Integrand size = 125, antiderivative size = 33 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{x+\frac {\log \left (e^{4+x}+\frac {1}{4} \left (x-4 x \left (\frac {3}{2 x}+x\right )\right )\right )}{x}} \] Output:
exp(x+ln(-(x+3/2/x)*x+1/4*x+exp(4+x))/x)
\[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=\int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx \] Input:
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)*L og[(-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]
Output:
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)*L og[(-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]
Time = 4.95 (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 = {7257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (-4 x^4+x^3-14 x^2+e^{x+4} \left (4 x^2+4 x\right )+\left (4 x^2-x-4 e^{x+4}+6\right ) \log \left (\frac {1}{4} \left (-4 x^2+x+4 e^{x+4}-6\right )\right )+x\right ) \exp \left (\frac {x^2+\log \left (\frac {1}{4} \left (-4 x^2+x+4 e^{x+4}-6\right )\right )}{x}\right )}{-4 x^4+x^3+4 e^{x+4} x^2-6 x^2} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle 4^{-1/x} e^x \left (-4 x^2+x+4 e^{x+4}-6\right )^{\frac {1}{x}}\) |
Input:
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]
Output:
(E^x*(-6 + 4*E^(4 + x) + x - 4*x^2)^x^(-1))/4^x^(-1)
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 5.54 (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 | \({\mathrm e}^{\frac {\ln \left ({\mathrm e}^{4+x}-x^{2}+\frac {x}{4}-\frac {3}{2}\right )+x^{2}}{x}}\) | \(25\) |
Input:
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)
Output:
(exp(4+x)-x^2+1/4*x-3/2)^(1/x)*exp(x)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (\frac {x^{2} + \log \left (-x^{2} + \frac {1}{4} \, x + e^{\left (x + 4\right )} - \frac {3}{2}\right )}{x}\right )} \] Input:
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((log(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")
Output:
e^((x^2 + log(-x^2 + 1/4*x + e^(x + 4) - 3/2))/x)
Timed out. \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=\text {Timed out} \] Input:
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)*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)
Output:
Timed out
Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (x - \frac {2 \, \log \left (2\right )}{x} + \frac {\log \left (-4 \, x^{2} + x + 4 \, e^{\left (x + 4\right )} - 6\right )}{x}\right )} \] Input:
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((log(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")
Output:
e^(x - 2*log(2)/x + log(-4*x^2 + x + 4*e^(x + 4) - 6)/x)
Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (x + \frac {\log \left (-x^{2} + \frac {1}{4} \, x + e^{\left (x + 4\right )} - \frac {3}{2}\right )}{x}\right )} \] Input:
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((log(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")
Output:
e^(x + log(-x^2 + 1/4*x + e^(x + 4) - 3/2)/x)
Time = 7.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx={\mathrm {e}}^x\,{\left (\frac {x}{4}+{\mathrm {e}}^{x+4}-x^2-\frac {3}{2}\right )}^{1/x} \] Input:
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)
Output:
exp(x)*(x/4 + exp(x + 4) - x^2 - 3/2)^(1/x)
Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\frac {\mathrm {log}\left (e^{x} e^{4}-x^{2}+\frac {x}{4}-\frac {3}{2}\right )+x^{2}}{x}} \] Input:
int(((-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((log(exp(4+x)-x^2+1/4*x-3/2)+x^2)/x)/(4*x^2*ex p(4+x)-4*x^4+x^3-6*x^2),x)
Output:
e**((log((4*e**x*e**4 - 4*x**2 + x - 6)/4) + x**2)/x)