3.16.10 \(\int \frac {e^{-8-4 x} (-64-32 x-4 x^2+(-224 x-120 x^2-16 x^3) \log (x)+(-96-48 x-6 x^2+(-224 x-120 x^2-16 x^3) \log (x)) \log (\log (x))+(-32-16 x-2 x^2+(-56 x-30 x^2-4 x^3) \log (x)) \log ^2(\log (x)))}{x \log ^3(x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {e^{-8-4 x} (4+x)^2 (2+\log (\log (x)))^2}{\log ^2(x)} \]

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Rubi [F]  time = 8.77, 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^{-8-4 x} \left (-64-32 x-4 x^2+\left (-224 x-120 x^2-16 x^3\right ) \log (x)+\left (-96-48 x-6 x^2+\left (-224 x-120 x^2-16 x^3\right ) \log (x)\right ) \log (\log (x))+\left (-32-16 x-2 x^2+\left (-56 x-30 x^2-4 x^3\right ) \log (x)\right ) \log ^2(\log (x))\right )}{x \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-8 - 4*x)*(-64 - 32*x - 4*x^2 + (-224*x - 120*x^2 - 16*x^3)*Log[x] + (-96 - 48*x - 6*x^2 + (-224*x - 1
20*x^2 - 16*x^3)*Log[x])*Log[Log[x]] + (-32 - 16*x - 2*x^2 + (-56*x - 30*x^2 - 4*x^3)*Log[x])*Log[Log[x]]^2))/
(x*Log[x]^3),x]

[Out]

-32*Defer[Int][E^(-8 - 4*x)/Log[x]^3, x] - 64*Defer[Int][E^(-8 - 4*x)/(x*Log[x]^3), x] - 4*Defer[Int][(E^(-8 -
4*x)*x)/Log[x]^3, x] - 224*Defer[Int][E^(-8 - 4*x)/Log[x]^2, x] - 120*Defer[Int][(E^(-8 - 4*x)*x)/Log[x]^2, x
] - 16*Defer[Int][(E^(-8 - 4*x)*x^2)/Log[x]^2, x] - 48*Defer[Int][(E^(-8 - 4*x)*Log[Log[x]])/Log[x]^3, x] - 96
*Defer[Int][(E^(-8 - 4*x)*Log[Log[x]])/(x*Log[x]^3), x] - 6*Defer[Int][(E^(-8 - 4*x)*x*Log[Log[x]])/Log[x]^3,
x] - 224*Defer[Int][(E^(-8 - 4*x)*Log[Log[x]])/Log[x]^2, x] - 120*Defer[Int][(E^(-8 - 4*x)*x*Log[Log[x]])/Log[
x]^2, x] - 16*Defer[Int][(E^(-8 - 4*x)*x^2*Log[Log[x]])/Log[x]^2, x] - 16*Defer[Int][(E^(-8 - 4*x)*Log[Log[x]]
^2)/Log[x]^3, x] - 32*Defer[Int][(E^(-8 - 4*x)*Log[Log[x]]^2)/(x*Log[x]^3), x] - 2*Defer[Int][(E^(-8 - 4*x)*x*
Log[Log[x]]^2)/Log[x]^3, x] - 56*Defer[Int][(E^(-8 - 4*x)*Log[Log[x]]^2)/Log[x]^2, x] - 30*Defer[Int][(E^(-8 -
4*x)*x*Log[Log[x]]^2)/Log[x]^2, x] - 4*Defer[Int][(E^(-8 - 4*x)*x^2*Log[Log[x]]^2)/Log[x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{-8-4 x} (4+x) (2+\log (\log (x))) (-((4+x) (1+\log (\log (x))))-x (7+2 x) \log (x) (2+\log (\log (x))))}{x \log ^3(x)} \, dx\\ &=2 \int \frac {e^{-8-4 x} (4+x) (2+\log (\log (x))) (-((4+x) (1+\log (\log (x))))-x (7+2 x) \log (x) (2+\log (\log (x))))}{x \log ^3(x)} \, dx\\ &=2 \int \left (-\frac {2 e^{-8-4 x} (4+x) \left (4+x+14 x \log (x)+4 x^2 \log (x)\right )}{x \log ^3(x)}-\frac {e^{-8-4 x} (4+x) \left (12+3 x+28 x \log (x)+8 x^2 \log (x)\right ) \log (\log (x))}{x \log ^3(x)}-\frac {e^{-8-4 x} (4+x) \left (4+x+7 x \log (x)+2 x^2 \log (x)\right ) \log ^2(\log (x))}{x \log ^3(x)}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-8-4 x} (4+x) \left (12+3 x+28 x \log (x)+8 x^2 \log (x)\right ) \log (\log (x))}{x \log ^3(x)} \, dx\right )-2 \int \frac {e^{-8-4 x} (4+x) \left (4+x+7 x \log (x)+2 x^2 \log (x)\right ) \log ^2(\log (x))}{x \log ^3(x)} \, dx-4 \int \frac {e^{-8-4 x} (4+x) \left (4+x+14 x \log (x)+4 x^2 \log (x)\right )}{x \log ^3(x)} \, dx\\ &=-\left (2 \int \left (\frac {e^{-8-4 x} \left (12+3 x+28 x \log (x)+8 x^2 \log (x)\right ) \log (\log (x))}{\log ^3(x)}+\frac {4 e^{-8-4 x} \left (12+3 x+28 x \log (x)+8 x^2 \log (x)\right ) \log (\log (x))}{x \log ^3(x)}\right ) \, dx\right )-2 \int \left (\frac {e^{-8-4 x} \left (4+x+7 x \log (x)+2 x^2 \log (x)\right ) \log ^2(\log (x))}{\log ^3(x)}+\frac {4 e^{-8-4 x} \left (4+x+7 x \log (x)+2 x^2 \log (x)\right ) \log ^2(\log (x))}{x \log ^3(x)}\right ) \, dx-4 \int \left (\frac {e^{-8-4 x} (4+x)^2}{x \log ^3(x)}+\frac {2 e^{-8-4 x} \left (28+15 x+2 x^2\right )}{\log ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-8-4 x} \left (12+3 x+28 x \log (x)+8 x^2 \log (x)\right ) \log (\log (x))}{\log ^3(x)} \, dx\right )-2 \int \frac {e^{-8-4 x} \left (4+x+7 x \log (x)+2 x^2 \log (x)\right ) \log ^2(\log (x))}{\log ^3(x)} \, dx-4 \int \frac {e^{-8-4 x} (4+x)^2}{x \log ^3(x)} \, dx-8 \int \frac {e^{-8-4 x} \left (28+15 x+2 x^2\right )}{\log ^2(x)} \, dx-8 \int \frac {e^{-8-4 x} \left (12+3 x+28 x \log (x)+8 x^2 \log (x)\right ) \log (\log (x))}{x \log ^3(x)} \, dx-8 \int \frac {e^{-8-4 x} \left (4+x+7 x \log (x)+2 x^2 \log (x)\right ) \log ^2(\log (x))}{x \log ^3(x)} \, dx\\ &=-\left (2 \int \left (\frac {12 e^{-8-4 x} \log (\log (x))}{\log ^3(x)}+\frac {3 e^{-8-4 x} x \log (\log (x))}{\log ^3(x)}+\frac {28 e^{-8-4 x} x \log (\log (x))}{\log ^2(x)}+\frac {8 e^{-8-4 x} x^2 \log (\log (x))}{\log ^2(x)}\right ) \, dx\right )-2 \int \left (\frac {4 e^{-8-4 x} \log ^2(\log (x))}{\log ^3(x)}+\frac {e^{-8-4 x} x \log ^2(\log (x))}{\log ^3(x)}+\frac {7 e^{-8-4 x} x \log ^2(\log (x))}{\log ^2(x)}+\frac {2 e^{-8-4 x} x^2 \log ^2(\log (x))}{\log ^2(x)}\right ) \, dx-4 \int \left (\frac {8 e^{-8-4 x}}{\log ^3(x)}+\frac {16 e^{-8-4 x}}{x \log ^3(x)}+\frac {e^{-8-4 x} x}{\log ^3(x)}\right ) \, dx-8 \int \left (\frac {28 e^{-8-4 x}}{\log ^2(x)}+\frac {15 e^{-8-4 x} x}{\log ^2(x)}+\frac {2 e^{-8-4 x} x^2}{\log ^2(x)}\right ) \, dx-8 \int \left (\frac {3 e^{-8-4 x} \log (\log (x))}{\log ^3(x)}+\frac {12 e^{-8-4 x} \log (\log (x))}{x \log ^3(x)}+\frac {28 e^{-8-4 x} \log (\log (x))}{\log ^2(x)}+\frac {8 e^{-8-4 x} x \log (\log (x))}{\log ^2(x)}\right ) \, dx-8 \int \left (\frac {e^{-8-4 x} \log ^2(\log (x))}{\log ^3(x)}+\frac {4 e^{-8-4 x} \log ^2(\log (x))}{x \log ^3(x)}+\frac {7 e^{-8-4 x} \log ^2(\log (x))}{\log ^2(x)}+\frac {2 e^{-8-4 x} x \log ^2(\log (x))}{\log ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-8-4 x} x \log ^2(\log (x))}{\log ^3(x)} \, dx\right )-4 \int \frac {e^{-8-4 x} x}{\log ^3(x)} \, dx-4 \int \frac {e^{-8-4 x} x^2 \log ^2(\log (x))}{\log ^2(x)} \, dx-6 \int \frac {e^{-8-4 x} x \log (\log (x))}{\log ^3(x)} \, dx-2 \left (8 \int \frac {e^{-8-4 x} \log ^2(\log (x))}{\log ^3(x)} \, dx\right )-14 \int \frac {e^{-8-4 x} x \log ^2(\log (x))}{\log ^2(x)} \, dx-16 \int \frac {e^{-8-4 x} x^2}{\log ^2(x)} \, dx-16 \int \frac {e^{-8-4 x} x^2 \log (\log (x))}{\log ^2(x)} \, dx-16 \int \frac {e^{-8-4 x} x \log ^2(\log (x))}{\log ^2(x)} \, dx-2 \left (24 \int \frac {e^{-8-4 x} \log (\log (x))}{\log ^3(x)} \, dx\right )-32 \int \frac {e^{-8-4 x}}{\log ^3(x)} \, dx-32 \int \frac {e^{-8-4 x} \log ^2(\log (x))}{x \log ^3(x)} \, dx-56 \int \frac {e^{-8-4 x} x \log (\log (x))}{\log ^2(x)} \, dx-56 \int \frac {e^{-8-4 x} \log ^2(\log (x))}{\log ^2(x)} \, dx-64 \int \frac {e^{-8-4 x}}{x \log ^3(x)} \, dx-64 \int \frac {e^{-8-4 x} x \log (\log (x))}{\log ^2(x)} \, dx-96 \int \frac {e^{-8-4 x} \log (\log (x))}{x \log ^3(x)} \, dx-120 \int \frac {e^{-8-4 x} x}{\log ^2(x)} \, dx-224 \int \frac {e^{-8-4 x}}{\log ^2(x)} \, dx-224 \int \frac {e^{-8-4 x} \log (\log (x))}{\log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.75, size = 24, normalized size = 1.00 \begin {gather*} \frac {e^{-4 (2+x)} (4+x)^2 (2+\log (\log (x)))^2}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-8 - 4*x)*(-64 - 32*x - 4*x^2 + (-224*x - 120*x^2 - 16*x^3)*Log[x] + (-96 - 48*x - 6*x^2 + (-224
*x - 120*x^2 - 16*x^3)*Log[x])*Log[Log[x]] + (-32 - 16*x - 2*x^2 + (-56*x - 30*x^2 - 4*x^3)*Log[x])*Log[Log[x]
]^2))/(x*Log[x]^3),x]

[Out]

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

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fricas [B]  time = 0.59, size = 61, normalized size = 2.54 \begin {gather*} \frac {{\left (x^{2} + 8 \, x + 16\right )} e^{\left (-4 \, x - 8\right )} \log \left (\log \relax (x)\right )^{2} + 4 \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (-4 \, x - 8\right )} \log \left (\log \relax (x)\right ) + 4 \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (-4 \, x - 8\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-30*x^2-56*x)*log(x)-2*x^2-16*x-32)*log(log(x))^2+((-16*x^3-120*x^2-224*x)*log(x)-6*x^2-48*
x-96)*log(log(x))+(-16*x^3-120*x^2-224*x)*log(x)-4*x^2-32*x-64)/x/exp(2*x+4)^2/log(x)^3,x, algorithm="fricas")

[Out]

((x^2 + 8*x + 16)*e^(-4*x - 8)*log(log(x))^2 + 4*(x^2 + 8*x + 16)*e^(-4*x - 8)*log(log(x)) + 4*(x^2 + 8*x + 16
)*e^(-4*x - 8))/log(x)^2

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giac [B]  time = 0.35, size = 97, normalized size = 4.04 \begin {gather*} \frac {{\left (x^{2} e^{\left (-4 \, x\right )} \log \left (\log \relax (x)\right )^{2} + 4 \, x^{2} e^{\left (-4 \, x\right )} \log \left (\log \relax (x)\right ) + 8 \, x e^{\left (-4 \, x\right )} \log \left (\log \relax (x)\right )^{2} + 4 \, x^{2} e^{\left (-4 \, x\right )} + 32 \, x e^{\left (-4 \, x\right )} \log \left (\log \relax (x)\right ) + 16 \, e^{\left (-4 \, x\right )} \log \left (\log \relax (x)\right )^{2} + 32 \, x e^{\left (-4 \, x\right )} + 64 \, e^{\left (-4 \, x\right )} \log \left (\log \relax (x)\right ) + 64 \, e^{\left (-4 \, x\right )}\right )} e^{\left (-8\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-30*x^2-56*x)*log(x)-2*x^2-16*x-32)*log(log(x))^2+((-16*x^3-120*x^2-224*x)*log(x)-6*x^2-48*
x-96)*log(log(x))+(-16*x^3-120*x^2-224*x)*log(x)-4*x^2-32*x-64)/x/exp(2*x+4)^2/log(x)^3,x, algorithm="giac")

[Out]

(x^2*e^(-4*x)*log(log(x))^2 + 4*x^2*e^(-4*x)*log(log(x)) + 8*x*e^(-4*x)*log(log(x))^2 + 4*x^2*e^(-4*x) + 32*x*
e^(-4*x)*log(log(x)) + 16*e^(-4*x)*log(log(x))^2 + 32*x*e^(-4*x) + 64*e^(-4*x)*log(log(x)) + 64*e^(-4*x))*e^(-
8)/log(x)^2

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maple [B]  time = 0.03, size = 69, normalized size = 2.88

method	result	size
risch	\(\frac {\left (x^{2}+8 x +16\right ) {\mathrm e}^{-4 x -8} \ln \left (\ln \relax (x )\right )^{2}}{\ln \relax (x )^{2}}+\frac {4 \left (x^{2}+8 x +16\right ) {\mathrm e}^{-4 x -8} \ln \left (\ln \relax (x )\right )}{\ln \relax (x )^{2}}+\frac {4 \left (x^{2}+8 x +16\right ) {\mathrm e}^{-4 x -8}}{\ln \relax (x )^{2}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^3-30*x^2-56*x)*ln(x)-2*x^2-16*x-32)*ln(ln(x))^2+((-16*x^3-120*x^2-224*x)*ln(x)-6*x^2-48*x-96)*ln(l
n(x))+(-16*x^3-120*x^2-224*x)*ln(x)-4*x^2-32*x-64)/x/exp(2*x+4)^2/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

(x^2+8*x+16)*exp(-4*x-8)/ln(x)^2*ln(ln(x))^2+4*(x^2+8*x+16)*exp(-4*x-8)/ln(x)^2*ln(ln(x))+4*(x^2+8*x+16)*exp(-
4*x-8)/ln(x)^2

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maxima [B]  time = 0.57, size = 57, normalized size = 2.38 \begin {gather*} \frac {{\left ({\left (x^{2} + 8 \, x + 16\right )} e^{\left (-4 \, x\right )} \log \left (\log \relax (x)\right )^{2} + 4 \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (-4 \, x\right )} \log \left (\log \relax (x)\right ) + 4 \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (-4 \, x\right )}\right )} e^{\left (-8\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-30*x^2-56*x)*log(x)-2*x^2-16*x-32)*log(log(x))^2+((-16*x^3-120*x^2-224*x)*log(x)-6*x^2-48*
x-96)*log(log(x))+(-16*x^3-120*x^2-224*x)*log(x)-4*x^2-32*x-64)/x/exp(2*x+4)^2/log(x)^3,x, algorithm="maxima")

[Out]

((x^2 + 8*x + 16)*e^(-4*x)*log(log(x))^2 + 4*(x^2 + 8*x + 16)*e^(-4*x)*log(log(x)) + 4*(x^2 + 8*x + 16)*e^(-4*
x))*e^(-8)/log(x)^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{-4\,x-8}\,\left (32\,x+\ln \left (\ln \relax (x)\right )\,\left (48\,x+6\,x^2+\ln \relax (x)\,\left (16\,x^3+120\,x^2+224\,x\right )+96\right )+{\ln \left (\ln \relax (x)\right )}^2\,\left (16\,x+2\,x^2+\ln \relax (x)\,\left (4\,x^3+30\,x^2+56\,x\right )+32\right )+4\,x^2+\ln \relax (x)\,\left (16\,x^3+120\,x^2+224\,x\right )+64\right )}{x\,{\ln \relax (x)}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- 4*x - 8)*(32*x + log(log(x))*(48*x + 6*x^2 + log(x)*(224*x + 120*x^2 + 16*x^3) + 96) + log(log(x))
^2*(16*x + 2*x^2 + log(x)*(56*x + 30*x^2 + 4*x^3) + 32) + 4*x^2 + log(x)*(224*x + 120*x^2 + 16*x^3) + 64))/(x*
log(x)^3),x)

[Out]

int(-(exp(- 4*x - 8)*(32*x + log(log(x))*(48*x + 6*x^2 + log(x)*(224*x + 120*x^2 + 16*x^3) + 96) + log(log(x))
^2*(16*x + 2*x^2 + log(x)*(56*x + 30*x^2 + 4*x^3) + 32) + 4*x^2 + log(x)*(224*x + 120*x^2 + 16*x^3) + 64))/(x*
log(x)^3), x)

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sympy [B]  time = 0.54, size = 76, normalized size = 3.17 \begin {gather*} \frac {\left (x^{2} \log {\left (\log {\relax (x )} \right )}^{2} + 4 x^{2} \log {\left (\log {\relax (x )} \right )} + 4 x^{2} + 8 x \log {\left (\log {\relax (x )} \right )}^{2} + 32 x \log {\left (\log {\relax (x )} \right )} + 32 x + 16 \log {\left (\log {\relax (x )} \right )}^{2} + 64 \log {\left (\log {\relax (x )} \right )} + 64\right ) e^{- 4 x - 8}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**3-30*x**2-56*x)*ln(x)-2*x**2-16*x-32)*ln(ln(x))**2+((-16*x**3-120*x**2-224*x)*ln(x)-6*x**2-
48*x-96)*ln(ln(x))+(-16*x**3-120*x**2-224*x)*ln(x)-4*x**2-32*x-64)/x/exp(2*x+4)**2/ln(x)**3,x)

[Out]

(x**2*log(log(x))**2 + 4*x**2*log(log(x)) + 4*x**2 + 8*x*log(log(x))**2 + 32*x*log(log(x)) + 32*x + 16*log(log
(x))**2 + 64*log(log(x)) + 64)*exp(-4*x - 8)/log(x)**2

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