3.69.7 \(\int \frac {(48600 x+20250 e^{2+2 x} x) \log (\log (x))+(48600 x+e^{2+2 x} (20250 x-81000 x^2)) \log (x) \log ^2(\log (x))}{(248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}) \log (x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {25 x^2 \log ^2(\log (x))}{\left (4+\frac {5}{3} e^{2+2 x}\right )^4} \]

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Rubi [F]  time = 2.85, 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 {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((48600*x + 20250*E^(2 + 2*x)*x)*Log[Log[x]] + (48600*x + E^(2 + 2*x)*(20250*x - 81000*x^2))*Log[x]*Log[Lo
g[x]]^2)/((248832 + 518400*E^(2 + 2*x) + 432000*E^(4 + 4*x) + 180000*E^(6 + 6*x) + 37500*E^(8 + 8*x) + 3125*E^
(10 + 10*x))*Log[x]),x]

[Out]

4050*Defer[Int][(x*Log[Log[x]])/((12 + 5*E^(2 + 2*x))^4*Log[x]), x] + 4050*Defer[Int][(x*Log[Log[x]]^2)/(12 +
5*E^(2 + 2*x))^4, x] + 194400*Defer[Int][(x^2*Log[Log[x]]^2)/(12 + 5*E^(2 + 2*x))^5, x] - 16200*Defer[Int][(x^
2*Log[Log[x]]^2)/(12 + 5*E^(2 + 2*x))^4, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4050 x \log (\log (x)) \left (12+5 e^{2+2 x}-\left (-12+5 e^{2+2 x} (-1+4 x)\right ) \log (x) \log (\log (x))\right )}{\left (12+5 e^{2+2 x}\right )^5 \log (x)} \, dx\\ &=4050 \int \frac {x \log (\log (x)) \left (12+5 e^{2+2 x}-\left (-12+5 e^{2+2 x} (-1+4 x)\right ) \log (x) \log (\log (x))\right )}{\left (12+5 e^{2+2 x}\right )^5 \log (x)} \, dx\\ &=4050 \int \left (\frac {48 x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^5}-\frac {x \log (\log (x)) (-1-\log (x) \log (\log (x))+4 x \log (x) \log (\log (x)))}{\left (12+5 e^{2+2 x}\right )^4 \log (x)}\right ) \, dx\\ &=-\left (4050 \int \frac {x \log (\log (x)) (-1-\log (x) \log (\log (x))+4 x \log (x) \log (\log (x)))}{\left (12+5 e^{2+2 x}\right )^4 \log (x)} \, dx\right )+194400 \int \frac {x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^5} \, dx\\ &=-\left (4050 \int \left (-\frac {x \log (\log (x))}{\left (12+5 e^{2+2 x}\right )^4 \log (x)}-\frac {x \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4}+\frac {4 x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4}\right ) \, dx\right )+194400 \int \frac {x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^5} \, dx\\ &=4050 \int \frac {x \log (\log (x))}{\left (12+5 e^{2+2 x}\right )^4 \log (x)} \, dx+4050 \int \frac {x \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4} \, dx-16200 \int \frac {x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4} \, dx+194400 \int \frac {x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^5} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 23, normalized size = 0.92 \begin {gather*} \frac {2025 x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((48600*x + 20250*E^(2 + 2*x)*x)*Log[Log[x]] + (48600*x + E^(2 + 2*x)*(20250*x - 81000*x^2))*Log[x]*
Log[Log[x]]^2)/((248832 + 518400*E^(2 + 2*x) + 432000*E^(4 + 4*x) + 180000*E^(6 + 6*x) + 37500*E^(8 + 8*x) + 3
125*E^(10 + 10*x))*Log[x]),x]

[Out]

(2025*x^2*Log[Log[x]]^2)/(12 + 5*E^(2 + 2*x))^4

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fricas [B]  time = 0.68, size = 46, normalized size = 1.84 \begin {gather*} \frac {2025 \, x^{2} \log \left (\log \relax (x)\right )^{2}}{625 \, e^{\left (8 \, x + 8\right )} + 6000 \, e^{\left (6 \, x + 6\right )} + 21600 \, e^{\left (4 \, x + 4\right )} + 34560 \, e^{\left (2 \, x + 2\right )} + 20736} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-81000*x^2+20250*x)*exp(x+1)^2+48600*x)*log(x)*log(log(x))^2+(20250*x*exp(x+1)^2+48600*x)*log(log
(x)))/(3125*exp(x+1)^10+37500*exp(x+1)^8+180000*exp(x+1)^6+432000*exp(x+1)^4+518400*exp(x+1)^2+248832)/log(x),
x, algorithm="fricas")

[Out]

2025*x^2*log(log(x))^2/(625*e^(8*x + 8) + 6000*e^(6*x + 6) + 21600*e^(4*x + 4) + 34560*e^(2*x + 2) + 20736)

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giac [B]  time = 0.17, size = 46, normalized size = 1.84 \begin {gather*} \frac {2025 \, x^{2} \log \left (\log \relax (x)\right )^{2}}{625 \, e^{\left (8 \, x + 8\right )} + 6000 \, e^{\left (6 \, x + 6\right )} + 21600 \, e^{\left (4 \, x + 4\right )} + 34560 \, e^{\left (2 \, x + 2\right )} + 20736} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-81000*x^2+20250*x)*exp(x+1)^2+48600*x)*log(x)*log(log(x))^2+(20250*x*exp(x+1)^2+48600*x)*log(log
(x)))/(3125*exp(x+1)^10+37500*exp(x+1)^8+180000*exp(x+1)^6+432000*exp(x+1)^4+518400*exp(x+1)^2+248832)/log(x),
x, algorithm="giac")

[Out]

2025*x^2*log(log(x))^2/(625*e^(8*x + 8) + 6000*e^(6*x + 6) + 21600*e^(4*x + 4) + 34560*e^(2*x + 2) + 20736)

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maple [A]  time = 0.04, size = 23, normalized size = 0.92

method	result	size
risch	\(\frac {2025 x^{2} \ln \left (\ln \relax (x )\right )^{2}}{\left (5 \,{\mathrm e}^{2 x +2}+12\right )^{4}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-81000*x^2+20250*x)*exp(x+1)^2+48600*x)*ln(x)*ln(ln(x))^2+(20250*x*exp(x+1)^2+48600*x)*ln(ln(x)))/(3125
*exp(x+1)^10+37500*exp(x+1)^8+180000*exp(x+1)^6+432000*exp(x+1)^4+518400*exp(x+1)^2+248832)/ln(x),x,method=_RE
TURNVERBOSE)

[Out]

2025*x^2/(5*exp(2*x+2)+12)^4*ln(ln(x))^2

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maxima [B]  time = 0.47, size = 46, normalized size = 1.84 \begin {gather*} \frac {2025 \, x^{2} \log \left (\log \relax (x)\right )^{2}}{625 \, e^{\left (8 \, x + 8\right )} + 6000 \, e^{\left (6 \, x + 6\right )} + 21600 \, e^{\left (4 \, x + 4\right )} + 34560 \, e^{\left (2 \, x + 2\right )} + 20736} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-81000*x^2+20250*x)*exp(x+1)^2+48600*x)*log(x)*log(log(x))^2+(20250*x*exp(x+1)^2+48600*x)*log(log
(x)))/(3125*exp(x+1)^10+37500*exp(x+1)^8+180000*exp(x+1)^6+432000*exp(x+1)^4+518400*exp(x+1)^2+248832)/log(x),
x, algorithm="maxima")

[Out]

2025*x^2*log(log(x))^2/(625*e^(8*x + 8) + 6000*e^(6*x + 6) + 21600*e^(4*x + 4) + 34560*e^(2*x + 2) + 20736)

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mupad [B]  time = 4.36, size = 46, normalized size = 1.84 \begin {gather*} \frac {81\,x^2\,{\ln \left (\ln \relax (x)\right )}^2}{25\,\left (\frac {6912\,{\mathrm {e}}^{2\,x+2}}{125}+\frac {864\,{\mathrm {e}}^{4\,x+4}}{25}+\frac {48\,{\mathrm {e}}^{6\,x+6}}{5}+{\mathrm {e}}^{8\,x+8}+\frac {20736}{625}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x))*(48600*x + 20250*x*exp(2*x + 2)) + log(log(x))^2*log(x)*(48600*x + exp(2*x + 2)*(20250*x - 81
000*x^2)))/(log(x)*(518400*exp(2*x + 2) + 432000*exp(4*x + 4) + 180000*exp(6*x + 6) + 37500*exp(8*x + 8) + 312
5*exp(10*x + 10) + 248832)),x)

[Out]

(81*x^2*log(log(x))^2)/(25*((6912*exp(2*x + 2))/125 + (864*exp(4*x + 4))/25 + (48*exp(6*x + 6))/5 + exp(8*x +
8) + 20736/625))

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sympy [B]  time = 0.37, size = 49, normalized size = 1.96 \begin {gather*} \frac {81 x^{2} \log {\left (\log {\relax (x )} \right )}^{2}}{\frac {6912 e^{2 x + 2}}{5} + 864 e^{4 x + 4} + 240 e^{6 x + 6} + 25 e^{8 x + 8} + \frac {20736}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-81000*x**2+20250*x)*exp(x+1)**2+48600*x)*ln(x)*ln(ln(x))**2+(20250*x*exp(x+1)**2+48600*x)*ln(ln(
x)))/(3125*exp(x+1)**10+37500*exp(x+1)**8+180000*exp(x+1)**6+432000*exp(x+1)**4+518400*exp(x+1)**2+248832)/ln(
x),x)

[Out]

81*x**2*log(log(x))**2/(6912*exp(2*x + 2)/5 + 864*exp(4*x + 4) + 240*exp(6*x + 6) + 25*exp(8*x + 8) + 20736/25
)

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