3.31.75 \(\int \frac {e^{\frac {-1+2 x}{2 x}} (-5+10 x)-10 x \log (3)}{2 e^{\frac {-1+2 x}{x}} x-4 e^{\frac {-1+2 x}{2 x}} x \log (3)+2 x \log ^2(3)} \, dx\)

Optimal. Leaf size=23 \[ 19+\frac {5 x}{e^{1-\frac {1}{2 x}}-\log (3)} \]

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Rubi [F]  time = 2.77, 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 {e^{\frac {-1+2 x}{2 x}} (-5+10 x)-10 x \log (3)}{2 e^{\frac {-1+2 x}{x}} x-4 e^{\frac {-1+2 x}{2 x}} x \log (3)+2 x \log ^2(3)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-1 + 2*x)/(2*x))*(-5 + 10*x) - 10*x*Log[3])/(2*E^((-1 + 2*x)/x)*x - 4*E^((-1 + 2*x)/(2*x))*x*Log[3] +
2*x*Log[3]^2),x]

[Out]

5*E^(-1 + 1/(2*x))*x - (5*Log[x])/(2*Log[3]) + 5*Log[3]*Defer[Int][E^(-1 + x^(-1))/(E - E^(1/(2*x))*Log[3]), x
] + (5*E^2*Defer[Subst][Defer[Int][1/(x*(-E + E^(x/2)*Log[3])^2), x], x, x^(-1)])/(2*Log[3]) + (5*E*Defer[Subs
t][Defer[Int][1/(x*(-E + E^(x/2)*Log[3])), x], x, x^(-1)])/Log[3] + (5*Log[3]*Defer[Subst][Defer[Int][E^(-1 +
x)/(E*x - E^(x/2)*x*Log[3]), x], x, x^(-1)])/2

Rubi steps

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

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Mathematica [A]  time = 0.40, size = 34, normalized size = 1.48 \begin {gather*} -\frac {5 x \left (\log (9)-\frac {e \log (9)}{e-e^{\left .\frac {1}{2}\right /x} \log (3)}\right )}{2 \log ^2(3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-1 + 2*x)/(2*x))*(-5 + 10*x) - 10*x*Log[3])/(2*E^((-1 + 2*x)/x)*x - 4*E^((-1 + 2*x)/(2*x))*x*Lo
g[3] + 2*x*Log[3]^2),x]

[Out]

(-5*x*(Log[9] - (E*Log[9])/(E - E^(1/(2*x))*Log[3])))/(2*Log[3]^2)

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fricas [A]  time = 0.86, size = 21, normalized size = 0.91 \begin {gather*} \frac {5 \, x}{e^{\left (\frac {2 \, x - 1}{2 \, x}\right )} - \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-5)*exp(1/2*(2*x-1)/x)-10*x*log(3))/(2*x*exp(1/2*(2*x-1)/x)^2-4*x*log(3)*exp(1/2*(2*x-1)/x)+2*
x*log(3)^2),x, algorithm="fricas")

[Out]

5*x/(e^(1/2*(2*x - 1)/x) - log(3))

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giac [A]  time = 0.35, size = 21, normalized size = 0.91 \begin {gather*} \frac {5 \, x}{e^{\left (\frac {2 \, x - 1}{2 \, x}\right )} - \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-5)*exp(1/2*(2*x-1)/x)-10*x*log(3))/(2*x*exp(1/2*(2*x-1)/x)^2-4*x*log(3)*exp(1/2*(2*x-1)/x)+2*
x*log(3)^2),x, algorithm="giac")

[Out]

5*x/(e^(1/2*(2*x - 1)/x) - log(3))

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maple [A]  time = 0.76, size = 22, normalized size = 0.96

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x-5)*exp(1/2*(2*x-1)/x)-10*x*ln(3))/(2*x*exp(1/2*(2*x-1)/x)^2-4*x*ln(3)*exp(1/2*(2*x-1)/x)+2*x*ln(3)^
2),x,method=_RETURNVERBOSE)

[Out]

-5*x/(ln(3)-exp(1/2*(2*x-1)/x))

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maxima [A]  time = 0.72, size = 25, normalized size = 1.09 \begin {gather*} -\frac {5 \, x e^{\left (\frac {1}{2 \, x}\right )}}{e^{\left (\frac {1}{2 \, x}\right )} \log \relax (3) - e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-5)*exp(1/2*(2*x-1)/x)-10*x*log(3))/(2*x*exp(1/2*(2*x-1)/x)^2-4*x*log(3)*exp(1/2*(2*x-1)/x)+2*
x*log(3)^2),x, algorithm="maxima")

[Out]

-5*x*e^(1/2/x)/(e^(1/2/x)*log(3) - e)

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mupad [B]  time = 1.88, size = 18, normalized size = 0.78 \begin {gather*} -\frac {10\,x}{\ln \relax (9)-2\,{\mathrm {e}}^{1-\frac {1}{2\,x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x - 1/2)/x)*(10*x - 5) - 10*x*log(3))/(2*x*exp((2*(x - 1/2))/x) + 2*x*log(3)^2 - 4*x*exp((x - 1/2)/x
)*log(3)),x)

[Out]

-(10*x)/(log(9) - 2*exp(1 - 1/(2*x)))

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sympy [A]  time = 0.13, size = 14, normalized size = 0.61 \begin {gather*} \frac {5 x}{e^{\frac {x - \frac {1}{2}}{x}} - \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-5)*exp(1/2*(2*x-1)/x)-10*x*ln(3))/(2*x*exp(1/2*(2*x-1)/x)**2-4*x*ln(3)*exp(1/2*(2*x-1)/x)+2*x
*ln(3)**2),x)

[Out]

5*x/(exp((x - 1/2)/x) - log(3))

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