\(\int \frac {e^{e^{-4+4 x^2}} (-8+8 x-2 x^2+e^{-4+4 x^2} (64 x^2-64 x^3+16 x^4))+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx\) [848]

Optimal result

Integrand size = 74, antiderivative size = 26 \[ \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx=\frac {e^{e^{-4+4 x^2}}}{x}+\frac {\log (3)}{4-2 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1608, 27, 12, 6820, 2326} \[ \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx=\frac {e^{e^{4 x^2-4}}}{x}+\frac {\log (3)}{2 (2-x)} \]

[In]

Int[(E^E^(-4 + 4*x^2)*(-8 + 8*x - 2*x^2 + E^(-4 + 4*x^2)*(64*x^2 - 64*x^3 + 16*x^4)) + x^2*Log[3])/(8*x^2 - 8*
x^3 + 2*x^4),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 1608

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 2326

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]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{x^2 \left (8-8 x+2 x^2\right )} \, dx \\ & = \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{2 (-2+x)^2 x^2} \, dx \\ & = \frac {1}{2} \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{(-2+x)^2 x^2} \, dx \\ & = \frac {1}{2} \int \left (-\frac {2 e^{-4+e^{-4+4 x^2}} \left (e^4-8 e^{4 x^2} x^2\right )}{x^2}+\frac {\log (3)}{(-2+x)^2}\right ) \, dx \\ & = \frac {\log (3)}{2 (2-x)}-\int \frac {e^{-4+e^{-4+4 x^2}} \left (e^4-8 e^{4 x^2} x^2\right )}{x^2} \, dx \\ & = \frac {e^{e^{-4+4 x^2}}}{x}+\frac {\log (3)}{2 (2-x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx=\frac {\frac {2 e^{4+e^{-4+4 x^2}}}{x}+\frac {e^4 \log (3)}{2-x}}{2 e^4} \]

[In]

Integrate[(E^E^(-4 + 4*x^2)*(-8 + 8*x - 2*x^2 + E^(-4 + 4*x^2)*(64*x^2 - 64*x^3 + 16*x^4)) + x^2*Log[3])/(8*x^
2 - 8*x^3 + 2*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {\ln \left (3\right )}{2 \left (-2+x \right )}+\frac {{\mathrm e}^{{\mathrm e}^{4 \left (-1+x \right ) \left (1+x \right )}}}{x}\)	\(25\)
parallelrisch	\(-\frac {x \ln \left (3\right )-2 \,{\mathrm e}^{{\mathrm e}^{4 x^{2}-4}} x +4 \,{\mathrm e}^{{\mathrm e}^{4 x^{2}-4}}}{2 x \left (-2+x \right )}\)	\(39\)

[In]

int((((16*x^4-64*x^3+64*x^2)*exp(x^2-1)^4-2*x^2+8*x-8)*exp(exp(x^2-1)^4)+x^2*ln(3))/(2*x^4-8*x^3+8*x^2),x,meth
od=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx=\frac {2 \, {\left (x - 2\right )} e^{\left (e^{\left (4 \, x^{2} - 4\right )}\right )} - x \log \left (3\right )}{2 \, {\left (x^{2} - 2 \, x\right )}} \]

[In]

integrate((((16*x^4-64*x^3+64*x^2)*exp(x^2-1)^4-2*x^2+8*x-8)*exp(exp(x^2-1)^4)+x^2*log(3))/(2*x^4-8*x^3+8*x^2)
,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx=- \frac {\log {\left (3 \right )}}{2 x - 4} + \frac {e^{e^{4 x^{2} - 4}}}{x} \]

[In]

integrate((((16*x**4-64*x**3+64*x**2)*exp(x**2-1)**4-2*x**2+8*x-8)*exp(exp(x**2-1)**4)+x**2*ln(3))/(2*x**4-8*x
**3+8*x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx=\frac {e^{\left (e^{\left (4 \, x^{2} - 4\right )}\right )}}{x} - \frac {\log \left (3\right )}{2 \, {\left (x - 2\right )}} \]

[In]

integrate((((16*x^4-64*x^3+64*x^2)*exp(x^2-1)^4-2*x^2+8*x-8)*exp(exp(x^2-1)^4)+x^2*log(3))/(2*x^4-8*x^3+8*x^2)
,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx=\int { \frac {x^{2} \log \left (3\right ) - 2 \, {\left (x^{2} - 8 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )} e^{\left (4 \, x^{2} - 4\right )} - 4 \, x + 4\right )} e^{\left (e^{\left (4 \, x^{2} - 4\right )}\right )}}{2 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )}} \,d x } \]

[In]

integrate((((16*x^4-64*x^3+64*x^2)*exp(x^2-1)^4-2*x^2+8*x-8)*exp(exp(x^2-1)^4)+x^2*log(3))/(2*x^4-8*x^3+8*x^2)
,x, algorithm="giac")

[Out]

integrate(1/2*(x^2*log(3) - 2*(x^2 - 8*(x^4 - 4*x^3 + 4*x^2)*e^(4*x^2 - 4) - 4*x + 4)*e^(e^(4*x^2 - 4)))/(x^4
- 4*x^3 + 4*x^2), x)

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{-4+4 x^2}} \left (-8+8 x-2 x^2+e^{-4+4 x^2} \left (64 x^2-64 x^3+16 x^4\right )\right )+x^2 \log (3)}{8 x^2-8 x^3+2 x^4} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{-4}\,{\mathrm {e}}^{4\,x^2}}}{x}-\frac {\ln \left (3\right )}{2\,x-4} \]

[In]

int((exp(exp(4*x^2 - 4))*(8*x + exp(4*x^2 - 4)*(64*x^2 - 64*x^3 + 16*x^4) - 2*x^2 - 8) + x^2*log(3))/(8*x^2 -
8*x^3 + 2*x^4),x)

[Out]

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