\(\int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} (14 x^3+4 x^4+(-22 x^2-6 x^3) \log (4+x)+(8 x+2 x^2) \log ^2(4+x))))}{4+x} \, dx\) [9449]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 161, antiderivative size = 36 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=2+e^{3-e^{3-e^{x^2 (-x+\log (4+x))^2}-x}-x} \]

[Out]

2+exp(3-exp(3-exp(x^2*(ln(4+x)-x)^2)-x)-x)

Rubi [A] (verified)

Time = 4.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.006, Rules used = {6838} \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=\exp \left (-\exp \left ((x+4)^{-2 x^3} \left (-e^{x^4+x^2 \log ^2(x+4)}\right )-x+3\right )-x+3\right ) \]

[In]

Int[(E^(3 - E^(3 - E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2) - x) - x)*(-4 - x + E^(3 - E^(x^4 - 2*x^3*Log
[4 + x] + x^2*Log[4 + x]^2) - x)*(4 + x + E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2)*(14*x^3 + 4*x^4 + (-22
*x^2 - 6*x^3)*Log[4 + x] + (8*x + 2*x^2)*Log[4 + x]^2))))/(4 + x),x]

[Out]

E^(3 - E^(3 - x - E^(x^4 + x^2*Log[4 + x]^2)/(4 + x)^(2*x^3)) - 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}& = \exp \left (3-\exp \left (3-x-e^{x^4+x^2 \log ^2(4+x)} (4+x)^{-2 x^3}\right )-x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{3-e^{3-x-e^{x^4+x^2 \log ^2(4+x)} (4+x)^{-2 x^3}}-x} \]

[In]

Integrate[(E^(3 - E^(3 - E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2) - x) - x)*(-4 - x + E^(3 - E^(x^4 - 2*x
^3*Log[4 + x] + x^2*Log[4 + x]^2) - x)*(4 + x + E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2)*(14*x^3 + 4*x^4
+ (-22*x^2 - 6*x^3)*Log[4 + x] + (8*x + 2*x^2)*Log[4 + x]^2))))/(4 + x),x]

[Out]

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

Maple [A] (verified)

Time = 4.73 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14

method result size
risch \({\mathrm e}^{-{\mathrm e}^{-\left (4+x \right )^{-2 x^{3}} {\mathrm e}^{x^{2} \left (\ln \left (4+x \right )^{2}+x^{2}\right )}+3-x}+3-x}\) \(41\)
parallelrisch \({\mathrm e}^{-{\mathrm e}^{-{\mathrm e}^{x^{2} \ln \left (4+x \right )^{2}-2 x^{3} \ln \left (4+x \right )+x^{4}}+3-x}+3-x}\) \(41\)

[In]

int(((((2*x^2+8*x)*ln(4+x)^2+(-6*x^3-22*x^2)*ln(4+x)+4*x^4+14*x^3)*exp(x^2*ln(4+x)^2-2*x^3*ln(4+x)+x^4)+4+x)*e
xp(-exp(x^2*ln(4+x)^2-2*x^3*ln(4+x)+x^4)+3-x)-x-4)*exp(-exp(-exp(x^2*ln(4+x)^2-2*x^3*ln(4+x)+x^4)+3-x)+3-x)/(4
+x),x,method=_RETURNVERBOSE)

[Out]

exp(-exp(-(4+x)^(-2*x^3)*exp(x^2*(ln(4+x)^2+x^2))+3-x)+3-x)

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{\left (-x - e^{\left (-x - e^{\left (x^{4} - 2 \, x^{3} \log \left (x + 4\right ) + x^{2} \log \left (x + 4\right )^{2}\right )} + 3\right )} + 3\right )} \]

[In]

integrate(((((2*x^2+8*x)*log(4+x)^2+(-6*x^3-22*x^2)*log(4+x)+4*x^4+14*x^3)*exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x
^4)+4+x)*exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+3-x)-x-4)*exp(-exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4
)+3-x)+3-x)/(4+x),x, algorithm="fricas")

[Out]

e^(-x - e^(-x - e^(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) + 3) + 3)

Sympy [A] (verification not implemented)

Time = 166.79 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{- x - e^{- x - e^{x^{4} - 2 x^{3} \log {\left (x + 4 \right )} + x^{2} \log {\left (x + 4 \right )}^{2}} + 3} + 3} \]

[In]

integrate(((((2*x**2+8*x)*ln(4+x)**2+(-6*x**3-22*x**2)*ln(4+x)+4*x**4+14*x**3)*exp(x**2*ln(4+x)**2-2*x**3*ln(4
+x)+x**4)+4+x)*exp(-exp(x**2*ln(4+x)**2-2*x**3*ln(4+x)+x**4)+3-x)-x-4)*exp(-exp(-exp(x**2*ln(4+x)**2-2*x**3*ln
(4+x)+x**4)+3-x)+3-x)/(4+x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{\left (-x - e^{\left (-x - e^{\left (x^{4} - 2 \, x^{3} \log \left (x + 4\right ) + x^{2} \log \left (x + 4\right )^{2}\right )} + 3\right )} + 3\right )} \]

[In]

integrate(((((2*x^2+8*x)*log(4+x)^2+(-6*x^3-22*x^2)*log(4+x)+4*x^4+14*x^3)*exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x
^4)+4+x)*exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+3-x)-x-4)*exp(-exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4
)+3-x)+3-x)/(4+x),x, algorithm="maxima")

[Out]

e^(-x - e^(-x - e^(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) + 3) + 3)

Giac [A] (verification not implemented)

none

Time = 23.50 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{\left (-x - e^{\left (-x - e^{\left (x^{4} - 2 \, x^{3} \log \left (x + 4\right ) + x^{2} \log \left (x + 4\right )^{2}\right )} + 3\right )} + 3\right )} \]

[In]

integrate(((((2*x^2+8*x)*log(4+x)^2+(-6*x^3-22*x^2)*log(4+x)+4*x^4+14*x^3)*exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x
^4)+4+x)*exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+3-x)-x-4)*exp(-exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4
)+3-x)+3-x)/(4+x),x, algorithm="giac")

[Out]

e^(-x - e^(-x - e^(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) + 3) + 3)

Mupad [B] (verification not implemented)

Time = 14.80 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx={\mathrm {e}}^{-{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{x^2\,{\ln \left (x+4\right )}^2}}{{\left (x+4\right )}^{2\,x^3}}}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3 \]

[In]

int(-(exp(3 - exp(3 - exp(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) - x) - x)*(x - exp(3 - exp(x^4 - 2*x^3*lo
g(x + 4) + x^2*log(x + 4)^2) - x)*(x + exp(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2)*(log(x + 4)^2*(8*x + 2*x
^2) - log(x + 4)*(22*x^2 + 6*x^3) + 14*x^3 + 4*x^4) + 4) + 4))/(x + 4),x)

[Out]

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