3.75.77 \(\int \frac {1+e^{e^2+x} (-2 e-2 x)-x}{-x+e^{e^2+x} (e^2+2 e x+x^2)} \, dx\)

Optimal. Leaf size=21 \[ x-\log \left (x-e^{e^2+x} (e+x)^2\right ) \]

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Rubi [F]  time = 0.95, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1+e^{e^2+x} (-2 e-2 x)-x}{-x+e^{e^2+x} \left (e^2+2 e x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-2*Log[E + x] - Defer[Int][(E^(2 + E^2 + x) - x + 2*E^(1 + E^2 + x)*x + E^(E^2 + x)*x^2)^(-1), x] - Defer[Int]
[x/(E^(2 + E^2 + x) - x + 2*E^(1 + E^2 + x)*x + E^(E^2 + x)*x^2), x] + 2*E*Defer[Int][1/((E + x)*(E^(2 + E^2 +
x) - x + 2*E^(1 + E^2 + x)*x + E^(E^2 + x)*x^2)), x]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 39, normalized size = 1.86 \begin {gather*} x-\log \left (e^{2+e^2+x}-x+2 e^{1+e^2+x} x+e^{e^2+x} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - Log[E^(2 + E^2 + x) - x + 2*E^(1 + E^2 + x)*x + E^(E^2 + x)*x^2]

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fricas [B]  time = 0.94, size = 47, normalized size = 2.24 \begin {gather*} x - 2 \, \log \left (x + e\right ) - \log \left (\frac {{\left (x^{2} + 2 \, x e + e^{2}\right )} e^{\left (x + e^{2}\right )} - x}{x^{2} + 2 \, x e + e^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 2*log(x + e) - log(((x^2 + 2*x*e + e^2)*e^(x + e^2) - x)/(x^2 + 2*x*e + e^2))

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giac [A]  time = 0.25, size = 36, normalized size = 1.71 \begin {gather*} x - \log \left (x^{2} e^{\left (x + e^{2} + 1\right )} - x e + 2 \, x e^{\left (x + e^{2} + 2\right )} + e^{\left (x + e^{2} + 3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 37, normalized size = 1.76

method	result	size
risch	\(-2 \ln \left (x +{\mathrm e}\right )+{\mathrm e}^{2}+x -\ln \left ({\mathrm e}^{x +{\mathrm e}^{2}}-\frac {x}{{\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}}\right )\)	\(37\)
norman	\(x -\ln \left ({\mathrm e}^{2} {\mathrm e}^{x +{\mathrm e}^{2}}+2 \,{\mathrm e} \,{\mathrm e}^{x +{\mathrm e}^{2}} x +{\mathrm e}^{x +{\mathrm e}^{2}} x^{2}-x \right )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 64, normalized size = 3.05 \begin {gather*} x - 2 \, \log \left (x + e\right ) - \log \left (\frac {{\left (x^{2} e^{\left (e^{2}\right )} + 2 \, x e^{\left (e^{2} + 1\right )} + e^{\left (e^{2} + 2\right )}\right )} e^{x} - x}{x^{2} e^{\left (e^{2}\right )} + 2 \, x e^{\left (e^{2} + 1\right )} + e^{\left (e^{2} + 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.31, size = 33, normalized size = 1.57 \begin {gather*} x-\ln \left ({\mathrm {e}}^{x+{\mathrm {e}}^2+2}-x+2\,x\,{\mathrm {e}}^{x+{\mathrm {e}}^2+1}+x^2\,{\mathrm {e}}^{x+{\mathrm {e}}^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - log(exp(x + exp(2) + 2) - x + 2*x*exp(x + exp(2) + 1) + x^2*exp(x + exp(2)))

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sympy [A]  time = 0.39, size = 32, normalized size = 1.52 \begin {gather*} x - 2 \log {\left (x + e \right )} - \log {\left (- \frac {x}{x^{2} + 2 e x + e^{2}} + e^{x + e^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 2*log(x + E) - log(-x/(x**2 + 2*E*x + exp(2)) + exp(x + exp(2)))

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