3.45.89 \(\int \frac {e^{e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}}} (3 x^2+9 x^4+e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}} (-18 x^2+60 x^4-18 x^6+(54-162 x^2) \log (x)+(-54+162 x^2) \log ^2(x)))}{x^2-6 x^4+9 x^6} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^{e^{\left (-x+\frac {3 \log (x)}{x}\right )^2}}}{\frac {1}{3 x}-x} \]

________________________________________________________________________________________

Rubi [B]  time = 1.01, antiderivative size = 125, normalized size of antiderivative = 4.03, number of steps used = 3, number of rules used = 3, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {1594, 28, 2288} \begin {gather*} \frac {3 e^{\frac {e^{\frac {x^4+9 \log ^2(x)}{x^2}}}{x^6}} \left (3 x^6-10 x^4+3 x^2+9 \left (1-3 x^2\right ) \log ^2(x)-9 \left (1-3 x^2\right ) \log (x)\right )}{x^2 \left (1-3 x^2\right )^2 \left (\frac {-2 x^3+3 x+6 x \log (x)-\frac {9 \log (x)}{x}}{x^2}+\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^((x^4 - 6*x^2*Log[x] + 9*Log[x]^2)/x^2)*(3*x^2 + 9*x^4 + E^((x^4 - 6*x^2*Log[x] + 9*Log[x]^2)/x^2)*(-
18*x^2 + 60*x^4 - 18*x^6 + (54 - 162*x^2)*Log[x] + (-54 + 162*x^2)*Log[x]^2)))/(x^2 - 6*x^4 + 9*x^6),x]

[Out]

(3*E^(E^((x^4 + 9*Log[x]^2)/x^2)/x^6)*(3*x^2 - 10*x^4 + 3*x^6 - 9*(1 - 3*x^2)*Log[x] + 9*(1 - 3*x^2)*Log[x]^2)
)/(x^2*(1 - 3*x^2)^2*((3*x - 2*x^3 - (9*Log[x])/x + 6*x*Log[x])/x^2 + (x^4 - 6*x^2*Log[x] + 9*Log[x]^2)/x^3))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && 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^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}}} \left (3 x^2+9 x^4+e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}} \left (-18 x^2+60 x^4-18 x^6+\left (54-162 x^2\right ) \log (x)+\left (-54+162 x^2\right ) \log ^2(x)\right )\right )}{x^2 \left (1-6 x^2+9 x^4\right )} \, dx\\ &=9 \int \frac {e^{e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}}} \left (3 x^2+9 x^4+e^{\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^2}} \left (-18 x^2+60 x^4-18 x^6+\left (54-162 x^2\right ) \log (x)+\left (-54+162 x^2\right ) \log ^2(x)\right )\right )}{x^2 \left (-3+9 x^2\right )^2} \, dx\\ &=\frac {3 e^{\frac {e^{\frac {x^4+9 \log ^2(x)}{x^2}}}{x^6}} \left (3 x^2-10 x^4+3 x^6-9 \left (1-3 x^2\right ) \log (x)+9 \left (1-3 x^2\right ) \log ^2(x)\right )}{x^2 \left (1-3 x^2\right )^2 \left (\frac {3 x-2 x^3-\frac {9 \log (x)}{x}+6 x \log (x)}{x^2}+\frac {x^4-6 x^2 \log (x)+9 \log ^2(x)}{x^3}\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 33, normalized size = 1.06 \begin {gather*} -\frac {3 e^{\frac {e^{x^2+\frac {9 \log ^2(x)}{x^2}}}{x^6}} x}{-1+3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^((x^4 - 6*x^2*Log[x] + 9*Log[x]^2)/x^2)*(3*x^2 + 9*x^4 + E^((x^4 - 6*x^2*Log[x] + 9*Log[x]^2)/x
^2)*(-18*x^2 + 60*x^4 - 18*x^6 + (54 - 162*x^2)*Log[x] + (-54 + 162*x^2)*Log[x]^2)))/(x^2 - 6*x^4 + 9*x^6),x]

[Out]

(-3*E^(E^(x^2 + (9*Log[x]^2)/x^2)/x^6)*x)/(-1 + 3*x^2)

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 35, normalized size = 1.13 \begin {gather*} -\frac {3 \, x e^{\left (e^{\left (\frac {x^{4} - 6 \, x^{2} \log \relax (x) + 9 \, \log \relax (x)^{2}}{x^{2}}\right )}\right )}}{3 \, x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((162*x^2-54)*log(x)^2+(-162*x^2+54)*log(x)-18*x^6+60*x^4-18*x^2)*exp((9*log(x)^2-6*x^2*log(x)+x^4)
/x^2)+9*x^4+3*x^2)*exp(exp((9*log(x)^2-6*x^2*log(x)+x^4)/x^2))/(9*x^6-6*x^4+x^2),x, algorithm="fricas")

[Out]

-3*x*e^(e^((x^4 - 6*x^2*log(x) + 9*log(x)^2)/x^2))/(3*x^2 - 1)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 \, {\left (3 \, x^{4} + x^{2} - 2 \, {\left (3 \, x^{6} - 10 \, x^{4} - 9 \, {\left (3 \, x^{2} - 1\right )} \log \relax (x)^{2} + 3 \, x^{2} + 9 \, {\left (3 \, x^{2} - 1\right )} \log \relax (x)\right )} e^{\left (\frac {x^{4} - 6 \, x^{2} \log \relax (x) + 9 \, \log \relax (x)^{2}}{x^{2}}\right )}\right )} e^{\left (e^{\left (\frac {x^{4} - 6 \, x^{2} \log \relax (x) + 9 \, \log \relax (x)^{2}}{x^{2}}\right )}\right )}}{9 \, x^{6} - 6 \, x^{4} + x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((162*x^2-54)*log(x)^2+(-162*x^2+54)*log(x)-18*x^6+60*x^4-18*x^2)*exp((9*log(x)^2-6*x^2*log(x)+x^4)
/x^2)+9*x^4+3*x^2)*exp(exp((9*log(x)^2-6*x^2*log(x)+x^4)/x^2))/(9*x^6-6*x^4+x^2),x, algorithm="giac")

[Out]

integrate(3*(3*x^4 + x^2 - 2*(3*x^6 - 10*x^4 - 9*(3*x^2 - 1)*log(x)^2 + 3*x^2 + 9*(3*x^2 - 1)*log(x))*e^((x^4
- 6*x^2*log(x) + 9*log(x)^2)/x^2))*e^(e^((x^4 - 6*x^2*log(x) + 9*log(x)^2)/x^2))/(9*x^6 - 6*x^4 + x^2), x)

________________________________________________________________________________________

maple [A]  time = 0.11, size = 31, normalized size = 1.00

method	result	size
risch	\(-\frac {3 x \,{\mathrm e}^{{\mathrm e}^{\frac {\left (-x^{2}+3 \ln \relax (x )\right )^{2}}{x^{2}}}}}{3 x^{2}-1}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((162*x^2-54)*ln(x)^2+(-162*x^2+54)*ln(x)-18*x^6+60*x^4-18*x^2)*exp((9*ln(x)^2-6*x^2*ln(x)+x^4)/x^2)+9*x^
4+3*x^2)*exp(exp((9*ln(x)^2-6*x^2*ln(x)+x^4)/x^2))/(9*x^6-6*x^4+x^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 31, normalized size = 1.00 \begin {gather*} -\frac {3 \, x e^{\left (\frac {e^{\left (x^{2} + \frac {9 \, \log \relax (x)^{2}}{x^{2}}\right )}}{x^{6}}\right )}}{3 \, x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((162*x^2-54)*log(x)^2+(-162*x^2+54)*log(x)-18*x^6+60*x^4-18*x^2)*exp((9*log(x)^2-6*x^2*log(x)+x^4)
/x^2)+9*x^4+3*x^2)*exp(exp((9*log(x)^2-6*x^2*log(x)+x^4)/x^2))/(9*x^6-6*x^4+x^2),x, algorithm="maxima")

[Out]

-3*x*e^(e^(x^2 + 9*log(x)^2/x^2)/x^6)/(3*x^2 - 1)

________________________________________________________________________________________

mupad [B]  time = 3.44, size = 29, normalized size = 0.94 \begin {gather*} -\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\frac {9\,{\ln \relax (x)}^2}{x^2}}}{x^6}}}{x^2-\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp((9*log(x)^2 - 6*x^2*log(x) + x^4)/x^2))*(3*x^2 - exp((9*log(x)^2 - 6*x^2*log(x) + x^4)/x^2)*(18*x
^2 - log(x)^2*(162*x^2 - 54) - 60*x^4 + 18*x^6 + log(x)*(162*x^2 - 54)) + 9*x^4))/(x^2 - 6*x^4 + 9*x^6),x)

[Out]

-(x*exp((exp(x^2)*exp((9*log(x)^2)/x^2))/x^6))/(x^2 - 1/3)

________________________________________________________________________________________

sympy [A]  time = 0.76, size = 36, normalized size = 1.16 \begin {gather*} - \frac {3 x e^{e^{\frac {x^{4} - 6 x^{2} \log {\relax (x )} + 9 \log {\relax (x )}^{2}}{x^{2}}}}}{3 x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((162*x**2-54)*ln(x)**2+(-162*x**2+54)*ln(x)-18*x**6+60*x**4-18*x**2)*exp((9*ln(x)**2-6*x**2*ln(x)+
x**4)/x**2)+9*x**4+3*x**2)*exp(exp((9*ln(x)**2-6*x**2*ln(x)+x**4)/x**2))/(9*x**6-6*x**4+x**2),x)

[Out]

-3*x*exp(exp((x**4 - 6*x**2*log(x) + 9*log(x)**2)/x**2))/(3*x**2 - 1)

________________________________________________________________________________________