3.83.20 \(\int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} (16+8 x+2 e^2 x+x^2)}{80+40 x+5 x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{5} e^{e+\frac {e^2 (2+x)}{4+x}} x \]

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Rubi [F]  time = 0.86, 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 {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+8 x+2 e^2 x+x^2\right )}{80+40 x+5 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^2*(2 + x) + E*(4 + x))/(4 + x))*(16 + 8*x + 2*E^2*x + x^2))/(80 + 40*x + 5*x^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+\left (8+2 e^2\right ) x+x^2\right )}{80+40 x+5 x^2} \, dx\\ &=\int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+\left (8+2 e^2\right ) x+x^2\right )}{5 (4+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+\left (8+2 e^2\right ) x+x^2\right )}{(4+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {2 e (2+e)+e (1+e) x}{4+x}} \left (16+2 \left (4+e^2\right ) x+x^2\right )}{(4+x)^2} \, dx\\ &=\frac {1}{5} \int \left (e^{\frac {2 e (2+e)+e (1+e) x}{4+x}}-\frac {8 e^{2+\frac {2 e (2+e)+e (1+e) x}{4+x}}}{(4+x)^2}+\frac {2 e^{2+\frac {2 e (2+e)+e (1+e) x}{4+x}}}{4+x}\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {2 e (2+e)+e (1+e) x}{4+x}} \, dx+\frac {2}{5} \int \frac {e^{2+\frac {2 e (2+e)+e (1+e) x}{4+x}}}{4+x} \, dx-\frac {8}{5} \int \frac {e^{2+\frac {2 e (2+e)+e (1+e) x}{4+x}}}{(4+x)^2} \, dx\\ &=\frac {1}{5} \int e^{\frac {2 e (2+e)+e (1+e) x}{4+x}} \, dx+\frac {2}{5} \int \frac {\exp \left (2+e (1+e)-\frac {4 e (1+e)-2 e (2+e)}{4+x}\right )}{4+x} \, dx-\frac {8}{5} \int \frac {\exp \left (2+e (1+e)-\frac {4 e (1+e)-2 e (2+e)}{4+x}\right )}{(4+x)^2} \, dx\\ &=-\frac {4}{5} e^{e (1+e)-\frac {2 e^2}{4+x}}-\frac {2}{5} e^{2+e+e^2} \text {Ei}\left (-\frac {2 e^2}{4+x}\right )+\frac {1}{5} \int e^{\frac {2 e (2+e)+e (1+e) x}{4+x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 22, normalized size = 1.05 \begin {gather*} \frac {1}{5} e^{\frac {e (4+x+e (2+x))}{4+x}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^2*(2 + x) + E*(4 + x))/(4 + x))*(16 + 8*x + 2*E^2*x + x^2))/(80 + 40*x + 5*x^2),x]

[Out]

(E^((E*(4 + x + E*(2 + x)))/(4 + x))*x)/5

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fricas [A]  time = 0.90, size = 23, normalized size = 1.10 \begin {gather*} \frac {1}{5} \, x e^{\left (\frac {{\left (x + 2\right )} e^{2} + {\left (x + 4\right )} e}{x + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*x+x^2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/(5*x^2+40*x+80),x, algorithm="fricas"
)

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} + 2 \, x e^{2} + 8 \, x + 16\right )} e^{\left (\frac {{\left (x + 2\right )} e^{2} + {\left (x + 4\right )} e}{x + 4}\right )}}{5 \, {\left (x^{2} + 8 \, x + 16\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*x+x^2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/(5*x^2+40*x+80),x, algorithm="giac")

[Out]

integrate(1/5*(x^2 + 2*x*e^2 + 8*x + 16)*e^(((x + 2)*e^2 + (x + 4)*e)/(x + 4))/(x^2 + 8*x + 16), x)

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maple [A]  time = 0.35, size = 28, normalized size = 1.33

method	result	size
gosper	\(\frac {{\mathrm e}^{\frac {{\mathrm e}^{2} x +x \,{\mathrm e}+2 \,{\mathrm e}^{2}+4 \,{\mathrm e}}{4+x}} x}{5}\)	\(28\)
risch	\(\frac {{\mathrm e}^{\frac {{\mathrm e}^{2} x +x \,{\mathrm e}+2 \,{\mathrm e}^{2}+4 \,{\mathrm e}}{4+x}} x}{5}\)	\(28\)
derivativedivides	\(\frac {2 \,{\mathrm e}^{2} \left (\frac {{\mathrm e}^{{\mathrm e}^{2}+{\mathrm e}-\frac {2 \,{\mathrm e}^{2}}{4+x}} {\mathrm e}^{-2} \left (4+x \right )}{2}-2 \,{\mathrm e}^{{\mathrm e}^{2}+{\mathrm e}-\frac {2 \,{\mathrm e}^{2}}{4+x}} {\mathrm e}^{-2}\right )}{5}\)	\(51\)
default	\(\frac {2 \,{\mathrm e}^{2} \left (\frac {{\mathrm e}^{{\mathrm e}^{2}+{\mathrm e}-\frac {2 \,{\mathrm e}^{2}}{4+x}} {\mathrm e}^{-2} \left (4+x \right )}{2}-2 \,{\mathrm e}^{{\mathrm e}^{2}+{\mathrm e}-\frac {2 \,{\mathrm e}^{2}}{4+x}} {\mathrm e}^{-2}\right )}{5}\)	\(51\)
norman	\(\frac {\frac {4 x \,{\mathrm e}^{\frac {\left (2+x \right ) {\mathrm e}^{2}+\left (4+x \right ) {\mathrm e}}{4+x}}}{5}+\frac {x^{2} {\mathrm e}^{\frac {\left (2+x \right ) {\mathrm e}^{2}+\left (4+x \right ) {\mathrm e}}{4+x}}}{5}}{4+x}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(2)*x+x^2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/(5*x^2+40*x+80),x,method=_RETURNVERBOSE)

[Out]

1/5*exp((exp(2)*x+x*exp(1)+2*exp(2)+4*exp(1))/(4+x))*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{5} \, x e^{\left (-\frac {2 \, e^{2}}{x + 4} + e^{2} + e\right )} + \frac {8}{5} \, e^{\left (-\frac {2 \, e^{2}}{x + 4} + e^{2} + e - 2\right )} - \frac {16}{5} \, \int \frac {e^{\left (-\frac {2 \, e^{2}}{x + 4} + e^{2} + e\right )}}{x^{2} + 8 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*x+x^2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/(5*x^2+40*x+80),x, algorithm="maxima"
)

[Out]

1/5*x*e^(-2*e^2/(x + 4) + e^2 + e) + 8/5*e^(-2*e^2/(x + 4) + e^2 + e - 2) - 16/5*integrate(e^(-2*e^2/(x + 4) +
e^2 + e)/(x^2 + 8*x + 16), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^2\,\left (x+2\right )+\mathrm {e}\,\left (x+4\right )}{x+4}}\,\left (8\,x+2\,x\,{\mathrm {e}}^2+x^2+16\right )}{5\,x^2+40\,x+80} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(2)*(x + 2) + exp(1)*(x + 4))/(x + 4))*(8*x + 2*x*exp(2) + x^2 + 16))/(40*x + 5*x^2 + 80),x)

[Out]

int((exp((exp(2)*(x + 2) + exp(1)*(x + 4))/(x + 4))*(8*x + 2*x*exp(2) + x^2 + 16))/(40*x + 5*x^2 + 80), x)

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sympy [A]  time = 0.23, size = 20, normalized size = 0.95 \begin {gather*} \frac {x e^{\frac {\left (x + 2\right ) e^{2} + e \left (x + 4\right )}{x + 4}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*x+x**2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/(5*x**2+40*x+80),x)

[Out]

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

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