\(\int \frac {e^{\frac {1+e^3}{4+e^x-x}} (e^x (-1-e^3) \log (4)+(1+e^3) \log (4))}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx\) [6866]

Optimal result

Integrand size = 69, antiderivative size = 24 \[ \int \frac {e^{\frac {1+e^3}{4+e^x-x}} \left (e^x \left (-1-e^3\right ) \log (4)+\left (1+e^3\right ) \log (4)\right )}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx=\frac {1}{2} e^{\frac {1+e^3}{4+e^x-x}} \log (4) \]

[Out]

ln(2)*exp((exp(3)+1)/(exp(x)-x+4))

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6820, 12, 6838} \[ \int \frac {e^{\frac {1+e^3}{4+e^x-x}} \left (e^x \left (-1-e^3\right ) \log (4)+\left (1+e^3\right ) \log (4)\right )}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx=\frac {1}{2} e^{\frac {1+e^3}{-x+e^x+4}} \log (4) \]

[In]

Int[(E^((1 + E^3)/(4 + E^x - x))*(E^x*(-1 - E^3)*Log[4] + (1 + E^3)*Log[4]))/(32 + 2*E^(2*x) + E^x*(16 - 4*x)
- 16*x + 2*x^2),x]

[Out]

(E^((1 + E^3)/(4 + E^x - x))*Log[4])/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6820

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1+e^3}{4+e^x-x}} \left (e^x \left (-1-e^3\right ) \log (4)+\left (1+e^3\right ) \log (4)\right )}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx=\frac {1}{2} e^{\frac {1+e^3}{4+e^x-x}} \log (4) \]

[In]

Integrate[(E^((1 + E^3)/(4 + E^x - x))*(E^x*(-1 - E^3)*Log[4] + (1 + E^3)*Log[4]))/(32 + 2*E^(2*x) + E^x*(16 -
4*x) - 16*x + 2*x^2),x]

[Out]

(E^((1 + E^3)/(4 + E^x - x))*Log[4])/2

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
risch	\(\ln \left (2\right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+1}{{\mathrm e}^{x}-x +4}}\)	\(19\)
parallelrisch	\(\ln \left (2\right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+1}{{\mathrm e}^{x}-x +4}}\)	\(19\)
norman	\(\frac {x \ln \left (2\right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+1}{{\mathrm e}^{x}-x +4}}-4 \ln \left (2\right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+1}{{\mathrm e}^{x}-x +4}}-{\mathrm e}^{x} \ln \left (2\right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+1}{{\mathrm e}^{x}-x +4}}}{x -{\mathrm e}^{x}-4}\)	\(71\)

[In]

int((2*(-exp(3)-1)*ln(2)*exp(x)+2*(exp(3)+1)*ln(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)^2+(-4*x+16)*exp(x)+
2*x^2-16*x+32),x,method=_RETURNVERBOSE)

[Out]

ln(2)*exp((exp(3)+1)/(exp(x)-x+4))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {1+e^3}{4+e^x-x}} \left (e^x \left (-1-e^3\right ) \log (4)+\left (1+e^3\right ) \log (4)\right )}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx=e^{\left (-\frac {e^{3} + 1}{x - e^{x} - 4}\right )} \log \left (2\right ) \]

[In]

integrate((2*(-exp(3)-1)*log(2)*exp(x)+2*(exp(3)+1)*log(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)^2+(-4*x+16)
*exp(x)+2*x^2-16*x+32),x, algorithm="fricas")

[Out]

e^(-(e^3 + 1)/(x - e^x - 4))*log(2)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {e^{\frac {1+e^3}{4+e^x-x}} \left (e^x \left (-1-e^3\right ) \log (4)+\left (1+e^3\right ) \log (4)\right )}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx=e^{\frac {1 + e^{3}}{- x + e^{x} + 4}} \log {\left (2 \right )} \]

[In]

integrate((2*(-exp(3)-1)*ln(2)*exp(x)+2*(exp(3)+1)*ln(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)**2+(-4*x+16)*
exp(x)+2*x**2-16*x+32),x)

[Out]

exp((1 + exp(3))/(-x + exp(x) + 4))*log(2)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {1+e^3}{4+e^x-x}} \left (e^x \left (-1-e^3\right ) \log (4)+\left (1+e^3\right ) \log (4)\right )}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx=e^{\left (-\frac {e^{3}}{x - e^{x} - 4} - \frac {1}{x - e^{x} - 4}\right )} \log \left (2\right ) \]

[In]

integrate((2*(-exp(3)-1)*log(2)*exp(x)+2*(exp(3)+1)*log(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)^2+(-4*x+16)
*exp(x)+2*x^2-16*x+32),x, algorithm="maxima")

[Out]

e^(-e^3/(x - e^x - 4) - 1/(x - e^x - 4))*log(2)

Giac [F]

\[ \int \frac {e^{\frac {1+e^3}{4+e^x-x}} \left (e^x \left (-1-e^3\right ) \log (4)+\left (1+e^3\right ) \log (4)\right )}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx=\int { -\frac {{\left ({\left (e^{3} + 1\right )} e^{x} \log \left (2\right ) - {\left (e^{3} + 1\right )} \log \left (2\right )\right )} e^{\left (-\frac {e^{3} + 1}{x - e^{x} - 4}\right )}}{x^{2} - 2 \, {\left (x - 4\right )} e^{x} - 8 \, x + e^{\left (2 \, x\right )} + 16} \,d x } \]

[In]

integrate((2*(-exp(3)-1)*log(2)*exp(x)+2*(exp(3)+1)*log(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)^2+(-4*x+16)
*exp(x)+2*x^2-16*x+32),x, algorithm="giac")

[Out]

integrate(-((e^3 + 1)*e^x*log(2) - (e^3 + 1)*log(2))*e^(-(e^3 + 1)/(x - e^x - 4))/(x^2 - 2*(x - 4)*e^x - 8*x +
e^(2*x) + 16), x)

Mupad [B] (verification not implemented)

Time = 12.59 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {e^{\frac {1+e^3}{4+e^x-x}} \left (e^x \left (-1-e^3\right ) \log (4)+\left (1+e^3\right ) \log (4)\right )}{32+2 e^{2 x}+e^x (16-4 x)-16 x+2 x^2} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^3+1}{{\mathrm {e}}^x-x+4}}\,\ln \left (2\right ) \]

[In]

int((exp((exp(3) + 1)/(exp(x) - x + 4))*(2*log(2)*(exp(3) + 1) - 2*exp(x)*log(2)*(exp(3) + 1)))/(2*exp(2*x) -
16*x - exp(x)*(4*x - 16) + 2*x^2 + 32),x)

[Out]

exp((exp(3) + 1)/(exp(x) - x + 4))*log(2)