3.38.73 \(\int \frac {e^{\frac {1}{2} (-1-(-10 x^2+2 x^3) \log (\log (\log (x))))} (5 x-x^2+(10 x-3 x^2) \log (x) \log (\log (x)) \log (\log (\log (x))))}{20 \log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=22 \[ \frac {1}{20} e^{-\frac {1}{2}-(-5+x) x^2 \log (\log (\log (x)))} \]

________________________________________________________________________________________

Rubi [F]  time = 1.37, 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 {\exp \left (\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )\right ) \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-1 - (-10*x^2 + 2*x^3)*Log[Log[Log[x]]])/2)*(5*x - x^2 + (10*x - 3*x^2)*Log[x]*Log[Log[x]]*Log[Log[Lo
g[x]]]))/(20*Log[x]*Log[Log[x]]),x]

[Out]

Defer[Int][(x*Log[Log[x]]^(-1 + 5*x^2 - x^3))/Log[x], x]/(4*Sqrt[E]) - Defer[Int][(x^2*Log[Log[x]]^(-1 + 5*x^2
- x^3))/Log[x], x]/(20*Sqrt[E]) + Defer[Int][x*Log[Log[x]]^((5 - x)*x^2)*Log[Log[Log[x]]], x]/(2*Sqrt[E]) - (
3*Defer[Int][x^2*Log[Log[x]]^((5 - x)*x^2)*Log[Log[Log[x]]], x])/(20*Sqrt[E])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{20} \int \frac {\exp \left (\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )\right ) \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{\log (x) \log (\log (x))} \, dx\\ &=\frac {1}{20} \int \frac {x \log ^{-1+5 x^2-x^3}(\log (x)) (5-x-(-10+3 x) \log (x) \log (\log (x)) \log (\log (\log (x))))}{\sqrt {e} \log (x)} \, dx\\ &=\frac {\int \frac {x \log ^{-1+5 x^2-x^3}(\log (x)) (5-x-(-10+3 x) \log (x) \log (\log (x)) \log (\log (\log (x))))}{\log (x)} \, dx}{20 \sqrt {e}}\\ &=\frac {\int \left (-\frac {(-5+x) x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)}-x (-10+3 x) \log ^{5 x^2-x^3}(\log (x)) \log (\log (\log (x)))\right ) \, dx}{20 \sqrt {e}}\\ &=-\frac {\int \frac {(-5+x) x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{20 \sqrt {e}}-\frac {\int x (-10+3 x) \log ^{5 x^2-x^3}(\log (x)) \log (\log (\log (x))) \, dx}{20 \sqrt {e}}\\ &=-\frac {\int \left (-\frac {5 x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)}+\frac {x^2 \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)}\right ) \, dx}{20 \sqrt {e}}-\frac {\int x (-10+3 x) \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x))) \, dx}{20 \sqrt {e}}\\ &=-\frac {\int \frac {x^2 \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{20 \sqrt {e}}-\frac {\int \left (-10 x \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x)))+3 x^2 \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x)))\right ) \, dx}{20 \sqrt {e}}+\frac {\int \frac {x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{4 \sqrt {e}}\\ &=-\frac {\int \frac {x^2 \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{20 \sqrt {e}}-\frac {3 \int x^2 \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x))) \, dx}{20 \sqrt {e}}+\frac {\int \frac {x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{4 \sqrt {e}}+\frac {\int x \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x))) \, dx}{2 \sqrt {e}}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 21, normalized size = 0.95 \begin {gather*} \frac {\log ^{-\left ((-5+x) x^2\right )}(\log (x))}{20 \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-1 - (-10*x^2 + 2*x^3)*Log[Log[Log[x]]])/2)*(5*x - x^2 + (10*x - 3*x^2)*Log[x]*Log[Log[x]]*Log[
Log[Log[x]]]))/(20*Log[x]*Log[Log[x]]),x]

[Out]

1/(20*Sqrt[E]*Log[Log[x]]^((-5 + x)*x^2))

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 20, normalized size = 0.91 \begin {gather*} \frac {1}{20} \, e^{\left (-{\left (x^{3} - 5 \, x^{2}\right )} \log \left (\log \left (\log \relax (x)\right )\right ) - \frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-3*x^2+10*x)*log(x)*log(log(x))*log(log(log(x)))-x^2+5*x)/log(x)/log(log(x))/exp(1/2*(2*x^3-1
0*x^2)*log(log(log(x)))+1/2),x, algorithm="fricas")

[Out]

1/20*e^(-(x^3 - 5*x^2)*log(log(log(x))) - 1/2)

________________________________________________________________________________________

giac [A]  time = 2.65, size = 23, normalized size = 1.05 \begin {gather*} \frac {1}{20} \, e^{\left (-x^{3} \log \left (\log \left (\log \relax (x)\right )\right ) + 5 \, x^{2} \log \left (\log \left (\log \relax (x)\right )\right ) - \frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-3*x^2+10*x)*log(x)*log(log(x))*log(log(log(x)))-x^2+5*x)/log(x)/log(log(x))/exp(1/2*(2*x^3-1
0*x^2)*log(log(log(x)))+1/2),x, algorithm="giac")

[Out]

1/20*e^(-x^3*log(log(log(x))) + 5*x^2*log(log(log(x))) - 1/2)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 18, normalized size = 0.82

method	result	size
risch	\(\frac {\ln \left (\ln \relax (x )\right )^{-x^{2} \left (x -5\right )} {\mathrm e}^{-\frac {1}{2}}}{20}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/20*((-3*x^2+10*x)*ln(x)*ln(ln(x))*ln(ln(ln(x)))-x^2+5*x)/ln(x)/ln(ln(x))/exp(1/2*(2*x^3-10*x^2)*ln(ln(ln
(x)))+1/2),x,method=_RETURNVERBOSE)

[Out]

1/20/(ln(ln(x))^(x^2*(x-5)))*exp(-1/2)

________________________________________________________________________________________

maxima [A]  time = 0.67, size = 23, normalized size = 1.05 \begin {gather*} \frac {1}{20} \, e^{\left (-x^{3} \log \left (\log \left (\log \relax (x)\right )\right ) + 5 \, x^{2} \log \left (\log \left (\log \relax (x)\right )\right ) - \frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-3*x^2+10*x)*log(x)*log(log(x))*log(log(log(x)))-x^2+5*x)/log(x)/log(log(x))/exp(1/2*(2*x^3-1
0*x^2)*log(log(log(x)))+1/2),x, algorithm="maxima")

[Out]

1/20*e^(-x^3*log(log(log(x))) + 5*x^2*log(log(log(x))) - 1/2)

________________________________________________________________________________________

mupad [B]  time = 2.28, size = 19, normalized size = 0.86 \begin {gather*} \frac {{\ln \left (\ln \relax (x)\right )}^{5\,x^2-x^3}\,{\mathrm {e}}^{-\frac {1}{2}}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((log(log(log(x)))*(10*x^2 - 2*x^3))/2 - 1/2)*(x/4 - x^2/20 + (log(log(x))*log(log(log(x)))*log(x)*(10
*x - 3*x^2))/20))/(log(log(x))*log(x)),x)

[Out]

(log(log(x))^(5*x^2 - x^3)*exp(-1/2))/20

________________________________________________________________________________________

sympy [A]  time = 1.27, size = 22, normalized size = 1.00 \begin {gather*} \frac {e^{- \left (x^{3} - 5 x^{2}\right ) \log {\left (\log {\left (\log {\relax (x )} \right )} \right )} - \frac {1}{2}}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-3*x**2+10*x)*ln(x)*ln(ln(x))*ln(ln(ln(x)))-x**2+5*x)/ln(x)/ln(ln(x))/exp(1/2*(2*x**3-10*x**2
)*ln(ln(ln(x)))+1/2),x)

[Out]

exp(-(x**3 - 5*x**2)*log(log(log(x))) - 1/2)/20

________________________________________________________________________________________