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

Optimal. Leaf size=23 \[ -2+4 x+\frac {4 x^2}{e^{1-x}-x} \]

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Rubi [F]  time = 0.33, 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 {4 e^2+4 e^{1+x} x^2}{e^2-2 e^{1+x} x+e^{2 x} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

4*E^2*Defer[Int][(E - E^x*x)^(-2), x] + 4*E^2*Defer[Int][x/(-E + E^x*x)^2, x] + 4*E*Defer[Int][x/(-E + E^x*x),
x]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 15, normalized size = 0.65 \begin {gather*} -\frac {4 e x}{-e+e^x x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*E*x)/(-E + E^x*x)

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fricas [A]  time = 0.55, size = 18, normalized size = 0.78 \begin {gather*} -\frac {4 \, x e^{2}}{x e^{\left (x + 1\right )} - e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*exp(1)*exp(x)+4*exp(1)^2)/(exp(x)^2*x^2-2*x*exp(1)*exp(x)+exp(1)^2),x, algorithm="fricas")

[Out]

-4*x*e^2/(x*e^(x + 1) - e^2)

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giac [A]  time = 0.38, size = 16, normalized size = 0.70 \begin {gather*} -\frac {4 \, x e}{x e^{x} - e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*exp(1)*exp(x)+4*exp(1)^2)/(exp(x)^2*x^2-2*x*exp(1)*exp(x)+exp(1)^2),x, algorithm="giac")

[Out]

-4*x*e/(x*e^x - e)

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maple [A]  time = 0.06, size = 16, normalized size = 0.70

method	result	size
norman	\(\frac {4 x \,{\mathrm e}}{-{\mathrm e}^{x} x +{\mathrm e}}\)	\(16\)
risch	\(\frac {4 x \,{\mathrm e}}{-{\mathrm e}^{x} x +{\mathrm e}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2*exp(1)*exp(x)+4*exp(1)^2)/(exp(x)^2*x^2-2*x*exp(1)*exp(x)+exp(1)^2),x,method=_RETURNVERBOSE)

[Out]

4*x*exp(1)/(-exp(x)*x+exp(1))

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maxima [A]  time = 0.50, size = 16, normalized size = 0.70 \begin {gather*} -\frac {4 \, x e}{x e^{x} - e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*exp(1)*exp(x)+4*exp(1)^2)/(exp(x)^2*x^2-2*x*exp(1)*exp(x)+exp(1)^2),x, algorithm="maxima")

[Out]

-4*x*e/(x*e^x - e)

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mupad [B]  time = 1.00, size = 15, normalized size = 0.65 \begin {gather*} \frac {4\,x\,\mathrm {e}}{\mathrm {e}-x\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x*exp(1))/(exp(1) - x*exp(x))

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sympy [A]  time = 0.10, size = 15, normalized size = 0.65 \begin {gather*} - \frac {4 e x}{x e^{x} - e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*E*x/(x*exp(x) - E)

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