\(\int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx\) [8110]

Optimal result

Integrand size = 52, antiderivative size = 30 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\frac {1}{5} \log \left (\left (2 x-4 e^{-x+x \log (2)} \left (-e^x+x\right )\right )^2\right ) \]

[Out]

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

Rubi [F(-1)]

Timed out. \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\text {\$Aborted} \]

[In]

Int[(-4 + 2*E^(x - x*Log[2]) + 4*x + 4*E^x*Log[2] - 4*x*Log[2])/(10*E^x - 10*x + 5*E^(x - x*Log[2])*x),x]

[Out]

$Aborted

Rubi steps Aborted

Mathematica [A] (verified)

Time = 3.52 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=-\frac {2 x}{5}+\frac {2}{5} \log \left (2^{1+x} e^x-2^{1+x} x+e^x x\right ) \]

[In]

Integrate[(-4 + 2*E^(x - x*Log[2]) + 4*x + 4*E^x*Log[2] - 4*x*Log[2])/(10*E^x - 10*x + 5*E^(x - x*Log[2])*x),x
]

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\frac {2 x \ln \left (2\right )}{5}-\frac {2 x}{5}+\frac {2 \ln \left (x \,{\mathrm e}^{-x \ln \left (2\right )+x}+2 \,{\mathrm e}^{x}-2 x \right )}{5}\)	\(31\)
norman	\(\left (\frac {2 \ln \left (2\right )}{5}-\frac {2}{5}\right ) x +\frac {2 \ln \left (10 \,{\mathrm e}^{x}+5 x \,{\mathrm e}^{-x \ln \left (2\right )+x}-10 x \right )}{5}\)	\(32\)
risch	\(\frac {2 \ln \left (x \right )}{5}+\frac {2 x \ln \left (2\right )}{5}-\frac {2 x}{5}+\frac {2 \ln \left (\left (\frac {1}{2}\right )^{x} {\mathrm e}^{x}-\frac {2 \left (x -{\mathrm e}^{x}\right )}{x}\right )}{5}\)	\(35\)

[In]

int((4*exp(x)*ln(2)+2*exp(-x*ln(2)+x)-4*x*ln(2)+4*x-4)/(10*exp(x)+5*x*exp(-x*ln(2)+x)-10*x),x,method=_RETURNVE
RBOSE)

[Out]

2/5*x*ln(2)-2/5*x+2/5*ln(x*exp(-x*ln(2)+x)+2*exp(x)-2*x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\frac {2}{5} \, x \log \left (2\right ) - \frac {2}{5} \, x + \frac {2}{5} \, \log \left (x e^{\left (-x \log \left (2\right ) + x\right )} - 2 \, x + 2 \, e^{x}\right ) \]

[In]

integrate((4*exp(x)*log(2)+2*exp(-x*log(2)+x)-4*x*log(2)+4*x-4)/(10*exp(x)+5*x*exp(-x*log(2)+x)-10*x),x, algor
ithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=- \frac {2 x}{5} + \frac {2 x \log {\left (2 \right )}}{5} + \frac {2 \log {\left (x e^{x} e^{- x \log {\left (2 \right )}} - 2 x + 2 e^{x} \right )}}{5} \]

[In]

integrate((4*exp(x)*ln(2)+2*exp(-x*ln(2)+x)-4*x*ln(2)+4*x-4)/(10*exp(x)+5*x*exp(-x*ln(2)+x)-10*x),x)

[Out]

-2*x/5 + 2*x*log(2)/5 + 2*log(x*exp(x)*exp(-x*log(2)) - 2*x + 2*exp(x))/5

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=-\frac {2}{5} \, x + \frac {2}{5} \, \log \left (-x + e^{x}\right ) + \frac {2}{5} \, \log \left (\frac {2 \cdot 2^{x} {\left (x - e^{x}\right )} - x e^{x}}{2 \, {\left (x - e^{x}\right )}}\right ) \]

[In]

integrate((4*exp(x)*log(2)+2*exp(-x*log(2)+x)-4*x*log(2)+4*x-4)/(10*exp(x)+5*x*exp(-x*log(2)+x)-10*x),x, algor
ithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\frac {2}{5} \, x \log \left (2\right ) - \frac {2}{5} \, x + \frac {2}{5} \, \log \left (x e^{\left (-x \log \left (2\right ) + x\right )} - 2 \, x + 2 \, e^{x}\right ) \]

[In]

integrate((4*exp(x)*log(2)+2*exp(-x*log(2)+x)-4*x*log(2)+4*x-4)/(10*exp(x)+5*x*exp(-x*log(2)+x)-10*x),x, algor
ithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\frac {2\,\ln \left (2\,{\mathrm {e}}^x-2\,x+\frac {x\,{\mathrm {e}}^x}{2^x}\right )}{5}+x\,\left (\frac {\ln \left (4\right )}{5}-\frac {2}{5}\right ) \]

[In]

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

[Out]

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