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

Optimal. Leaf size=15 \[ e^{-e^{\frac {1}{5+x}}} (-3+x) \]

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Rubi [A]  time = 0.09, antiderivative size = 18, normalized size of antiderivative = 1.20, number of steps used = 2, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {27, 2288} \begin {gather*} -e^{-e^{\frac {1}{x+5}}} (3-x) \end {gather*}

Antiderivative was successfully verified.

[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 2288

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

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} e^{-e^{\frac {1}{5+x}}} (-3+x) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 1.05, size = 13, normalized size = 0.87 \begin {gather*} {\left (x - 3\right )} e^{\left (-e^{\left (\frac {1}{x + 5}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*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)))

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giac [A]  time = 0.19, size = 23, normalized size = 1.53 \begin {gather*} x e^{\left (-e^{\left (\frac {1}{x + 5}\right )}\right )} - 3 \, e^{\left (-e^{\left (\frac {1}{x + 5}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*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)))

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maple [A]  time = 0.45, size = 14, normalized size = 0.93

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.56, size = 13, normalized size = 0.87 \begin {gather*} {\left (x - 3\right )} e^{\left (-e^{\left (\frac {1}{x + 5}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*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)))

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mupad [B]  time = 0.18, size = 13, normalized size = 0.87 \begin {gather*} {\mathrm {e}}^{-{\mathrm {e}}^{\frac {1}{x+5}}}\,\left (x-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 5.30, size = 10, normalized size = 0.67 \begin {gather*} \left (x - 3\right ) e^{- e^{\frac {1}{x + 5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*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)))

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