3.28.58 \(\int \frac {e^{\frac {-2+(-14+2 x) \log (\log (x))}{\log (\log (x))}} (1+x \log (x) \log ^2(\log (x)))}{2 x \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=20 \[ 8+\frac {1}{4} e^{2 \left (-7+x-\frac {1}{\log (\log (x))}\right )} \]

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Rubi [A]  time = 1.03, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 6688, 6706} \begin {gather*} \frac {1}{4} e^{-2 \left (-x+\frac {1}{\log (\log (x))}+7\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-2 + (-14 + 2*x)*Log[Log[x]])/Log[Log[x]])*(1 + x*Log[x]*Log[Log[x]]^2))/(2*x*Log[x]*Log[Log[x]]^2),x
]

[Out]

1/(4*E^(2*(7 - x + Log[Log[x]]^(-1))))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {e^{\frac {-2+(-14+2 x) \log (\log (x))}{\log (\log (x))}} \left (1+x \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))} \, dx\\ &=\frac {1}{2} \int \frac {e^{-2 \left (7-x+\frac {1}{\log (\log (x))}\right )} \left (1+x \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))} \, dx\\ &=\frac {1}{4} e^{-2 \left (7-x+\frac {1}{\log (\log (x))}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 18, normalized size = 0.90 \begin {gather*} \frac {1}{4} e^{-14+2 x-\frac {2}{\log (\log (x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-2 + (-14 + 2*x)*Log[Log[x]])/Log[Log[x]])*(1 + x*Log[x]*Log[Log[x]]^2))/(2*x*Log[x]*Log[Log[x]
]^2),x]

[Out]

E^(-14 + 2*x - 2/Log[Log[x]])/4

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fricas [A]  time = 0.62, size = 19, normalized size = 0.95 \begin {gather*} \frac {1}{4} \, e^{\left (\frac {2 \, {\left ({\left (x - 7\right )} \log \left (\log \relax (x)\right ) - 1\right )}}{\log \left (\log \relax (x)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(x*log(x)*log(log(x))^2+1)*exp(((2*x-14)*log(log(x))-2)/log(log(x)))/x/log(x)/log(log(x))^2,x, a
lgorithm="fricas")

[Out]

1/4*e^(2*((x - 7)*log(log(x)) - 1)/log(log(x)))

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giac [A]  time = 0.23, size = 15, normalized size = 0.75 \begin {gather*} \frac {1}{4} \, e^{\left (2 \, x - \frac {2}{\log \left (\log \relax (x)\right )} - 14\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(x*log(x)*log(log(x))^2+1)*exp(((2*x-14)*log(log(x))-2)/log(log(x)))/x/log(x)/log(log(x))^2,x, a
lgorithm="giac")

[Out]

1/4*e^(2*x - 2/log(log(x)) - 14)

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maple [A]  time = 0.02, size = 23, normalized size = 1.15

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {2 x \ln \left (\ln \relax (x )\right )-14 \ln \left (\ln \relax (x )\right )-2}{\ln \left (\ln \relax (x )\right )}}}{4}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(x*ln(x)*ln(ln(x))^2+1)*exp(((2*x-14)*ln(ln(x))-2)/ln(ln(x)))/x/ln(x)/ln(ln(x))^2,x,method=_RETURNVERB
OSE)

[Out]

1/4*exp(2*(x*ln(ln(x))-7*ln(ln(x))-1)/ln(ln(x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{2} \, \int \frac {{\left (x \log \relax (x) \log \left (\log \relax (x)\right )^{2} + 1\right )} e^{\left (\frac {2 \, {\left ({\left (x - 7\right )} \log \left (\log \relax (x)\right ) - 1\right )}}{\log \left (\log \relax (x)\right )}\right )}}{x \log \relax (x) \log \left (\log \relax (x)\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(x*log(x)*log(log(x))^2+1)*exp(((2*x-14)*log(log(x))-2)/log(log(x)))/x/log(x)/log(log(x))^2,x, a
lgorithm="maxima")

[Out]

1/2*integrate((x*log(x)*log(log(x))^2 + 1)*e^(2*((x - 7)*log(log(x)) - 1)/log(log(x)))/(x*log(x)*log(log(x))^2
), x)

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mupad [B]  time = 1.83, size = 16, normalized size = 0.80 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-14}\,{\mathrm {e}}^{-\frac {2}{\ln \left (\ln \relax (x)\right )}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((log(log(x))*(2*x - 14) - 2)/log(log(x)))*(x*log(log(x))^2*log(x) + 1))/(2*x*log(log(x))^2*log(x)),x)

[Out]

(exp(2*x)*exp(-14)*exp(-2/log(log(x))))/4

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sympy [A]  time = 0.40, size = 19, normalized size = 0.95 \begin {gather*} \frac {e^{\frac {\left (2 x - 14\right ) \log {\left (\log {\relax (x )} \right )} - 2}{\log {\left (\log {\relax (x )} \right )}}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(x*ln(x)*ln(ln(x))**2+1)*exp(((2*x-14)*ln(ln(x))-2)/ln(ln(x)))/x/ln(x)/ln(ln(x))**2,x)

[Out]

exp(((2*x - 14)*log(log(x)) - 2)/log(log(x)))/4

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