrm e}^{-\frac {1}{5} x^{4}+\frac {4}{5} x^{3}+\frac {4}{5} x^{2}-\frac {16}{5} x -\frac {14}{5}}}\)	\(27\)
parallelrisch	\(x +{\mathrm e}^{{\mathrm e}^{-\frac {1}{5} x^{4}+\frac {4}{5} x^{3}+\frac {4}{5} x^{2}-\frac {16}{5} x -\frac {14}{5}} \ln \left (\ln \left (x \right )\right )}\)	\(29\)

Input:

int(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-1 
4/5)*ln(x)*ln(ln(x))+5*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5))*exp(exp( 
-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*ln(ln(x)))+5*x*ln(x))/x/ln(x),x,meth 
od=_RETURNVERBOSE)
 

Output:

x+ln(x)^exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx=x + \log \left (x\right )^{e^{\left (-\frac {1}{5} \, x^{4} + \frac {4}{5} \, x^{3} + \frac {4}{5} \, x^{2} - \frac {16}{5} \, x - \frac {14}{5}\right )}} \] Input:

integrate(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16 
/5*x-14/5)*log(x)*log(log(x))+5*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)) 
*exp(exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(log(x)))+5*x*log(x))/x/ 
log(x),x, algorithm="fricas")
 

Output:

x + log(x)^e^(-1/5*x^4 + 4/5*x^3 + 4/5*x^2 - 16/5*x - 14/5)
 

Sympy [A] (verification not implemented)

Time = 5.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx=x + e^{e^{- \frac {x^{4}}{5} + \frac {4 x^{3}}{5} + \frac {4 x^{2}}{5} - \frac {16 x}{5} - \frac {14}{5}} \log {\left (\log {\left (x \right )} \right )}} \] Input:

integrate(1/5*(((-4*x**4+12*x**3+8*x**2-16*x)*exp(-1/5*x**4+4/5*x**3+4/5*x 
**2-16/5*x-14/5)*ln(x)*ln(ln(x))+5*exp(-1/5*x**4+4/5*x**3+4/5*x**2-16/5*x- 
14/5))*exp(exp(-1/5*x**4+4/5*x**3+4/5*x**2-16/5*x-14/5)*ln(ln(x)))+5*x*ln( 
x))/x/ln(x),x)
 

Output:

x + exp(exp(-x**4/5 + 4*x**3/5 + 4*x**2/5 - 16*x/5 - 14/5)*log(log(x)))
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx=x + \log \left (x\right )^{e^{\left (-\frac {1}{5} \, x^{4} + \frac {4}{5} \, x^{3} + \frac {4}{5} \, x^{2} - \frac {16}{5} \, x - \frac {14}{5}\right )}} \] Input:

integrate(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16 
/5*x-14/5)*log(x)*log(log(x))+5*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)) 
*exp(exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(log(x)))+5*x*log(x))/x/ 
log(x),x, algorithm="maxima")
 

Output:

x + log(x)^e^(-1/5*x^4 + 4/5*x^3 + 4/5*x^2 - 16/5*x - 14/5)
 

Giac [A] (verification not implemented)

Time = 7.63 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx=x + \log \left (x\right )^{e^{\left (-\frac {1}{5} \, x^{4} + \frac {4}{5} \, x^{3} + \frac {4}{5} \, x^{2} - \frac {16}{5} \, x - \frac {14}{5}\right )}} \] Input:

integrate(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16 
/5*x-14/5)*log(x)*log(log(x))+5*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)) 
*exp(exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(log(x)))+5*x*log(x))/x/ 
log(x),x, algorithm="giac")
 

Output:

x + log(x)^e^(-1/5*x^4 + 4/5*x^3 + 4/5*x^2 - 16/5*x - 14/5)
 

Mupad [B] (verification not implemented)

Time = 1.86 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx=x+{\ln \left (x\right )}^{{\mathrm {e}}^{-\frac {x^4}{5}+\frac {4\,x^3}{5}+\frac {4\,x^2}{5}-\frac {16\,x}{5}-\frac {14}{5}}} \] Input:

int(((exp(log(log(x))*exp((4*x^2)/5 - (16*x)/5 + (4*x^3)/5 - x^4/5 - 14/5) 
)*(5*exp((4*x^2)/5 - (16*x)/5 + (4*x^3)/5 - x^4/5 - 14/5) - log(log(x))*ex 
p((4*x^2)/5 - (16*x)/5 + (4*x^3)/5 - x^4/5 - 14/5)*log(x)*(16*x - 8*x^2 - 
12*x^3 + 4*x^4)))/5 + x*log(x))/(x*log(x)),x)
 

Output:

x + log(x)^exp((4*x^2)/5 - (16*x)/5 + (4*x^3)/5 - x^4/5 - 14/5)
 

Reduce [F]

\[ \int \frac {5 x \log (x)+\log ^{e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}}(x) \left (5 e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )}+e^{\frac {1}{5} \left (-14-16 x+4 x^2+4 x^3-x^4\right )} \left (-16 x+8 x^2+12 x^3-4 x^4\right ) \log (x) \log (\log (x))\right )}{5 x \log (x)} \, dx =\text {Too large to display} \] Input:

int(1/5*(((-4*x^4+12*x^3+8*x^2-16*x)*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-1 
4/5)*log(x)*log(log(x))+5*exp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5))*exp(e 
xp(-1/5*x^4+4/5*x^3+4/5*x^2-16/5*x-14/5)*log(log(x)))+5*x*log(x))/x/log(x) 
,x)
 

Output:

(5*int(e**((4*e**((x**4 + 16*x + 4)/5)*e**2*x**3 + 4*e**((x**4 + 16*x + 4) 
/5)*e**2*x**2 + 30*e**((x**4 + 16*x + 4)/5)*e**2*x + e**((x**4 + 16*x + 4) 
/5)*e**2 + 5*e**((4*x**3 + 4*x**2)/5)*log(log(x)))/(5*e**((x**4 + 16*x + 4 
)/5)*e**2))/(e**((x**4 + 46*x)/5)*log(x)*x),x) - 4*int((e**((4*e**((x**4 + 
 16*x + 4)/5)*e**2*x**3 + 4*e**((x**4 + 16*x + 4)/5)*e**2*x**2 + 30*e**((x 
**4 + 16*x + 4)/5)*e**2*x + e**((x**4 + 16*x + 4)/5)*e**2 + 5*e**((4*x**3 
+ 4*x**2)/5)*log(log(x)))/(5*e**((x**4 + 16*x + 4)/5)*e**2))*log(log(x))*x 
**3)/e**((x**4 + 46*x)/5),x) + 12*int((e**((4*e**((x**4 + 16*x + 4)/5)*e** 
2*x**3 + 4*e**((x**4 + 16*x + 4)/5)*e**2*x**2 + 30*e**((x**4 + 16*x + 4)/5 
)*e**2*x + e**((x**4 + 16*x + 4)/5)*e**2 + 5*e**((4*x**3 + 4*x**2)/5)*log( 
log(x)))/(5*e**((x**4 + 16*x + 4)/5)*e**2))*log(log(x))*x**2)/e**((x**4 + 
46*x)/5),x) + 8*int((e**((4*e**((x**4 + 16*x + 4)/5)*e**2*x**3 + 4*e**((x* 
*4 + 16*x + 4)/5)*e**2*x**2 + 30*e**((x**4 + 16*x + 4)/5)*e**2*x + e**((x* 
*4 + 16*x + 4)/5)*e**2 + 5*e**((4*x**3 + 4*x**2)/5)*log(log(x)))/(5*e**((x 
**4 + 16*x + 4)/5)*e**2))*log(log(x))*x)/e**((x**4 + 46*x)/5),x) - 16*int( 
(e**((4*e**((x**4 + 16*x + 4)/5)*e**2*x**3 + 4*e**((x**4 + 16*x + 4)/5)*e* 
*2*x**2 + 30*e**((x**4 + 16*x + 4)/5)*e**2*x + e**((x**4 + 16*x + 4)/5)*e* 
*2 + 5*e**((4*x**3 + 4*x**2)/5)*log(log(x)))/(5*e**((x**4 + 16*x + 4)/5)*e 
**2))*log(log(x)))/e**((x**4 + 46*x)/5),x) + 5*e**3*x)/(5*e**3)