Integrand size = 64, antiderivative size = 34 \[ \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx=-e^x+3 (-5+x)-2 x-\frac {-2-x^2}{2 (x-\log (\log (2)))} \] Output:
x-15-(-x^2-2)/(2*x-2*ln(ln(2)))-exp(x)
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx=\frac {1}{2} \left (-2 e^x+3 x+\frac {2}{x-\log (\log (2))}+\frac {\log ^2(\log (2))}{x-\log (\log (2))}\right ) \] Input:
Integrate[(-2 + 3*x^2 - 2*E^x*x^2 + (-6*x + 4*E^x*x)*Log[Log[2]] + (2 - 2* E^x)*Log[Log[2]]^2)/(2*x^2 - 4*x*Log[Log[2]] + 2*Log[Log[2]]^2),x]
Output:
(-2*E^x + 3*x + 2/(x - Log[Log[2]]) + Log[Log[2]]^2/(x - Log[Log[2]]))/2
Time = 0.71 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {7277, 27, 7293, 2009}
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.
\(\displaystyle \int \frac {-2 e^x x^2+3 x^2+\left (2-2 e^x\right ) \log ^2(\log (2))+\left (4 e^x x-6 x\right ) \log (\log (2))-2}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 8 \int -\frac {2 e^x x^2-3 x^2-2 \left (1-e^x\right ) \log ^2(\log (2))+2 \left (3 x-2 e^x x\right ) \log (\log (2))+2}{16 (x-\log (\log (2)))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{2} \int \frac {2 e^x x^2-3 x^2-2 \left (1-e^x\right ) \log ^2(\log (2))+2 \left (3 x-2 e^x x\right ) \log (\log (2))+2}{(x-\log (\log (2)))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (-\frac {3 x^2}{(x-\log (\log (2)))^2}+\frac {6 \log (\log (2)) x}{(x-\log (\log (2)))^2}+2 e^x+\frac {2 \left (1-\log ^2(\log (2))\right )}{(x-\log (\log (2)))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (3 x-2 e^x+\frac {2 \left (1-\log ^2(\log (2))\right )}{x-\log (\log (2))}+\frac {3 \log ^2(\log (2))}{x-\log (\log (2))}\right )\) |
Input:
Int[(-2 + 3*x^2 - 2*E^x*x^2 + (-6*x + 4*E^x*x)*Log[Log[2]] + (2 - 2*E^x)*L og[Log[2]]^2)/(2*x^2 - 4*x*Log[Log[2]] + 2*Log[Log[2]]^2),x]
Output:
(-2*E^x + 3*x + (3*Log[Log[2]]^2)/(x - Log[Log[2]]) + (2*(1 - Log[Log[2]]^ 2))/(x - Log[Log[2]]))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.69 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85
method | result | size |
parts | \(\frac {3 x}{2}-\frac {-\ln \left (\ln \left (2\right )\right )^{2}-2}{2 \left (-\ln \left (\ln \left (2\right )\right )+x \right )}-{\mathrm e}^{x}\) | \(29\) |
norman | \(\frac {{\mathrm e}^{x} x -\frac {3 x^{2}}{2}-{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right )-1+\ln \left (\ln \left (2\right )\right )^{2}}{\ln \left (\ln \left (2\right )\right )-x}\) | \(34\) |
risch | \(\frac {3 x}{2}-\frac {\ln \left (\ln \left (2\right )\right )^{2}}{2 \left (\ln \left (\ln \left (2\right )\right )-x \right )}-\frac {1}{\ln \left (\ln \left (2\right )\right )-x}-{\mathrm e}^{x}\) | \(36\) |
parallelrisch | \(\frac {-2-3 x^{2}+2 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right )+2 \ln \left (\ln \left (2\right )\right )^{2}}{2 \ln \left (\ln \left (2\right )\right )-2 x}\) | \(38\) |
default | \(\frac {1}{-\ln \left (\ln \left (2\right )\right )+x}+\frac {3 x}{2}+\frac {\ln \left (\ln \left (2\right )\right )^{2}}{2 x -2 \ln \left (\ln \left (2\right )\right )}-{\mathrm e}^{x}-2 \ln \left (\ln \left (2\right )\right )^{2} \left (-\frac {{\mathrm e}^{x}}{-\ln \left (\ln \left (2\right )\right )+x}-\ln \left (2\right ) \operatorname {expIntegral}_{1}\left (\ln \left (\ln \left (2\right )\right )-x \right )\right )+2 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right ) \operatorname {expIntegral}_{1}\left (\ln \left (\ln \left (2\right )\right )-x \right )+2 \ln \left (\ln \left (2\right )\right ) \left (\ln \left (\ln \left (2\right )\right ) \left (-\frac {{\mathrm e}^{x}}{-\ln \left (\ln \left (2\right )\right )+x}-\ln \left (2\right ) \operatorname {expIntegral}_{1}\left (\ln \left (\ln \left (2\right )\right )-x \right )\right )-\ln \left (2\right ) \operatorname {expIntegral}_{1}\left (\ln \left (\ln \left (2\right )\right )-x \right )\right )\) | \(134\) |
Input:
int(((-2*exp(x)+2)*ln(ln(2))^2+(4*exp(x)*x-6*x)*ln(ln(2))-2*exp(x)*x^2+3*x ^2-2)/(2*ln(ln(2))^2-4*x*ln(ln(2))+2*x^2),x,method=_RETURNVERBOSE)
Output:
3/2*x-1/2*(-ln(ln(2))^2-2)/(-ln(ln(2))+x)-exp(x)
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx=\frac {3 \, x^{2} - 2 \, x e^{x} - {\left (3 \, x - 2 \, e^{x}\right )} \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2} + 2}{2 \, {\left (x - \log \left (\log \left (2\right )\right )\right )}} \] Input:
integrate(((-2*exp(x)+2)*log(log(2))^2+(4*exp(x)*x-6*x)*log(log(2))-2*exp( x)*x^2+3*x^2-2)/(2*log(log(2))^2-4*x*log(log(2))+2*x^2),x, algorithm="fric as")
Output:
1/2*(3*x^2 - 2*x*e^x - (3*x - 2*e^x)*log(log(2)) + log(log(2))^2 + 2)/(x - log(log(2)))
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx=\frac {3 x}{2} - e^{x} + \frac {\log {\left (\log {\left (2 \right )} \right )}^{2} + 2}{2 x - 2 \log {\left (\log {\left (2 \right )} \right )}} \] Input:
integrate(((-2*exp(x)+2)*ln(ln(2))**2+(4*exp(x)*x-6*x)*ln(ln(2))-2*exp(x)* x**2+3*x**2-2)/(2*ln(ln(2))**2-4*x*ln(ln(2))+2*x**2),x)
Output:
3*x/2 - exp(x) + (log(log(2))**2 + 2)/(2*x - 2*log(log(2)))
\[ \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx=\int { -\frac {2 \, x^{2} e^{x} + 2 \, {\left (e^{x} - 1\right )} \log \left (\log \left (2\right )\right )^{2} - 3 \, x^{2} - 2 \, {\left (2 \, x e^{x} - 3 \, x\right )} \log \left (\log \left (2\right )\right ) + 2}{2 \, {\left (x^{2} - 2 \, x \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2}\right )}} \,d x } \] Input:
integrate(((-2*exp(x)+2)*log(log(2))^2+(4*exp(x)*x-6*x)*log(log(2))-2*exp( x)*x^2+3*x^2-2)/(2*log(log(2))^2-4*x*log(log(2))+2*x^2),x, algorithm="maxi ma")
Output:
2*integrate(e^x/(x^3 - 3*x^2*log(log(2)) + 3*x*log(log(2))^2 - log(log(2)) ^3), x)*log(log(2))^2 + exp_integral_e(2, -x + log(log(2)))*log(2)*log(log (2))^2/(x - log(log(2))) + 3*(log(log(2))/(x - log(log(2))) - log(x - log( log(2))))*log(log(2)) + 3*log(x - log(log(2)))*log(log(2)) + 3/2*x - (x^2 - 2*x*log(log(2)))*e^x/(x^2 - 2*x*log(log(2)) + log(log(2))^2) - 5/2*log(l og(2))^2/(x - log(log(2))) + 1/(x - log(log(2)))
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx=\frac {3 \, x^{2} - 2 \, x e^{x} - 3 \, x \log \left (\log \left (2\right )\right ) + 2 \, e^{x} \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2} + 2}{2 \, {\left (x - \log \left (\log \left (2\right )\right )\right )}} \] Input:
integrate(((-2*exp(x)+2)*log(log(2))^2+(4*exp(x)*x-6*x)*log(log(2))-2*exp( x)*x^2+3*x^2-2)/(2*log(log(2))^2-4*x*log(log(2))+2*x^2),x, algorithm="giac ")
Output:
1/2*(3*x^2 - 2*x*e^x - 3*x*log(log(2)) + 2*e^x*log(log(2)) + log(log(2))^2 + 2)/(x - log(log(2)))
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx=\frac {\frac {{\ln \left (\ln \left (2\right )\right )}^2}{2}+1}{x-\ln \left (\ln \left (2\right )\right )}-\frac {2\,{\mathrm {e}}^x\,\ln \left (\ln \left (2\right )\right )-3\,x\,\ln \left (\ln \left (2\right )\right )}{2\,\ln \left (\ln \left (2\right )\right )} \] Input:
int(-(2*x^2*exp(x) + log(log(2))*(6*x - 4*x*exp(x)) + log(log(2))^2*(2*exp (x) - 2) - 3*x^2 + 2)/(2*log(log(2))^2 - 4*x*log(log(2)) + 2*x^2),x)
Output:
(log(log(2))^2/2 + 1)/(x - log(log(2))) - (2*exp(x)*log(log(2)) - 3*x*log( log(2)))/(2*log(log(2)))
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.62 \[ \int \frac {-2+3 x^2-2 e^x x^2+\left (-6 x+4 e^x x\right ) \log (\log (2))+\left (2-2 e^x\right ) \log ^2(\log (2))}{2 x^2-4 x \log (\log (2))+2 \log ^2(\log (2))} \, dx=\frac {-2 e^{x} \mathrm {log}\left (\mathrm {log}\left (2\right )\right )^{2}+2 e^{x} \mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x +2 \mathrm {log}\left (\mathrm {log}\left (2\right )\right )^{2} x -3 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x^{2}-2 x}{2 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) \left (\mathrm {log}\left (\mathrm {log}\left (2\right )\right )-x \right )} \] Input:
int(((-2*exp(x)+2)*log(log(2))^2+(4*exp(x)*x-6*x)*log(log(2))-2*exp(x)*x^2 +3*x^2-2)/(2*log(log(2))^2-4*x*log(log(2))+2*x^2),x)
Output:
( - 2*e**x*log(log(2))**2 + 2*e**x*log(log(2))*x + 2*log(log(2))**2*x - 3* log(log(2))*x**2 - 2*x)/(2*log(log(2))*(log(log(2)) - x))