\(\int \frac {e^{2 e^{x^2}} (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x))}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx\) [7292]

Optimal result

Integrand size = 67, antiderivative size = 18 \[ \int \frac {e^{2 e^{x^2}} \left (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x)\right )}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{2 e^{x^2}}}{(-x+\log (x))^2} \]

[Out]

exp(exp(x^2))^2/(ln(x)-x)^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(18)=36\).

Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.89, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2326} \[ \int \frac {e^{2 e^{x^2}} \left (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x)\right )}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{2 e^{x^2}-x^2} \left (e^{x^2} x^3-e^{x^2} x^2 \log (x)\right )}{x \left (x^4-3 x^3 \log (x)+3 x^2 \log ^2(x)-x \log ^3(x)\right )} \]

[In]

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

[Out]

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{2 e^{x^2}-x^2} \left (e^{x^2} x^3-e^{x^2} x^2 \log (x)\right )}{x \left (x^4-3 x^3 \log (x)+3 x^2 \log ^2(x)-x \log ^3(x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 e^{x^2}} \left (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x)\right )}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{2 e^{x^2}}}{(x-\log (x))^2} \]

[In]

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

[Out]

E^(2*E^x^2)/(x - Log[x])^2

Maple [A] (verified)

Time = 1.61 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{x^{2}}}}{\ln \left (x \right )^{2}-2 x \ln \left (x \right )+x^{2}}\)	\(24\)

[In]

int((4*x^2*exp(x^2)*ln(x)-4*x^3*exp(x^2)+2*x-2)*exp(exp(x^2))^2/(x*ln(x)^3-3*x^2*ln(x)^2+3*x^3*ln(x)-x^4),x,me
thod=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {e^{2 e^{x^2}} \left (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x)\right )}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{\left (2 \, e^{\left (x^{2}\right )}\right )}}{x^{2} - 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}} \]

[In]

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

[Out]

e^(2*e^(x^2))/(x^2 - 2*x*log(x) + log(x)^2)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {e^{2 e^{x^2}} \left (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x)\right )}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{2 e^{x^{2}}}}{x^{2} - 2 x \log {\left (x \right )} + \log {\left (x \right )}^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {e^{2 e^{x^2}} \left (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x)\right )}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{\left (2 \, e^{\left (x^{2}\right )}\right )}}{x^{2} - 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}} \]

[In]

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

[Out]

e^(2*e^(x^2))/(x^2 - 2*x*log(x) + log(x)^2)

Giac [F]

\[ \int \frac {e^{2 e^{x^2}} \left (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x)\right )}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx=\int { \frac {2 \, {\left (2 \, x^{3} e^{\left (x^{2}\right )} - 2 \, x^{2} e^{\left (x^{2}\right )} \log \left (x\right ) - x + 1\right )} e^{\left (2 \, e^{\left (x^{2}\right )}\right )}}{x^{4} - 3 \, x^{3} \log \left (x\right ) + 3 \, x^{2} \log \left (x\right )^{2} - x \log \left (x\right )^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 e^{x^2}} \left (-2+2 x-4 e^{x^2} x^3+4 e^{x^2} x^2 \log (x)\right )}{-x^4+3 x^3 \log (x)-3 x^2 \log ^2(x)+x \log ^3(x)} \, dx=\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}}}{{\left (x-\ln \left (x\right )\right )}^2} \]

[In]

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

[Out]

exp(2*exp(x^2))/(x - log(x))^2