3.23.92 \(\int \frac {-9+1048576 e^{5-5 e^x-5 x} (-1-5 x-5 e^x x)+\log (x)}{2 x^2} \, dx\)

Optimal. Leaf size=36 \[ \frac {4}{x}+\frac {e^{5+5 \left (-e^x-x+\log (16)\right )}+x-\log (x)}{2 x} \]

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Rubi [F]  time = 0.41, 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 {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-9 + 1048576*E^(5 - 5*E^x - 5*x)*(-1 - 5*x - 5*E^x*x) + Log[x])/(2*x^2),x]

[Out]

-1/2*1/x + (9 - Log[x])/(2*x) - 524288*Defer[Int][1/(E^(5*(-1 + E^x + x))*x^2), x] - 2621440*Defer[Int][E^(5 -
5*E^x - 4*x)/x, x] - 2621440*Defer[Int][1/(E^(5*(-1 + E^x + x))*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-9+1048576 e^{5-5 e^x-5 x} \left (-1-5 x-5 e^x x\right )+\log (x)}{x^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {1048576 e^{-5 \left (-1+e^x+x\right )} (-1-5 x)}{x^2}-\frac {5242880 e^{5-5 e^x-4 x}}{x}+\frac {-9+\log (x)}{x^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-9+\log (x)}{x^2} \, dx+524288 \int \frac {e^{-5 \left (-1+e^x+x\right )} (-1-5 x)}{x^2} \, dx-2621440 \int \frac {e^{5-5 e^x-4 x}}{x} \, dx\\ &=-\frac {1}{2 x}+\frac {9-\log (x)}{2 x}+524288 \int \left (-\frac {e^{-5 \left (-1+e^x+x\right )}}{x^2}-\frac {5 e^{-5 \left (-1+e^x+x\right )}}{x}\right ) \, dx-2621440 \int \frac {e^{5-5 e^x-4 x}}{x} \, dx\\ &=-\frac {1}{2 x}+\frac {9-\log (x)}{2 x}-524288 \int \frac {e^{-5 \left (-1+e^x+x\right )}}{x^2} \, dx-2621440 \int \frac {e^{5-5 e^x-4 x}}{x} \, dx-2621440 \int \frac {e^{-5 \left (-1+e^x+x\right )}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 25, normalized size = 0.69 \begin {gather*} \frac {8+1048576 e^{-5 \left (-1+e^x+x\right )}-\log (x)}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9 + 1048576*E^(5 - 5*E^x - 5*x)*(-1 - 5*x - 5*E^x*x) + Log[x])/(2*x^2),x]

[Out]

(8 + 1048576/E^(5*(-1 + E^x + x)) - Log[x])/(2*x)

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fricas [A]  time = 0.86, size = 25, normalized size = 0.69 \begin {gather*} \frac {e^{\left (-5 \, x - 5 \, e^{x} + 20 \, \log \relax (2) + 5\right )} - \log \relax (x) + 8}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*log(2)-5*x+5)+log(x)-9)/x^2,x, algorithm="fricas")

[Out]

1/2*(e^(-5*x - 5*e^x + 20*log(2) + 5) - log(x) + 8)/x

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giac [A]  time = 0.25, size = 31, normalized size = 0.86 \begin {gather*} -\frac {{\left (e^{x} \log \relax (x) - 8 \, e^{x} - 1048576 \, e^{\left (-4 \, x - 5 \, e^{x} + 5\right )}\right )} e^{\left (-x\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*log(2)-5*x+5)+log(x)-9)/x^2,x, algorithm="giac")

[Out]

-1/2*(e^x*log(x) - 8*e^x - 1048576*e^(-4*x - 5*e^x + 5))*e^(-x)/x

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maple [A]  time = 0.09, size = 29, normalized size = 0.81

method	result	size
risch	\(-\frac {\ln \relax (x )}{2 x}+\frac {4}{x}+\frac {524288 \,{\mathrm e}^{-5 \,{\mathrm e}^{x}+5-5 x}}{x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*ln(2)-5*x+5)+ln(x)-9)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/2*ln(x)/x+4/x+524288*exp(-5*exp(x)+5-5*x)/x

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maxima [A]  time = 0.98, size = 34, normalized size = 0.94 \begin {gather*} -\frac {{\left ({\left (\log \relax (x) + 1\right )} e^{\left (5 \, x\right )} - 1048576 \, e^{\left (-5 \, e^{x} + 5\right )}\right )} e^{\left (-5 \, x\right )}}{2 \, x} + \frac {9}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*log(2)-5*x+5)+log(x)-9)/x^2,x, algorithm="maxima")

[Out]

-1/2*((log(x) + 1)*e^(5*x) - 1048576*e^(-5*e^x + 5))*e^(-5*x)/x + 9/2/x

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mupad [B]  time = 1.48, size = 29, normalized size = 0.81 \begin {gather*} \frac {4}{x}-\frac {\ln \relax (x)}{2\,x}+\frac {524288\,{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-5\,{\mathrm {e}}^x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(20*log(2) - 5*x - 5*exp(x) + 5)*(5*x + 5*x*exp(x) + 1))/2 - log(x)/2 + 9/2)/x^2,x)

[Out]

4/x - log(x)/(2*x) + (524288*exp(-5*x)*exp(5)*exp(-5*exp(x)))/x

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sympy [A]  time = 0.33, size = 24, normalized size = 0.67 \begin {gather*} \frac {524288 e^{- 5 x - 5 e^{x} + 5}}{x} - \frac {\log {\relax (x )}}{2 x} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-5*exp(x)*x-5*x-1)*exp(-5*exp(x)+20*ln(2)-5*x+5)+ln(x)-9)/x**2,x)

[Out]

524288*exp(-5*x - 5*exp(x) + 5)/x - log(x)/(2*x) + 4/x

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