Integrand size = 61, antiderivative size = 33 \[ \int \frac {e^{-6+2 x} \left (e^{6-2 x} \left (-10+x+x^3\right ) \log ^2(2)+4^{2 e^{-6+2 x}} \left (-2 e^{6-2 x}+4 x \log (4)\right )\right )}{x^3 \log ^2(2)} \, dx=\frac {x^2+\frac {5-x+\frac {4^{2 e^{-6+2 x}}}{\log ^2(2)}}{x}}{x} \] Output:
(x^2+(exp(2*ln(2)/exp(3-x)^2)^2/ln(2)^2-x+5)/x)/x
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-6+2 x} \left (e^{6-2 x} \left (-10+x+x^3\right ) \log ^2(2)+4^{2 e^{-6+2 x}} \left (-2 e^{6-2 x}+4 x \log (4)\right )\right )}{x^3 \log ^2(2)} \, dx=\frac {5}{x^2}-\frac {1}{x}+x+\frac {2^{-1+4 e^{-6+2 x}} \log (4)}{x^2 \log ^3(2)} \] Input:
Integrate[(E^(-6 + 2*x)*(E^(6 - 2*x)*(-10 + x + x^3)*Log[2]^2 + 4^(2*E^(-6 + 2*x))*(-2*E^(6 - 2*x) + 4*x*Log[4])))/(x^3*Log[2]^2),x]
Output:
5/x^2 - x^(-1) + x + (2^(-1 + 4*E^(-6 + 2*x))*Log[4])/(x^2*Log[2]^3)
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.
\(\displaystyle \int \frac {e^{2 x-6} \left (e^{6-2 x} \left (x^3+x-10\right ) \log ^2(2)+4^{2 e^{2 x-6}} \left (4 x \log (4)-2 e^{6-2 x}\right )\right )}{x^3 \log ^2(2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^{2 x-6} \left (e^{6-2 x} \log ^2(2) \left (-x^3-x+10\right )+2^{1+4 e^{2 x-6}} \left (e^{6-2 x}-2 x \log (4)\right )\right )}{x^3}dx}{\log ^2(2)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{2 x-6} \left (e^{6-2 x} \log ^2(2) \left (-x^3-x+10\right )+2^{1+4 e^{2 x-6}} \left (e^{6-2 x}-2 x \log (4)\right )\right )}{x^3}dx}{\log ^2(2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (-\frac {\log ^2(2) x^3+\log ^2(2) x-2^{1+4 e^{2 x-6}}-10 \log ^2(2)}{x^3}-\frac {4^{1+2 e^{2 x-6}} e^{2 x-6} \log (4)}{x^2}\right )dx}{\log ^2(2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\int \frac {2^{1+4 e^{2 x-6}}}{x^3}dx-\log (4) \int \frac {4^{1+2 e^{2 x-6}} e^{2 x-6}}{x^2}dx-\frac {5 \log ^2(2)}{x^2}+x \left (-\log ^2(2)\right )+\frac {\log ^2(2)}{x}}{\log ^2(2)}\) |
Input:
Int[(E^(-6 + 2*x)*(E^(6 - 2*x)*(-10 + x + x^3)*Log[2]^2 + 4^(2*E^(-6 + 2*x ))*(-2*E^(6 - 2*x) + 4*x*Log[4])))/(x^3*Log[2]^2),x]
Output:
$Aborted
Time = 0.64 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30
method | result | size |
risch | \(x +\frac {-x \ln \left (2\right )^{2}+5 \ln \left (2\right )^{2}}{\ln \left (2\right )^{2} x^{2}}+\frac {4^{2 \,{\mathrm e}^{2 x -6}}}{\ln \left (2\right )^{2} x^{2}}\) | \(43\) |
parallelrisch | \(\frac {x^{3} \ln \left (2\right )^{2}-x \ln \left (2\right )^{2}+5 \ln \left (2\right )^{2}+{\mathrm e}^{4 \ln \left (2\right ) {\mathrm e}^{2 x -6}}}{\ln \left (2\right )^{2} x^{2}}\) | \(46\) |
Input:
int(((-2*exp(-x+3)^2+8*x*ln(2))*exp(2*ln(2)/exp(-x+3)^2)^2+(x^3+x-10)*ln(2 )^2*exp(-x+3)^2)/x^3/ln(2)^2/exp(-x+3)^2,x,method=_RETURNVERBOSE)
Output:
x+1/ln(2)^2*(-x*ln(2)^2+5*ln(2)^2)/x^2+1/ln(2)^2/x^2*(4^exp(2*x-6))^2
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-6+2 x} \left (e^{6-2 x} \left (-10+x+x^3\right ) \log ^2(2)+4^{2 e^{-6+2 x}} \left (-2 e^{6-2 x}+4 x \log (4)\right )\right )}{x^3 \log ^2(2)} \, dx=\frac {{\left (x^{3} - x + 5\right )} \log \left (2\right )^{2} + 2^{4 \, e^{\left (2 \, x - 6\right )}}}{x^{2} \log \left (2\right )^{2}} \] Input:
integrate(((-2*exp(3-x)^2+8*x*log(2))*exp(2*log(2)/exp(3-x)^2)^2+(x^3+x-10 )*log(2)^2*exp(3-x)^2)/x^3/log(2)^2/exp(3-x)^2,x, algorithm="fricas")
Output:
((x^3 - x + 5)*log(2)^2 + 2^(4*e^(2*x - 6)))/(x^2*log(2)^2)
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-6+2 x} \left (e^{6-2 x} \left (-10+x+x^3\right ) \log ^2(2)+4^{2 e^{-6+2 x}} \left (-2 e^{6-2 x}+4 x \log (4)\right )\right )}{x^3 \log ^2(2)} \, dx=x + \frac {5 - x}{x^{2}} + \frac {e^{4 e^{2 x - 6} \log {\left (2 \right )}}}{x^{2} \log {\left (2 \right )}^{2}} \] Input:
integrate(((-2*exp(3-x)**2+8*x*ln(2))*exp(2*ln(2)/exp(3-x)**2)**2+(x**3+x- 10)*ln(2)**2*exp(3-x)**2)/x**3/ln(2)**2/exp(3-x)**2,x)
Output:
x + (5 - x)/x**2 + exp(4*exp(2*x - 6)*log(2))/(x**2*log(2)**2)
Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-6+2 x} \left (e^{6-2 x} \left (-10+x+x^3\right ) \log ^2(2)+4^{2 e^{-6+2 x}} \left (-2 e^{6-2 x}+4 x \log (4)\right )\right )}{x^3 \log ^2(2)} \, dx=\frac {x \log \left (2\right )^{2} - \frac {\log \left (2\right )^{2}}{x} + \frac {5 \, \log \left (2\right )^{2}}{x^{2}} + \frac {2^{4 \, e^{\left (2 \, x - 6\right )}}}{x^{2}}}{\log \left (2\right )^{2}} \] Input:
integrate(((-2*exp(3-x)^2+8*x*log(2))*exp(2*log(2)/exp(3-x)^2)^2+(x^3+x-10 )*log(2)^2*exp(3-x)^2)/x^3/log(2)^2/exp(3-x)^2,x, algorithm="maxima")
Output:
(x*log(2)^2 - log(2)^2/x + 5*log(2)^2/x^2 + 2^(4*e^(2*x - 6))/x^2)/log(2)^ 2
\[ \int \frac {e^{-6+2 x} \left (e^{6-2 x} \left (-10+x+x^3\right ) \log ^2(2)+4^{2 e^{-6+2 x}} \left (-2 e^{6-2 x}+4 x \log (4)\right )\right )}{x^3 \log ^2(2)} \, dx=\int { \frac {{\left ({\left (x^{3} + x - 10\right )} e^{\left (-2 \, x + 6\right )} \log \left (2\right )^{2} + 2 \, {\left (4 \, x \log \left (2\right ) - e^{\left (-2 \, x + 6\right )}\right )} 2^{4 \, e^{\left (2 \, x - 6\right )}}\right )} e^{\left (2 \, x - 6\right )}}{x^{3} \log \left (2\right )^{2}} \,d x } \] Input:
integrate(((-2*exp(3-x)^2+8*x*log(2))*exp(2*log(2)/exp(3-x)^2)^2+(x^3+x-10 )*log(2)^2*exp(3-x)^2)/x^3/log(2)^2/exp(3-x)^2,x, algorithm="giac")
Output:
integrate(((x^3 + x - 10)*e^(-2*x + 6)*log(2)^2 + 2*(4*x*log(2) - e^(-2*x + 6))*2^(4*e^(2*x - 6)))*e^(2*x - 6)/(x^3*log(2)^2), x)
Time = 2.79 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-6+2 x} \left (e^{6-2 x} \left (-10+x+x^3\right ) \log ^2(2)+4^{2 e^{-6+2 x}} \left (-2 e^{6-2 x}+4 x \log (4)\right )\right )}{x^3 \log ^2(2)} \, dx=x+\frac {2^{4\,{\mathrm {e}}^{2\,x-6}}-x\,{\ln \left (2\right )}^2+5\,{\ln \left (2\right )}^2}{x^2\,{\ln \left (2\right )}^2} \] Input:
int(-(exp(2*x - 6)*(exp(4*exp(2*x - 6)*log(2))*(2*exp(6 - 2*x) - 8*x*log(2 )) - exp(6 - 2*x)*log(2)^2*(x + x^3 - 10)))/(x^3*log(2)^2),x)
Output:
x + (2^(4*exp(2*x - 6)) - x*log(2)^2 + 5*log(2)^2)/(x^2*log(2)^2)
Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-6+2 x} \left (e^{6-2 x} \left (-10+x+x^3\right ) \log ^2(2)+4^{2 e^{-6+2 x}} \left (-2 e^{6-2 x}+4 x \log (4)\right )\right )}{x^3 \log ^2(2)} \, dx=\frac {e^{\frac {4 e^{2 x} \mathrm {log}\left (2\right )}{e^{6}}}+\mathrm {log}\left (2\right )^{2} x^{3}-\mathrm {log}\left (2\right )^{2} x +5 \mathrm {log}\left (2\right )^{2}}{\mathrm {log}\left (2\right )^{2} x^{2}} \] Input:
int(((-2*exp(3-x)^2+8*x*log(2))*exp(2*log(2)/exp(3-x)^2)^2+(x^3+x-10)*log( 2)^2*exp(3-x)^2)/x^3/log(2)^2/exp(3-x)^2,x)
Output:
(e**((4*e**(2*x)*log(2))/e**6) + log(2)**2*x**3 - log(2)**2*x + 5*log(2)** 2)/(log(2)**2*x**2)