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

Optimal. Leaf size=29 \[ -\log \left (\frac {5}{2}\right )+e^{e^x} \left (2-\log \left (4+e^{x-x^2}\right )\right ) \]

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Rubi [A]  time = 2.03, antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 5, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6688, 6742, 2282, 2194, 2554} \begin {gather*} 2 e^{e^x}-e^{e^x} \log \left (e^{x-x^2}+4\right ) \end {gather*}

Antiderivative was successfully verified.

[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 2194

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 2282

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 2554

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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 1.13, size = 21, normalized size = 0.72 \begin {gather*} -e^{e^x} \left (-2+\log \left (4+e^{x-x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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)]))

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fricas [A]  time = 0.61, size = 31, normalized size = 1.07 \begin {gather*} -{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)+(2*x-1)*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)

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giac [A]  time = 0.19, size = 39, normalized size = 1.34 \begin {gather*} {\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)+(2*x-1)*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)

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maple [A]  time = 0.07, size = 22, normalized size = 0.76

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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maxima [A]  time = 0.84, size = 25, normalized size = 0.86 \begin {gather*} {\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)+(2*x-1)*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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)+(2*x-1)*exp(-x**
2+x))*exp(exp(x)))/(exp(-x**2+x)+4),x)

[Out]

Timed out

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