Integrand size = 121, antiderivative size = 34 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=e^{\frac {5}{\log (2)}}+2 x-\frac {\log (2)}{\log \left (e^{\frac {x}{4+e^2}}-x\right )} \] Output:
2*x+exp(5/ln(2))-ln(2)/ln(exp(x/(4+exp(2)))-x)
Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=2 x-\frac {\log (2)}{\log \left (e^{\frac {x}{4+e^2}}-x\right )} \] Input:
Integrate[(E^(x/(4 + E^2))*Log[2] + (-4 - E^2)*Log[2] + (E^(x/(4 + E^2))*( 8 + 2*E^2) - 8*x - 2*E^2*x)*Log[E^(x/(4 + E^2)) - x]^2)/((E^(x/(4 + E^2))* (4 + E^2) - 4*x - E^2*x)*Log[E^(x/(4 + E^2)) - x]^2),x]
Output:
2*x - Log[2]/Log[E^(x/(4 + E^2)) - x]
Time = 0.65 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6, 7239, 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.
\(\displaystyle \int \frac {\left (-2 e^2 x-8 x+\left (8+2 e^2\right ) e^{\frac {x}{4+e^2}}\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )+e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)}{\left (-e^2 x-4 x+\left (4+e^2\right ) e^{\frac {x}{4+e^2}}\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (-2 e^2 x-8 x+\left (8+2 e^2\right ) e^{\frac {x}{4+e^2}}\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )+e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)}{\left (\left (-4-e^2\right ) x+\left (4+e^2\right ) e^{\frac {x}{4+e^2}}\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {\left (e^{\frac {x}{4+e^2}}-4 \left (1+\frac {e^2}{4}\right )\right ) \log (2)}{\left (4+e^2\right ) \left (e^{\frac {x}{4+e^2}}-x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}+2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 x-\frac {\log (2)}{\log \left (e^{\frac {x}{4+e^2}}-x\right )}\) |
Input:
Int[(E^(x/(4 + E^2))*Log[2] + (-4 - E^2)*Log[2] + (E^(x/(4 + E^2))*(8 + 2* E^2) - 8*x - 2*E^2*x)*Log[E^(x/(4 + E^2)) - x]^2)/((E^(x/(4 + E^2))*(4 + E ^2) - 4*x - E^2*x)*Log[E^(x/(4 + E^2)) - x]^2),x]
Output:
2*x - Log[2]/Log[E^(x/(4 + E^2)) - x]
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74
method | result | size |
risch | \(2 x -\frac {\ln \left (2\right )}{\ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right )}\) | \(25\) |
norman | \(\frac {2 x \ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right )-\ln \left (2\right )}{\ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right )}\) | \(40\) |
parallelrisch | \(-\frac {-2 \,{\mathrm e}^{2} x \ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right )+{\mathrm e}^{2} \ln \left (2\right )-8 x \ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right )+4 \ln \left (2\right )}{\ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right ) \left (4+{\mathrm e}^{2}\right )}\) | \(71\) |
Input:
int((((2*exp(2)+8)*exp(x/(4+exp(2)))-2*exp(2)*x-8*x)*ln(exp(x/(4+exp(2)))- x)^2+ln(2)*exp(x/(4+exp(2)))+(-exp(2)-4)*ln(2))/((4+exp(2))*exp(x/(4+exp(2 )))-exp(2)*x-4*x)/ln(exp(x/(4+exp(2)))-x)^2,x,method=_RETURNVERBOSE)
Output:
2*x-ln(2)/ln(exp(x/(4+exp(2)))-x)
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=\frac {2 \, x \log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right ) - \log \left (2\right )}{\log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right )} \] Input:
integrate((((2*exp(2)+8)*exp(x/(4+exp(2)))-2*exp(2)*x-8*x)*log(exp(x/(4+ex p(2)))-x)^2+log(2)*exp(x/(4+exp(2)))+(-exp(2)-4)*log(2))/((4+exp(2))*exp(x /(4+exp(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x, algorithm="fricas ")
Output:
(2*x*log(-x + e^(x/(e^2 + 4))) - log(2))/log(-x + e^(x/(e^2 + 4)))
Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.50 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=2 x - \frac {\log {\left (2 \right )}}{\log {\left (- x + e^{\frac {x}{4 + e^{2}}} \right )}} \] Input:
integrate((((2*exp(2)+8)*exp(x/(4+exp(2)))-2*exp(2)*x-8*x)*ln(exp(x/(4+exp (2)))-x)**2+ln(2)*exp(x/(4+exp(2)))+(-exp(2)-4)*ln(2))/((4+exp(2))*exp(x/( 4+exp(2)))-exp(2)*x-4*x)/ln(exp(x/(4+exp(2)))-x)**2,x)
Output:
2*x - log(2)/log(-x + exp(x/(4 + exp(2))))
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=\frac {2 \, x \log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right ) - \log \left (2\right )}{\log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right )} \] Input:
integrate((((2*exp(2)+8)*exp(x/(4+exp(2)))-2*exp(2)*x-8*x)*log(exp(x/(4+ex p(2)))-x)^2+log(2)*exp(x/(4+exp(2)))+(-exp(2)-4)*log(2))/((4+exp(2))*exp(x /(4+exp(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x, algorithm="maxima ")
Output:
(2*x*log(-x + e^(x/(e^2 + 4))) - log(2))/log(-x + e^(x/(e^2 + 4)))
Time = 0.44 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=\frac {2 \, x \log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right ) - \log \left (2\right )}{\log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right )} \] Input:
integrate((((2*exp(2)+8)*exp(x/(4+exp(2)))-2*exp(2)*x-8*x)*log(exp(x/(4+ex p(2)))-x)^2+log(2)*exp(x/(4+exp(2)))+(-exp(2)-4)*log(2))/((4+exp(2))*exp(x /(4+exp(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x, algorithm="giac")
Output:
(2*x*log(-x + e^(x/(e^2 + 4))) - log(2))/log(-x + e^(x/(e^2 + 4)))
Time = 2.67 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=2\,x-\frac {\ln \left (2\right )}{\ln \left ({\mathrm {e}}^{\frac {x}{{\mathrm {e}}^2+4}}-x\right )} \] Input:
int((log(exp(x/(exp(2) + 4)) - x)^2*(8*x + 2*x*exp(2) - exp(x/(exp(2) + 4) )*(2*exp(2) + 8)) - exp(x/(exp(2) + 4))*log(2) + log(2)*(exp(2) + 4))/(log (exp(x/(exp(2) + 4)) - x)^2*(4*x + x*exp(2) - exp(x/(exp(2) + 4))*(exp(2) + 4))),x)
Output:
2*x - log(2)/log(exp(x/(exp(2) + 4)) - x)
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=\frac {2 \,\mathrm {log}\left (e^{\frac {x}{e^{2}+4}}-x \right ) x -\mathrm {log}\left (2\right )}{\mathrm {log}\left (e^{\frac {x}{e^{2}+4}}-x \right )} \] Input:
int((((2*exp(2)+8)*exp(x/(4+exp(2)))-2*exp(2)*x-8*x)*log(exp(x/(4+exp(2))) -x)^2+log(2)*exp(x/(4+exp(2)))+(-exp(2)-4)*log(2))/((4+exp(2))*exp(x/(4+ex p(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x)
Output:
(2*log(e**(x/(e**2 + 4)) - x)*x - log(2))/log(e**(x/(e**2 + 4)) - x)