\(\int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} (x+x^2+(-2-3 x-2 x^2+2 x^3) \log (2))}{(-2 x-3 x^2+x^4) \log (2)} \, dx\) [4623]

Optimal result

Integrand size = 68, antiderivative size = 32 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {e^{x \left (2+\frac {\log (2 x)+\frac {\log (-4+2 x)}{\log (2)}}{x}\right )}}{1+x} \]

[Out]

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

Rubi [F]

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

[In]

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]

[Out]

(2*E^4*(-4 + 2*x)^Log[2]^(-1)*Gamma[1 + Log[2]^(-1), 4 - 2*x])/(4 - 2*x)^Log[2]^(-1) - (2*E^4*(-4 + 2*x)^Log[2
]^(-1)*Gamma[Log[2]^(-1), 4 - 2*x]*(1 - Log[4]))/((4 - 2*x)^Log[2]^(-1)*Log[2]) - (4*Log[8]*Defer[Int][(E^(2*x
)*(-4 + 2*x)^(-1 + Log[2]^(-1)))/(1 + x)^2, x])/Log[2] - (4*(1 - Log[128])*Defer[Int][(E^(2*x)*(-4 + 2*x)^(-1
+ Log[2]^(-1)))/(1 + x), x])/Log[2]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\exp \left (\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}\right ) \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2 x-3 x^2+x^4} \, dx}{\log (2)} \\ & = \frac {\int \frac {\exp \left (\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}\right ) \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{x \left (-2-3 x+x^3\right )} \, dx}{\log (2)} \\ & = \frac {\int \frac {2 e^{\frac {2 x \log (2)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2-3 x+x^3} \, dx}{\log (2)} \\ & = \frac {2 \int \frac {e^{\frac {2 x \log (2)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2-3 x+x^3} \, dx}{\log (2)} \\ & = \frac {2 \int \frac {e^{2 x} (-4+2 x)^{\frac {1}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2-3 x+x^3} \, dx}{\log (2)} \\ & = \frac {2 \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\frac {1}{2}+x+\frac {x^2}{2}} \, dx}{\log (2)} \\ & = \frac {2 \int \frac {2 e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{(1+x)^2} \, dx}{\log (2)} \\ & = \frac {4 \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{(1+x)^2} \, dx}{\log (2)} \\ & = \frac {4 \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \left (2 x^3 \log (2)+x^2 (1-\log (4))-\log (4)+x (1-\log (8))\right )}{(1+x)^2} \, dx}{\log (2)} \\ & = \frac {4 \int \left (e^{2 x} (-4+2 x)^{\frac {1}{\log (2)}} \log (2)+e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} (1-\log (4))-\frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \log (8)}{(1+x)^2}+\frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} (-1+\log (128))}{1+x}\right ) \, dx}{\log (2)} \\ & = 4 \int e^{2 x} (-4+2 x)^{\frac {1}{\log (2)}} \, dx+\frac {(4 (1-\log (4))) \int e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \, dx}{\log (2)}-\frac {(4 \log (8)) \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}}}{(1+x)^2} \, dx}{\log (2)}-\frac {(4 (1-\log (128))) \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}}}{1+x} \, dx}{\log (2)} \\ & = 2 e^4 (4-2 x)^{-\frac {1}{\log (2)}} (-4+2 x)^{\frac {1}{\log (2)}} \Gamma \left (1+\frac {1}{\log (2)},4-2 x\right )-\frac {2 e^4 (4-2 x)^{-\frac {1}{\log (2)}} (-4+2 x)^{\frac {1}{\log (2)}} \Gamma \left (\frac {1}{\log (2)},4-2 x\right ) (1-\log (4))}{\log (2)}-\frac {(4 \log (8)) \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}}}{(1+x)^2} \, dx}{\log (2)}-\frac {(4 (1-\log (128))) \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}}}{1+x} \, dx}{\log (2)} \\ \end{align*}

Mathematica [F]

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

[In]

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]

[Out]

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]

Maple [A] (verified)

Time = 2.45 (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	\(\frac {{\mathrm e}^{\frac {\ln \left (2\right ) \ln \left (2 x \right )+\ln \left (2 x -4\right )+2 x \ln \left (2\right )}{\ln \left (2\right )}}}{1+x}\)	\(32\)

[In]

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,m
ethod=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {e^{\left (\frac {2 \, x \log \left (2\right ) + \log \left (2\right ) \log \left (2 \, x\right ) + \log \left (2 \, x - 4\right )}{\log \left (2\right )}\right )}}{x + 1} \]

[In]

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")

[Out]

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

Sympy [F(-1)]

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

[In]

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)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {2 \, x e^{\left (2 \, x + \frac {\log \left (x - 2\right )}{\log \left (2\right )} + 1\right )}}{x + 1} \]

[In]

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")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\frac {x e^{\left (\frac {2 \, x \log \left (2\right ) + \log \left (2\right )^{2} + \log \left (2\right ) + \log \left (x - 2\right )}{\log \left (2\right )}\right )}}{x + 1} \]

[In]

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")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx=\int -\frac {{\mathrm {e}}^{\frac {\ln \left (2\,x-4\right )+\ln \left (2\,x\right )\,\ln \left (2\right )+2\,x\,\ln \left (2\right )}{\ln \left (2\right )}}\,\left (x-\ln \left (2\right )\,\left (-2\,x^3+2\,x^2+3\,x+2\right )+x^2\right )}{\ln \left (2\right )\,\left (-x^4+3\,x^2+2\,x\right )} \,d x \]

[In]

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)

[Out]

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)