3.60.51 \(\int \frac {e^{-\frac {2 (e^{9+6 x+x^2}-x)}{x \log (\log (x))}} (8 e^{17+6 x+x^2}-8 e^8 x+e^{17+6 x+x^2} (8-48 x-16 x^2) \log (x) \log (\log (x))-8 e^8 x \log (x) \log ^2(\log (x)))}{x^4 \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=30 \[ \frac {4 e^{8-\frac {2 \left (e^{(3+x)^2}-x\right )}{x \log (\log (x))}}}{x^2} \]

________________________________________________________________________________________

Rubi [B]  time = 1.58, antiderivative size = 162, normalized size of antiderivative = 5.40, number of steps used = 1, number of rules used = 1, integrand size = 97, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {2288} \begin {gather*} \frac {4 e^{-\frac {2 \left (e^{x^2+6 x+9}-x\right )}{x \log (\log (x))}} \left (e^{x^2+6 x+17}+e^{x^2+6 x+17} \left (-2 x^2-6 x+1\right ) \log (x) \log (\log (x))-e^8 x\right )}{x^4 \log (x) \left (\frac {e^{x^2+6 x+9}-x}{x^2 \log (x) \log ^2(\log (x))}+\frac {e^{x^2+6 x+9}-x}{x^2 \log (\log (x))}+\frac {1-2 e^{x^2+6 x+9} (x+3)}{x \log (\log (x))}\right ) \log ^2(\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*E^(17 + 6*x + x^2) - 8*E^8*x + E^(17 + 6*x + x^2)*(8 - 48*x - 16*x^2)*Log[x]*Log[Log[x]] - 8*E^8*x*Log[
x]*Log[Log[x]]^2)/(E^((2*(E^(9 + 6*x + x^2) - x))/(x*Log[Log[x]]))*x^4*Log[x]*Log[Log[x]]^2),x]

[Out]

(4*(E^(17 + 6*x + x^2) - E^8*x + E^(17 + 6*x + x^2)*(1 - 6*x - 2*x^2)*Log[x]*Log[Log[x]]))/(E^((2*(E^(9 + 6*x
+ x^2) - x))/(x*Log[Log[x]]))*x^4*Log[x]*((E^(9 + 6*x + x^2) - x)/(x^2*Log[x]*Log[Log[x]]^2) + (E^(9 + 6*x + x
^2) - x)/(x^2*Log[Log[x]]) + (1 - 2*E^(9 + 6*x + x^2)*(3 + x))/(x*Log[Log[x]]))*Log[Log[x]]^2)

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {4 e^{-\frac {2 \left (e^{9+6 x+x^2}-x\right )}{x \log (\log (x))}} \left (e^{17+6 x+x^2}-e^8 x+e^{17+6 x+x^2} \left (1-6 x-2 x^2\right ) \log (x) \log (\log (x))\right )}{x^4 \log (x) \left (\frac {e^{9+6 x+x^2}-x}{x^2 \log (x) \log ^2(\log (x))}+\frac {e^{9+6 x+x^2}-x}{x^2 \log (\log (x))}+\frac {1-2 e^{9+6 x+x^2} (3+x)}{x \log (\log (x))}\right ) \log ^2(\log (x))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 29, normalized size = 0.97 \begin {gather*} \frac {4 e^{8+\frac {2-\frac {2 e^{(3+x)^2}}{x}}{\log (\log (x))}}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^(17 + 6*x + x^2) - 8*E^8*x + E^(17 + 6*x + x^2)*(8 - 48*x - 16*x^2)*Log[x]*Log[Log[x]] - 8*E^8*
x*Log[x]*Log[Log[x]]^2)/(E^((2*(E^(9 + 6*x + x^2) - x))/(x*Log[Log[x]]))*x^4*Log[x]*Log[Log[x]]^2),x]

[Out]

(4*E^(8 + (2 - (2*E^(3 + x)^2)/x)/Log[Log[x]]))/x^2

________________________________________________________________________________________

fricas [A]  time = 0.81, size = 36, normalized size = 1.20 \begin {gather*} \frac {4 \, e^{\left (\frac {2 \, {\left (x e^{8} - e^{\left (x^{2} + 6 \, x + 17\right )}\right )} e^{\left (-8\right )}}{x \log \left (\log \relax (x)\right )} + 8\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*exp(4)^2*log(x)*log(log(x))^2+(-16*x^2-48*x+8)*exp(4)^2*exp(x^2+6*x+9)*log(x)*log(log(x))+8*ex
p(4)^2*exp(x^2+6*x+9)-8*x*exp(4)^2)/x^4/log(x)/log(log(x))^2/exp((exp(x^2+6*x+9)-x)/x/log(log(x)))^2,x, algori
thm="fricas")

[Out]

4*e^(2*(x*e^8 - e^(x^2 + 6*x + 17))*e^(-8)/(x*log(log(x))) + 8)/x^2

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*exp(4)^2*log(x)*log(log(x))^2+(-16*x^2-48*x+8)*exp(4)^2*exp(x^2+6*x+9)*log(x)*log(log(x))+8*ex
p(4)^2*exp(x^2+6*x+9)-8*x*exp(4)^2)/x^4/log(x)/log(log(x))^2/exp((exp(x^2+6*x+9)-x)/x/log(log(x)))^2,x, algori
thm="giac")

[Out]

undef

________________________________________________________________________________________

maple [A]  time = 0.06, size = 33, normalized size = 1.10

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{\frac {8 x \ln \left (\ln \relax (x )\right )-2 \,{\mathrm e}^{\left (3+x \right )^{2}}+2 x}{x \ln \left (\ln \relax (x )\right )}}}{x^{2}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x*exp(4)^2*ln(x)*ln(ln(x))^2+(-16*x^2-48*x+8)*exp(4)^2*exp(x^2+6*x+9)*ln(x)*ln(ln(x))+8*exp(4)^2*exp(x
^2+6*x+9)-8*x*exp(4)^2)/x^4/ln(x)/ln(ln(x))^2/exp((exp(x^2+6*x+9)-x)/x/ln(ln(x)))^2,x,method=_RETURNVERBOSE)

[Out]

4/x^2*exp(2*(4*x*ln(ln(x))-exp((3+x)^2)+x)/ln(ln(x))/x)

________________________________________________________________________________________

maxima [A]  time = 0.47, size = 34, normalized size = 1.13 \begin {gather*} \frac {4 \, e^{\left (-\frac {2 \, e^{\left (x^{2} + 6 \, x + 9\right )}}{x \log \left (\log \relax (x)\right )} + \frac {2}{\log \left (\log \relax (x)\right )} + 8\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*exp(4)^2*log(x)*log(log(x))^2+(-16*x^2-48*x+8)*exp(4)^2*exp(x^2+6*x+9)*log(x)*log(log(x))+8*ex
p(4)^2*exp(x^2+6*x+9)-8*x*exp(4)^2)/x^4/log(x)/log(log(x))^2/exp((exp(x^2+6*x+9)-x)/x/log(log(x)))^2,x, algori
thm="maxima")

[Out]

4*e^(-2*e^(x^2 + 6*x + 9)/(x*log(log(x))) + 2/log(log(x)) + 8)/x^2

________________________________________________________________________________________

mupad [B]  time = 4.50, size = 36, normalized size = 1.20 \begin {gather*} \frac {4\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{6\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^9}{x\,\ln \left (\ln \relax (x)\right )}}\,{\mathrm {e}}^8\,{\mathrm {e}}^{\frac {2}{\ln \left (\ln \relax (x)\right )}}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*(x - exp(6*x + x^2 + 9)))/(x*log(log(x))))*(8*x*exp(8) - 8*exp(8)*exp(6*x + x^2 + 9) + 8*x*log(lo
g(x))^2*exp(8)*log(x) + log(log(x))*exp(8)*exp(6*x + x^2 + 9)*log(x)*(48*x + 16*x^2 - 8)))/(x^4*log(log(x))^2*
log(x)),x)

[Out]

(4*exp(-(2*exp(6*x)*exp(x^2)*exp(9))/(x*log(log(x))))*exp(8)*exp(2/log(log(x))))/x^2

________________________________________________________________________________________

sympy [A]  time = 1.95, size = 31, normalized size = 1.03 \begin {gather*} \frac {4 e^{8} e^{- \frac {2 \left (- x + e^{x^{2} + 6 x + 9}\right )}{x \log {\left (\log {\relax (x )} \right )}}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*exp(4)**2*ln(x)*ln(ln(x))**2+(-16*x**2-48*x+8)*exp(4)**2*exp(x**2+6*x+9)*ln(x)*ln(ln(x))+8*exp
(4)**2*exp(x**2+6*x+9)-8*x*exp(4)**2)/x**4/ln(x)/ln(ln(x))**2/exp((exp(x**2+6*x+9)-x)/x/ln(ln(x)))**2,x)

[Out]

4*exp(8)*exp(-2*(-x + exp(x**2 + 6*x + 9))/(x*log(log(x))))/x**2

________________________________________________________________________________________