\(\int \frac {e^{-1-x+x \log (2)+x^2 \log (\frac {1}{-4 x+x^2})} (4+3 x-2 x^2+(-4+x) \log (2)+(-8 x+2 x^2) \log (\frac {1}{-4 x+x^2}))}{-16+4 x} \, dx\) [6264]

Optimal result

Integrand size = 69, antiderivative size = 32 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} e^{-1-x+x \log (2)+x^2 \log \left (-\frac {1}{(4-x) x}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=2^{x-2} e^{-x-1} \left (\frac {1}{x^2-4 x}\right )^{x^2} \]

[In]

Int[(E^(-1 - x + x*Log[2] + x^2*Log[(-4*x + x^2)^(-1)])*(4 + 3*x - 2*x^2 + (-4 + x)*Log[2] + (-8*x + 2*x^2)*Lo
g[(-4*x + x^2)^(-1)]))/(-16 + 4*x),x]

[Out]

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

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}& = 2^{-2+x} e^{-1-x} \left (\frac {1}{-4 x+x^2}\right )^{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=2^{-2+x} e^{-1-x} \left (\frac {1}{-4 x+x^2}\right )^{x^2} \]

[In]

Integrate[(E^(-1 - x + x*Log[2] + x^2*Log[(-4*x + x^2)^(-1)])*(4 + 3*x - 2*x^2 + (-4 + x)*Log[2] + (-8*x + 2*x
^2)*Log[(-4*x + x^2)^(-1)]))/(-16 + 4*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {\left (\frac {1}{x^{2}-4 x}\right )^{x^{2}} 2^{x} {\mathrm e}^{-1-x}}{4}\)	\(25\)
default	\(\frac {{\mathrm e}^{x^{2} \ln \left (\frac {1}{x^{2}-4 x}\right )+x \ln \left (2\right )-x -1}}{4}\)	\(27\)
norman	\(\frac {{\mathrm e}^{x^{2} \ln \left (\frac {1}{x^{2}-4 x}\right )+x \ln \left (2\right )-x -1}}{4}\)	\(27\)
parallelrisch	\(\frac {{\mathrm e}^{x^{2} \ln \left (\frac {1}{\left (x -4\right ) x}\right )+x \ln \left (2\right )-x -1}}{4}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (x^{2} \log \left (\frac {1}{x^{2} - 4 \, x}\right ) + x \log \left (2\right ) - x - 1\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {e^{x^{2} \log {\left (\frac {1}{x^{2} - 4 x} \right )} - x + x \log {\left (2 \right )} - 1}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (-x^{2} \log \left (x - 4\right ) - x^{2} \log \left (x\right ) + x \log \left (2\right ) - x - 1\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (-x^{2} \log \left (x^{2} - 4 \, x\right ) + x \log \left (2\right ) - x - 1\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {2^x\,{\mathrm {e}}^{-x-1}\,{\left (-\frac {1}{4\,x-x^2}\right )}^{x^2}}{4} \]

[In]

int((exp(x*log(2) - x + x^2*log(-1/(4*x - x^2)) - 1)*(3*x - log(-1/(4*x - x^2))*(8*x - 2*x^2) + log(2)*(x - 4)
- 2*x^2 + 4))/(4*x - 16),x)

[Out]

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