3.28.11 \(\int \frac {6+e^{-5+e^{\frac {1}{4} (1-4 x-x^2)}} (2+e^{\frac {1}{4} (1-4 x-x^2)} (2+x))}{6+2 e^{-5+e^{\frac {1}{4} (1-4 x-x^2)}}} \, dx\)

Optimal. Leaf size=28 \[ x-\log \left (3+e^{-5+e^{\frac {1}{4} (1-x-x (3+x))}}\right ) \]

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Rubi [A]  time = 0.79, antiderivative size = 29, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 2, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6742, 6684} \begin {gather*} x-\log \left (e^{e^{-\frac {x^2}{4}-x+\frac {1}{4}}}+3 e^5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + E^(-5 + E^((1 - 4*x - x^2)/4))*(2 + E^((1 - 4*x - x^2)/4)*(2 + x)))/(6 + 2*E^(-5 + E^((1 - 4*x - x^2)
/4))),x]

[Out]

x - Log[3*E^5 + E^E^(1/4 - x - x^2/4)]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 \left (1+\frac {e^{\frac {1}{4}+e^{\frac {1}{4}-x-\frac {x^2}{4}}-x-\frac {x^2}{4}} (2+x)}{2 \left (3 e^5+e^{e^{\frac {1}{4}-x-\frac {x^2}{4}}}\right )}\right ) \, dx\\ &=x+\frac {1}{2} \int \frac {e^{\frac {1}{4}+e^{\frac {1}{4}-x-\frac {x^2}{4}}-x-\frac {x^2}{4}} (2+x)}{3 e^5+e^{e^{\frac {1}{4}-x-\frac {x^2}{4}}}} \, dx\\ &=x-\log \left (3 e^5+e^{e^{\frac {1}{4}-x-\frac {x^2}{4}}}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.70, size = 29, normalized size = 1.04 \begin {gather*} x-\log \left (3 e^5+e^{e^{\frac {1}{4}-x-\frac {x^2}{4}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + E^(-5 + E^((1 - 4*x - x^2)/4))*(2 + E^((1 - 4*x - x^2)/4)*(2 + x)))/(6 + 2*E^(-5 + E^((1 - 4*x
- x^2)/4))),x]

[Out]

x - Log[3*E^5 + E^E^(1/4 - x - x^2/4)]

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fricas [A]  time = 0.50, size = 21, normalized size = 0.75 \begin {gather*} x - \log \left (e^{\left (e^{\left (-\frac {1}{4} \, x^{2} - x + \frac {1}{4}\right )} - 5\right )} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*exp(-1/4*x^2-x+1/4)+2)*exp(exp(-1/4*x^2-x+1/4)-5)+6)/(2*exp(exp(-1/4*x^2-x+1/4)-5)+6),x, alg
orithm="fricas")

[Out]

x - log(e^(e^(-1/4*x^2 - x + 1/4) - 5) + 3)

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giac [B]  time = 0.55, size = 87, normalized size = 3.11 \begin {gather*} -\frac {1}{4} \, x^{2} e^{\left (-\frac {1}{4} \, x^{2} - x + \frac {1}{4}\right )} - \frac {1}{4} \, x^{2} - x e^{\left (-\frac {1}{4} \, x^{2} - x + \frac {1}{4}\right )} - e^{\left (-\frac {1}{4} \, x^{2} - x + \frac {1}{4}\right )} - \log \left (e^{\left (-\frac {1}{4} \, x^{2} - x + e^{\left (-\frac {1}{4} \, x^{2} - x + \frac {1}{4}\right )}\right )} + 3 \, e^{\left (-\frac {1}{4} \, x^{2} - x + 5\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*exp(-1/4*x^2-x+1/4)+2)*exp(exp(-1/4*x^2-x+1/4)-5)+6)/(2*exp(exp(-1/4*x^2-x+1/4)-5)+6),x, alg
orithm="giac")

[Out]

-1/4*x^2*e^(-1/4*x^2 - x + 1/4) - 1/4*x^2 - x*e^(-1/4*x^2 - x + 1/4) - e^(-1/4*x^2 - x + 1/4) - log(e^(-1/4*x^
2 - x + e^(-1/4*x^2 - x + 1/4)) + 3*e^(-1/4*x^2 - x + 5))

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maple [A]  time = 0.10, size = 23, normalized size = 0.82

method	result	size
risch	\(x -5-\ln \left ({\mathrm e}^{{\mathrm e}^{-\frac {1}{4} x^{2}-x +\frac {1}{4}}-5}+3\right )\)	\(23\)
norman	\(x -\ln \left (2 \,{\mathrm e}^{{\mathrm e}^{-\frac {1}{4} x^{2}-x +\frac {1}{4}}-5}+6\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2+x)*exp(-1/4*x^2-x+1/4)+2)*exp(exp(-1/4*x^2-x+1/4)-5)+6)/(2*exp(exp(-1/4*x^2-x+1/4)-5)+6),x,method=_RE
TURNVERBOSE)

[Out]

x-5-ln(exp(exp(-1/4*x^2-x+1/4)-5)+3)

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maxima [A]  time = 0.76, size = 22, normalized size = 0.79 \begin {gather*} x - \log \left (3 \, e^{5} + e^{\left (e^{\left (-\frac {1}{4} \, x^{2} - x + \frac {1}{4}\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*exp(-1/4*x^2-x+1/4)+2)*exp(exp(-1/4*x^2-x+1/4)-5)+6)/(2*exp(exp(-1/4*x^2-x+1/4)-5)+6),x, alg
orithm="maxima")

[Out]

x - log(3*e^5 + e^(e^(-1/4*x^2 - x + 1/4)))

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mupad [B]  time = 0.17, size = 24, normalized size = 0.86 \begin {gather*} x-\ln \left ({\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^{1/4}\,{\mathrm {e}}^{-\frac {x^2}{4}}}\,{\mathrm {e}}^{-5}+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(1/4 - x^2/4 - x) - 5)*(exp(1/4 - x^2/4 - x)*(x + 2) + 2) + 6)/(2*exp(exp(1/4 - x^2/4 - x) - 5) +
6),x)

[Out]

x - log(exp(exp(-x)*exp(1/4)*exp(-x^2/4))*exp(-5) + 3)

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sympy [A]  time = 0.35, size = 34, normalized size = 1.21 \begin {gather*} x - \frac {e^{- \frac {x^{2}}{4} - x + \frac {1}{4}}}{2} - \frac {\log {\left (e^{e^{- \frac {x^{2}}{4} - x + \frac {1}{4}} - 5} + 3 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*exp(-1/4*x**2-x+1/4)+2)*exp(exp(-1/4*x**2-x+1/4)-5)+6)/(2*exp(exp(-1/4*x**2-x+1/4)-5)+6),x)

[Out]

x - exp(-x**2/4 - x + 1/4)/2 - log(exp(exp(-x**2/4 - x + 1/4) - 5) + 3)/2

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