3.85.7 \(\int \frac {-2 x^2+e^x (4 x-2 x^2)+e^{1+x} (e^x-x^2)}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx\)

Optimal. Leaf size=28 \[ \frac {-e^x+x}{5 e^3 x \left (e^{1+x}+2 x\right )} \]

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Rubi [F]  time = 0.70, 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 {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x^2 + E^x*(4*x - 2*x^2) + E^(1 + x)*(E^x - x^2))/(5*E^(5 + 2*x)*x^2 + 20*E^(4 + x)*x^3 + 20*E^3*x^4),x
]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 28, normalized size = 1.00 \begin {gather*} \frac {-e^x+x}{5 e^3 x \left (e^{1+x}+2 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^2 + E^x*(4*x - 2*x^2) + E^(1 + x)*(E^x - x^2))/(5*E^(5 + 2*x)*x^2 + 20*E^(4 + x)*x^3 + 20*E^3*
x^4),x]

[Out]

(-E^x + x)/(5*E^3*x*(E^(1 + x) + 2*x))

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fricas [A]  time = 0.88, size = 29, normalized size = 1.04 \begin {gather*} \frac {x e^{4} - e^{\left (x + 4\right )}}{5 \, {\left (2 \, x^{2} e^{7} + x e^{\left (x + 8\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-x^2)*exp(x+1)+(-2*x^2+4*x)*exp(x)-2*x^2)/(5*x^2*exp(3)*exp(x+1)^2+20*x^3*exp(3)*exp(x+1)+20
*x^4*exp(3)),x, algorithm="fricas")

[Out]

1/5*(x*e^4 - e^(x + 4))/(2*x^2*e^7 + x*e^(x + 8))

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giac [A]  time = 0.15, size = 24, normalized size = 0.86 \begin {gather*} \frac {x - e^{x}}{5 \, {\left (2 \, x^{2} e^{3} + x e^{\left (x + 4\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-x^2)*exp(x+1)+(-2*x^2+4*x)*exp(x)-2*x^2)/(5*x^2*exp(3)*exp(x+1)^2+20*x^3*exp(3)*exp(x+1)+20
*x^4*exp(3)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.16, size = 32, normalized size = 1.14

method	result	size
norman	\(\frac {\frac {x \,{\mathrm e}^{-3}}{5}-\frac {{\mathrm e}^{-3} {\mathrm e}^{x}}{5}}{x \left ({\mathrm e} \,{\mathrm e}^{x}+2 x \right )}\)	\(32\)
risch	\(-\frac {{\mathrm e}^{-4}}{5 x}+\frac {{\mathrm e}^{-4} {\mathrm e}}{10 x +5 \,{\mathrm e}^{x +1}}+\frac {2 \,{\mathrm e}^{-4}}{5 \left (2 x +{\mathrm e}^{x +1}\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)-x^2)*exp(x+1)+(-2*x^2+4*x)*exp(x)-2*x^2)/(5*x^2*exp(3)*exp(x+1)^2+20*x^3*exp(3)*exp(x+1)+20*x^4*e
xp(3)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 24, normalized size = 0.86 \begin {gather*} \frac {x - e^{x}}{5 \, {\left (2 \, x^{2} e^{3} + x e^{\left (x + 4\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-x^2)*exp(x+1)+(-2*x^2+4*x)*exp(x)-2*x^2)/(5*x^2*exp(3)*exp(x+1)^2+20*x^3*exp(3)*exp(x+1)+20
*x^4*exp(3)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.28, size = 23, normalized size = 0.82 \begin {gather*} \frac {x-{\mathrm {e}}^x}{5\,x\,\left ({\mathrm {e}}^{x+4}+2\,x\,{\mathrm {e}}^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + 1)*(exp(x) - x^2) + exp(x)*(4*x - 2*x^2) - 2*x^2)/(20*x^4*exp(3) + 5*x^2*exp(3)*exp(2*x + 2) + 20
*x^3*exp(x + 1)*exp(3)),x)

[Out]

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

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sympy [A]  time = 0.15, size = 27, normalized size = 0.96 \begin {gather*} \frac {2 + e}{10 x e^{4} + 5 e^{5} e^{x}} - \frac {1}{5 x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-x**2)*exp(x+1)+(-2*x**2+4*x)*exp(x)-2*x**2)/(5*x**2*exp(3)*exp(x+1)**2+20*x**3*exp(3)*exp(x
+1)+20*x**4*exp(3)),x)

[Out]

(2 + E)/(10*x*exp(4) + 5*exp(5)*exp(x)) - exp(-4)/(5*x)

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