3.54.40 \(\int \frac {e^{\frac {4-2 \log (\frac {x^2-\log (4)+\log (x)}{6+x})}{\log (\frac {x^2-\log (4)+\log (x)}{6+x})}} (-24-4 x-48 x^2-4 x^3-4 x \log (4)+4 x \log (x))}{(6 x^3+x^4+(-6 x-x^2) \log (4)+(6 x+x^2) \log (x)) \log ^2(\frac {x^2-\log (4)+\log (x)}{6+x})} \, dx\)

Optimal. Leaf size=27 \[ 2+e^{-2+\frac {4}{\log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}} \]

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Rubi [F]  time = 17.92, 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 {4-2 \log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}{\log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}\right ) \left (-24-4 x-48 x^2-4 x^3-4 x \log (4)+4 x \log (x)\right )}{\left (6 x^3+x^4+\left (-6 x-x^2\right ) \log (4)+\left (6 x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((4 - 2*Log[(x^2 - Log[4] + Log[x])/(6 + x)])/Log[(x^2 - Log[4] + Log[x])/(6 + x)])*(-24 - 4*x - 48*x^2
- 4*x^3 - 4*x*Log[4] + 4*x*Log[x]))/((6*x^3 + x^4 + (-6*x - x^2)*Log[4] + (6*x + x^2)*Log[x])*Log[(x^2 - Log[
4] + Log[x])/(6 + x)]^2),x]

[Out]

-4*Defer[Int][E^(-2 + 4/Log[(x^2 + Log[x/4])/(6 + x)])/(x*(x^2 + Log[x/4])*Log[(x^2 + Log[x/4])/(6 + x)]^2), x
] - 4*Defer[Int][(E^(-2 + 4/Log[(x^2 + Log[x/4])/(6 + x)])*x)/((x^2 + Log[x/4])*Log[(x^2 + Log[x/4])/(6 + x)]^
2), x] - 96*Defer[Subst][Defer[Int][E^(-2 + 4/Log[(16*x^2 + Log[x])/(6 + 4*x)])/((16*x^2 + Log[x])*Log[(16*x^2
+ Log[x])/(6 + 4*x)]^2), x], x, x/4] + 592*Defer[Subst][Defer[Int][E^(-2 + 4/Log[(16*x^2 + Log[x])/(6 + 4*x)]
)/((6 + 4*x)*(16*x^2 + Log[x])*Log[(16*x^2 + Log[x])/(6 + 4*x)]^2), x], x, x/4] - 16*(1 + Log[4])*Defer[Subst]
[Defer[Int][E^(-2 + 4/Log[(16*x^2 + Log[x])/(6 + 4*x)])/((6 + 4*x)*(16*x^2 + Log[x])*Log[(16*x^2 + Log[x])/(6
+ 4*x)]^2), x], x, x/4] + 16*Defer[Subst][Defer[Int][(E^(-2 + 4/Log[(16*x^2 + Log[x])/(6 + 4*x)])*Log[4*x])/((
6 + 4*x)*(16*x^2 + Log[x])*Log[(16*x^2 + Log[x])/(6 + 4*x)]^2), x], x, x/4]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {4-2 \log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}{\log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}\right ) \left (-24-48 x^2-4 x^3+x (-4-4 \log (4))+4 x \log (x)\right )}{\left (6 x^3+x^4+\left (-6 x-x^2\right ) \log (4)+\left (6 x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )} \, dx\\ &=\int \frac {4 e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (-6-12 x^2-x^3-x (1+\log (4))+x \log (x)\right )}{x (6+x) \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )} \, dx\\ &=4 \int \frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (-6-12 x^2-x^3-x (1+\log (4))+x \log (x)\right )}{x (6+x) \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )} \, dx\\ &=4 \int \left (\frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (6+12 x^2+x^3+x (1+\log (4))-x \log (x)\right )}{6 (6+x) \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}+\frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (-6-12 x^2-x^3-x (1+\log (4))+x \log (x)\right )}{6 x \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}\right ) \, dx\\ &=\frac {2}{3} \int \frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (6+12 x^2+x^3+x (1+\log (4))-x \log (x)\right )}{(6+x) \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )} \, dx+\frac {2}{3} \int \frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (-6-12 x^2-x^3-x (1+\log (4))+x \log (x)\right )}{x \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.28, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {4-2 \log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}{\log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}} \left (-24-4 x-48 x^2-4 x^3-4 x \log (4)+4 x \log (x)\right )}{\left (6 x^3+x^4+\left (-6 x-x^2\right ) \log (4)+\left (6 x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((4 - 2*Log[(x^2 - Log[4] + Log[x])/(6 + x)])/Log[(x^2 - Log[4] + Log[x])/(6 + x)])*(-24 - 4*x -
48*x^2 - 4*x^3 - 4*x*Log[4] + 4*x*Log[x]))/((6*x^3 + x^4 + (-6*x - x^2)*Log[4] + (6*x + x^2)*Log[x])*Log[(x^2
- Log[4] + Log[x])/(6 + x)]^2),x]

[Out]

Integrate[(E^((4 - 2*Log[(x^2 - Log[4] + Log[x])/(6 + x)])/Log[(x^2 - Log[4] + Log[x])/(6 + x)])*(-24 - 4*x -
48*x^2 - 4*x^3 - 4*x*Log[4] + 4*x*Log[x]))/((6*x^3 + x^4 + (-6*x - x^2)*Log[4] + (6*x + x^2)*Log[x])*Log[(x^2
- Log[4] + Log[x])/(6 + x)]^2), x]

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fricas [A]  time = 1.07, size = 41, normalized size = 1.52 \begin {gather*} e^{\left (-\frac {2 \, {\left (\log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right ) - 2\right )}}{\log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)-8*x*log(2)-4*x^3-48*x^2-4*x-24)*exp((-2*log((log(x)-2*log(2)+x^2)/(x+6))+4)/log((log(x)-
2*log(2)+x^2)/(x+6)))/((x^2+6*x)*log(x)+2*(-x^2-6*x)*log(2)+x^4+6*x^3)/log((log(x)-2*log(2)+x^2)/(x+6))^2,x, a
lgorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, {\left (x^{3} + 12 \, x^{2} + 2 \, x \log \relax (2) - x \log \relax (x) + x + 6\right )} e^{\left (-\frac {2 \, {\left (\log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right ) - 2\right )}}{\log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right )}\right )}}{{\left (x^{4} + 6 \, x^{3} - 2 \, {\left (x^{2} + 6 \, x\right )} \log \relax (2) + {\left (x^{2} + 6 \, x\right )} \log \relax (x)\right )} \log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)-8*x*log(2)-4*x^3-48*x^2-4*x-24)*exp((-2*log((log(x)-2*log(2)+x^2)/(x+6))+4)/log((log(x)-
2*log(2)+x^2)/(x+6)))/((x^2+6*x)*log(x)+2*(-x^2-6*x)*log(2)+x^4+6*x^3)/log((log(x)-2*log(2)+x^2)/(x+6))^2,x, a
lgorithm="giac")

[Out]

integrate(-4*(x^3 + 12*x^2 + 2*x*log(2) - x*log(x) + x + 6)*e^(-2*(log((x^2 - 2*log(2) + log(x))/(x + 6)) - 2)
/log((x^2 - 2*log(2) + log(x))/(x + 6)))/((x^4 + 6*x^3 - 2*(x^2 + 6*x)*log(2) + (x^2 + 6*x)*log(x))*log((x^2 -
2*log(2) + log(x))/(x + 6))^2), x)

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maple [C]  time = 12.63, size = 433, normalized size = 16.04

method	result	size
risch	\({\mathrm e}^{-\frac {2 \left (i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +6}\right )+i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2} \mathrm {csgn}\left (i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right ) \mathrm {csgn}\left (\frac {i}{x +6}\right ) \mathrm {csgn}\left (i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )\right )-2 i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2}+2 i \pi -2 \ln \left (x +6\right )+2 \ln \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )+2 \ln \relax (2)-4\right )}{i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +6}\right )+i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2} \mathrm {csgn}\left (i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right ) \mathrm {csgn}\left (\frac {i}{x +6}\right ) \mathrm {csgn}\left (i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )\right )-2 i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2}+2 i \pi -2 \ln \left (x +6\right )+2 \ln \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )+2 \ln \relax (2)}}\)	\(433\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(x)-8*x*ln(2)-4*x^3-48*x^2-4*x-24)*exp((-2*ln((ln(x)-2*ln(2)+x^2)/(x+6))+4)/ln((ln(x)-2*ln(2)+x^2)/
(x+6)))/((x^2+6*x)*ln(x)+2*(-x^2-6*x)*ln(2)+x^4+6*x^3)/ln((ln(x)-2*ln(2)+x^2)/(x+6))^2,x,method=_RETURNVERBOSE
)

[Out]

exp(-2*(I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2)-1/2*ln(x)))^3+I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2)-1/2*ln(x)))^2*csgn(I
/(x+6))+I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2)-1/2*ln(x)))^2*csgn(I*(-1/2*x^2+ln(2)-1/2*ln(x)))-I*Pi*csgn(I/(x+6)*(
-1/2*x^2+ln(2)-1/2*ln(x)))*csgn(I/(x+6))*csgn(I*(-1/2*x^2+ln(2)-1/2*ln(x)))-2*I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2
)-1/2*ln(x)))^2+2*I*Pi-2*ln(x+6)+2*ln(-1/2*x^2+ln(2)-1/2*ln(x))+2*ln(2)-4)/(I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2)-
1/2*ln(x)))^3+I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2)-1/2*ln(x)))^2*csgn(I/(x+6))+I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2)-
1/2*ln(x)))^2*csgn(I*(-1/2*x^2+ln(2)-1/2*ln(x)))-I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2)-1/2*ln(x)))*csgn(I/(x+6))*c
sgn(I*(-1/2*x^2+ln(2)-1/2*ln(x)))-2*I*Pi*csgn(I/(x+6)*(-1/2*x^2+ln(2)-1/2*ln(x)))^2+2*I*Pi-2*ln(x+6)+2*ln(-1/2
*x^2+ln(2)-1/2*ln(x))+2*ln(2)))

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maxima [B]  time = 0.96, size = 384, normalized size = 14.22 \begin {gather*} \frac {x^{3} e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )}}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} + \frac {12 \, x^{2} e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )}}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} + \frac {2 \, x e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )} \log \relax (2)}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} - \frac {x e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )} \log \relax (x)}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} + \frac {x e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )}}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} + \frac {6 \, e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )}}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)-8*x*log(2)-4*x^3-48*x^2-4*x-24)*exp((-2*log((log(x)-2*log(2)+x^2)/(x+6))+4)/log((log(x)-
2*log(2)+x^2)/(x+6)))/((x^2+6*x)*log(x)+2*(-x^2-6*x)*log(2)+x^4+6*x^3)/log((log(x)-2*log(2)+x^2)/(x+6))^2,x, a
lgorithm="maxima")

[Out]

x^3*e^(4/(log(x^2 - 2*log(2) + log(x)) - log(x + 6)))/(x^3*e^2 + 12*x^2*e^2 + x*(2*log(2) + 1)*e^2 - x*e^2*log
(x) + 6*e^2) + 12*x^2*e^(4/(log(x^2 - 2*log(2) + log(x)) - log(x + 6)))/(x^3*e^2 + 12*x^2*e^2 + x*(2*log(2) +
1)*e^2 - x*e^2*log(x) + 6*e^2) + 2*x*e^(4/(log(x^2 - 2*log(2) + log(x)) - log(x + 6)))*log(2)/(x^3*e^2 + 12*x^
2*e^2 + x*(2*log(2) + 1)*e^2 - x*e^2*log(x) + 6*e^2) - x*e^(4/(log(x^2 - 2*log(2) + log(x)) - log(x + 6)))*log
(x)/(x^3*e^2 + 12*x^2*e^2 + x*(2*log(2) + 1)*e^2 - x*e^2*log(x) + 6*e^2) + x*e^(4/(log(x^2 - 2*log(2) + log(x)
) - log(x + 6)))/(x^3*e^2 + 12*x^2*e^2 + x*(2*log(2) + 1)*e^2 - x*e^2*log(x) + 6*e^2) + 6*e^(4/(log(x^2 - 2*lo
g(2) + log(x)) - log(x + 6)))/(x^3*e^2 + 12*x^2*e^2 + x*(2*log(2) + 1)*e^2 - x*e^2*log(x) + 6*e^2)

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mupad [B]  time = 4.05, size = 23, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{-2}\,{\mathrm {e}}^{\frac {4}{\ln \left (\frac {\ln \left (\frac {x}{4}\right )+x^2}{x+6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*log((log(x) - 2*log(2) + x^2)/(x + 6)) - 4)/log((log(x) - 2*log(2) + x^2)/(x + 6)))*(4*x + 8*x*l
og(2) - 4*x*log(x) + 48*x^2 + 4*x^3 + 24))/(log((log(x) - 2*log(2) + x^2)/(x + 6))^2*(log(x)*(6*x + x^2) + 6*x
^3 + x^4 - 2*log(2)*(6*x + x^2))),x)

[Out]

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

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sympy [A]  time = 62.22, size = 37, normalized size = 1.37 \begin {gather*} e^{\frac {4 - 2 \log {\left (\frac {x^{2} + \log {\relax (x )} - 2 \log {\relax (2 )}}{x + 6} \right )}}{\log {\left (\frac {x^{2} + \log {\relax (x )} - 2 \log {\relax (2 )}}{x + 6} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(x)-8*x*ln(2)-4*x**3-48*x**2-4*x-24)*exp((-2*ln((ln(x)-2*ln(2)+x**2)/(x+6))+4)/ln((ln(x)-2*ln
(2)+x**2)/(x+6)))/((x**2+6*x)*ln(x)+2*(-x**2-6*x)*ln(2)+x**4+6*x**3)/ln((ln(x)-2*ln(2)+x**2)/(x+6))**2,x)

[Out]

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

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