3.37.23 \(\int \frac {-x^4+e^{\frac {5-x+2 x^2-2 x \log (2)}{-x^2+x \log (2)}} (10 x-x^2-5 \log (2))+2 x^3 \log (2)-x^2 \log ^2(2)}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx\)

Optimal. Leaf size=25 \[ e^{-2-\frac {-1+\frac {5}{x}}{x-\log (2)}}-x \]

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Rubi [F]  time = 3.86, 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*} \int \frac {-x^4+e^{\frac {5-x+2 x^2-2 x \log (2)}{-x^2+x \log (2)}} \left (10 x-x^2-5 \log (2)\right )+2 x^3 \log (2)-x^2 \log ^2(2)}{x^4-2 x^3 \log (2)+x^2 \log ^2(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-x^4 + E^((5 - x + 2*x^2 - 2*x*Log[2])/(-x^2 + x*Log[2]))*(10*x - x^2 - 5*Log[2]) + 2*x^3*Log[2] - x^2*Lo
g[2]^2)/(x^4 - 2*x^3*Log[2] + x^2*Log[2]^2),x]

[Out]

-x - (Log[32]*Defer[Int][(4^(x - Log[2])^(-1)*E^((-5 + x - 2*x^2)/(x*(x - Log[2]))))/x^2, x])/Log[2]^2 - (1 -
Log[32]/Log[2]^2)*Defer[Int][(4^(x - Log[2])^(-1)*E^((-5 + x - 2*x^2)/(x*(x - Log[2]))))/(x - Log[2])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x^4+e^{\frac {5-x+2 x^2-2 x \log (2)}{-x^2+x \log (2)}} \left (10 x-x^2-5 \log (2)\right )+2 x^3 \log (2)-x^2 \log ^2(2)}{x^2 \left (x^2-2 x \log (2)+\log ^2(2)\right )} \, dx\\ &=\int \frac {-x^4+e^{\frac {5-x+2 x^2-2 x \log (2)}{-x^2+x \log (2)}} \left (10 x-x^2-5 \log (2)\right )+2 x^3 \log (2)-x^2 \log ^2(2)}{x^2 (x-\log (2))^2} \, dx\\ &=\int \left (-1+\frac {4^{\frac {1}{x-\log (2)}} e^{\frac {-5+x-2 x^2}{x (x-\log (2))}} \left (10 x-x^2-\log (32)\right )}{x^2 (x-\log (2))^2}\right ) \, dx\\ &=-x+\int \frac {4^{\frac {1}{x-\log (2)}} e^{\frac {-5+x-2 x^2}{x (x-\log (2))}} \left (10 x-x^2-\log (32)\right )}{x^2 (x-\log (2))^2} \, dx\\ &=-x+\int \left (-\frac {4^{\frac {1}{x-\log (2)}} e^{\frac {-5+x-2 x^2}{x (x-\log (2))}} \log (32)}{x^2 \log ^2(2)}+\frac {4^{\frac {1}{x-\log (2)}} e^{\frac {-5+x-2 x^2}{x (x-\log (2))}} \left (-\log ^2(2)+\log (32)\right )}{(x-\log (2))^2 \log ^2(2)}\right ) \, dx\\ &=-x-\frac {\log (32) \int \frac {4^{\frac {1}{x-\log (2)}} e^{\frac {-5+x-2 x^2}{x (x-\log (2))}}}{x^2} \, dx}{\log ^2(2)}+\left (-1+\frac {\log (32)}{\log ^2(2)}\right ) \int \frac {4^{\frac {1}{x-\log (2)}} e^{\frac {-5+x-2 x^2}{x (x-\log (2))}}}{(x-\log (2))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 3.10, size = 30, normalized size = 1.20 \begin {gather*} e^{\frac {-5+x-2 x^2+x \log (4)}{x (x-\log (2))}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^4 + E^((5 - x + 2*x^2 - 2*x*Log[2])/(-x^2 + x*Log[2]))*(10*x - x^2 - 5*Log[2]) + 2*x^3*Log[2] -
x^2*Log[2]^2)/(x^4 - 2*x^3*Log[2] + x^2*Log[2]^2),x]

[Out]

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

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fricas [A]  time = 0.67, size = 33, normalized size = 1.32 \begin {gather*} -x + e^{\left (-\frac {2 \, x^{2} - 2 \, x \log \relax (2) - x + 5}{x^{2} - x \log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*log(2)-x^2+10*x)*exp((-2*x*log(2)+2*x^2-x+5)/(x*log(2)-x^2))-x^2*log(2)^2+2*x^3*log(2)-x^4)/(x^
2*log(2)^2-2*x^3*log(2)+x^4),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.28, size = 64, normalized size = 2.56 \begin {gather*} -x + e^{\left (-\frac {2 \, x^{2}}{x^{2} - x \log \relax (2)} + \frac {2 \, x \log \relax (2)}{x^{2} - x \log \relax (2)} + \frac {x}{x^{2} - x \log \relax (2)} - \frac {5}{x^{2} - x \log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*log(2)-x^2+10*x)*exp((-2*x*log(2)+2*x^2-x+5)/(x*log(2)-x^2))-x^2*log(2)^2+2*x^3*log(2)-x^4)/(x^
2*log(2)^2-2*x^3*log(2)+x^4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 32, normalized size = 1.28

method	result	size
risch	\(-x +{\mathrm e}^{-\frac {2 x \ln \relax (2)-2 x^{2}+x -5}{x \left (\ln \relax (2)-x \right )}}\)	\(32\)
norman	\(\frac {x^{3}-x \ln \relax (2)^{2}+x \ln \relax (2) {\mathrm e}^{\frac {-2 x \ln \relax (2)+2 x^{2}-x +5}{x \ln \relax (2)-x^{2}}}-x^{2} {\mathrm e}^{\frac {-2 x \ln \relax (2)+2 x^{2}-x +5}{x \ln \relax (2)-x^{2}}}}{\left (\ln \relax (2)-x \right ) x}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*ln(2)-x^2+10*x)*exp((-2*x*ln(2)+2*x^2-x+5)/(x*ln(2)-x^2))-x^2*ln(2)^2+2*x^3*ln(2)-x^4)/(x^2*ln(2)^2-2
*x^3*ln(2)+x^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.95, size = 89, normalized size = 3.56 \begin {gather*} -2 \, {\left (\frac {\log \relax (2)}{x - \log \relax (2)} - \log \left (x - \log \relax (2)\right )\right )} \log \relax (2) - 2 \, \log \relax (2) \log \left (x - \log \relax (2)\right ) - x + \frac {2 \, \log \relax (2)^{2}}{x - \log \relax (2)} + e^{\left (-\frac {5}{x \log \relax (2) - \log \relax (2)^{2}} + \frac {1}{x - \log \relax (2)} + \frac {5}{x \log \relax (2)} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*log(2)-x^2+10*x)*exp((-2*x*log(2)+2*x^2-x+5)/(x*log(2)-x^2))-x^2*log(2)^2+2*x^3*log(2)-x^4)/(x^
2*log(2)^2-2*x^3*log(2)+x^4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.48, size = 73, normalized size = 2.92 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {x}{x\,\ln \relax (2)-x^2}}\,{\mathrm {e}}^{\frac {2\,x^2}{x\,\ln \relax (2)-x^2}}\,{\mathrm {e}}^{\frac {5}{x\,\ln \relax (2)-x^2}}}{2^{\frac {2\,x}{x\,\ln \relax (2)-x^2}}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2*log(2)^2 + exp(-(x + 2*x*log(2) - 2*x^2 - 5)/(x*log(2) - x^2))*(5*log(2) - 10*x + x^2) - 2*x^3*log(2
) + x^4)/(x^2*log(2)^2 - 2*x^3*log(2) + x^4),x)

[Out]

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

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sympy [A]  time = 0.42, size = 26, normalized size = 1.04 \begin {gather*} - x + e^{\frac {2 x^{2} - 2 x \log {\relax (2 )} - x + 5}{- x^{2} + x \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*ln(2)-x**2+10*x)*exp((-2*x*ln(2)+2*x**2-x+5)/(x*ln(2)-x**2))-x**2*ln(2)**2+2*x**3*ln(2)-x**4)/(
x**2*ln(2)**2-2*x**3*ln(2)+x**4),x)

[Out]

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

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