∫ e1 _2 (exx3 log(2)+x4 log(2))(−4x3 log(2)+ex(−3x2−x3) log(2))(iπ+log(− log(log(2))))2 ___________________________________________________________________________________________________________________________ 18−12e1 _2 (exx3 log(2)+x4 log(2))+2eexx3 log(2)+x4 log(2) dx [3107]

Optimal. Leaf size=33 \[ \frac {(i \pi +\log (-\log (\log (2))))^2}{-3+2^{\frac {1}{2} x^3 \left (e^x+x\right )}} \]

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Rubi [A]
time = 0.95, antiderivative size = 36, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, integrand size = 110, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {12, 6820, 6818} \begin {gather*} -\frac {(\log (-\log (\log (2)))+i \pi )^2}{3-2^{\frac {1}{2} x^3 \left (x+e^x\right )}} \end {gather*}

Int[(E^((E^x*x^3*Log[2] + x^4*Log[2])/2)*(-4*x^3*Log[2] + E^x*(-3*x^2 - x^3)*Log[2])*(I*Pi + Log[-Log[Log[2]]]

)^2)/(18 - 12*E^((E^x*x^3*Log[2] + x^4*Log[2])/2) + 2*E^(E^x*x^3*Log[2] + x^4*Log[2])),x]

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /; !F

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

\begin {gather*} \begin {aligned} \text {integral} &=(i \pi +\log (-\log (\log (2))))^2 \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right )}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx\\ &=(i \pi +\log (-\log (\log (2))))^2 \int \frac {2^{\frac {1}{2} \left (-2+e^x x^3+x^4\right )} x^2 \left (-4 x-e^x (3+x)\right ) \log (2)}{\left (3-2^{\frac {1}{2} x^3 \left (e^x+x\right )}\right )^2} \, dx\\ &=\left (\log (2) (i \pi +\log (-\log (\log (2))))^2\right ) \int \frac {2^{\frac {1}{2} \left (-2+e^x x^3+x^4\right )} x^2 \left (-4 x-e^x (3+x)\right )}{\left (3-2^{\frac {1}{2} x^3 \left (e^x+x\right )}\right )^2} \, dx\\ &=-\frac {(i \pi +\log (-\log (\log (2))))^2}{3-2^{\frac {1}{2} x^3 \left (e^x+x\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.14, size = 34, normalized size = 1.03 \begin {gather*} -\frac {(\pi -i \log (-\log (\log (2))))^2}{-3+2^{\frac {1}{2} x^3 \left (e^x+x\right )}} \end {gather*}

Integrate[(E^((E^x*x^3*Log[2] + x^4*Log[2])/2)*(-4*x^3*Log[2] + E^x*(-3*x^2 - x^3)*Log[2])*(I*Pi + Log[-Log[Lo

g[2]]])^2)/(18 - 12*E^((E^x*x^3*Log[2] + x^4*Log[2])/2) + 2*E^(E^x*x^3*Log[2] + x^4*Log[2])),x]

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method	result	size

risch	\(\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{2^{\frac {{\mathrm e}^{x} x^{3}}{2}} 2^{\frac {x^{4}}{2}}-3}\)	\(29\)
norman	\(\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{{\mathrm e}^{\frac {x^{3} \ln \left (2\right ) {\mathrm e}^{x}}{2}+\frac {x^{4} \ln \left (2\right )}{2}}-3}\)	\(30\)

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

*x^3*ln(2)*exp(x)+1/2*x^4*ln(2))^2-12*exp(1/2*x^3*ln(2)*exp(x)+1/2*x^4*ln(2))+18),x,method=_RETURNVERBOSE)

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Maxima [A]
time = 0.55, size = 29, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \end {gather*}

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

^2/(2*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))^2-12*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))+18),x, algorith

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Fricas [A]
time = 0.39, size = 29, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \end {gather*}

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

^2/(2*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))^2-12*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))+18),x, algorith

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Sympy [A]
time = 0.35, size = 51, normalized size = 1.55 \begin {gather*} \frac {- \pi ^{2} + \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}^{2} + 2 i \pi \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}}{e^{\frac {x^{4} \log {\left (2 \right )}}{2}} e^{\frac {x^{3} e^{x} \log {\left (2 \right )}}{2}} - 3} \end {gather*}

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

2/(2*exp(1/2*x**3*ln(2)*exp(x)+1/2*x**4*ln(2))**2-12*exp(1/2*x**3*ln(2)*exp(x)+1/2*x**4*ln(2))+18),x)

(-pi**2 + log(-log(log(2)))**2 + 2*I*pi*log(-log(log(2))))/(exp(x**4*log(2)/2)*exp(x**3*exp(x)*log(2)/2) - 3)

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Giac [A]
time = 0.40, size = 29, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \end {gather*}

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

^2/(2*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))^2-12*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))+18),x, algorith

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\ln \left (\ln \left (\ln \left (2\right )\right )\right )}^2\,{\mathrm {e}}^{\frac {x^4\,\ln \left (2\right )}{2}+\frac {x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}{2}}\,\left (4\,x^3\,\ln \left (2\right )+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x^3+3\,x^2\right )\right )}{2\,{\mathrm {e}}^{x^4\,\ln \left (2\right )+x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}-12\,{\mathrm {e}}^{\frac {x^4\,\ln \left (2\right )}{2}+\frac {x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}{2}}+18} \,d x \end {gather*}

int(-(log(log(log(2)))^2*exp((x^4*log(2))/2 + (x^3*exp(x)*log(2))/2)*(4*x^3*log(2) + exp(x)*log(2)*(3*x^2 + x^

3)))/(2*exp(x^4*log(2) + x^3*exp(x)*log(2)) - 12*exp((x^4*log(2))/2 + (x^3*exp(x)*log(2))/2) + 18),x)

int(-(log(log(log(2)))^2*exp((x^4*log(2))/2 + (x^3*exp(x)*log(2))/2)*(4*x^3*log(2) + exp(x)*log(2)*(3*x^2 + x^

3)))/(2*exp(x^4*log(2) + x^3*exp(x)*log(2)) - 12*exp((x^4*log(2))/2 + (x^3*exp(x)*log(2))/2) + 18), x)

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3.32.7 \(\int \frac {e^{\frac {1}{2} (e^x x^3 \log (2)+x^4 \log (2))} (-4 x^3 \log (2)+e^x (-3 x^2-x^3) \log (2)) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} (e^x x^3 \log (2)+x^4 \log (2))}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx\) [3107]