3.21.55 \(\int \frac {\log ^{\frac {2 x}{e^4 (-8 x-2 x^2)+(5+2 x) \log (\log (6))}}(6) (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6)))}{e^8 (64 x^2+32 x^3+4 x^4)+e^4 (-80 x-52 x^2-8 x^3) \log (\log (6))+(25+20 x+4 x^2) \log ^2(\log (6))} \, dx\)

Optimal. Leaf size=25 \[ e^{\frac {x}{\frac {5}{2}+x-\frac {e^4 x (4+x)}{\log (\log (6))}}} \]

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Rubi [A]  time = 0.78, antiderivative size = 31, normalized size of antiderivative = 1.24, number of steps used = 1, number of rules used = 1, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6706} \begin {gather*} \log ^{-\frac {2 x}{2 e^4 \left (x^2+4 x\right )-(2 x+5) \log (\log (6))}}(6) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Log[6]^((2*x)/(E^4*(-8*x - 2*x^2) + (5 + 2*x)*Log[Log[6]]))*(4*E^4*x^2*Log[Log[6]] + 10*Log[Log[6]]^2))/(
E^8*(64*x^2 + 32*x^3 + 4*x^4) + E^4*(-80*x - 52*x^2 - 8*x^3)*Log[Log[6]] + (25 + 20*x + 4*x^2)*Log[Log[6]]^2),
x]

[Out]

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

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} &=\log ^{-\frac {2 x}{2 e^4 \left (4 x+x^2\right )-(5+2 x) \log (\log (6))}}(6)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.16, size = 28, normalized size = 1.12 \begin {gather*} \log ^{-\frac {2 x}{2 e^4 x (4+x)-(5+2 x) \log (\log (6))}}(6) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[6]^((2*x)/(E^4*(-8*x - 2*x^2) + (5 + 2*x)*Log[Log[6]]))*(4*E^4*x^2*Log[Log[6]] + 10*Log[Log[6]]
^2))/(E^8*(64*x^2 + 32*x^3 + 4*x^4) + E^4*(-80*x - 52*x^2 - 8*x^3)*Log[Log[6]] + (25 + 20*x + 4*x^2)*Log[Log[6
]]^2),x]

[Out]

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

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fricas [A]  time = 0.63, size = 30, normalized size = 1.20 \begin {gather*} \log \relax (6)^{-\frac {2 \, x}{2 \, {\left (x^{2} + 4 \, x\right )} e^{4} - {\left (2 \, x + 5\right )} \log \left (\log \relax (6)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(log(6))^2+4*x^2*exp(4)*log(log(6)))*exp(2*x*log(log(6))/((5+2*x)*log(log(6))+(-2*x^2-8*x)*ex
p(4)))/((4*x^2+20*x+25)*log(log(6))^2+(-8*x^3-52*x^2-80*x)*exp(4)*log(log(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),
x, algorithm="fricas")

[Out]

log(6)^(-2*x/(2*(x^2 + 4*x)*e^4 - (2*x + 5)*log(log(6))))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (2 \, x^{2} e^{4} \log \left (\log \relax (6)\right ) + 5 \, \log \left (\log \relax (6)\right )^{2}\right )} \log \relax (6)^{-\frac {2 \, x}{2 \, {\left (x^{2} + 4 \, x\right )} e^{4} - {\left (2 \, x + 5\right )} \log \left (\log \relax (6)\right )}}}{4 \, {\left (2 \, x^{3} + 13 \, x^{2} + 20 \, x\right )} e^{4} \log \left (\log \relax (6)\right ) - {\left (4 \, x^{2} + 20 \, x + 25\right )} \log \left (\log \relax (6)\right )^{2} - 4 \, {\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )} e^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(log(6))^2+4*x^2*exp(4)*log(log(6)))*exp(2*x*log(log(6))/((5+2*x)*log(log(6))+(-2*x^2-8*x)*ex
p(4)))/((4*x^2+20*x+25)*log(log(6))^2+(-8*x^3-52*x^2-80*x)*exp(4)*log(log(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),
x, algorithm="giac")

[Out]

integrate(-2*(2*x^2*e^4*log(log(6)) + 5*log(log(6))^2)*log(6)^(-2*x/(2*(x^2 + 4*x)*e^4 - (2*x + 5)*log(log(6))
))/(4*(2*x^3 + 13*x^2 + 20*x)*e^4*log(log(6)) - (4*x^2 + 20*x + 25)*log(log(6))^2 - 4*(x^4 + 8*x^3 + 16*x^2)*e
^8), x)

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maple [A]  time = 0.29, size = 34, normalized size = 1.36

method	result	size
gosper	\({\mathrm e}^{-\frac {2 \ln \left (\ln \relax (6)\right ) x}{2 x^{2} {\mathrm e}^{4}+8 x \,{\mathrm e}^{4}-2 \ln \left (\ln \relax (6)\right ) x -5 \ln \left (\ln \relax (6)\right )}}\)	\(34\)
risch	\(\left (\ln \relax (2)+\ln \relax (3)\right )^{\frac {2 x}{-2 x^{2} {\mathrm e}^{4}+2 \ln \left (\ln \relax (2)+\ln \relax (3)\right ) x -8 x \,{\mathrm e}^{4}+5 \ln \left (\ln \relax (2)+\ln \relax (3)\right )}}\)	\(42\)
norman	\(\frac {\left (8 \,{\mathrm e}^{4}-2 \ln \left (\ln \relax (6)\right )\right ) x \,{\mathrm e}^{\frac {2 x \ln \left (\ln \relax (6)\right )}{\left (5+2 x \right ) \ln \left (\ln \relax (6)\right )+\left (-2 x^{2}-8 x \right ) {\mathrm e}^{4}}}-5 \ln \left (\ln \relax (6)\right ) {\mathrm e}^{\frac {2 x \ln \left (\ln \relax (6)\right )}{\left (5+2 x \right ) \ln \left (\ln \relax (6)\right )+\left (-2 x^{2}-8 x \right ) {\mathrm e}^{4}}}+2 x^{2} {\mathrm e}^{4} {\mathrm e}^{\frac {2 x \ln \left (\ln \relax (6)\right )}{\left (5+2 x \right ) \ln \left (\ln \relax (6)\right )+\left (-2 x^{2}-8 x \right ) {\mathrm e}^{4}}}}{2 x^{2} {\mathrm e}^{4}+8 x \,{\mathrm e}^{4}-2 \ln \left (\ln \relax (6)\right ) x -5 \ln \left (\ln \relax (6)\right )}\)	\(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*ln(ln(6))^2+4*x^2*exp(4)*ln(ln(6)))*exp(2*x*ln(ln(6))/((5+2*x)*ln(ln(6))+(-2*x^2-8*x)*exp(4)))/((4*x^2
+20*x+25)*ln(ln(6))^2+(-8*x^3-52*x^2-80*x)*exp(4)*ln(ln(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),x,method=_RETURNVE
RBOSE)

[Out]

exp(-2*ln(ln(6))*x/(2*x^2*exp(4)+8*x*exp(4)-2*ln(ln(6))*x-5*ln(ln(6))))

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maxima [A]  time = 0.72, size = 45, normalized size = 1.80 \begin {gather*} \frac {1}{{\left (\log \relax (3) + \log \relax (2)\right )}^{\frac {2 \, x}{2 \, x^{2} e^{4} + 2 \, x {\left (4 \, e^{4} - \log \left (\log \relax (3) + \log \relax (2)\right )\right )} - 5 \, \log \left (\log \relax (3) + \log \relax (2)\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(log(6))^2+4*x^2*exp(4)*log(log(6)))*exp(2*x*log(log(6))/((5+2*x)*log(log(6))+(-2*x^2-8*x)*ex
p(4)))/((4*x^2+20*x+25)*log(log(6))^2+(-8*x^3-52*x^2-80*x)*exp(4)*log(log(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),
x, algorithm="maxima")

[Out]

1/((log(3) + log(2))^(2*x/(2*x^2*e^4 + 2*x*(4*e^4 - log(log(3) + log(2))) - 5*log(log(3) + log(2)))))

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mupad [B]  time = 3.73, size = 33, normalized size = 1.32 \begin {gather*} {\mathrm {e}}^{\frac {2\,x\,\ln \left (\ln \relax (6)\right )}{5\,\ln \left (\ln \relax (6)\right )-8\,x\,{\mathrm {e}}^4-2\,x^2\,{\mathrm {e}}^4+2\,x\,\ln \left (\ln \relax (6)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(2*x*log(log(6)))/(exp(4)*(8*x + 2*x^2) - log(log(6))*(2*x + 5)))*(10*log(log(6))^2 + 4*x^2*exp(4)*l
og(log(6))))/(log(log(6))^2*(20*x + 4*x^2 + 25) + exp(8)*(64*x^2 + 32*x^3 + 4*x^4) - exp(4)*log(log(6))*(80*x
+ 52*x^2 + 8*x^3)),x)

[Out]

exp((2*x*log(log(6)))/(5*log(log(6)) - 8*x*exp(4) - 2*x^2*exp(4) + 2*x*log(log(6))))

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sympy [A]  time = 0.97, size = 32, normalized size = 1.28 \begin {gather*} e^{\frac {2 x \log {\left (\log {\relax (6 )} \right )}}{\left (2 x + 5\right ) \log {\left (\log {\relax (6 )} \right )} + \left (- 2 x^{2} - 8 x\right ) e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*ln(ln(6))**2+4*x**2*exp(4)*ln(ln(6)))*exp(2*x*ln(ln(6))/((5+2*x)*ln(ln(6))+(-2*x**2-8*x)*exp(4))
)/((4*x**2+20*x+25)*ln(ln(6))**2+(-8*x**3-52*x**2-80*x)*exp(4)*ln(ln(6))+(4*x**4+32*x**3+64*x**2)*exp(4)**2),x
)

[Out]

exp(2*x*log(log(6))/((2*x + 5)*log(log(6)) + (-2*x**2 - 8*x)*exp(4)))

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