3.66.89 \(\int \frac {e^{e^{e^{2 x^2-4 x \log (\frac {4}{x})+2 \log ^2(\frac {4}{x})}}} (e^{\frac {8}{-3+x}} (-9-2 x-x^2)+e^{e^{2 x^2-4 x \log (\frac {4}{x})+2 \log ^2(\frac {4}{x})}+2 x^2-4 x \log (\frac {4}{x})+2 \log ^2(\frac {4}{x})} (e^{\frac {8}{-3+x}} (36 x+12 x^2-20 x^3+4 x^4)+e^{\frac {8}{-3+x}} (-36-12 x+20 x^2-4 x^3) \log (\frac {4}{x})))}{9 x^2-6 x^3+x^4} \, dx\)

Optimal. Leaf size=32 \[ \frac {e^{e^{e^{2 \left (-x+\log \left (\frac {4}{x}\right )\right )^2}}+\frac {8}{-3+x}}}{x} \]

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Rubi [B]  time = 7.92, antiderivative size = 132, normalized size of antiderivative = 4.12, number of steps used = 3, number of rules used = 3, integrand size = 185, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {1594, 27, 2288} \begin {gather*} \frac {\left (e^{-\frac {8}{3-x}} \left (x^4-5 x^3+3 x^2+9 x\right )-e^{-\frac {8}{3-x}} \left (x^3-5 x^2+3 x+9\right ) \log \left (\frac {4}{x}\right )\right ) \exp \left (\exp \left (4^{-4 x} \left (\frac {1}{x}\right )^{-4 x} e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )}\right )\right )}{(3-x)^2 x^2 \left (x-\log \left (\frac {4}{x}\right )-\frac {\log \left (\frac {4}{x}\right )}{x}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(-9 - 2*x - x^2) + E^(E^(2*x^2 - 4*x*Log[4/x]
+ 2*Log[4/x]^2) + 2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(36*x + 12*x^2 - 20*x^3 + 4*x^4) + E^(
8/(-3 + x))*(-36 - 12*x + 20*x^2 - 4*x^3)*Log[4/x])))/(9*x^2 - 6*x^3 + x^4),x]

[Out]

(E^E^(E^(2*x^2 + 2*Log[4/x]^2)/(4^(4*x)*(x^(-1))^(4*x)))*((9*x + 3*x^2 - 5*x^3 + x^4)/E^(8/(3 - x)) - ((9 + 3*
x - 5*x^2 + x^3)*Log[4/x])/E^(8/(3 - x))))/((3 - x)^2*x^2*(1 + x - Log[4/x] - Log[4/x]/x))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+\exp \left (e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )\right ) \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{x^2 \left (9-6 x+x^2\right )} \, dx\\ &=\int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+\exp \left (e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )\right ) \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{(-3+x)^2 x^2} \, dx\\ &=\frac {\exp \left (\exp \left (4^{-4 x} e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )} \left (\frac {1}{x}\right )^{-4 x}\right )\right ) \left (e^{-\frac {8}{3-x}} \left (9 x+3 x^2-5 x^3+x^4\right )-e^{-\frac {8}{3-x}} \left (9+3 x-5 x^2+x^3\right ) \log \left (\frac {4}{x}\right )\right )}{(3-x)^2 x^2 \left (1+x-\log \left (\frac {4}{x}\right )-\frac {\log \left (\frac {4}{x}\right )}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 10.19, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^E^E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(-9 - 2*x - x^2) + E^(E^(2*x^2 - 4*x*Lo
g[4/x] + 2*Log[4/x]^2) + 2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(36*x + 12*x^2 - 20*x^3 + 4*x^4)
+ E^(8/(-3 + x))*(-36 - 12*x + 20*x^2 - 4*x^3)*Log[4/x])))/(9*x^2 - 6*x^3 + x^4),x]

[Out]

Integrate[(E^E^E^(2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(-9 - 2*x - x^2) + E^(E^(2*x^2 - 4*x*Lo
g[4/x] + 2*Log[4/x]^2) + 2*x^2 - 4*x*Log[4/x] + 2*Log[4/x]^2)*(E^(8/(-3 + x))*(36*x + 12*x^2 - 20*x^3 + 4*x^4)
+ E^(8/(-3 + x))*(-36 - 12*x + 20*x^2 - 4*x^3)*Log[4/x])))/(9*x^2 - 6*x^3 + x^4), x]

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fricas [A]  time = 0.58, size = 40, normalized size = 1.25 \begin {gather*} \frac {e^{\left (\frac {8}{x - 3} + e^{\left (e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+20*x^2-12*x-36)*exp(4/(x-3))^2*log(4/x)+(4*x^4-20*x^3+12*x^2+36*x)*exp(4/(x-3))^2)*exp(log
(4/x)^2-2*x*log(4/x)+x^2)^2*exp(exp(log(4/x)^2-2*x*log(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(x-3))^2)*exp(exp(exp(l
og(4/x)^2-2*x*log(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (4 \, {\left ({\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )} e^{\left (\frac {8}{x - 3}\right )} \log \left (\frac {4}{x}\right ) - {\left (x^{4} - 5 \, x^{3} + 3 \, x^{2} + 9 \, x\right )} e^{\left (\frac {8}{x - 3}\right )}\right )} e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2} + e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )} + {\left (x^{2} + 2 \, x + 9\right )} e^{\left (\frac {8}{x - 3}\right )}\right )} e^{\left (e^{\left (e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )}\right )}}{x^{4} - 6 \, x^{3} + 9 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+20*x^2-12*x-36)*exp(4/(x-3))^2*log(4/x)+(4*x^4-20*x^3+12*x^2+36*x)*exp(4/(x-3))^2)*exp(log
(4/x)^2-2*x*log(4/x)+x^2)^2*exp(exp(log(4/x)^2-2*x*log(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(x-3))^2)*exp(exp(exp(l
og(4/x)^2-2*x*log(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x, algorithm="giac")

[Out]

integrate(-(4*((x^3 - 5*x^2 + 3*x + 9)*e^(8/(x - 3))*log(4/x) - (x^4 - 5*x^3 + 3*x^2 + 9*x)*e^(8/(x - 3)))*e^(
2*x^2 - 4*x*log(4/x) + 2*log(4/x)^2 + e^(2*x^2 - 4*x*log(4/x) + 2*log(4/x)^2)) + (x^2 + 2*x + 9)*e^(8/(x - 3))
)*e^(e^(e^(2*x^2 - 4*x*log(4/x) + 2*log(4/x)^2)))/(x^4 - 6*x^3 + 9*x^2), x)

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (-4 x^{3}+20 x^{2}-12 x -36\right ) {\mathrm e}^{\frac {8}{x -3}} \ln \left (\frac {4}{x}\right )+\left (4 x^{4}-20 x^{3}+12 x^{2}+36 x \right ) {\mathrm e}^{\frac {8}{x -3}}\right ) {\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}} {\mathrm e}^{{\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}}}+\left (-x^{2}-2 x -9\right ) {\mathrm e}^{\frac {8}{x -3}}\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}}}}}{x^{4}-6 x^{3}+9 x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^3+20*x^2-12*x-36)*exp(4/(x-3))^2*ln(4/x)+(4*x^4-20*x^3+12*x^2+36*x)*exp(4/(x-3))^2)*exp(ln(4/x)^2-
2*x*ln(4/x)+x^2)^2*exp(exp(ln(4/x)^2-2*x*ln(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(x-3))^2)*exp(exp(exp(ln(4/x)^2-2*
x*ln(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x)

[Out]

int((((-4*x^3+20*x^2-12*x-36)*exp(4/(x-3))^2*ln(4/x)+(4*x^4-20*x^3+12*x^2+36*x)*exp(4/(x-3))^2)*exp(ln(4/x)^2-
2*x*ln(4/x)+x^2)^2*exp(exp(ln(4/x)^2-2*x*ln(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(x-3))^2)*exp(exp(exp(ln(4/x)^2-2*
x*ln(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x)

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maxima [A]  time = 0.93, size = 49, normalized size = 1.53 \begin {gather*} \frac {e^{\left (\frac {8}{x - 3} + e^{\left (e^{\left (2 \, x^{2} - 8 \, x \log \relax (2) + 8 \, \log \relax (2)^{2} + 4 \, x \log \relax (x) - 8 \, \log \relax (2) \log \relax (x) + 2 \, \log \relax (x)^{2}\right )}\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+20*x^2-12*x-36)*exp(4/(x-3))^2*log(4/x)+(4*x^4-20*x^3+12*x^2+36*x)*exp(4/(x-3))^2)*exp(log
(4/x)^2-2*x*log(4/x)+x^2)^2*exp(exp(log(4/x)^2-2*x*log(4/x)+x^2)^2)+(-x^2-2*x-9)*exp(4/(x-3))^2)*exp(exp(exp(l
og(4/x)^2-2*x*log(4/x)+x^2)^2))/(x^4-6*x^3+9*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.61, size = 52, normalized size = 1.62 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{{\left (\frac {1}{256}\right )}^x\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x}\right )}^2}\,{\mathrm {e}}^{8\,{\ln \relax (2)}^2}\,{\mathrm {e}}^{2\,x^2}\,{\left (\frac {1}{x}\right )}^{8\,\ln \relax (2)-4\,x}}}\,{\mathrm {e}}^{\frac {8}{x-3}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(exp(2*log(4/x)^2 - 4*x*log(4/x) + 2*x^2)))*(exp(8/(x - 3))*(2*x + x^2 + 9) - exp(exp(2*log(4/x)^
2 - 4*x*log(4/x) + 2*x^2))*exp(2*log(4/x)^2 - 4*x*log(4/x) + 2*x^2)*(exp(8/(x - 3))*(36*x + 12*x^2 - 20*x^3 +
4*x^4) - exp(8/(x - 3))*log(4/x)*(12*x - 20*x^2 + 4*x^3 + 36))))/(9*x^2 - 6*x^3 + x^4),x)

[Out]

(exp(exp((1/256)^x*exp(2*log(1/x)^2)*exp(8*log(2)^2)*exp(2*x^2)*(1/x)^(8*log(2) - 4*x)))*exp(8/(x - 3)))/x

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sympy [A]  time = 87.09, size = 34, normalized size = 1.06 \begin {gather*} \frac {e^{\frac {8}{x - 3}} e^{e^{e^{2 x^{2} - 4 x \log {\left (\frac {4}{x} \right )} + 2 \log {\left (\frac {4}{x} \right )}^{2}}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**3+20*x**2-12*x-36)*exp(4/(x-3))**2*ln(4/x)+(4*x**4-20*x**3+12*x**2+36*x)*exp(4/(x-3))**2)*e
xp(ln(4/x)**2-2*x*ln(4/x)+x**2)**2*exp(exp(ln(4/x)**2-2*x*ln(4/x)+x**2)**2)+(-x**2-2*x-9)*exp(4/(x-3))**2)*exp
(exp(exp(ln(4/x)**2-2*x*ln(4/x)+x**2)**2))/(x**4-6*x**3+9*x**2),x)

[Out]

exp(8/(x - 3))*exp(exp(exp(2*x**2 - 4*x*log(4/x) + 2*log(4/x)**2)))/x

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