Integrand size = 68, antiderivative size = 32 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {e^{x \left (2+\frac {\log (2 x)+\frac {\log (-4+2 x)}{\log (2)}}{x}\right )}}{1+x} \] Output:
exp(x*(2+(ln(2*x-4)/ln(2)+ln(2*x))/x))/(1+x)
\[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx \] Input:
Integrate[(E^((2*x*Log[2] + Log[2]*Log[2*x] + Log[-4 + 2*x])/Log[2])*(x + x^2 + (-2 - 3*x - 2*x^2 + 2*x^3)*Log[2]))/((-2*x - 3*x^2 + x^4)*Log[2]),x]
Output:
Integrate[(E^((2*x*Log[2] + Log[2]*Log[2*x] + Log[-4 + 2*x])/Log[2])*(x + x^2 + (-2 - 3*x - 2*x^2 + 2*x^3)*Log[2]))/(-2*x - 3*x^2 + x^4), x]/Log[2]
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.
\(\displaystyle \int \frac {\left (x^2+\left (2 x^3-2 x^2-3 x-2\right ) \log (2)+x\right ) \exp \left (\frac {2 x \log (2)+\log (2) \log (2 x)+\log (2 x-4)}{\log (2)}\right )}{\left (x^4-3 x^2-2 x\right ) \log (2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {2 e^{2 x} x (2 x-4)^{\frac {1}{\log (2)}} \left (x^2+x-\left (-2 x^3+2 x^2+3 x+2\right ) \log (2)\right )}{-x^4+3 x^2+2 x}dx}{\log (2)}\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \frac {\int -\frac {2 e^{2 x} (2 x-4)^{\frac {1}{\log (2)}} \left (x^2+x-\left (-2 x^3+2 x^2+3 x+2\right ) \log (2)\right )}{-x^3+3 x+2}dx}{\log (2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \int \frac {e^{2 x} (2 x-4)^{\frac {1}{\log (2)}} \left (x^2+x-\left (-2 x^3+2 x^2+3 x+2\right ) \log (2)\right )}{-x^3+3 x+2}dx}{\log (2)}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle -\frac {2 \int \frac {e^{2 x} (2 x-4)^{-1+\frac {1}{\log (2)}} \left (x^2+x-\left (-2 x^3+2 x^2+3 x+2\right ) \log (2)\right )}{-\frac {x^2}{2}-x-\frac {1}{2}}dx}{\log (2)}\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle -\frac {2 \int \frac {e^{2 x} (2 x-4)^{-1+\frac {1}{\log (2)}} \left (x^2+x-\left (-2 x^3+2 x^2+3 x+2\right ) \log (2)\right )}{\left (\frac {i x}{\sqrt {2}}+\frac {i}{\sqrt {2}}\right )^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {2 \int \frac {e^{2 x} (2 x-4)^{-1+\frac {1}{\log (2)}} \left (2 \log (2) x^3+(1-\log (4)) x^2+(1-\log (8)) x-\log (4)\right )}{\left (\frac {i x}{\sqrt {2}}+\frac {i}{\sqrt {2}}\right )^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \int \left (-e^{2 x} \log (4) (2 x-4)^{\frac {1}{\log (2)}}-\frac {2 e^{2 x} (-1+\log (128)) (2 x-4)^{-1+\frac {1}{\log (2)}}}{x+1}+\frac {e^{2 x} \log (64) (2 x-4)^{-1+\frac {1}{\log (2)}}}{(x+1)^2}+e^{2 x} (-2+\log (16)) (2 x-4)^{-1+\frac {1}{\log (2)}}\right )dx}{\log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (\log (64) \int \frac {e^{2 x} (2 x-4)^{-1+\frac {1}{\log (2)}}}{(x+1)^2}dx+2 (1-\log (128)) \int \frac {e^{2 x} (2 x-4)^{-1+\frac {1}{\log (2)}}}{x+1}dx+\frac {1}{2} e^4 (2-\log (16)) (4-2 x)^{-\frac {1}{\log (2)}} (2 x-4)^{\frac {1}{\log (2)}} \Gamma \left (\frac {1}{\log (2)},4-2 x\right )-\frac {1}{2} e^4 \log (4) (4-2 x)^{-\frac {1}{\log (2)}} (2 x-4)^{\frac {1}{\log (2)}} \Gamma \left (1+\frac {1}{\log (2)},4-2 x\right )\right )}{\log (2)}\) |
Input:
Int[(E^((2*x*Log[2] + Log[2]*Log[2*x] + Log[-4 + 2*x])/Log[2])*(x + x^2 + (-2 - 3*x - 2*x^2 + 2*x^3)*Log[2]))/((-2*x - 3*x^2 + x^4)*Log[2]),x]
Output:
$Aborted
Time = 2.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {2 x \left (2 x -4\right )^{\frac {1}{\ln \left (2\right )}} {\mathrm e}^{2 x}}{1+x}\) | \(23\) |
gosper | \(\frac {{\mathrm e}^{\frac {\ln \left (2\right ) \ln \left (2 x \right )+\ln \left (2 x -4\right )+2 x \ln \left (2\right )}{\ln \left (2\right )}}}{1+x}\) | \(32\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {\ln \left (2\right ) \ln \left (2 x \right )+\ln \left (2 x -4\right )+2 x \ln \left (2\right )}{\ln \left (2\right )}}}{1+x}\) | \(32\) |
Input:
int(((2*x^3-2*x^2-3*x-2)*ln(2)+x^2+x)*exp((ln(2)*ln(2*x)+ln(2*x-4)+2*x*ln( 2))/ln(2))/(x^4-3*x^2-2*x)/ln(2),x,method=_RETURNVERBOSE)
Output:
2*x*(2*x-4)^(1/ln(2))*exp(2*x)/(1+x)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {e^{\left (\frac {2 \, x \log \left (2\right ) + \log \left (2\right ) \log \left (2 \, x\right ) + \log \left (2 \, x - 4\right )}{\log \left (2\right )}\right )}}{x + 1} \] Input:
integrate(((2*x^3-2*x^2-3*x-2)*log(2)+x^2+x)*exp((log(2)*log(2*x)+log(2*x- 4)+2*x*log(2))/log(2))/(x^4-3*x^2-2*x)/log(2),x, algorithm="fricas")
Output:
e^((2*x*log(2) + log(2)*log(2*x) + log(2*x - 4))/log(2))/(x + 1)
Timed out. \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\text {Timed out} \] Input:
integrate(((2*x**3-2*x**2-3*x-2)*ln(2)+x**2+x)*exp((ln(2)*ln(2*x)+ln(2*x-4 )+2*x*ln(2))/ln(2))/(x**4-3*x**2-2*x)/ln(2),x)
Output:
Timed out
Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {2 \, x e^{\left (2 \, x + \frac {\log \left (x - 2\right )}{\log \left (2\right )} + 1\right )}}{x + 1} \] Input:
integrate(((2*x^3-2*x^2-3*x-2)*log(2)+x^2+x)*exp((log(2)*log(2*x)+log(2*x- 4)+2*x*log(2))/log(2))/(x^4-3*x^2-2*x)/log(2),x, algorithm="maxima")
Output:
2*x*e^(2*x + log(x - 2)/log(2) + 1)/(x + 1)
Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {x e^{\left (\frac {2 \, x \log \left (2\right ) + \log \left (2\right )^{2} + \log \left (2\right ) + \log \left (x - 2\right )}{\log \left (2\right )}\right )}}{x + 1} \] Input:
integrate(((2*x^3-2*x^2-3*x-2)*log(2)+x^2+x)*exp((log(2)*log(2*x)+log(2*x- 4)+2*x*log(2))/log(2))/(x^4-3*x^2-2*x)/log(2),x, algorithm="giac")
Output:
x*e^((2*x*log(2) + log(2)^2 + log(2) + log(x - 2))/log(2))/(x + 1)
Timed out. \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\int -\frac {{\mathrm {e}}^{\frac {\ln \left (2\,x-4\right )+\ln \left (2\,x\right )\,\ln \left (2\right )+2\,x\,\ln \left (2\right )}{\ln \left (2\right )}}\,\left (x-\ln \left (2\right )\,\left (-2\,x^3+2\,x^2+3\,x+2\right )+x^2\right )}{\ln \left (2\right )\,\left (-x^4+3\,x^2+2\,x\right )} \,d x \] Input:
int(-(exp((log(2*x - 4) + log(2*x)*log(2) + 2*x*log(2))/log(2))*(x - log(2 )*(3*x + 2*x^2 - 2*x^3 + 2) + x^2))/(log(2)*(2*x + 3*x^2 - x^4)),x)
Output:
int(-(exp((log(2*x - 4) + log(2*x)*log(2) + 2*x*log(2))/log(2))*(x - log(2 )*(3*x + 2*x^2 - 2*x^3 + 2) + x^2))/(log(2)*(2*x + 3*x^2 - x^4)), x)
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {2 e^{\frac {\mathrm {log}\left (2 x -4\right )+2 \,\mathrm {log}\left (2\right ) x}{\mathrm {log}\left (2\right )}} x}{x +1} \] Input:
int(((2*x^3-2*x^2-3*x-2)*log(2)+x^2+x)*exp((log(2)*log(2*x)+log(2*x-4)+2*x *log(2))/log(2))/(x^4-3*x^2-2*x)/log(2),x)
Output:
(2*e**((log(2*x - 4) + 2*log(2)*x)/log(2))*x)/(x + 1)