3.59.4 \(\int \frac {e^{-e^x} (-32 x^2+8 x^4+8 x^5+(-16 x^2+4 x^5) \log (5)+e^x (8 x^5+8 x^6+4 x^6 \log (5))+(64 x+64 x^2+32 x^2 \log (5)+e^x (-32 x^2-32 x^3-16 x^3 \log (5))) \log (-2-2 x-x \log (5)))}{2 x^6+2 x^7+x^7 \log (5)+(-16 x^3-16 x^4-8 x^4 \log (5)) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx\)

Optimal. Leaf size=34 \[ \frac {e^{-e^x} x}{-\frac {x^2}{4}+\frac {\log (-2+x-x (3+\log (5)))}{x}} \]

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Rubi [F]  time = 11.13, 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 {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-32*x^2 + 8*x^4 + 8*x^5 + (-16*x^2 + 4*x^5)*Log[5] + E^x*(8*x^5 + 8*x^6 + 4*x^6*Log[5]) + (64*x + 64*x^2
+ 32*x^2*Log[5] + E^x*(-32*x^2 - 32*x^3 - 16*x^3*Log[5]))*Log[-2 - 2*x - x*Log[5]])/(E^E^x*(2*x^6 + 2*x^7 + x^
7*Log[5] + (-16*x^3 - 16*x^4 - 8*x^4*Log[5])*Log[-2 - 2*x - x*Log[5]] + (32 + 32*x + 16*x*Log[5])*Log[-2 - 2*x
- x*Log[5]]^2)),x]

[Out]

(32*Defer[Int][1/(E^E^x*(x^3 - 4*Log[-2 - x*(2 + Log[5])])^2), x])/(2 + Log[5]) - 16*Defer[Int][x/(E^E^x*(x^3
- 4*Log[-2 - x*(2 + Log[5])])^2), x] + 12*Defer[Int][x^4/(E^E^x*(x^3 - 4*Log[-2 - x*(2 + Log[5])])^2), x] - (6
4*Defer[Int][1/(E^E^x*(2 + x*(2 + Log[5]))*(x^3 - 4*Log[-2 - x*(2 + Log[5])])^2), x])/(2 + Log[5]) - 8*Defer[I
nt][x/(E^E^x*(x^3 - 4*Log[-2 - x*(2 + Log[5])])), x] + 4*Defer[Int][(E^(-E^x + x)*x^2)/(x^3 - 4*Log[-2 - x*(2
+ Log[5])]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+x^7 (2+\log (5))+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx\\ &=\int \frac {4 e^{-e^x} x \left (2 x^3-4 x (2+\log (5))+e^x x^5 (2+\log (5))+x^4 \left (2+2 e^x+\log (5)\right )-4 \left (-2+e^x x\right ) (2+x (2+\log (5))) \log (-2-x (2+\log (5)))\right )}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx\\ &=4 \int \frac {e^{-e^x} x \left (2 x^3-4 x (2+\log (5))+e^x x^5 (2+\log (5))+x^4 \left (2+2 e^x+\log (5)\right )-4 \left (-2+e^x x\right ) (2+x (2+\log (5))) \log (-2-x (2+\log (5)))\right )}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx\\ &=4 \int \left (\frac {2 e^{-e^x} x^4}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {4 e^{-e^x} x^2 (-2-\log (5))}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {2 e^{-e^x} x^5 \left (1+\frac {\log (5)}{2}\right )}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {e^{-e^x+x} x^2}{x^3-4 \log (-2-x (2+\log (5)))}+\frac {8 e^{-e^x} x \log (-2-x (2+\log (5)))}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}\right ) \, dx\\ &=4 \int \frac {e^{-e^x+x} x^2}{x^3-4 \log (-2-x (2+\log (5)))} \, dx+8 \int \frac {e^{-e^x} x^4}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx+32 \int \frac {e^{-e^x} x \log (-2-x (2+\log (5)))}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx+(4 (2+\log (5))) \int \frac {e^{-e^x} x^5}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx-(16 (2+\log (5))) \int \frac {e^{-e^x} x^2}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx\\ &=4 \int \frac {e^{-e^x+x} x^2}{x^3-4 \log (-2-x (2+\log (5)))} \, dx+8 \int \left (-\frac {8 e^{-e^x}}{(2+\log (5))^4 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {4 e^{-e^x} x}{(2+\log (5))^3 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}-\frac {2 e^{-e^x} x^2}{(2+\log (5))^2 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {e^{-e^x} x^3}{(2+\log (5)) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {16 e^{-e^x}}{(2+\log (5))^4 (2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}\right ) \, dx+32 \int \left (\frac {e^{-e^x} x^4}{4 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}-\frac {e^{-e^x} x}{4 \left (x^3-4 \log (-2-x (2+\log (5)))\right )}\right ) \, dx+(4 (2+\log (5))) \int \left (\frac {16 e^{-e^x}}{(2+\log (5))^5 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}-\frac {8 e^{-e^x} x}{(2+\log (5))^4 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {4 e^{-e^x} x^2}{(2+\log (5))^3 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}-\frac {2 e^{-e^x} x^3}{(2+\log (5))^2 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {e^{-e^x} x^4}{(2+\log (5)) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {32 e^{-e^x}}{(-2-\log (5))^5 (2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}\right ) \, dx-(16 (2+\log (5))) \int \left (-\frac {2 e^{-e^x}}{(2+\log (5))^2 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {e^{-e^x} x}{(2+\log (5)) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {4 e^{-e^x}}{(2+\log (5))^2 (2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}\right ) \, dx\\ &=4 \int \frac {e^{-e^x} x^4}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx+4 \int \frac {e^{-e^x+x} x^2}{x^3-4 \log (-2-x (2+\log (5)))} \, dx+8 \int \frac {e^{-e^x} x^4}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx-8 \int \frac {e^{-e^x} x}{x^3-4 \log (-2-x (2+\log (5)))} \, dx-16 \int \frac {e^{-e^x} x}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx+\frac {32 \int \frac {e^{-e^x}}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx}{2+\log (5)}-\frac {64 \int \frac {e^{-e^x}}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx}{2+\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 32, normalized size = 0.94 \begin {gather*} \frac {4 e^{-e^x} x^2}{-x^3+4 \log (-2-x (2+\log (5)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*x^2 + 8*x^4 + 8*x^5 + (-16*x^2 + 4*x^5)*Log[5] + E^x*(8*x^5 + 8*x^6 + 4*x^6*Log[5]) + (64*x + 6
4*x^2 + 32*x^2*Log[5] + E^x*(-32*x^2 - 32*x^3 - 16*x^3*Log[5]))*Log[-2 - 2*x - x*Log[5]])/(E^E^x*(2*x^6 + 2*x^
7 + x^7*Log[5] + (-16*x^3 - 16*x^4 - 8*x^4*Log[5])*Log[-2 - 2*x - x*Log[5]] + (32 + 32*x + 16*x*Log[5])*Log[-2
- 2*x - x*Log[5]]^2)),x]

[Out]

(4*x^2)/(E^E^x*(-x^3 + 4*Log[-2 - x*(2 + Log[5])]))

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fricas [A]  time = 0.64, size = 29, normalized size = 0.85 \begin {gather*} -\frac {4 \, x^{2} e^{\left (-e^{x}\right )}}{x^{3} - 4 \, \log \left (-x \log \relax (5) - 2 \, x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^3*log(5)-32*x^3-32*x^2)*exp(x)+32*x^2*log(5)+64*x^2+64*x)*log(-x*log(5)-2*x-2)+(4*x^6*log(5
)+8*x^6+8*x^5)*exp(x)+(4*x^5-16*x^2)*log(5)+8*x^5+8*x^4-32*x^2)/((16*x*log(5)+32*x+32)*log(-x*log(5)-2*x-2)^2+
(-8*x^4*log(5)-16*x^4-16*x^3)*log(-x*log(5)-2*x-2)+x^7*log(5)+2*x^7+2*x^6)/exp(exp(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 29, normalized size = 0.85 \begin {gather*} -\frac {4 \, x^{2} e^{\left (-e^{x}\right )}}{x^{3} - 4 \, \log \left (-x \log \relax (5) - 2 \, x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^3*log(5)-32*x^3-32*x^2)*exp(x)+32*x^2*log(5)+64*x^2+64*x)*log(-x*log(5)-2*x-2)+(4*x^6*log(5
)+8*x^6+8*x^5)*exp(x)+(4*x^5-16*x^2)*log(5)+8*x^5+8*x^4-32*x^2)/((16*x*log(5)+32*x+32)*log(-x*log(5)-2*x-2)^2+
(-8*x^4*log(5)-16*x^4-16*x^3)*log(-x*log(5)-2*x-2)+x^7*log(5)+2*x^7+2*x^6)/exp(exp(x)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (-16 x^{3} \ln \relax (5)-32 x^{3}-32 x^{2}\right ) {\mathrm e}^{x}+32 x^{2} \ln \relax (5)+64 x^{2}+64 x \right ) \ln \left (-x \ln \relax (5)-2 x -2\right )+\left (4 x^{6} \ln \relax (5)+8 x^{6}+8 x^{5}\right ) {\mathrm e}^{x}+\left (4 x^{5}-16 x^{2}\right ) \ln \relax (5)+8 x^{5}+8 x^{4}-32 x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{x}}}{\left (16 x \ln \relax (5)+32 x +32\right ) \ln \left (-x \ln \relax (5)-2 x -2\right )^{2}+\left (-8 x^{4} \ln \relax (5)-16 x^{4}-16 x^{3}\right ) \ln \left (-x \ln \relax (5)-2 x -2\right )+x^{7} \ln \relax (5)+2 x^{7}+2 x^{6}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16*x^3*ln(5)-32*x^3-32*x^2)*exp(x)+32*x^2*ln(5)+64*x^2+64*x)*ln(-x*ln(5)-2*x-2)+(4*x^6*ln(5)+8*x^6+8*x
^5)*exp(x)+(4*x^5-16*x^2)*ln(5)+8*x^5+8*x^4-32*x^2)/((16*x*ln(5)+32*x+32)*ln(-x*ln(5)-2*x-2)^2+(-8*x^4*ln(5)-1
6*x^4-16*x^3)*ln(-x*ln(5)-2*x-2)+x^7*ln(5)+2*x^7+2*x^6)/exp(exp(x)),x)

[Out]

int((((-16*x^3*ln(5)-32*x^3-32*x^2)*exp(x)+32*x^2*ln(5)+64*x^2+64*x)*ln(-x*ln(5)-2*x-2)+(4*x^6*ln(5)+8*x^6+8*x
^5)*exp(x)+(4*x^5-16*x^2)*ln(5)+8*x^5+8*x^4-32*x^2)/((16*x*ln(5)+32*x+32)*ln(-x*ln(5)-2*x-2)^2+(-8*x^4*ln(5)-1
6*x^4-16*x^3)*ln(-x*ln(5)-2*x-2)+x^7*ln(5)+2*x^7+2*x^6)/exp(exp(x)),x)

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maxima [A]  time = 0.54, size = 30, normalized size = 0.88 \begin {gather*} -\frac {4 \, x^{2}}{x^{3} e^{\left (e^{x}\right )} - 4 \, e^{\left (e^{x}\right )} \log \left (-x {\left (\log \relax (5) + 2\right )} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^3*log(5)-32*x^3-32*x^2)*exp(x)+32*x^2*log(5)+64*x^2+64*x)*log(-x*log(5)-2*x-2)+(4*x^6*log(5
)+8*x^6+8*x^5)*exp(x)+(4*x^5-16*x^2)*log(5)+8*x^5+8*x^4-32*x^2)/((16*x*log(5)+32*x+32)*log(-x*log(5)-2*x-2)^2+
(-8*x^4*log(5)-16*x^4-16*x^3)*log(-x*log(5)-2*x-2)+x^7*log(5)+2*x^7+2*x^6)/exp(exp(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.02, size = 254, normalized size = 7.47 \begin {gather*} \frac {\frac {4\,x^2\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,\left (2\,x^3\,{\mathrm {e}}^x-4\,\ln \relax (5)+2\,x^4\,{\mathrm {e}}^x+x^3\,\ln \relax (5)+2\,x^2+2\,x^3+x^4\,{\mathrm {e}}^x\,\ln \relax (5)-8\right )}{3\,x^3\,\ln \relax (5)-4\,\ln \relax (5)+6\,x^2+6\,x^3-8}-\frac {16\,x\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,\ln \left (-2\,x-x\,\ln \relax (5)-2\right )\,\left (x\,{\mathrm {e}}^x-2\right )\,\left (2\,x+x\,\ln \relax (5)+2\right )}{3\,x^3\,\ln \relax (5)-4\,\ln \relax (5)+6\,x^2+6\,x^3-8}}{4\,\ln \left (-2\,x-x\,\ln \relax (5)-2\right )-x^3}-\frac {{\mathrm {e}}^{-{\mathrm {e}}^x}\,\left (\frac {16\,x}{\ln \left (125\right )+6}-\frac {8\,x^2\,{\mathrm {e}}^x}{\ln \left (125\right )+6}+\frac {x^2\,\left (8\,\ln \relax (5)+16\right )}{\ln \left (125\right )+6}-\frac {x^3\,{\mathrm {e}}^x\,\left (\ln \left (625\right )+8\right )}{\ln \left (125\right )+6}\right )}{x^3+\frac {6\,x^2}{\ln \left (125\right )+6}-\frac {\ln \left (625\right )+8}{\ln \left (125\right )+6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(x))*(log(- 2*x - x*log(5) - 2)*(64*x + 32*x^2*log(5) - exp(x)*(16*x^3*log(5) + 32*x^2 + 32*x^3)
+ 64*x^2) - log(5)*(16*x^2 - 4*x^5) + exp(x)*(4*x^6*log(5) + 8*x^5 + 8*x^6) - 32*x^2 + 8*x^4 + 8*x^5))/(log(-
2*x - x*log(5) - 2)^2*(32*x + 16*x*log(5) + 32) + x^7*log(5) - log(- 2*x - x*log(5) - 2)*(8*x^4*log(5) + 16*x^
3 + 16*x^4) + 2*x^6 + 2*x^7),x)

[Out]

((4*x^2*exp(-exp(x))*(2*x^3*exp(x) - 4*log(5) + 2*x^4*exp(x) + x^3*log(5) + 2*x^2 + 2*x^3 + x^4*exp(x)*log(5)
- 8))/(3*x^3*log(5) - 4*log(5) + 6*x^2 + 6*x^3 - 8) - (16*x*exp(-exp(x))*log(- 2*x - x*log(5) - 2)*(x*exp(x) -
2)*(2*x + x*log(5) + 2))/(3*x^3*log(5) - 4*log(5) + 6*x^2 + 6*x^3 - 8))/(4*log(- 2*x - x*log(5) - 2) - x^3) -
(exp(-exp(x))*((16*x)/(log(125) + 6) - (8*x^2*exp(x))/(log(125) + 6) + (x^2*(8*log(5) + 16))/(log(125) + 6) -
(x^3*exp(x)*(log(625) + 8))/(log(125) + 6)))/((6*x^2)/(log(125) + 6) - (log(625) + 8)/(log(125) + 6) + x^3)

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sympy [A]  time = 0.62, size = 29, normalized size = 0.85 \begin {gather*} - \frac {4 x^{2} e^{- e^{x}}}{x^{3} - 4 \log {\left (- 2 x - x \log {\relax (5 )} - 2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x**3*ln(5)-32*x**3-32*x**2)*exp(x)+32*x**2*ln(5)+64*x**2+64*x)*ln(-x*ln(5)-2*x-2)+(4*x**6*ln(
5)+8*x**6+8*x**5)*exp(x)+(4*x**5-16*x**2)*ln(5)+8*x**5+8*x**4-32*x**2)/((16*x*ln(5)+32*x+32)*ln(-x*ln(5)-2*x-2
)**2+(-8*x**4*ln(5)-16*x**4-16*x**3)*ln(-x*ln(5)-2*x-2)+x**7*ln(5)+2*x**7+2*x**6)/exp(exp(x)),x)

[Out]

-4*x**2*exp(-exp(x))/(x**3 - 4*log(-2*x - x*log(5) - 2))

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