3.5.31 \(\int \frac {(-36 x^2+36 x \log (x)) \log (x-\log (x))+\frac {e^x (-15+15 x+(-15 x-15 x^2+(15+15 x) \log (x)) \log (x-\log (x)))}{\log (x-\log (x))}}{e^x (15 x^2-60 x^4+(-15 x+60 x^3) \log (x))+\frac {e^{2 x} (-25 x^3+25 x^2 \log (x))}{\log (x-\log (x))}+(18 x^3-36 x^5+(-18 x^2+36 x^4) \log (x)) \log (x-\log (x))} \, dx\) [431]

3.5.31.1 Optimal result

Integrand size = 155, antiderivative size = 29 \[ \int \frac {\left (-36 x^2+36 x \log (x)\right ) \log (x-\log (x))+\frac {e^x \left (-15+15 x+\left (-15 x-15 x^2+(15+15 x) \log (x)\right ) \log (x-\log (x))\right )}{\log (x-\log (x))}}{e^x \left (15 x^2-60 x^4+\left (-15 x+60 x^3\right ) \log (x)\right )+\frac {e^{2 x} \left (-25 x^3+25 x^2 \log (x)\right )}{\log (x-\log (x))}+\left (18 x^3-36 x^5+\left (-18 x^2+36 x^4\right ) \log (x)\right ) \log (x-\log (x))} \, dx=\log \left (5-\frac {3}{x \left (\frac {6 x}{5}+\frac {e^x}{\log (x-\log (x))}\right )}\right ) \]

output

ln(5-3/x/(exp(-ln(ln(x-ln(x)))+x)+6/5*x))
 

3.5.31.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(29)=58\).

Time = 0.35 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {\left (-36 x^2+36 x \log (x)\right ) \log (x-\log (x))+\frac {e^x \left (-15+15 x+\left (-15 x-15 x^2+(15+15 x) \log (x)\right ) \log (x-\log (x))\right )}{\log (x-\log (x))}}{e^x \left (15 x^2-60 x^4+\left (-15 x+60 x^3\right ) \log (x)\right )+\frac {e^{2 x} \left (-25 x^3+25 x^2 \log (x)\right )}{\log (x-\log (x))}+\left (18 x^3-36 x^5+\left (-18 x^2+36 x^4\right ) \log (x)\right ) \log (x-\log (x))} \, dx=3 \left (-\frac {\log (x)}{3}-\frac {1}{3} \log \left (5 e^x+6 x \log (x-\log (x))\right )+\frac {1}{3} \log \left (5 e^x x-3 \log (x-\log (x))+6 x^2 \log (x-\log (x))\right )\right ) \]

input

Integrate[((-36*x^2 + 36*x*Log[x])*Log[x - Log[x]] + (E^x*(-15 + 15*x + (- 
15*x - 15*x^2 + (15 + 15*x)*Log[x])*Log[x - Log[x]]))/Log[x - Log[x]])/(E^ 
x*(15*x^2 - 60*x^4 + (-15*x + 60*x^3)*Log[x]) + (E^(2*x)*(-25*x^3 + 25*x^2 
*Log[x]))/Log[x - Log[x]] + (18*x^3 - 36*x^5 + (-18*x^2 + 36*x^4)*Log[x])* 
Log[x - Log[x]]),x]
 

output

3*(-1/3*Log[x] - Log[5*E^x + 6*x*Log[x - Log[x]]]/3 + Log[5*E^x*x - 3*Log[ 
x - Log[x]] + 6*x^2*Log[x - Log[x]]]/3)
 

3.5.31.3 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.

\(\displaystyle \int \frac {\left (36 x \log (x)-36 x^2\right ) \log (x-\log (x))+\frac {e^x \left (\left (-15 x^2-15 x+(15 x+15) \log (x)\right ) \log (x-\log (x))+15 x-15\right )}{\log (x-\log (x))}}{\frac {e^{2 x} \left (25 x^2 \log (x)-25 x^3\right )}{\log (x-\log (x))}+e^x \left (-60 x^4+\left (60 x^3-15 x\right ) \log (x)+15 x^2\right )+\left (-36 x^5+18 x^3+\left (36 x^4-18 x^2\right ) \log (x)\right ) \log (x-\log (x))} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\log (x-\log (x)) \left (-\left (\left (36 x \log (x)-36 x^2\right ) \log (x-\log (x))\right )-\frac {e^x \left (\left (-15 x^2-15 x+(15 x+15) \log (x)\right ) \log (x-\log (x))+15 x-15\right )}{\log (x-\log (x))}\right )}{x (x-\log (x)) \left (36 x^3 \log ^2(x-\log (x))+60 e^x x^2 \log (x-\log (x))+25 e^{2 x} x-18 x \log ^2(x-\log (x))-15 e^x \log (x-\log (x))\right )}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

Int[((-36*x^2 + 36*x*Log[x])*Log[x - Log[x]] + (E^x*(-15 + 15*x + (-15*x - 
 15*x^2 + (15 + 15*x)*Log[x])*Log[x - Log[x]]))/Log[x - Log[x]])/(E^x*(15* 
x^2 - 60*x^4 + (-15*x + 60*x^3)*Log[x]) + (E^(2*x)*(-25*x^3 + 25*x^2*Log[x 
]))/Log[x - Log[x]] + (18*x^3 - 36*x^5 + (-18*x^2 + 36*x^4)*Log[x])*Log[x 
- Log[x]]),x]
 

output

$Aborted
 

3.5.31.3.1 Defintions of rubi rules used

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] 
]
 

3.5.31.4 Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62

\[\ln \left (\frac {{\mathrm e}^{x}}{\ln \left (x -\ln \left (x \right )\right )}+\frac {\frac {6 x^{2}}{5}-\frac {3}{5}}{x}\right )-\ln \left (\frac {{\mathrm e}^{x}}{\ln \left (x -\ln \left (x \right )\right )}+\frac {6 x}{5}\right )\]

input

int(((((15*x+15)*ln(x)-15*x^2-15*x)*ln(x-ln(x))+15*x-15)*exp(-ln(ln(x-ln(x 
)))+x)+(36*x*ln(x)-36*x^2)*ln(x-ln(x)))/((25*x^2*ln(x)-25*x^3)*ln(x-ln(x)) 
*exp(-ln(ln(x-ln(x)))+x)^2+((60*x^3-15*x)*ln(x)-60*x^4+15*x^2)*ln(x-ln(x)) 
*exp(-ln(ln(x-ln(x)))+x)+((36*x^4-18*x^2)*ln(x)-36*x^5+18*x^3)*ln(x-ln(x)) 
),x)
 

output

ln(1/ln(x-ln(x))*exp(x)+3/5*(2*x^2-1)/x)-ln(1/ln(x-ln(x))*exp(x)+6/5*x)
 

3.5.31.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {\left (-36 x^2+36 x \log (x)\right ) \log (x-\log (x))+\frac {e^x \left (-15+15 x+\left (-15 x-15 x^2+(15+15 x) \log (x)\right ) \log (x-\log (x))\right )}{\log (x-\log (x))}}{e^x \left (15 x^2-60 x^4+\left (-15 x+60 x^3\right ) \log (x)\right )+\frac {e^{2 x} \left (-25 x^3+25 x^2 \log (x)\right )}{\log (x-\log (x))}+\left (18 x^3-36 x^5+\left (-18 x^2+36 x^4\right ) \log (x)\right ) \log (x-\log (x))} \, dx=-\log \left (6 \, x + 5 \, e^{\left (x - \log \left (\log \left (x - \log \left (x\right )\right )\right )\right )}\right ) + \log \left (\frac {6 \, x^{2} + 5 \, x e^{\left (x - \log \left (\log \left (x - \log \left (x\right )\right )\right )\right )} - 3}{x}\right ) \]

input

integrate(((((15*x+15)*log(x)-15*x^2-15*x)*log(x-log(x))+15*x-15)*exp(-log 
(log(x-log(x)))+x)+(36*x*log(x)-36*x^2)*log(x-log(x)))/((25*x^2*log(x)-25* 
x^3)*log(x-log(x))*exp(-log(log(x-log(x)))+x)^2+((60*x^3-15*x)*log(x)-60*x 
^4+15*x^2)*log(x-log(x))*exp(-log(log(x-log(x)))+x)+((36*x^4-18*x^2)*log(x 
)-36*x^5+18*x^3)*log(x-log(x))),x, algorithm=\
 

output

-log(6*x + 5*e^(x - log(log(x - log(x))))) + log((6*x^2 + 5*x*e^(x - log(l 
og(x - log(x)))) - 3)/x)
 

3.5.31.6 Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (-36 x^2+36 x \log (x)\right ) \log (x-\log (x))+\frac {e^x \left (-15+15 x+\left (-15 x-15 x^2+(15+15 x) \log (x)\right ) \log (x-\log (x))\right )}{\log (x-\log (x))}}{e^x \left (15 x^2-60 x^4+\left (-15 x+60 x^3\right ) \log (x)\right )+\frac {e^{2 x} \left (-25 x^3+25 x^2 \log (x)\right )}{\log (x-\log (x))}+\left (18 x^3-36 x^5+\left (-18 x^2+36 x^4\right ) \log (x)\right ) \log (x-\log (x))} \, dx=\text {Exception raised: PolynomialError} \]

input

integrate(((((15*x+15)*ln(x)-15*x**2-15*x)*ln(x-ln(x))+15*x-15)*exp(-ln(ln 
(x-ln(x)))+x)+(36*x*ln(x)-36*x**2)*ln(x-ln(x)))/((25*x**2*ln(x)-25*x**3)*l 
n(x-ln(x))*exp(-ln(ln(x-ln(x)))+x)**2+((60*x**3-15*x)*ln(x)-60*x**4+15*x** 
2)*ln(x-ln(x))*exp(-ln(ln(x-ln(x)))+x)+((36*x**4-18*x**2)*ln(x)-36*x**5+18 
*x**3)*ln(x-ln(x))),x)
 

output

Exception raised: PolynomialError >> 1/(-5*_t0*x**2 + 5*x**3) contains an 
element of the set of generators.
 

3.5.31.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (29) = 58\).

Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.41 \[ \int \frac {\left (-36 x^2+36 x \log (x)\right ) \log (x-\log (x))+\frac {e^x \left (-15+15 x+\left (-15 x-15 x^2+(15+15 x) \log (x)\right ) \log (x-\log (x))\right )}{\log (x-\log (x))}}{e^x \left (15 x^2-60 x^4+\left (-15 x+60 x^3\right ) \log (x)\right )+\frac {e^{2 x} \left (-25 x^3+25 x^2 \log (x)\right )}{\log (x-\log (x))}+\left (18 x^3-36 x^5+\left (-18 x^2+36 x^4\right ) \log (x)\right ) \log (x-\log (x))} \, dx=\log \left (2 \, x^{2} - 1\right ) - 2 \, \log \left (x\right ) + \log \left (\frac {5 \, x e^{x} + 3 \, {\left (2 \, x^{2} - 1\right )} \log \left (x - \log \left (x\right )\right )}{3 \, {\left (2 \, x^{2} - 1\right )}}\right ) - \log \left (\frac {6 \, x \log \left (x - \log \left (x\right )\right ) + 5 \, e^{x}}{6 \, x}\right ) \]

input

integrate(((((15*x+15)*log(x)-15*x^2-15*x)*log(x-log(x))+15*x-15)*exp(-log 
(log(x-log(x)))+x)+(36*x*log(x)-36*x^2)*log(x-log(x)))/((25*x^2*log(x)-25* 
x^3)*log(x-log(x))*exp(-log(log(x-log(x)))+x)^2+((60*x^3-15*x)*log(x)-60*x 
^4+15*x^2)*log(x-log(x))*exp(-log(log(x-log(x)))+x)+((36*x^4-18*x^2)*log(x 
)-36*x^5+18*x^3)*log(x-log(x))),x, algorithm=\
 

output

log(2*x^2 - 1) - 2*log(x) + log(1/3*(5*x*e^x + 3*(2*x^2 - 1)*log(x - log(x 
)))/(2*x^2 - 1)) - log(1/6*(6*x*log(x - log(x)) + 5*e^x)/x)
 

3.5.31.8 Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {\left (-36 x^2+36 x \log (x)\right ) \log (x-\log (x))+\frac {e^x \left (-15+15 x+\left (-15 x-15 x^2+(15+15 x) \log (x)\right ) \log (x-\log (x))\right )}{\log (x-\log (x))}}{e^x \left (15 x^2-60 x^4+\left (-15 x+60 x^3\right ) \log (x)\right )+\frac {e^{2 x} \left (-25 x^3+25 x^2 \log (x)\right )}{\log (x-\log (x))}+\left (18 x^3-36 x^5+\left (-18 x^2+36 x^4\right ) \log (x)\right ) \log (x-\log (x))} \, dx=\log \left (6 \, x^{2} \log \left (x - \log \left (x\right )\right ) + 5 \, x e^{x} - 3 \, \log \left (x - \log \left (x\right )\right )\right ) - \log \left (6 \, x \log \left (x - \log \left (x\right )\right ) + 5 \, e^{x}\right ) - \log \left (x\right ) \]

input

integrate(((((15*x+15)*log(x)-15*x^2-15*x)*log(x-log(x))+15*x-15)*exp(-log 
(log(x-log(x)))+x)+(36*x*log(x)-36*x^2)*log(x-log(x)))/((25*x^2*log(x)-25* 
x^3)*log(x-log(x))*exp(-log(log(x-log(x)))+x)^2+((60*x^3-15*x)*log(x)-60*x 
^4+15*x^2)*log(x-log(x))*exp(-log(log(x-log(x)))+x)+((36*x^4-18*x^2)*log(x 
)-36*x^5+18*x^3)*log(x-log(x))),x, algorithm=\
 

output

log(6*x^2*log(x - log(x)) + 5*x*e^x - 3*log(x - log(x))) - log(6*x*log(x - 
 log(x)) + 5*e^x) - log(x)
 

3.5.31.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-36 x^2+36 x \log (x)\right ) \log (x-\log (x))+\frac {e^x \left (-15+15 x+\left (-15 x-15 x^2+(15+15 x) \log (x)\right ) \log (x-\log (x))\right )}{\log (x-\log (x))}}{e^x \left (15 x^2-60 x^4+\left (-15 x+60 x^3\right ) \log (x)\right )+\frac {e^{2 x} \left (-25 x^3+25 x^2 \log (x)\right )}{\log (x-\log (x))}+\left (18 x^3-36 x^5+\left (-18 x^2+36 x^4\right ) \log (x)\right ) \log (x-\log (x))} \, dx=-\int \frac {\ln \left (x-\ln \left (x\right )\right )\,\left (36\,x\,\ln \left (x\right )-36\,x^2\right )-{\mathrm {e}}^{x-\ln \left (\ln \left (x-\ln \left (x\right )\right )\right )}\,\left (\ln \left (x-\ln \left (x\right )\right )\,\left (15\,x-\ln \left (x\right )\,\left (15\,x+15\right )+15\,x^2\right )-15\,x+15\right )}{\ln \left (x-\ln \left (x\right )\right )\,\left (\ln \left (x\right )\,\left (18\,x^2-36\,x^4\right )-18\,x^3+36\,x^5\right )-{\mathrm {e}}^{2\,x-2\,\ln \left (\ln \left (x-\ln \left (x\right )\right )\right )}\,\ln \left (x-\ln \left (x\right )\right )\,\left (25\,x^2\,\ln \left (x\right )-25\,x^3\right )+{\mathrm {e}}^{x-\ln \left (\ln \left (x-\ln \left (x\right )\right )\right )}\,\ln \left (x-\ln \left (x\right )\right )\,\left (\ln \left (x\right )\,\left (15\,x-60\,x^3\right )-15\,x^2+60\,x^4\right )} \,d x \]

input

int(-(log(x - log(x))*(36*x*log(x) - 36*x^2) - exp(x - log(log(x - log(x)) 
))*(log(x - log(x))*(15*x - log(x)*(15*x + 15) + 15*x^2) - 15*x + 15))/(lo 
g(x - log(x))*(log(x)*(18*x^2 - 36*x^4) - 18*x^3 + 36*x^5) - exp(2*x - 2*l 
og(log(x - log(x))))*log(x - log(x))*(25*x^2*log(x) - 25*x^3) + exp(x - lo 
g(log(x - log(x))))*log(x - log(x))*(log(x)*(15*x - 60*x^3) - 15*x^2 + 60* 
x^4)),x)
 

output

-int((log(x - log(x))*(36*x*log(x) - 36*x^2) - exp(x - log(log(x - log(x)) 
))*(log(x - log(x))*(15*x - log(x)*(15*x + 15) + 15*x^2) - 15*x + 15))/(lo 
g(x - log(x))*(log(x)*(18*x^2 - 36*x^4) - 18*x^3 + 36*x^5) - exp(2*x - 2*l 
og(log(x - log(x))))*log(x - log(x))*(25*x^2*log(x) - 25*x^3) + exp(x - lo 
g(log(x - log(x))))*log(x - log(x))*(log(x)*(15*x - 60*x^3) - 15*x^2 + 60* 
x^4)), x)