\(\int \frac {e^{e^x} (e^x (8+2 e^{x-x^2})+e^{x-x^2} (-1+2 x))+e^{e^x+x} (-4-e^{x-x^2}) \log (4+e^{x-x^2})}{4+e^{x-x^2}} \, dx\) [2360]

Optimal result

Integrand size = 87, antiderivative size = 29 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=-\log \left (\frac {5}{2}\right )+e^{e^x} \left (2-\log \left (4+e^{x-x^2}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6820, 6874, 2320, 2225, 2634} \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=2 e^{e^x}-e^{e^x} \log \left (e^{x-x^2}+4\right ) \]

[In]

Int[(E^E^x*(E^x*(8 + 2*E^(x - x^2)) + E^(x - x^2)*(-1 + 2*x)) + E^(E^x + x)*(-4 - E^(x - x^2))*Log[4 + E^(x -
x^2)])/(4 + E^(x - x^2)),x]

[Out]

2*E^E^x - E^E^x*Log[4 + E^(x - x^2)]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{e^x+x} \left (-1+2 e^x+8 e^{x^2}+2 x-\left (e^x+4 e^{x^2}\right ) \log \left (4+e^{x-x^2}\right )\right )}{e^x+4 e^{x^2}} \, dx \\ & = \int \left (2 e^{e^x+x}+\frac {e^{e^x+x} (-1+2 x)}{e^x+4 e^{x^2}}-e^{e^x+x} \log \left (4+e^{x-x^2}\right )\right ) \, dx \\ & = 2 \int e^{e^x+x} \, dx+\int \frac {e^{e^x+x} (-1+2 x)}{e^x+4 e^{x^2}} \, dx-\int e^{e^x+x} \log \left (4+e^{x-x^2}\right ) \, dx \\ & = -e^{e^x} \log \left (4+e^{x-x^2}\right )+2 \text {Subst}\left (\int e^x \, dx,x,e^x\right )+\int \frac {e^{e^x+x} (1-2 x)}{e^x+4 e^{x^2}} \, dx+\int \left (-\frac {e^{e^x+x}}{e^x+4 e^{x^2}}+\frac {2 e^{e^x+x} x}{e^x+4 e^{x^2}}\right ) \, dx \\ & = 2 e^{e^x}-e^{e^x} \log \left (4+e^{x-x^2}\right )+2 \int \frac {e^{e^x+x} x}{e^x+4 e^{x^2}} \, dx-\int \frac {e^{e^x+x}}{e^x+4 e^{x^2}} \, dx+\int \left (\frac {e^{e^x+x}}{e^x+4 e^{x^2}}-\frac {2 e^{e^x+x} x}{e^x+4 e^{x^2}}\right ) \, dx \\ & = 2 e^{e^x}-e^{e^x} \log \left (4+e^{x-x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 5.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=-e^{e^x} \left (-2+\log \left (4+e^{x-x^2}\right )\right ) \]

[In]

Integrate[(E^E^x*(E^x*(8 + 2*E^(x - x^2)) + E^(x - x^2)*(-1 + 2*x)) + E^(E^x + x)*(-4 - E^(x - x^2))*Log[4 + E
^(x - x^2)])/(4 + E^(x - x^2)),x]

[Out]

-(E^E^x*(-2 + Log[4 + E^(x - x^2)]))

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76

method	result	size
risch	\(-{\mathrm e}^{{\mathrm e}^{x}} \ln \left ({\mathrm e}^{-x \left (-1+x \right )}+4\right )+2 \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(22\)
parallelrisch	\(-{\mathrm e}^{{\mathrm e}^{x}} \ln \left ({\mathrm e}^{-x^{2}+x}+4\right )+2 \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(23\)

[In]

int(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*ln(exp(-x^2+x)+4)+((2*exp(-x^2+x)+8)*exp(x)+(-1+2*x)*exp(-x^2+x))*exp
(exp(x)))/(exp(-x^2+x)+4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=-{\left (e^{\left (x + e^{x}\right )} \log \left (e^{\left (-x^{2} + x\right )} + 4\right ) - 2 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \]

[In]

integrate(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*log(exp(-x^2+x)+4)+((2*exp(-x^2+x)+8)*exp(x)+(-1+2*x)*exp(-x^2+
x))*exp(exp(x)))/(exp(-x^2+x)+4),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=\text {Timed out} \]

[In]

integrate(((-exp(-x**2+x)-4)*exp(x)*exp(exp(x))*ln(exp(-x**2+x)+4)+((2*exp(-x**2+x)+8)*exp(x)+(-1+2*x)*exp(-x*
*2+x))*exp(exp(x)))/(exp(-x**2+x)+4),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx={\left (x^{2} + 2\right )} e^{\left (e^{x}\right )} - e^{\left (e^{x}\right )} \log \left (4 \, e^{\left (x^{2}\right )} + e^{x}\right ) \]

[In]

integrate(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*log(exp(-x^2+x)+4)+((2*exp(-x^2+x)+8)*exp(x)+(-1+2*x)*exp(-x^2+
x))*exp(exp(x)))/(exp(-x^2+x)+4),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx={\left (x^{2} e^{\left (x + e^{x}\right )} - e^{\left (x + e^{x}\right )} \log \left (4 \, e^{\left (x^{2}\right )} + e^{x}\right ) + 2 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \]

[In]

integrate(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*log(exp(-x^2+x)+4)+((2*exp(-x^2+x)+8)*exp(x)+(-1+2*x)*exp(-x^2+
x))*exp(exp(x)))/(exp(-x^2+x)+4),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=\int \frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{x-x^2}\,\left (2\,x-1\right )+{\mathrm {e}}^x\,\left (2\,{\mathrm {e}}^{x-x^2}+8\right )\right )-\ln \left ({\mathrm {e}}^{x-x^2}+4\right )\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{x-x^2}+4\right )}{{\mathrm {e}}^{x-x^2}+4} \,d x \]

[In]

int((exp(exp(x))*(exp(x - x^2)*(2*x - 1) + exp(x)*(2*exp(x - x^2) + 8)) - log(exp(x - x^2) + 4)*exp(exp(x))*ex
p(x)*(exp(x - x^2) + 4))/(exp(x - x^2) + 4),x)

[Out]

int((exp(exp(x))*(exp(x - x^2)*(2*x - 1) + exp(x)*(2*exp(x - x^2) + 8)) - log(exp(x - x^2) + 4)*exp(x + exp(x)
)*(exp(x - x^2) + 4))/(exp(x - x^2) + 4), x)