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

Optimal. Leaf size=30 \[ \frac {1}{5} \log \left (\left (2 x-4 e^{-x+x \log (2)} \left (-e^x+x\right )\right )^2\right ) \]

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Rubi [F]  time = 180.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[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

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Mathematica [A]  time = 0.82, size = 34, normalized size = 1.13 \begin {gather*} -\frac {2 x}{5}+\frac {2}{5} \log \left (2^{1+x} e^x-2^{1+x} x+e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.57, size = 30, normalized size = 1.00 \begin {gather*} \frac {2}{5} \, x \log \relax (2) - \frac {2}{5} \, x + \frac {2}{5} \, \log \left (x e^{\left (-x \log \relax (2) + x\right )} - 2 \, x + 2 \, e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.22, size = 30, normalized size = 1.00 \begin {gather*} \frac {2}{5} \, x \log \relax (2) - \frac {2}{5} \, x + \frac {2}{5} \, \log \left (x e^{\left (-x \log \relax (2) + x\right )} - 2 \, x + 2 \, e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.10, size = 32, normalized size = 1.07

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

Verification of antiderivative is not currently implemented for this CAS.

[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*ln(2)-2/5)*x+2/5*ln(10*exp(x)+5*x*exp(-x*ln(2)+x)-10*x)

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maxima [A]  time = 0.51, size = 43, normalized size = 1.43 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.24, size = 29, normalized size = 0.97 \begin {gather*} \frac {2\,\ln \left (2\,{\mathrm {e}}^x-2\,x+\frac {x\,{\mathrm {e}}^x}{2^x}\right )}{5}+x\,\left (\frac {\ln \relax (4)}{5}-\frac {2}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 2.33, size = 37, normalized size = 1.23 \begin {gather*} - \frac {2 x}{5} + \frac {2 x \log {\relax (2 )}}{5} + \frac {2 \log {\left (x e^{x} e^{- x \log {\relax (2 )}} - 2 x + 2 e^{x} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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