3.70.23 \(\int \frac {8 x^2 \log (4) \log (x)+(-8 e^{e^5}-2 x^2 \log (4)) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx\)

Optimal. Leaf size=27 \[ 3+\frac {4+e^{-e^5} x^2 \log (4)}{4+\log ^2(\log (x))} \]

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Rubi [F]  time = 2.28, 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 {8 x^2 \log (4) \log (x)+\left (-8 e^{e^5}-2 x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \log ^2(\log (x))}{16 e^{e^5} x \log (x)+8 e^{e^5} x \log (x) \log ^2(\log (x))+e^{e^5} x \log (x) \log ^4(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*x^2*Log[4]*Log[x] + (-8*E^E^5 - 2*x^2*Log[4])*Log[Log[x]] + 2*x^2*Log[4]*Log[x]*Log[Log[x]]^2)/(16*E^E^
5*x*Log[x] + 8*E^E^5*x*Log[x]*Log[Log[x]]^2 + E^E^5*x*Log[x]*Log[Log[x]]^4),x]

[Out]

4/(4 + Log[Log[x]]^2) - (2*Log[4]*Defer[Int][(x*Log[Log[x]])/(Log[x]*(4 + Log[Log[x]]^2)^2), x])/E^E^5 + (Log[
16]*Defer[Int][x/(4 + Log[Log[x]]^2), x])/E^E^5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-e^5} \left (-2 \left (4 e^{e^5}+x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \left (4+\log ^2(\log (x))\right )\right )}{x \log (x) \left (4+\log ^2(\log (x))\right )^2} \, dx\\ &=e^{-e^5} \int \frac {-2 \left (4 e^{e^5}+x^2 \log (4)\right ) \log (\log (x))+2 x^2 \log (4) \log (x) \left (4+\log ^2(\log (x))\right )}{x \log (x) \left (4+\log ^2(\log (x))\right )^2} \, dx\\ &=e^{-e^5} \int \left (-\frac {2 \left (4 e^{e^5}+x^2 \log (4)\right ) \log (\log (x))}{x \log (x) \left (4+\log ^2(\log (x))\right )^2}+\frac {x \log (16)}{4+\log ^2(\log (x))}\right ) \, dx\\ &=-\left (\left (2 e^{-e^5}\right ) \int \frac {\left (4 e^{e^5}+x^2 \log (4)\right ) \log (\log (x))}{x \log (x) \left (4+\log ^2(\log (x))\right )^2} \, dx\right )+\left (e^{-e^5} \log (16)\right ) \int \frac {x}{4+\log ^2(\log (x))} \, dx\\ &=-\left (\left (2 e^{-e^5}\right ) \int \left (\frac {4 e^{e^5} \log (\log (x))}{x \log (x) \left (4+\log ^2(\log (x))\right )^2}+\frac {x \log (4) \log (\log (x))}{\log (x) \left (4+\log ^2(\log (x))\right )^2}\right ) \, dx\right )+\left (e^{-e^5} \log (16)\right ) \int \frac {x}{4+\log ^2(\log (x))} \, dx\\ &=-\left (8 \int \frac {\log (\log (x))}{x \log (x) \left (4+\log ^2(\log (x))\right )^2} \, dx\right )-\left (2 e^{-e^5} \log (4)\right ) \int \frac {x \log (\log (x))}{\log (x) \left (4+\log ^2(\log (x))\right )^2} \, dx+\left (e^{-e^5} \log (16)\right ) \int \frac {x}{4+\log ^2(\log (x))} \, dx\\ &=-\left (8 \operatorname {Subst}\left (\int \frac {\log (x)}{x \left (4+\log ^2(x)\right )^2} \, dx,x,\log (x)\right )\right )-\left (2 e^{-e^5} \log (4)\right ) \int \frac {x \log (\log (x))}{\log (x) \left (4+\log ^2(\log (x))\right )^2} \, dx+\left (e^{-e^5} \log (16)\right ) \int \frac {x}{4+\log ^2(\log (x))} \, dx\\ &=-\left (8 \operatorname {Subst}\left (\int \frac {x}{\left (4+x^2\right )^2} \, dx,x,\log (\log (x))\right )\right )-\left (2 e^{-e^5} \log (4)\right ) \int \frac {x \log (\log (x))}{\log (x) \left (4+\log ^2(\log (x))\right )^2} \, dx+\left (e^{-e^5} \log (16)\right ) \int \frac {x}{4+\log ^2(\log (x))} \, dx\\ &=\frac {4}{4+\log ^2(\log (x))}-\left (2 e^{-e^5} \log (4)\right ) \int \frac {x \log (\log (x))}{\log (x) \left (4+\log ^2(\log (x))\right )^2} \, dx+\left (e^{-e^5} \log (16)\right ) \int \frac {x}{4+\log ^2(\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 31, normalized size = 1.15 \begin {gather*} \frac {e^{-e^5} \left (4 e^{e^5}+x^2 \log (4)\right )}{4+\log ^2(\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x^2*Log[4]*Log[x] + (-8*E^E^5 - 2*x^2*Log[4])*Log[Log[x]] + 2*x^2*Log[4]*Log[x]*Log[Log[x]]^2)/(1
6*E^E^5*x*Log[x] + 8*E^E^5*x*Log[x]*Log[Log[x]]^2 + E^E^5*x*Log[x]*Log[Log[x]]^4),x]

[Out]

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

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fricas [A]  time = 0.86, size = 31, normalized size = 1.15 \begin {gather*} \frac {2 \, {\left (x^{2} \log \relax (2) + 2 \, e^{\left (e^{5}\right )}\right )}}{e^{\left (e^{5}\right )} \log \left (\log \relax (x)\right )^{2} + 4 \, e^{\left (e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(2)*log(x)*log(log(x))^2+(-8*exp(exp(5))-4*x^2*log(2))*log(log(x))+16*x^2*log(2)*log(x))/(
x*exp(exp(5))*log(x)*log(log(x))^4+8*x*exp(exp(5))*log(x)*log(log(x))^2+16*x*exp(exp(5))*log(x)),x, algorithm=
"fricas")

[Out]

2*(x^2*log(2) + 2*e^(e^5))/(e^(e^5)*log(log(x))^2 + 4*e^(e^5))

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giac [A]  time = 0.25, size = 24, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (x^{2} e^{\left (-e^{5}\right )} \log \relax (2) + 2\right )}}{\log \left (\log \relax (x)\right )^{2} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(2)*log(x)*log(log(x))^2+(-8*exp(exp(5))-4*x^2*log(2))*log(log(x))+16*x^2*log(2)*log(x))/(
x*exp(exp(5))*log(x)*log(log(x))^4+8*x*exp(exp(5))*log(x)*log(log(x))^2+16*x*exp(exp(5))*log(x)),x, algorithm=
"giac")

[Out]

2*(x^2*e^(-e^5)*log(2) + 2)/(log(log(x))^2 + 4)

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maple [A]  time = 0.08, size = 29, normalized size = 1.07

method	result	size
risch	\(\frac {2 \left (x^{2} \ln \relax (2)+2 \,{\mathrm e}^{{\mathrm e}^{5}}\right ) {\mathrm e}^{-{\mathrm e}^{5}}}{4+\ln \left (\ln \relax (x )\right )^{2}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2*ln(2)*ln(x)*ln(ln(x))^2+(-8*exp(exp(5))-4*x^2*ln(2))*ln(ln(x))+16*x^2*ln(2)*ln(x))/(x*exp(exp(5))*l
n(x)*ln(ln(x))^4+8*x*exp(exp(5))*ln(x)*ln(ln(x))^2+16*x*exp(exp(5))*ln(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 31, normalized size = 1.15 \begin {gather*} \frac {2 \, {\left (x^{2} \log \relax (2) + 2 \, e^{\left (e^{5}\right )}\right )}}{e^{\left (e^{5}\right )} \log \left (\log \relax (x)\right )^{2} + 4 \, e^{\left (e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(2)*log(x)*log(log(x))^2+(-8*exp(exp(5))-4*x^2*log(2))*log(log(x))+16*x^2*log(2)*log(x))/(
x*exp(exp(5))*log(x)*log(log(x))^4+8*x*exp(exp(5))*log(x)*log(log(x))^2+16*x*exp(exp(5))*log(x)),x, algorithm=
"maxima")

[Out]

2*(x^2*log(2) + 2*e^(e^5))/(e^(e^5)*log(log(x))^2 + 4*e^(e^5))

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mupad [B]  time = 4.30, size = 28, normalized size = 1.04 \begin {gather*} \frac {2\,{\mathrm {e}}^{-{\mathrm {e}}^5}\,\left (\ln \relax (2)\,x^2+2\,{\mathrm {e}}^{{\mathrm {e}}^5}\right )}{{\ln \left (\ln \relax (x)\right )}^2+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^2*log(2)*log(x) - log(log(x))*(8*exp(exp(5)) + 4*x^2*log(2)) + 4*x^2*log(log(x))^2*log(2)*log(x))/(1
6*x*exp(exp(5))*log(x) + 8*x*log(log(x))^2*exp(exp(5))*log(x) + x*log(log(x))^4*exp(exp(5))*log(x)),x)

[Out]

(2*exp(-exp(5))*(2*exp(exp(5)) + x^2*log(2)))/(log(log(x))^2 + 4)

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sympy [A]  time = 0.26, size = 32, normalized size = 1.19 \begin {gather*} \frac {2 x^{2} \log {\relax (2 )} + 4 e^{e^{5}}}{e^{e^{5}} \log {\left (\log {\relax (x )} \right )}^{2} + 4 e^{e^{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2*ln(2)*ln(x)*ln(ln(x))**2+(-8*exp(exp(5))-4*x**2*ln(2))*ln(ln(x))+16*x**2*ln(2)*ln(x))/(x*exp
(exp(5))*ln(x)*ln(ln(x))**4+8*x*exp(exp(5))*ln(x)*ln(ln(x))**2+16*x*exp(exp(5))*ln(x)),x)

[Out]

(2*x**2*log(2) + 4*exp(exp(5)))/(exp(exp(5))*log(log(x))**2 + 4*exp(exp(5)))

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