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

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 = 0.88, 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, 6688, 6686} \begin {gather*} -\frac {(\log (-\log (\log (2)))+i \pi )^2}{3-2^{\frac {1}{2} x^3 \left (x+e^x\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

Rule 12

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

Rule 6686

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

Rule 6688

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

Rubi steps

\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 = 0.25, 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*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

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fricas [A]  time = 0.66, size = 29, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \left (\log \relax (2)\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \relax (2) + \frac {1}{2} \, x^{3} e^{x} \log \relax (2)\right )} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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
m="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 29, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \left (\log \relax (2)\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \relax (2) + \frac {1}{2} \, x^{3} e^{x} \log \relax (2)\right )} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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
m="giac")

[Out]

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

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maple [A]  time = 0.19, size = 29, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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

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maxima [A]  time = 1.11, size = 29, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \left (\log \relax (2)\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \relax (2) + \frac {1}{2} \, x^{3} e^{x} \log \relax (2)\right )} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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
m="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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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sympy [A]  time = 0.70, size = 51, normalized size = 1.55 \begin {gather*} \frac {- \pi ^{2} + \log {\left (- \log {\left (\log {\relax (2 )} \right )} \right )}^{2} + 2 i \pi \log {\left (- \log {\left (\log {\relax (2 )} \right )} \right )}}{e^{\frac {x^{4} \log {\relax (2 )}}{2}} e^{\frac {x^{3} e^{x} \log {\relax (2 )}}{2}} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

(-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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