3.27.50 \(\int \frac {-75-x-2 x^2+6 e^{-x+x \log (x)} x^2 \log (x)}{3 x^2} \, dx\) [2650]

3.27.50.1 Optimal result

Integrand size = 34, antiderivative size = 31 \[ \int \frac {-75-x-2 x^2+6 e^{-x+x \log (x)} x^2 \log (x)}{3 x^2} \, dx=\frac {25}{x}-\frac {1}{3} \log \left (e^{-6 e^{-x+x \log (x)}+2 x} x\right ) \]

output

25/x-1/3*ln(exp(x-3*exp(x*ln(x)-x))^2*x)
 

3.27.50.2 Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-75-x-2 x^2+6 e^{-x+x \log (x)} x^2 \log (x)}{3 x^2} \, dx=\frac {1}{3} \left (\frac {75}{x}-2 x+6 e^{-x} x^x-\log (x)\right ) \]

input

Integrate[(-75 - x - 2*x^2 + 6*E^(-x + x*Log[x])*x^2*Log[x])/(3*x^2),x]
 

output

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

3.27.50.3 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.

\(\displaystyle \int \frac {-2 x^2+6 x^2 e^{x \log (x)-x} \log (x)-x-75}{3 x^2} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int -\frac {-6 e^{-x} \log (x) x^{x+2}+2 x^2+x+75}{x^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{3} \int \frac {-6 e^{-x} \log (x) x^{x+2}+2 x^2+x+75}{x^2}dx\)

\(\Big \downarrow \) 2010

\(\displaystyle -\frac {1}{3} \int \left (\frac {2 x^2+x+75}{x^2}-6 e^{-x} x^x \log (x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (-6 \int \frac {\int e^{-x} x^xdx}{x}dx+6 \log (x) \int e^{-x} x^xdx-2 x+\frac {75}{x}-\log (x)\right )\)

input

Int[(-75 - x - 2*x^2 + 6*E^(-x + x*Log[x])*x^2*Log[x])/(3*x^2),x]
 

output

$Aborted
 

3.27.50.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

3.27.50.4 Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {2 x}{3}-\frac {\ln \left (x \right )}{3}+\frac {25}{x}+2 x^{x} {\mathrm e}^{-x}\)	\(23\)
default	\(-\frac {2 x}{3}-\frac {\ln \left (x \right )}{3}+\frac {25}{x}+2 \,{\mathrm e}^{x \ln \left (x \right )-x}\)	\(25\)
parts	\(-\frac {2 x}{3}-\frac {\ln \left (x \right )}{3}+\frac {25}{x}+2 \,{\mathrm e}^{x \ln \left (x \right )-x}\)	\(25\)
parallelrisch	\(-\frac {x \ln \left (x \right )+2 x^{2}-6 x \,{\mathrm e}^{\left (\ln \left (x \right )-1\right ) x}-75}{3 x}\)	\(27\)
norman	\(\frac {25-\frac {x \ln \left (x \right )}{3}-\frac {2 x^{2}}{3}+2 x \,{\mathrm e}^{x \ln \left (x \right )-x}}{x}\)	\(29\)

input

int(1/3*(6*x^2*ln(x)*exp(x*ln(x)-x)-2*x^2-x-75)/x^2,x,method=_RETURNVERBOS 
E)
 

output

-2/3*x-1/3*ln(x)+25/x+2*x^x*exp(-x)
 

3.27.50.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-75-x-2 x^2+6 e^{-x+x \log (x)} x^2 \log (x)}{3 x^2} \, dx=-\frac {2 \, x^{2} - 6 \, x e^{\left (x \log \left (x\right ) - x\right )} + x \log \left (x\right ) - 75}{3 \, x} \]

input

integrate(1/3*(6*x^2*log(x)*exp(x*log(x)-x)-2*x^2-x-75)/x^2,x, algorithm=\
 

output

-1/3*(2*x^2 - 6*x*e^(x*log(x) - x) + x*log(x) - 75)/x
 

3.27.50.6 Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-75-x-2 x^2+6 e^{-x+x \log (x)} x^2 \log (x)}{3 x^2} \, dx=- \frac {2 x}{3} + 2 e^{x \log {\left (x \right )} - x} - \frac {\log {\left (x \right )}}{3} + \frac {25}{x} \]

input

integrate(1/3*(6*x**2*ln(x)*exp(x*ln(x)-x)-2*x**2-x-75)/x**2,x)
 

output

-2*x/3 + 2*exp(x*log(x) - x) - log(x)/3 + 25/x
 

3.27.50.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-75-x-2 x^2+6 e^{-x+x \log (x)} x^2 \log (x)}{3 x^2} \, dx=-\frac {2}{3} \, x + \frac {25}{x} + 2 \, e^{\left (x \log \left (x\right ) - x\right )} - \frac {1}{3} \, \log \left (x\right ) \]

input

integrate(1/3*(6*x^2*log(x)*exp(x*log(x)-x)-2*x^2-x-75)/x^2,x, algorithm=\
 

output

-2/3*x + 25/x + 2*e^(x*log(x) - x) - 1/3*log(x)
 

3.27.50.8 Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-75-x-2 x^2+6 e^{-x+x \log (x)} x^2 \log (x)}{3 x^2} \, dx=-\frac {2 \, x^{2} - 6 \, x e^{\left (x \log \left (x\right ) - x\right )} + x \log \left (x\right ) - 75}{3 \, x} \]

input

integrate(1/3*(6*x^2*log(x)*exp(x*log(x)-x)-2*x^2-x-75)/x^2,x, algorithm=\
 

output

-1/3*(2*x^2 - 6*x*e^(x*log(x) - x) + x*log(x) - 75)/x
 

3.27.50.9 Mupad [B] (verification not implemented)

Time = 8.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-75-x-2 x^2+6 e^{-x+x \log (x)} x^2 \log (x)}{3 x^2} \, dx=\frac {25}{x}-\frac {\ln \left (x\right )}{3}-\frac {2\,x}{3}+2\,x^x\,{\mathrm {e}}^{-x} \]

input

int(-(x/3 + (2*x^2)/3 - 2*x^2*exp(x*log(x) - x)*log(x) + 25)/x^2,x)
 

output

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