3.45.98 \(\int \frac {e^{\frac {x+(3+3 e^x+12 x-3 x^2) \log (\frac {x+2 \log (\log (4))}{\log (\log (4))})}{3+3 e^x+12 x-3 x^2}} (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x (6+25 x-7 x^2)+(2+e^x (2-2 x)+2 x^2) \log (\log (4)))}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x (6 x+24 x^2-6 x^3)+(6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x (12+48 x-12 x^2)) \log (\log (4))} \, dx\)

Optimal. Leaf size=35 \[ e^{-\frac {x}{3 \left (-e^x+x \left (-4-\frac {1}{x}+x\right )\right )}} \left (2+\frac {x}{\log (\log (4))}\right ) \]

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Rubi [F]  time = 49.46, 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 {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}\right ) \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((x + (3 + 3*E^x + 12*x - 3*x^2)*Log[(x + 2*Log[Log[4]])/Log[Log[4]]])/(3 + 3*E^x + 12*x - 3*x^2))*(3 +
3*E^(2*x) + 25*x + 42*x^2 - 23*x^3 + 3*x^4 + E^x*(6 + 25*x - 7*x^2) + (2 + E^x*(2 - 2*x) + 2*x^2)*Log[Log[4]]
))/(3*x + 3*E^(2*x)*x + 24*x^2 + 42*x^3 - 24*x^4 + 3*x^5 + E^x*(6*x + 24*x^2 - 6*x^3) + (6 + 6*E^(2*x) + 48*x
+ 84*x^2 - 48*x^3 + 6*x^4 + E^x*(12 + 48*x - 12*x^2))*Log[Log[4]]),x]

[Out]

Defer[Int][E^(x/(3*(1 + E^x + 4*x - x^2))), x]/Log[Log[4]] + (2*Defer[Int][E^(x/(3*(1 + E^x + 4*x - x^2)))/(1
+ E^x + 4*x - x^2), x])/3 - 2*Defer[Int][(E^(x/(3*(1 + E^x + 4*x - x^2)))*x)/(-1 - E^x - 4*x + x^2)^2, x] + (4
- Log[Log[4]]^(-1))*Defer[Int][(E^(x/(3*(1 + E^x + 4*x - x^2)))*x^2)/(-1 - E^x - 4*x + x^2)^2, x] + (2*(3 - L
og[Log[4]])*Defer[Int][(E^(x/(3*(1 + E^x + 4*x - x^2)))*x^3)/(-1 - E^x - 4*x + x^2)^2, x])/(3*Log[Log[4]]) - D
efer[Int][(E^(x/(3*(1 + E^x + 4*x - x^2)))*x^4)/(-1 - E^x - 4*x + x^2)^2, x]/(3*Log[Log[4]]) + ((2 - Log[Log[4
]]^(-1))*Defer[Int][(E^(x/(3*(1 + E^x + 4*x - x^2)))*x)/(-1 - E^x - 4*x + x^2), x])/3 + Defer[Int][(E^(x/(3*(1
+ E^x + 4*x - x^2)))*x^2)/(-1 - E^x - 4*x + x^2), x]/(3*Log[Log[4]])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \left (3 e^{2 x}+25 x-23 x^3+3 x^4+3 \left (1+\frac {2}{3} \log (\log (4))\right )+2 x^2 (21+\log (\log (4)))+e^x \left (6+25 x-7 x^2+2 \log (\log (4))-2 x \log (\log (4))\right )\right )}{3 \left (1+e^x+4 x-x^2\right )^2 \log (\log (4))} \, dx\\ &=\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \left (3 e^{2 x}+25 x-23 x^3+3 x^4+3 \left (1+\frac {2}{3} \log (\log (4))\right )+2 x^2 (21+\log (\log (4)))+e^x \left (6+25 x-7 x^2+2 \log (\log (4))-2 x \log (\log (4))\right )\right )}{\left (1+e^x+4 x-x^2\right )^2} \, dx}{3 \log (\log (4))}\\ &=\frac {\int \left (3 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}}-\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x \left (3-6 x+x^2\right ) (x+2 \log (\log (4)))}{\left (-1-e^x-4 x+x^2\right )^2}+\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} (-1+x) (x+2 \log (\log (4)))}{-1-e^x-4 x+x^2}\right ) \, dx}{3 \log (\log (4))}\\ &=-\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x \left (3-6 x+x^2\right ) (x+2 \log (\log (4)))}{\left (-1-e^x-4 x+x^2\right )^2} \, dx}{3 \log (\log (4))}+\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} (-1+x) (x+2 \log (\log (4)))}{-1-e^x-4 x+x^2} \, dx}{3 \log (\log (4))}+\frac {\int e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \, dx}{\log (\log (4))}\\ &=\frac {\int \left (\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^2}{-1-e^x-4 x+x^2}+\frac {2 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \log (\log (4))}{1+e^x+4 x-x^2}+\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x (-1+2 \log (\log (4)))}{-1-e^x-4 x+x^2}\right ) \, dx}{3 \log (\log (4))}-\frac {\int \left (\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^4}{\left (-1-e^x-4 x+x^2\right )^2}+\frac {2 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^3 (-3+\log (\log (4)))}{\left (-1-e^x-4 x+x^2\right )^2}+\frac {6 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x \log (\log (4))}{\left (-1-e^x-4 x+x^2\right )^2}-\frac {3 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^2 (-1+4 \log (\log (4)))}{\left (-1-e^x-4 x+x^2\right )^2}\right ) \, dx}{3 \log (\log (4))}+\frac {\int e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \, dx}{\log (\log (4))}\\ &=\frac {2}{3} \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}}}{1+e^x+4 x-x^2} \, dx-2 \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x}{\left (-1-e^x-4 x+x^2\right )^2} \, dx-\left (-4+\frac {1}{\log (\log (4))}\right ) \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^2}{\left (-1-e^x-4 x+x^2\right )^2} \, dx-\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^4}{\left (-1-e^x-4 x+x^2\right )^2} \, dx}{3 \log (\log (4))}+\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^2}{-1-e^x-4 x+x^2} \, dx}{3 \log (\log (4))}+\frac {\int e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \, dx}{\log (\log (4))}+\frac {(2 (3-\log (\log (4)))) \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^3}{\left (-1-e^x-4 x+x^2\right )^2} \, dx}{3 \log (\log (4))}+\frac {(-1+2 \log (\log (4))) \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x}{-1-e^x-4 x+x^2} \, dx}{3 \log (\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 40, normalized size = 1.14 \begin {gather*} \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} (3 x+6 \log (\log (4)))}{3 \log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x + (3 + 3*E^x + 12*x - 3*x^2)*Log[(x + 2*Log[Log[4]])/Log[Log[4]]])/(3 + 3*E^x + 12*x - 3*x^2)
)*(3 + 3*E^(2*x) + 25*x + 42*x^2 - 23*x^3 + 3*x^4 + E^x*(6 + 25*x - 7*x^2) + (2 + E^x*(2 - 2*x) + 2*x^2)*Log[L
og[4]]))/(3*x + 3*E^(2*x)*x + 24*x^2 + 42*x^3 - 24*x^4 + 3*x^5 + E^x*(6*x + 24*x^2 - 6*x^3) + (6 + 6*E^(2*x) +
48*x + 84*x^2 - 48*x^3 + 6*x^4 + E^x*(12 + 48*x - 12*x^2))*Log[Log[4]]),x]

[Out]

(E^(x/(3*(1 + E^x + 4*x - x^2)))*(3*x + 6*Log[Log[4]]))/(3*Log[Log[4]])

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fricas [A]  time = 0.66, size = 53, normalized size = 1.51 \begin {gather*} e^{\left (\frac {3 \, {\left (x^{2} - 4 \, x - e^{x} - 1\right )} \log \left (\frac {x + 2 \, \log \left (2 \, \log \relax (2)\right )}{\log \left (2 \, \log \relax (2)\right )}\right ) - x}{3 \, {\left (x^{2} - 4 \, x - e^{x} - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x+2)*exp(x)+2*x^2+2)*log(2*log(2))+3*exp(x)^2+(-7*x^2+25*x+6)*exp(x)+3*x^4-23*x^3+42*x^2+25*x+
3)*exp(((3*exp(x)-3*x^2+12*x+3)*log((2*log(2*log(2))+x)/log(2*log(2)))+x)/(3*exp(x)-3*x^2+12*x+3))/((6*exp(x)^
2+(-12*x^2+48*x+12)*exp(x)+6*x^4-48*x^3+84*x^2+48*x+6)*log(2*log(2))+3*x*exp(x)^2+(-6*x^3+24*x^2+6*x)*exp(x)+3
*x^5-24*x^4+42*x^3+24*x^2+3*x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x+2)*exp(x)+2*x^2+2)*log(2*log(2))+3*exp(x)^2+(-7*x^2+25*x+6)*exp(x)+3*x^4-23*x^3+42*x^2+25*x+
3)*exp(((3*exp(x)-3*x^2+12*x+3)*log((2*log(2*log(2))+x)/log(2*log(2)))+x)/(3*exp(x)-3*x^2+12*x+3))/((6*exp(x)^
2+(-12*x^2+48*x+12)*exp(x)+6*x^4-48*x^3+84*x^2+48*x+6)*log(2*log(2))+3*x*exp(x)^2+(-6*x^3+24*x^2+6*x)*exp(x)+3
*x^5-24*x^4+42*x^3+24*x^2+3*x),x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.17, size = 118, normalized size = 3.37

method	result	size
risch	\({\mathrm e}^{\frac {-3 \ln \left (\frac {2 \ln \relax (2)+2 \ln \left (\ln \relax (2)\right )+x}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}\right ) x^{2}+3 \,{\mathrm e}^{x} \ln \left (\frac {2 \ln \relax (2)+2 \ln \left (\ln \relax (2)\right )+x}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}\right )+12 \ln \left (\frac {2 \ln \relax (2)+2 \ln \left (\ln \relax (2)\right )+x}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}\right ) x +3 \ln \left (\frac {2 \ln \relax (2)+2 \ln \left (\ln \relax (2)\right )+x}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}\right )+x}{3 \,{\mathrm e}^{x}-3 x^{2}+12 x +3}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x+2)*exp(x)+2*x^2+2)*ln(2*ln(2))+3*exp(x)^2+(-7*x^2+25*x+6)*exp(x)+3*x^4-23*x^3+42*x^2+25*x+3)*exp((
(3*exp(x)-3*x^2+12*x+3)*ln((2*ln(2*ln(2))+x)/ln(2*ln(2)))+x)/(3*exp(x)-3*x^2+12*x+3))/((6*exp(x)^2+(-12*x^2+48
*x+12)*exp(x)+6*x^4-48*x^3+84*x^2+48*x+6)*ln(2*ln(2))+3*x*exp(x)^2+(-6*x^3+24*x^2+6*x)*exp(x)+3*x^5-24*x^4+42*
x^3+24*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

exp(1/3*(-3*ln((2*ln(2)+2*ln(ln(2))+x)/(ln(2)+ln(ln(2))))*x^2+3*exp(x)*ln((2*ln(2)+2*ln(ln(2))+x)/(ln(2)+ln(ln
(2))))+12*ln((2*ln(2)+2*ln(ln(2))+x)/(ln(2)+ln(ln(2))))*x+3*ln((2*ln(2)+2*ln(ln(2))+x)/(ln(2)+ln(ln(2))))+x)/(
-x^2+exp(x)+4*x+1))

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maxima [A]  time = 0.96, size = 38, normalized size = 1.09 \begin {gather*} \frac {{\left (x + 2 \, \log \relax (2) + 2 \, \log \left (\log \relax (2)\right )\right )} e^{\left (-\frac {x}{3 \, {\left (x^{2} - 4 \, x - e^{x} - 1\right )}}\right )}}{\log \relax (2) + \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x+2)*exp(x)+2*x^2+2)*log(2*log(2))+3*exp(x)^2+(-7*x^2+25*x+6)*exp(x)+3*x^4-23*x^3+42*x^2+25*x+
3)*exp(((3*exp(x)-3*x^2+12*x+3)*log((2*log(2*log(2))+x)/log(2*log(2)))+x)/(3*exp(x)-3*x^2+12*x+3))/((6*exp(x)^
2+(-12*x^2+48*x+12)*exp(x)+6*x^4-48*x^3+84*x^2+48*x+6)*log(2*log(2))+3*x*exp(x)^2+(-6*x^3+24*x^2+6*x)*exp(x)+3
*x^5-24*x^4+42*x^3+24*x^2+3*x),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x+\ln \left (\frac {x+2\,\ln \left (2\,\ln \relax (2)\right )}{\ln \left (2\,\ln \relax (2)\right )}\right )\,\left (12\,x+3\,{\mathrm {e}}^x-3\,x^2+3\right )}{12\,x+3\,{\mathrm {e}}^x-3\,x^2+3}}\,\left (25\,x+3\,{\mathrm {e}}^{2\,x}+\ln \left (2\,\ln \relax (2)\right )\,\left (2\,x^2-{\mathrm {e}}^x\,\left (2\,x-2\right )+2\right )+{\mathrm {e}}^x\,\left (-7\,x^2+25\,x+6\right )+42\,x^2-23\,x^3+3\,x^4+3\right )}{3\,x+3\,x\,{\mathrm {e}}^{2\,x}+\ln \left (2\,\ln \relax (2)\right )\,\left (48\,x+6\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (-12\,x^2+48\,x+12\right )+84\,x^2-48\,x^3+6\,x^4+6\right )+24\,x^2+42\,x^3-24\,x^4+3\,x^5+{\mathrm {e}}^x\,\left (-6\,x^3+24\,x^2+6\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x + log((x + 2*log(2*log(2)))/log(2*log(2)))*(12*x + 3*exp(x) - 3*x^2 + 3))/(12*x + 3*exp(x) - 3*x^2
+ 3))*(25*x + 3*exp(2*x) + log(2*log(2))*(2*x^2 - exp(x)*(2*x - 2) + 2) + exp(x)*(25*x - 7*x^2 + 6) + 42*x^2
- 23*x^3 + 3*x^4 + 3))/(3*x + 3*x*exp(2*x) + log(2*log(2))*(48*x + 6*exp(2*x) + exp(x)*(48*x - 12*x^2 + 12) +
84*x^2 - 48*x^3 + 6*x^4 + 6) + 24*x^2 + 42*x^3 - 24*x^4 + 3*x^5 + exp(x)*(6*x + 24*x^2 - 6*x^3)),x)

[Out]

int((exp((x + log((x + 2*log(2*log(2)))/log(2*log(2)))*(12*x + 3*exp(x) - 3*x^2 + 3))/(12*x + 3*exp(x) - 3*x^2
+ 3))*(25*x + 3*exp(2*x) + log(2*log(2))*(2*x^2 - exp(x)*(2*x - 2) + 2) + exp(x)*(25*x - 7*x^2 + 6) + 42*x^2
- 23*x^3 + 3*x^4 + 3))/(3*x + 3*x*exp(2*x) + log(2*log(2))*(48*x + 6*exp(2*x) + exp(x)*(48*x - 12*x^2 + 12) +
84*x^2 - 48*x^3 + 6*x^4 + 6) + 24*x^2 + 42*x^3 - 24*x^4 + 3*x^5 + exp(x)*(6*x + 24*x^2 - 6*x^3)), x)

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sympy [A]  time = 3.75, size = 51, normalized size = 1.46 \begin {gather*} e^{\frac {x + \left (- 3 x^{2} + 12 x + 3 e^{x} + 3\right ) \log {\left (\frac {x + 2 \log {\left (2 \log {\relax (2 )} \right )}}{\log {\left (2 \log {\relax (2 )} \right )}} \right )}}{- 3 x^{2} + 12 x + 3 e^{x} + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x+2)*exp(x)+2*x**2+2)*ln(2*ln(2))+3*exp(x)**2+(-7*x**2+25*x+6)*exp(x)+3*x**4-23*x**3+42*x**2+2
5*x+3)*exp(((3*exp(x)-3*x**2+12*x+3)*ln((2*ln(2*ln(2))+x)/ln(2*ln(2)))+x)/(3*exp(x)-3*x**2+12*x+3))/((6*exp(x)
**2+(-12*x**2+48*x+12)*exp(x)+6*x**4-48*x**3+84*x**2+48*x+6)*ln(2*ln(2))+3*x*exp(x)**2+(-6*x**3+24*x**2+6*x)*e
xp(x)+3*x**5-24*x**4+42*x**3+24*x**2+3*x),x)

[Out]

exp((x + (-3*x**2 + 12*x + 3*exp(x) + 3)*log((x + 2*log(2*log(2)))/log(2*log(2))))/(-3*x**2 + 12*x + 3*exp(x)
+ 3))

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