3.4.4 \(\int \frac {e^{-16/x} (x-x^2+(16+16 e^5-16 x-x^2) \log (\frac {1}{5 x})+(-16+16 x+x^2) \log (\frac {1}{5 x}) \log (\log (\frac {1}{5 x})))}{x^2 \log (\frac {1}{5 x})} \, dx\)

Optimal. Leaf size=29 \[ e^{-16/x} \left (1+e^5-x+(-1+x) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 2.57, 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^{-16/x} \left (x-x^2+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )+\left (-16+16 x+x^2\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x - x^2 + (16 + 16*E^5 - 16*x - x^2)*Log[1/(5*x)] + (-16 + 16*x + x^2)*Log[1/(5*x)]*Log[Log[1/(5*x)]])/(E
^(16/x)*x^2*Log[1/(5*x)]),x]

[Out]

(1 + E^5 - x)/E^(16/x) + Defer[Int][1/(E^(16/x)*(Log[5] - Log[x^(-1)])), x] + Defer[Int][Log[Log[1/(5*x)]]/E^(
16/x), x] - Defer[Subst][Defer[Int][1/(E^(16*x)*x*Log[x/5]), x], x, x^(-1)] - 16*Defer[Subst][Defer[Int][Log[L
og[x/5]]/(E^(16*x)*x), x], x, x^(-1)] + 80*Defer[Subst][Defer[Int][Log[Log[x]]/E^(80*x), x], x, 1/(5*x)]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^{-16/x} \left (x-x^2+16 \left (1+e^5\right ) \log \left (\frac {1}{5 x}\right )-16 x \log \left (\frac {1}{5 x}\right )-x^2 \log \left (\frac {1}{5 x}\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )}+\frac {e^{-16/x} \left (-16+16 x+x^2\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )}{x^2}\right ) \, dx\\ &=\int \frac {e^{-16/x} \left (x-x^2+16 \left (1+e^5\right ) \log \left (\frac {1}{5 x}\right )-16 x \log \left (\frac {1}{5 x}\right )-x^2 \log \left (\frac {1}{5 x}\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx+\int \frac {e^{-16/x} \left (-16+16 x+x^2\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )}{x^2} \, dx\\ &=\int \frac {e^{-16/x} \left (-((-1+x) x)+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx+\int \left (e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right )-\frac {16 e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right )}{x^2}+\frac {16 e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right )}{x}\right ) \, dx\\ &=-\left (16 \int \frac {e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right )}{x^2} \, dx\right )+16 \int \frac {e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right )}{x} \, dx+\int \left (\frac {e^{-16/x} \left (16 \left (1+e^5\right )-16 x-x^2\right )}{x^2}+\frac {e^{-16/x} (1-x)}{x \log \left (\frac {1}{5 x}\right )}\right ) \, dx+\int e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right ) \, dx\\ &=16 \operatorname {Subst}\left (\int e^{-16 x} \log \left (\log \left (\frac {x}{5}\right )\right ) \, dx,x,\frac {1}{x}\right )-16 \operatorname {Subst}\left (\int \frac {e^{-16 x} \log \left (\log \left (\frac {x}{5}\right )\right )}{x} \, dx,x,\frac {1}{x}\right )+\int \frac {e^{-16/x} \left (16 \left (1+e^5\right )-16 x-x^2\right )}{x^2} \, dx+\int \frac {e^{-16/x} (1-x)}{x \log \left (\frac {1}{5 x}\right )} \, dx+\int e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right ) \, dx\\ &=e^{-16/x} \left (1+e^5-x\right )-16 \operatorname {Subst}\left (\int \frac {e^{-16 x} \log \left (\log \left (\frac {x}{5}\right )\right )}{x} \, dx,x,\frac {1}{x}\right )+80 \operatorname {Subst}\left (\int e^{-80 x} \log (\log (x)) \, dx,x,\frac {1}{5 x}\right )+\int \left (\frac {e^{-16/x}}{x \log \left (\frac {1}{5 x}\right )}+\frac {e^{-16/x}}{\log (5)-\log \left (\frac {1}{x}\right )}\right ) \, dx+\int e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right ) \, dx\\ &=e^{-16/x} \left (1+e^5-x\right )-16 \operatorname {Subst}\left (\int \frac {e^{-16 x} \log \left (\log \left (\frac {x}{5}\right )\right )}{x} \, dx,x,\frac {1}{x}\right )+80 \operatorname {Subst}\left (\int e^{-80 x} \log (\log (x)) \, dx,x,\frac {1}{5 x}\right )+\int \frac {e^{-16/x}}{x \log \left (\frac {1}{5 x}\right )} \, dx+\int \frac {e^{-16/x}}{\log (5)-\log \left (\frac {1}{x}\right )} \, dx+\int e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right ) \, dx\\ &=e^{-16/x} \left (1+e^5-x\right )-16 \operatorname {Subst}\left (\int \frac {e^{-16 x} \log \left (\log \left (\frac {x}{5}\right )\right )}{x} \, dx,x,\frac {1}{x}\right )+80 \operatorname {Subst}\left (\int e^{-80 x} \log (\log (x)) \, dx,x,\frac {1}{5 x}\right )+\int \frac {e^{-16/x}}{\log (5)-\log \left (\frac {1}{x}\right )} \, dx+\int e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right ) \, dx-\operatorname {Subst}\left (\int \frac {e^{-16 x}}{x \log \left (\frac {x}{5}\right )} \, dx,x,\frac {1}{x}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.63, size = 29, normalized size = 1.00 \begin {gather*} e^{-16/x} \left (1+e^5-x+(-1+x) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - x^2 + (16 + 16*E^5 - 16*x - x^2)*Log[1/(5*x)] + (-16 + 16*x + x^2)*Log[1/(5*x)]*Log[Log[1/(5*x)
]])/(E^(16/x)*x^2*Log[1/(5*x)]),x]

[Out]

(1 + E^5 - x + (-1 + x)*Log[Log[1/(5*x)]])/E^(16/x)

________________________________________________________________________________________

fricas [A]  time = 0.82, size = 33, normalized size = 1.14 \begin {gather*} {\left (x - 1\right )} e^{\left (-\frac {16}{x}\right )} \log \left (\log \left (\frac {1}{5 \, x}\right )\right ) - {\left (x - e^{5} - 1\right )} e^{\left (-\frac {16}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+16*x-16)*log(1/5/x)*log(log(1/5/x))+(16*exp(5)-x^2-16*x+16)*log(1/5/x)-x^2+x)/x^2/exp(4/x)^4/l
og(1/5/x),x, algorithm="fricas")

[Out]

(x - 1)*e^(-16/x)*log(log(1/5/x)) - (x - e^5 - 1)*e^(-16/x)

________________________________________________________________________________________

giac [B]  time = 0.47, size = 54, normalized size = 1.86 \begin {gather*} x e^{\left (-\frac {16}{x}\right )} \log \left (-\log \left (5 \, x\right )\right ) - x e^{\left (-\frac {16}{x}\right )} - e^{\left (-\frac {16}{x}\right )} \log \left (-\log \left (5 \, x\right )\right ) + e^{\left (-\frac {16}{x}\right )} + e^{\left (-\frac {16}{x} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+16*x-16)*log(1/5/x)*log(log(1/5/x))+(16*exp(5)-x^2-16*x+16)*log(1/5/x)-x^2+x)/x^2/exp(4/x)^4/l
og(1/5/x),x, algorithm="giac")

[Out]

x*e^(-16/x)*log(-log(5*x)) - x*e^(-16/x) - e^(-16/x)*log(-log(5*x)) + e^(-16/x) + e^(-16/x + 5)

________________________________________________________________________________________

maple [A]  time = 0.16, size = 36, normalized size = 1.24

method	result	size
risch	\(\left (x -1\right ) {\mathrm e}^{-\frac {16}{x}} \ln \left (-\ln \relax (5)-\ln \relax (x )\right )+\left ({\mathrm e}^{5}-x +1\right ) {\mathrm e}^{-\frac {16}{x}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+16*x-16)*ln(1/5/x)*ln(ln(1/5/x))+(16*exp(5)-x^2-16*x+16)*ln(1/5/x)-x^2+x)/x^2/exp(4/x)^4/ln(1/5/x),x
,method=_RETURNVERBOSE)

[Out]

(x-1)*exp(-16/x)*ln(-ln(5)-ln(x))+(exp(5)-x+1)*exp(-16/x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{\left (-\frac {16}{x} + 5\right )} - 16 \, \Gamma \left (-1, \frac {16}{x}\right ) + \int \frac {{\left (x^{2} - x {\left (16 \, \log \relax (5) + 1\right )} - 16 \, {\left (x - 1\right )} \log \relax (x) + {\left (x^{2} \log \relax (5) + 16 \, x \log \relax (5) + {\left (x^{2} + 16 \, x - 16\right )} \log \relax (x) - 16 \, \log \relax (5)\right )} \log \left (-\log \relax (5) - \log \relax (x)\right ) + 16 \, \log \relax (5)\right )} e^{\left (-\frac {16}{x}\right )}}{x^{2} \log \relax (5) + x^{2} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+16*x-16)*log(1/5/x)*log(log(1/5/x))+(16*exp(5)-x^2-16*x+16)*log(1/5/x)-x^2+x)/x^2/exp(4/x)^4/l
og(1/5/x),x, algorithm="maxima")

[Out]

e^(-16/x + 5) - 16*gamma(-1, 16/x) + integrate((x^2 - x*(16*log(5) + 1) - 16*(x - 1)*log(x) + (x^2*log(5) + 16
*x*log(5) + (x^2 + 16*x - 16)*log(x) - 16*log(5))*log(-log(5) - log(x)) + 16*log(5))*e^(-16/x)/(x^2*log(5) + x
^2*log(x)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {16}{x}}\,\left (x-\ln \left (\frac {1}{5\,x}\right )\,\left (x^2+16\,x-16\,{\mathrm {e}}^5-16\right )-x^2+\ln \left (\ln \left (\frac {1}{5\,x}\right )\right )\,\ln \left (\frac {1}{5\,x}\right )\,\left (x^2+16\,x-16\right )\right )}{x^2\,\ln \left (\frac {1}{5\,x}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-16/x)*(x - log(1/(5*x))*(16*x - 16*exp(5) + x^2 - 16) - x^2 + log(log(1/(5*x)))*log(1/(5*x))*(16*x +
x^2 - 16)))/(x^2*log(1/(5*x))),x)

[Out]

int((exp(-16/x)*(x - log(1/(5*x))*(16*x - 16*exp(5) + x^2 - 16) - x^2 + log(log(1/(5*x)))*log(1/(5*x))*(16*x +
x^2 - 16)))/(x^2*log(1/(5*x))), x)

________________________________________________________________________________________

sympy [A]  time = 99.14, size = 29, normalized size = 1.00 \begin {gather*} \left (x \log {\left (\log {\left (\frac {1}{5 x} \right )} \right )} - x - \log {\left (\log {\left (\frac {1}{5 x} \right )} \right )} + 1 + e^{5}\right ) e^{- \frac {16}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+16*x-16)*ln(1/5/x)*ln(ln(1/5/x))+(16*exp(5)-x**2-16*x+16)*ln(1/5/x)-x**2+x)/x**2/exp(4/x)**4/
ln(1/5/x),x)

[Out]

(x*log(log(1/(5*x))) - x - log(log(1/(5*x))) + 1 + exp(5))*exp(-16/x)

________________________________________________________________________________________