\(\int \frac {e^{\frac {x^2+\log (\frac {1}{4} (-6+4 e^{4+x}+x-4 x^2))}{x}} (x-14 x^2+x^3-4 x^4+e^{4+x} (4 x+4 x^2)+(6-4 e^{4+x}-x+4 x^2) \log (\frac {1}{4} (-6+4 e^{4+x}+x-4 x^2)))}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx\) [9777]

Optimal result

Integrand size = 125, antiderivative size = 33 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{x+\frac {\log \left (e^{4+x}+\frac {1}{4} \left (x-4 x \left (\frac {3}{2 x}+x\right )\right )\right )}{x}} \]

[Out]

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

Rubi [A] (verified)

Time = 4.36 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {6838} \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=4^{-1/x} e^x \left (-4 x^2+x+4 e^{x+4}-6\right )^{\frac {1}{x}} \]

[In]

Int[(E^((x^2 + Log[(-6 + 4*E^(4 + x) + x - 4*x^2)/4])/x)*(x - 14*x^2 + x^3 - 4*x^4 + E^(4 + x)*(4*x + 4*x^2) +
(6 - 4*E^(4 + x) - x + 4*x^2)*Log[(-6 + 4*E^(4 + x) + x - 4*x^2)/4]))/(-6*x^2 + 4*E^(4 + x)*x^2 + x^3 - 4*x^4
),x]

[Out]

(E^x*(-6 + 4*E^(4 + x) + x - 4*x^2)^x^(-1))/4^x^(-1)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = 4^{-1/x} e^x \left (-6+4 e^{4+x}+x-4 x^2\right )^{\frac {1}{x}} \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=\int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx \]

[In]

Integrate[(E^((x^2 + Log[(-6 + 4*E^(4 + x) + x - 4*x^2)/4])/x)*(x - 14*x^2 + x^3 - 4*x^4 + E^(4 + x)*(4*x + 4*
x^2) + (6 - 4*E^(4 + x) - x + 4*x^2)*Log[(-6 + 4*E^(4 + x) + x - 4*x^2)/4]))/(-6*x^2 + 4*E^(4 + x)*x^2 + x^3 -
4*x^4),x]

[Out]

Integrate[(E^((x^2 + Log[(-6 + 4*E^(4 + x) + x - 4*x^2)/4])/x)*(x - 14*x^2 + x^3 - 4*x^4 + E^(4 + x)*(4*x + 4*
x^2) + (6 - 4*E^(4 + x) - x + 4*x^2)*Log[(-6 + 4*E^(4 + x) + x - 4*x^2)/4]))/(-6*x^2 + 4*E^(4 + x)*x^2 + x^3 -
4*x^4), x]

Maple [A] (verified)

Time = 5.94 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67

method	result	size
risch	\(\left ({\mathrm e}^{4+x}-x^{2}+\frac {x}{4}-\frac {3}{2}\right )^{\frac {1}{x}} {\mathrm e}^{x}\)	\(22\)
parallelrisch	\({\mathrm e}^{\frac {\ln \left ({\mathrm e}^{4+x}-x^{2}+\frac {x}{4}-\frac {3}{2}\right )+x^{2}}{x}}\)	\(25\)

[In]

int(((-4*exp(4+x)+4*x^2-x+6)*ln(exp(4+x)-x^2+1/4*x-3/2)+(4*x^2+4*x)*exp(4+x)-4*x^4+x^3-14*x^2+x)*exp((ln(exp(4
+x)-x^2+1/4*x-3/2)+x^2)/x)/(4*x^2*exp(4+x)-4*x^4+x^3-6*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (\frac {x^{2} + \log \left (-x^{2} + \frac {1}{4} \, x + e^{\left (x + 4\right )} - \frac {3}{2}\right )}{x}\right )} \]

[In]

integrate(((-4*exp(4+x)+4*x^2-x+6)*log(exp(4+x)-x^2+1/4*x-3/2)+(4*x^2+4*x)*exp(4+x)-4*x^4+x^3-14*x^2+x)*exp((l
og(exp(4+x)-x^2+1/4*x-3/2)+x^2)/x)/(4*x^2*exp(4+x)-4*x^4+x^3-6*x^2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=\text {Timed out} \]

[In]

integrate(((-4*exp(4+x)+4*x**2-x+6)*ln(exp(4+x)-x**2+1/4*x-3/2)+(4*x**2+4*x)*exp(4+x)-4*x**4+x**3-14*x**2+x)*e
xp((ln(exp(4+x)-x**2+1/4*x-3/2)+x**2)/x)/(4*x**2*exp(4+x)-4*x**4+x**3-6*x**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (x - \frac {2 \, \log \left (2\right )}{x} + \frac {\log \left (-4 \, x^{2} + x + 4 \, e^{\left (x + 4\right )} - 6\right )}{x}\right )} \]

[In]

integrate(((-4*exp(4+x)+4*x^2-x+6)*log(exp(4+x)-x^2+1/4*x-3/2)+(4*x^2+4*x)*exp(4+x)-4*x^4+x^3-14*x^2+x)*exp((l
og(exp(4+x)-x^2+1/4*x-3/2)+x^2)/x)/(4*x^2*exp(4+x)-4*x^4+x^3-6*x^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (x + \frac {\log \left (-x^{2} + \frac {1}{4} \, x + e^{\left (x + 4\right )} - \frac {3}{2}\right )}{x}\right )} \]

[In]

integrate(((-4*exp(4+x)+4*x^2-x+6)*log(exp(4+x)-x^2+1/4*x-3/2)+(4*x^2+4*x)*exp(4+x)-4*x^4+x^3-14*x^2+x)*exp((l
og(exp(4+x)-x^2+1/4*x-3/2)+x^2)/x)/(4*x^2*exp(4+x)-4*x^4+x^3-6*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.66 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx={\mathrm {e}}^x\,{\left (\frac {x}{4}+{\mathrm {e}}^{x+4}-x^2-\frac {3}{2}\right )}^{1/x} \]

[In]

int((exp((log(x/4 + exp(x + 4) - x^2 - 3/2) + x^2)/x)*(x + exp(x + 4)*(4*x + 4*x^2) - log(x/4 + exp(x + 4) - x
^2 - 3/2)*(x + 4*exp(x + 4) - 4*x^2 - 6) - 14*x^2 + x^3 - 4*x^4))/(4*x^2*exp(x + 4) - 6*x^2 + x^3 - 4*x^4),x)

[Out]

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