3.101.75 \(\int \frac {2^{-1/x} (2^{\frac {1}{x}} e^{-5+x} (-x-x^2)+e^{-2+x} (x+x^2+\log (2)))}{x} \, dx\)

Optimal. Leaf size=23 \[ -e^{-5+x} x+2^{-1/x} e^{-2+x} x \]

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Rubi [F]  time = 0.53, 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 {2^{-1/x} \left (2^{\frac {1}{x}} e^{-5+x} \left (-x-x^2\right )+e^{-2+x} \left (x+x^2+\log (2)\right )\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.42, size = 20, normalized size = 0.87 \begin {gather*} e^{-5+x} \left (-1+2^{-1/x} e^3\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-5 + x)*(-1 + E^3/2^x^(-1))*x

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fricas [A]  time = 0.71, size = 30, normalized size = 1.30 \begin {gather*} -\frac {{\left (2^{\left (\frac {1}{x}\right )} x e^{\left (x - 2\right )} - x e^{\left (x + 1\right )}\right )} e^{\left (-3\right )}}{2^{\left (\frac {1}{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (x^{2} + x\right )} 2^{\left (\frac {1}{x}\right )} e^{\left (x - 5\right )} - {\left (x^{2} + x + \log \relax (2)\right )} e^{\left (x - 2\right )}}{2^{\left (\frac {1}{x}\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 22, normalized size = 0.96

method	result	size
risch	\(-x \,{\mathrm e}^{x -5}+2^{-\frac {1}{x}} x \,{\mathrm e}^{x -2}\)	\(22\)
norman	\(\left (x \,{\mathrm e}^{3} {\mathrm e}^{x -5}-{\mathrm e}^{x -5} {\mathrm e}^{\frac {\ln \relax (2)}{x}} x \right ) {\mathrm e}^{-\frac {\ln \relax (2)}{x}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 29, normalized size = 1.26 \begin {gather*} x e^{\left (x - \frac {\log \relax (2)}{x} - 2\right )} - {\left (x - 1\right )} e^{\left (x - 5\right )} - e^{\left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.24, size = 19, normalized size = 0.83 \begin {gather*} -x\,{\mathrm {e}}^x\,\left ({\mathrm {e}}^{-5}-\frac {{\mathrm {e}}^{-2}}{2^{1/x}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*exp(x)*(exp(-5) - exp(-2)/2^(1/x))

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sympy [A]  time = 10.64, size = 17, normalized size = 0.74 \begin {gather*} \left (- x + x e^{3} e^{- \frac {\log {\relax (2 )}}{x}}\right ) e^{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-x + x*exp(3)*exp(-log(2)/x))*exp(x - 5)

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