\(\int \frac {e^{-e^{\frac {1}{5+x}}} (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2)}{25+10 x+x^2} \, dx\) [2820]

Optimal result

Integrand size = 41, antiderivative size = 15 \[ \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{25+10 x+x^2} \, dx=e^{-e^{\frac {1}{5+x}}} (-3+x) \]

[Out]

(-3+x)/exp(exp(1/(5+x)))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {27, 2326} \[ \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{25+10 x+x^2} \, dx=-e^{-e^{\frac {1}{x+5}}} (3-x) \]

[In]

Int[(25 + E^(5 + x)^(-1)*(-3 + x) + 10*x + x^2)/(E^E^(5 + x)^(-1)*(25 + 10*x + x^2)),x]

[Out]

-((3 - x)/E^E^(5 + x)^(-1))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{(5+x)^2} \, dx \\ & = -e^{-e^{\frac {1}{5+x}}} (3-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{25+10 x+x^2} \, dx=e^{-e^{\frac {1}{5+x}}} (-3+x) \]

[In]

Integrate[(25 + E^(5 + x)^(-1)*(-3 + x) + 10*x + x^2)/(E^E^(5 + x)^(-1)*(25 + 10*x + x^2)),x]

[Out]

(-3 + x)/E^E^(5 + x)^(-1)

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
risch	\(\left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{\frac {1}{5+x}}}\)	\(14\)
parallelrisch	\(-\frac {\left (900-300 x \right ) {\mathrm e}^{-{\mathrm e}^{\frac {1}{5+x}}}}{300}\)	\(17\)
norman	\(\frac {\left (x^{2}+2 x -15\right ) {\mathrm e}^{-{\mathrm e}^{\frac {1}{5+x}}}}{5+x}\)	\(24\)

[In]

int(((-3+x)*exp(1/(5+x))+x^2+10*x+25)/(x^2+10*x+25)/exp(exp(1/(5+x))),x,method=_RETURNVERBOSE)

[Out]

(-3+x)*exp(-exp(1/(5+x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{25+10 x+x^2} \, dx={\left (x - 3\right )} e^{\left (-e^{\left (\frac {1}{x + 5}\right )}\right )} \]

[In]

integrate(((-3+x)*exp(1/(5+x))+x^2+10*x+25)/(x^2+10*x+25)/exp(exp(1/(5+x))),x, algorithm="fricas")

[Out]

(x - 3)*e^(-e^(1/(x + 5)))

Sympy [A] (verification not implemented)

Time = 2.42 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{25+10 x+x^2} \, dx=\left (x - 3\right ) e^{- e^{\frac {1}{x + 5}}} \]

[In]

integrate(((-3+x)*exp(1/(5+x))+x**2+10*x+25)/(x**2+10*x+25)/exp(exp(1/(5+x))),x)

[Out]

(x - 3)*exp(-exp(1/(x + 5)))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{25+10 x+x^2} \, dx={\left (x - 3\right )} e^{\left (-e^{\left (\frac {1}{x + 5}\right )}\right )} \]

[In]

integrate(((-3+x)*exp(1/(5+x))+x^2+10*x+25)/(x^2+10*x+25)/exp(exp(1/(5+x))),x, algorithm="maxima")

[Out]

(x - 3)*e^(-e^(1/(x + 5)))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.53 \[ \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{25+10 x+x^2} \, dx=x e^{\left (-e^{\left (\frac {1}{x + 5}\right )}\right )} - 3 \, e^{\left (-e^{\left (\frac {1}{x + 5}\right )}\right )} \]

[In]

integrate(((-3+x)*exp(1/(5+x))+x^2+10*x+25)/(x^2+10*x+25)/exp(exp(1/(5+x))),x, algorithm="giac")

[Out]

x*e^(-e^(1/(x + 5))) - 3*e^(-e^(1/(x + 5)))

Mupad [B] (verification not implemented)

Time = 8.88 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-e^{\frac {1}{5+x}}} \left (25+e^{\frac {1}{5+x}} (-3+x)+10 x+x^2\right )}{25+10 x+x^2} \, dx={\mathrm {e}}^{-{\mathrm {e}}^{\frac {1}{x+5}}}\,\left (x-3\right ) \]

[In]

int((exp(-exp(1/(x + 5)))*(10*x + exp(1/(x + 5))*(x - 3) + x^2 + 25))/(10*x + x^2 + 25),x)

[Out]

exp(-exp(1/(x + 5)))*(x - 3)