3.12.69 \(\int \frac {-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} (1+x)}{-x-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} x^2+e^{\frac {1}{\log (64)}} x \log (x)} \, dx\) [1169]

3.12.69.1 Optimal result

Integrand size = 60, antiderivative size = 22 \[ \int \frac {-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} (1+x)}{-x-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} x^2+e^{\frac {1}{\log (64)}} x \log (x)} \, dx=\log \left (2 e^x+e^{-\frac {1}{\log (64)}}-x-\log (x)\right ) \]

output

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

3.12.69.2 Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} (1+x)}{-x-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} x^2+e^{\frac {1}{\log (64)}} x \log (x)} \, dx=\log \left (1+2 e^{x+\frac {1}{\log (64)}}-e^{\frac {1}{\log (64)}} x-e^{\frac {1}{\log (64)}} \log (x)\right ) \]

input

Integrate[(-2*E^(x + Log[64]^(-1))*x + E^Log[64]^(-1)*(1 + x))/(-x - 2*E^( 
x + Log[64]^(-1))*x + E^Log[64]^(-1)*x^2 + E^Log[64]^(-1)*x*Log[x]),x]
 

output

Log[1 + 2*E^(x + Log[64]^(-1)) - E^Log[64]^(-1)*x - E^Log[64]^(-1)*Log[x]]
 

3.12.69.3 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.

\(\displaystyle \int \frac {(x+1) e^{\frac {1}{\log (64)}}-2 x e^{x+\frac {1}{\log (64)}}}{x^2 e^{\frac {1}{\log (64)}}-x-2 x e^{x+\frac {1}{\log (64)}}+x e^{\frac {1}{\log (64)}} \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (2 e^x x-x-1\right ) e^{\frac {1}{\log (64)}}}{x^2 \left (-e^{\frac {1}{\log (64)}}\right )+x+2 x e^{x+\frac {1}{\log (64)}}-x e^{\frac {1}{\log (64)}} \log (x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle e^{\frac {1}{\log (64)}} \int -\frac {-2 e^x x+x+1}{-e^{\frac {1}{\log (64)}} x^2+2 e^{x+\frac {1}{\log (64)}} x-e^{\frac {1}{\log (64)}} \log (x) x+x}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -e^{\frac {1}{\log (64)}} \int \frac {-2 e^x x+x+1}{-e^{\frac {1}{\log (64)}} x^2+2 e^{x+\frac {1}{\log (64)}} x-e^{\frac {1}{\log (64)}} \log (x) x+x}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -e^{\frac {1}{\log (64)}} \int \left (-\frac {1}{x \left (e^{\frac {1}{\log (64)}} x-2 e^{x+\frac {1}{\log (64)}}+e^{\frac {1}{\log (64)}} \log (x)-1\right )}-\frac {2 e^x}{-e^{\frac {1}{\log (64)}} x+2 e^{x+\frac {1}{\log (64)}}-e^{\frac {1}{\log (64)}} \log (x)+1}+\frac {1}{-e^{\frac {1}{\log (64)}} x+2 e^{x+\frac {1}{\log (64)}}-e^{\frac {1}{\log (64)}} \log (x)+1}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -e^{\frac {1}{\log (64)}} \left (\int \frac {1}{-e^{\frac {1}{\log (64)}} x+2 e^{x+\frac {1}{\log (64)}}-e^{\frac {1}{\log (64)}} \log (x)+1}dx-2 \int \frac {e^x}{-e^{\frac {1}{\log (64)}} x+2 e^{x+\frac {1}{\log (64)}}-e^{\frac {1}{\log (64)}} \log (x)+1}dx-\int \frac {1}{x \left (e^{\frac {1}{\log (64)}} x-2 e^{x+\frac {1}{\log (64)}}+e^{\frac {1}{\log (64)}} \log (x)-1\right )}dx\right )\)

input

Int[(-2*E^(x + Log[64]^(-1))*x + E^Log[64]^(-1)*(1 + x))/(-x - 2*E^(x + Lo 
g[64]^(-1))*x + E^Log[64]^(-1)*x^2 + E^Log[64]^(-1)*x*Log[x]),x]
 

output

$Aborted
 

3.12.69.3.1 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 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] 
]
 

3.12.69.4 Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
risch	\(\ln \left (x -{\mathrm e}^{-\frac {1}{6 \ln \left (2\right )}}-2 \,{\mathrm e}^{x}+\ln \left (x \right )\right )\)	\(19\)
norman	\(\ln \left ({\mathrm e}^{\frac {1}{6 \ln \left (2\right )}} \ln \left (x \right )-2 \,{\mathrm e}^{\frac {1}{6 \ln \left (2\right )}} {\mathrm e}^{x}+x \,{\mathrm e}^{\frac {1}{6 \ln \left (2\right )}}-1\right )\)	\(34\)
parallelrisch	\(\ln \left (\left ({\mathrm e}^{\frac {1}{6 \ln \left (2\right )}} \ln \left (x \right )-2 \,{\mathrm e}^{\frac {1}{6 \ln \left (2\right )}} {\mathrm e}^{x}+x \,{\mathrm e}^{\frac {1}{6 \ln \left (2\right )}}-1\right ) {\mathrm e}^{-\frac {1}{6 \ln \left (2\right )}}\right )\)	\(44\)

input

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

output

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

3.12.69.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} (1+x)}{-x-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} x^2+e^{\frac {1}{\log (64)}} x \log (x)} \, dx=\log \left (x e^{\left (\frac {1}{6 \, \log \left (2\right )}\right )} + e^{\left (\frac {1}{6 \, \log \left (2\right )}\right )} \log \left (x\right ) - 2 \, e^{\left (\frac {6 \, x \log \left (2\right ) + 1}{6 \, \log \left (2\right )}\right )} - 1\right ) \]

input

integrate((-2*x*exp(1/6/log(2))*exp(x)+(1+x)*exp(1/6/log(2)))/(x*exp(1/6/l 
og(2))*log(x)-2*x*exp(1/6/log(2))*exp(x)+x^2*exp(1/6/log(2))-x),x, algorit 
hm=\
 

output

log(x*e^(1/6/log(2)) + e^(1/6/log(2))*log(x) - 2*e^(1/6*(6*x*log(2) + 1)/l 
og(2)) - 1)
 

3.12.69.6 Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} (1+x)}{-x-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} x^2+e^{\frac {1}{\log (64)}} x \log (x)} \, dx=\log {\left (\frac {- x e^{\frac {1}{6 \log {\left (2 \right )}}} - e^{\frac {1}{6 \log {\left (2 \right )}}} \log {\left (x \right )} + 1}{2 e^{\frac {1}{6 \log {\left (2 \right )}}}} + e^{x} \right )} \]

input

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

output

log((-x*exp(1/(6*log(2))) - exp(1/(6*log(2)))*log(x) + 1)*exp(-1/(6*log(2) 
))/2 + exp(x))
 

3.12.69.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} (1+x)}{-x-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} x^2+e^{\frac {1}{\log (64)}} x \log (x)} \, dx=\log \left (-\frac {1}{2} \, {\left (x e^{\left (\frac {1}{6 \, \log \left (2\right )}\right )} + e^{\left (\frac {1}{6 \, \log \left (2\right )}\right )} \log \left (x\right ) - 2 \, e^{\left (x + \frac {1}{6 \, \log \left (2\right )}\right )} - 1\right )} e^{\left (-\frac {1}{6 \, \log \left (2\right )}\right )}\right ) \]

input

integrate((-2*x*exp(1/6/log(2))*exp(x)+(1+x)*exp(1/6/log(2)))/(x*exp(1/6/l 
og(2))*log(x)-2*x*exp(1/6/log(2))*exp(x)+x^2*exp(1/6/log(2))-x),x, algorit 
hm=\
 

output

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

3.12.69.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} (1+x)}{-x-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} x^2+e^{\frac {1}{\log (64)}} x \log (x)} \, dx=\log \left (-x e^{\left (\frac {1}{6 \, \log \left (2\right )}\right )} - e^{\left (\frac {1}{6 \, \log \left (2\right )}\right )} \log \left (x\right ) + 2 \, e^{\left (x + \frac {1}{6 \, \log \left (2\right )}\right )} + 1\right ) \]

input

integrate((-2*x*exp(1/6/log(2))*exp(x)+(1+x)*exp(1/6/log(2)))/(x*exp(1/6/l 
og(2))*log(x)-2*x*exp(1/6/log(2))*exp(x)+x^2*exp(1/6/log(2))-x),x, algorit 
hm=\
 

output

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

3.12.69.9 Mupad [B] (verification not implemented)

Time = 13.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} (1+x)}{-x-2 e^{x+\frac {1}{\log (64)}} x+e^{\frac {1}{\log (64)}} x^2+e^{\frac {1}{\log (64)}} x \log (x)} \, dx=\ln \left (x-{\mathrm {e}}^{-\frac {1}{6\,\ln \left (2\right )}}-2\,{\mathrm {e}}^x+\ln \left (x\right )\right ) \]

input

int(-(exp(1/(6*log(2)))*(x + 1) - 2*x*exp(1/(6*log(2)))*exp(x))/(x - x^2*e 
xp(1/(6*log(2))) + 2*x*exp(1/(6*log(2)))*exp(x) - x*exp(1/(6*log(2)))*log( 
x)),x)
 

output

log(x - exp(-1/(6*log(2))) - 2*exp(x) + log(x))