\(\int \frac {e^{\frac {x}{4+e^2}} \log (2)+(-4-e^2) \log (2)+(e^{\frac {x}{4+e^2}} (8+2 e^2)-8 x-2 e^2 x) \log ^2(e^{\frac {x}{4+e^2}}-x)}{(e^{\frac {x}{4+e^2}} (4+e^2)-4 x-e^2 x) \log ^2(e^{\frac {x}{4+e^2}}-x)} \, dx\) [3256]

Optimal result

Integrand size = 121, antiderivative size = 34 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=e^{\frac {5}{\log (2)}}+2 x-\frac {\log (2)}{\log \left (e^{\frac {x}{4+e^2}}-x\right )} \]

[Out]

2*x+exp(5/ln(2))-ln(2)/ln(exp(x/(4+exp(2)))-x)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6, 6820, 6818} \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=2 x-\frac {\log (2)}{\log \left (e^{\frac {x}{4+e^2}}-x\right )} \]

[In]

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]

[Out]

2*x - Log[2]/Log[E^(x/(4 + E^2)) - x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=2 x-\frac {\log (2)}{\log \left (e^{\frac {x}{4+e^2}}-x\right )} \]

[In]

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]

[Out]

2*x - Log[2]/Log[E^(x/(4 + E^2)) - x]

Maple [A] (verified)

Time = 0.88 (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	\(-\frac {-2 \,{\mathrm e}^{2} \ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right ) x +{\mathrm e}^{2} \ln \left (2\right )-8 x \ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right )+4 \ln \left (2\right )}{\ln \left ({\mathrm e}^{\frac {x}{4+{\mathrm e}^{2}}}-x \right ) \left (4+{\mathrm e}^{2}\right )}\)	\(71\)

[In]

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)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=\frac {2 \, x \log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right ) - \log \left (2\right )}{\log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right )} \]

[In]

integrate((((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+exp(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x, algorithm="fric
as")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.50 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=2 x - \frac {\log {\left (2 \right )}}{\log {\left (- x + e^{\frac {x}{4 + e^{2}}} \right )}} \]

[In]

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)

[Out]

2*x - log(2)/log(-x + exp(x/(4 + exp(2))))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=\frac {2 \, x \log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right ) - \log \left (2\right )}{\log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right )} \]

[In]

integrate((((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+exp(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x, algorithm="maxi
ma")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=\frac {2 \, x \log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right ) - \log \left (2\right )}{\log \left (-x + e^{\left (\frac {x}{e^{2} + 4}\right )}\right )} \]

[In]

integrate((((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+exp(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x, algorithm="giac
")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {x}{4+e^2}} \log (2)+\left (-4-e^2\right ) \log (2)+\left (e^{\frac {x}{4+e^2}} \left (8+2 e^2\right )-8 x-2 e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )}{\left (e^{\frac {x}{4+e^2}} \left (4+e^2\right )-4 x-e^2 x\right ) \log ^2\left (e^{\frac {x}{4+e^2}}-x\right )} \, dx=2\,x-\frac {\ln \left (2\right )}{\ln \left ({\mathrm {e}}^{\frac {x}{{\mathrm {e}}^2+4}}-x\right )} \]

[In]

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)

[Out]

2*x - log(2)/log(exp(x/(exp(2) + 4)) - x)