3.50.96 \(\int \frac {20^{\frac {1}{\log (\frac {1}{16} (64+16 x+x^2+(-64-8 x) \log (3+e^{e^x})+16 \log ^2(3+e^{e^x})))}} (6 \log (20)+e^{e^x} (2 \log (20)-8 e^x \log (20)))}{(-24+e^{e^x} (-8-x)-3 x+(12+4 e^{e^x}) \log (3+e^{e^x})) \log ^2(\frac {1}{16} (64+16 x+x^2+(-64-8 x) \log (3+e^{e^x})+16 \log ^2(3+e^{e^x})))} \, dx\)

Optimal. Leaf size=24 \[ 20^{\frac {1}{\log \left (\left (2+\frac {x}{4}-\log \left (3+e^{e^x}\right )\right )^2\right )}} \]

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Rubi [F]  time = 6.02, 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 {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(20^Log[(64 + 16*x + x^2 + (-64 - 8*x)*Log[3 + E^E^x] + 16*Log[3 + E^E^x]^2)/16]^(-1)*(6*Log[20] + E^E^x*(
2*Log[20] - 8*E^x*Log[20])))/((-24 + E^E^x*(-8 - x) - 3*x + (12 + 4*E^E^x)*Log[3 + E^E^x])*Log[(64 + 16*x + x^
2 + (-64 - 8*x)*Log[3 + E^E^x] + 16*Log[3 + E^E^x]^2)/16]^2),x]

[Out]

-(Log[20]*Defer[Int][(2^(1 + 2/Log[(8 + x - 4*Log[3 + E^E^x])^2/16])*5^Log[(8 + x - 4*Log[3 + E^E^x])^2/16]^(-
1))/((8 + x - 4*Log[3 + E^E^x])*Log[(8 + x - 4*Log[3 + E^E^x])^2/16]^2), x]) + Log[20]*Defer[Int][(2^(3 + 2/Lo
g[(8 + x - 4*Log[3 + E^E^x])^2/16])*5^Log[(8 + x - 4*Log[3 + E^E^x])^2/16]^(-1)*E^(E^x + x))/((3 + E^E^x)*(8 +
x - 4*Log[3 + E^E^x])*Log[(8 + x - 4*Log[3 + E^E^x])^2/16]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} \left (-3-e^{e^x}+4 e^{e^x+x}\right ) \log (20)}{\left (3+e^{e^x}\right ) \left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )} \, dx\\ &=\log (20) \int \frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} \left (-3-e^{e^x}+4 e^{e^x+x}\right )}{\left (3+e^{e^x}\right ) \left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )} \, dx\\ &=\log (20) \int \left (-\frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}}}{\left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}+\frac {2^{3+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} e^{e^x+x}}{\left (3+e^{e^x}\right ) \left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}\right ) \, dx\\ &=-\left (\log (20) \int \frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}}}{\left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )} \, dx\right )+\log (20) \int \frac {2^{3+\frac {2}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} e^{e^x+x}}{\left (3+e^{e^x}\right ) \left (8+x-4 \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 24, normalized size = 1.00 \begin {gather*} 20^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20^Log[(64 + 16*x + x^2 + (-64 - 8*x)*Log[3 + E^E^x] + 16*Log[3 + E^E^x]^2)/16]^(-1)*(6*Log[20] + E
^E^x*(2*Log[20] - 8*E^x*Log[20])))/((-24 + E^E^x*(-8 - x) - 3*x + (12 + 4*E^E^x)*Log[3 + E^E^x])*Log[(64 + 16*
x + x^2 + (-64 - 8*x)*Log[3 + E^E^x] + 16*Log[3 + E^E^x]^2)/16]^2),x]

[Out]

20^Log[(8 + x - 4*Log[3 + E^E^x])^2/16]^(-1)

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fricas [A]  time = 0.72, size = 32, normalized size = 1.33 \begin {gather*} 20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*log(20)*exp(x)+2*log(20))*exp(exp(x))+6*log(20))*exp(log(20)/log(log(exp(exp(x))+3)^2+1/16*(-8*
x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4))/((4*exp(exp(x))+12)*log(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/log(
log(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4)^2,x, algorithm="fricas")

[Out]

20^(1/log(1/16*x^2 - 1/2*(x + 8)*log(e^(e^x) + 3) + log(e^(e^x) + 3)^2 + x + 4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left ({\left (4 \, e^{x} \log \left (20\right ) - \log \left (20\right )\right )} e^{\left (e^{x}\right )} - 3 \, \log \left (20\right )\right )} 20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )}}{{\left ({\left (x + 8\right )} e^{\left (e^{x}\right )} - 4 \, {\left (e^{\left (e^{x}\right )} + 3\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + 3 \, x + 24\right )} \log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*log(20)*exp(x)+2*log(20))*exp(exp(x))+6*log(20))*exp(log(20)/log(log(exp(exp(x))+3)^2+1/16*(-8*
x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4))/((4*exp(exp(x))+12)*log(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/log(
log(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4)^2,x, algorithm="giac")

[Out]

integrate(2*((4*e^x*log(20) - log(20))*e^(e^x) - 3*log(20))*20^(1/log(1/16*x^2 - 1/2*(x + 8)*log(e^(e^x) + 3)
+ log(e^(e^x) + 3)^2 + x + 4))/(((x + 8)*e^(e^x) - 4*(e^(e^x) + 3)*log(e^(e^x) + 3) + 3*x + 24)*log(1/16*x^2 -
1/2*(x + 8)*log(e^(e^x) + 3) + log(e^(e^x) + 3)^2 + x + 4)^2), x)

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maple [C]  time = 1.52, size = 131, normalized size = 5.46

method	result	size
risch	\({\mathrm e}^{-\frac {2 \left (2 \ln \relax (2)+\ln \relax (5)\right )}{i \pi \mathrm {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )^{2}\right )^{3}-2 i \pi \mathrm {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )\right )+i \pi \,\mathrm {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )^{2}\right ) \mathrm {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )\right )^{2}+8 \ln \relax (2)-4 \ln \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*ln(20)*exp(x)+2*ln(20))*exp(exp(x))+6*ln(20))*exp(ln(20)/ln(ln(exp(exp(x))+3)^2+1/16*(-8*x-64)*ln(exp
(exp(x))+3)+1/16*x^2+x+4))/((4*exp(exp(x))+12)*ln(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/ln(ln(exp(exp(x))+
3)^2+1/16*(-8*x-64)*ln(exp(exp(x))+3)+1/16*x^2+x+4)^2,x,method=_RETURNVERBOSE)

[Out]

exp(-2*(2*ln(2)+ln(5))/(I*Pi*csgn(I*(-4*ln(exp(exp(x))+3)+x+8)^2)^3-2*I*Pi*csgn(I*(-4*ln(exp(exp(x))+3)+x+8)^2
)^2*csgn(I*(-4*ln(exp(exp(x))+3)+x+8))+I*Pi*csgn(I*(-4*ln(exp(exp(x))+3)+x+8)^2)*csgn(I*(-4*ln(exp(exp(x))+3)+
x+8))^2+8*ln(2)-4*ln(-4*ln(exp(exp(x))+3)+x+8)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, \int \frac {{\left ({\left (4 \, e^{x} \log \left (20\right ) - \log \left (20\right )\right )} e^{\left (e^{x}\right )} - 3 \, \log \left (20\right )\right )} 20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )}}{{\left ({\left (x + 8\right )} e^{\left (e^{x}\right )} - 4 \, {\left (e^{\left (e^{x}\right )} + 3\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + 3 \, x + 24\right )} \log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*log(20)*exp(x)+2*log(20))*exp(exp(x))+6*log(20))*exp(log(20)/log(log(exp(exp(x))+3)^2+1/16*(-8*
x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4))/((4*exp(exp(x))+12)*log(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/log(
log(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4)^2,x, algorithm="maxima")

[Out]

2*integrate(((4*e^x*log(20) - log(20))*e^(e^x) - 3*log(20))*20^(1/log(1/16*x^2 - 1/2*(x + 8)*log(e^(e^x) + 3)
+ log(e^(e^x) + 3)^2 + x + 4))/(((x + 8)*e^(e^x) - 4*(e^(e^x) + 3)*log(e^(e^x) + 3) + 3*x + 24)*log(1/16*x^2 -
1/2*(x + 8)*log(e^(e^x) + 3) + log(e^(e^x) + 3)^2 + x + 4)^2), x)

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mupad [B]  time = 3.83, size = 38, normalized size = 1.58 \begin {gather*} {20}^{\frac {1}{\ln \left (\frac {x^2}{16}-\frac {x\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )}{2}+x+{\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )}^2-4\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )+4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(20)/log(x + log(exp(exp(x)) + 3)^2 - (log(exp(exp(x)) + 3)*(8*x + 64))/16 + x^2/16 + 4))*(6*log(
20) + exp(exp(x))*(2*log(20) - 8*exp(x)*log(20))))/(log(x + log(exp(exp(x)) + 3)^2 - (log(exp(exp(x)) + 3)*(8*
x + 64))/16 + x^2/16 + 4)^2*(3*x - log(exp(exp(x)) + 3)*(4*exp(exp(x)) + 12) + exp(exp(x))*(x + 8) + 24)),x)

[Out]

20^(1/log(x - 4*log(exp(exp(x)) + 3) + log(exp(exp(x)) + 3)^2 + x^2/16 - (x*log(exp(exp(x)) + 3))/2 + 4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*ln(20)*exp(x)+2*ln(20))*exp(exp(x))+6*ln(20))*exp(ln(20)/ln(ln(exp(exp(x))+3)**2+1/16*(-8*x-64)
*ln(exp(exp(x))+3)+1/16*x**2+x+4))/((4*exp(exp(x))+12)*ln(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/ln(ln(exp(
exp(x))+3)**2+1/16*(-8*x-64)*ln(exp(exp(x))+3)+1/16*x**2+x+4)**2,x)

[Out]

Timed out

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