3.81.3 \(\int \frac {e^3+2 e^{5-x^2} x}{-3+e^{5-x^2}+e^3 (-2-x)} \, dx\)

Optimal. Leaf size=31 \[ -\frac {1}{4}-\log \left (\frac {1}{3} \left (3+e^3 \left (2-e^{2-x^2}+x\right )\right )\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.35, number of steps used = 1, number of rules used = 1, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6684} \begin {gather*} -\log \left (-e^{-x^2} \left (-e^{x^2+3} x-3 e^{x^2}-2 e^{x^2+3}+e^5\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6684

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\log \left (-e^{-x^2} \left (e^5-3 e^{x^2}-2 e^{3+x^2}-e^{3+x^2} x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.64, size = 22, normalized size = 0.71 \begin {gather*} -\log \left (-3+e^{5-x^2}-e^3 (2+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[-3 + E^(5 - x^2) - E^3*(2 + x)]

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fricas [A]  time = 0.52, size = 20, normalized size = 0.65 \begin {gather*} -\log \left (-{\left (x + 2\right )} e^{3} + e^{\left (-x^{2} + 5\right )} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-(x + 2)*e^3 + e^(-x^2 + 5) - 3)

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giac [A]  time = 0.17, size = 22, normalized size = 0.71 \begin {gather*} -\log \left (-x e^{3} - 2 \, e^{3} + e^{\left (-x^{2} + 5\right )} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 26, normalized size = 0.84

method	result	size
norman	\(-\ln \left (x \,{\mathrm e}^{3}-{\mathrm e}^{3} {\mathrm e}^{-x^{2}+2}+2 \,{\mathrm e}^{3}+3\right )\)	\(26\)
risch	\(2-\ln \left ({\mathrm e}^{-x^{2}+2}-\left (x \,{\mathrm e}^{3}+2 \,{\mathrm e}^{3}+3\right ) {\mathrm e}^{-3}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 21, normalized size = 0.68 \begin {gather*} -\log \left ({\left (x + 2\right )} e^{3} - e^{\left (-x^{2} + 5\right )} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log((x + 2)*e^3 - e^(-x^2 + 5) + 3)

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mupad [B]  time = 0.38, size = 20, normalized size = 0.65 \begin {gather*} -\ln \left (x+3\,{\mathrm {e}}^{-3}-{\mathrm {e}}^{2-x^2}+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3) + 2*x*exp(3)*exp(2 - x^2))/(exp(3)*(x + 2) - exp(3)*exp(2 - x^2) + 3),x)

[Out]

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

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sympy [A]  time = 0.17, size = 26, normalized size = 0.84 \begin {gather*} - \log {\left (\frac {- x e^{3} - 2 e^{3} - 3}{e^{3}} + e^{2 - x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(3)*exp(-x**2+2)+exp(3))/(exp(3)*exp(-x**2+2)+(-x-2)*exp(3)-3),x)

[Out]

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

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