Integrand size = 69, antiderivative size = 32 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} e^{-1-x+x \log (2)+x^2 \log \left (-\frac {1}{(4-x) x}\right )} \] Output:
1/4*exp(x*ln(2)+ln(-1/x/(4-x))*x^2-x-1)
Time = 5.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=2^{-2+x} e^{-1-x} \left (\frac {1}{-4 x+x^2}\right )^{x^2} \] Input:
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]
Output:
2^(-2 + x)*E^(-1 - x)*((-4*x + x^2)^(-1))^x^2
Time = 0.53 (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 = {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.
\(\displaystyle \int \frac {e^{x^2 \log \left (\frac {1}{x^2-4 x}\right )-x+x \log (2)-1} \left (-2 x^2+\left (2 x^2-8 x\right ) \log \left (\frac {1}{x^2-4 x}\right )+3 x+(x-4) \log (2)+4\right )}{4 x-16} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle 2^{x-2} e^{-x-1} \left (\frac {1}{x^2-4 x}\right )^{x^2}\) |
Input:
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)*Log[(-4*x + x^2)^(-1)]))/(-16 + 4*x),x]
Output:
2^(-2 + x)*E^(-1 - x)*((-4*x + x^2)^(-1))^x^2
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 = 0.61 (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 | \(\frac {{\mathrm e}^{x^{2} \ln \left (\frac {1}{\left (x -4\right ) x}\right )+x \ln \left (2\right )-x -1}}{4}\) | \(27\) |
Input:
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,method=_RETURNVERBOSE)
Output:
1/4*(1/(x^2-4*x))^(x^2)*2^x*exp(-1-x)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (x^{2} \log \left (\frac {1}{x^{2} - 4 \, x}\right ) + x \log \left (2\right ) - x - 1\right )} \] Input:
integrate(((2*x^2-8*x)*log(1/(x^2-4*x))+(-4+x)*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")
Output:
1/4*e^(x^2*log(1/(x^2 - 4*x)) + x*log(2) - x - 1)
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {e^{x^{2} \log {\left (\frac {1}{x^{2} - 4 x} \right )} - x + x \log {\left (2 \right )} - 1}}{4} \] Input:
integrate(((2*x**2-8*x)*ln(1/(x**2-4*x))+(-4+x)*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)
Output:
exp(x**2*log(1/(x**2 - 4*x)) - x + x*log(2) - 1)/4
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (-x^{2} \log \left (x - 4\right ) - x^{2} \log \left (x\right ) + x \log \left (2\right ) - x - 1\right )} \] Input:
integrate(((2*x^2-8*x)*log(1/(x^2-4*x))+(-4+x)*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")
Output:
1/4*e^(-x^2*log(x - 4) - x^2*log(x) + x*log(2) - x - 1)
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {1}{4} \, e^{\left (-x^{2} \log \left (x^{2} - 4 \, x\right ) + x \log \left (2\right ) - x - 1\right )} \] Input:
integrate(((2*x^2-8*x)*log(1/(x^2-4*x))+(-4+x)*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")
Output:
1/4*e^(-x^2*log(x^2 - 4*x) + x*log(2) - x - 1)
Time = 2.97 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {2^x\,{\mathrm {e}}^{-x-1}\,{\left (-\frac {1}{4\,x-x^2}\right )}^{x^2}}{4} \] Input:
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)
Output:
(2^x*exp(- x - 1)*(-1/(4*x - x^2))^(x^2))/4
\[ \int \frac {e^{-1-x+x \log (2)+x^2 \log \left (\frac {1}{-4 x+x^2}\right )} \left (4+3 x-2 x^2+(-4+x) \log (2)+\left (-8 x+2 x^2\right ) \log \left (\frac {1}{-4 x+x^2}\right )\right )}{-16+4 x} \, dx=\frac {-4 \left (\int \frac {2^{x}}{x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}} x -4 x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}}}d x \right ) \mathrm {log}\left (2\right )+4 \left (\int \frac {2^{x}}{x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}} x -4 x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}}}d x \right )-2 \left (\int \frac {2^{x} x^{2}}{x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}} x -4 x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}}}d x \right )-2 \left (\int \frac {2^{x} \mathrm {log}\left (x^{2}-4 x \right ) x^{2}}{x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}} x -4 x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}}}d x \right )+8 \left (\int \frac {2^{x} \mathrm {log}\left (x^{2}-4 x \right ) x}{x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}} x -4 x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}}}d x \right )+\left (\int \frac {2^{x} x}{x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}} x -4 x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}}}d x \right ) \mathrm {log}\left (2\right )+3 \left (\int \frac {2^{x} x}{x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}} x -4 x^{x^{2}} e^{x} \left (x -4\right )^{x^{2}}}d x \right )}{4 e} \] Input:
int(((2*x^2-8*x)*log(1/(x^2-4*x))+(-4+x)*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)
Output:
( - 4*int(2**x/(x**(x**2)*e**x*(x - 4)**(x**2)*x - 4*x**(x**2)*e**x*(x - 4 )**(x**2)),x)*log(2) + 4*int(2**x/(x**(x**2)*e**x*(x - 4)**(x**2)*x - 4*x* *(x**2)*e**x*(x - 4)**(x**2)),x) - 2*int((2**x*x**2)/(x**(x**2)*e**x*(x - 4)**(x**2)*x - 4*x**(x**2)*e**x*(x - 4)**(x**2)),x) - 2*int((2**x*log(x**2 - 4*x)*x**2)/(x**(x**2)*e**x*(x - 4)**(x**2)*x - 4*x**(x**2)*e**x*(x - 4) **(x**2)),x) + 8*int((2**x*log(x**2 - 4*x)*x)/(x**(x**2)*e**x*(x - 4)**(x* *2)*x - 4*x**(x**2)*e**x*(x - 4)**(x**2)),x) + int((2**x*x)/(x**(x**2)*e** x*(x - 4)**(x**2)*x - 4*x**(x**2)*e**x*(x - 4)**(x**2)),x)*log(2) + 3*int( (2**x*x)/(x**(x**2)*e**x*(x - 4)**(x**2)*x - 4*x**(x**2)*e**x*(x - 4)**(x* *2)),x))/(4*e)