3.98.85 \(\int \frac {e^{-2+x} (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} (-4-4 x-x^2)+e^{2-x} (32+16 x+2 x^2) (i \pi +\log (9-e))^2)}{(16+8 x+x^2) (i \pi +\log (9-e))^2} \, dx\)

Optimal. Leaf size=34 \[ -e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}}+2 x \]

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Rubi [F]  time = 5.44, 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^{-2+x} \left (\exp \left (\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right ) \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-2 + x)*(E^((E^(-2 + x)*x)/((4 + x)*(I*Pi + Log[9 - E])^2))*(-4 - 4*x - x^2) + E^(2 - x)*(32 + 16*x +
2*x^2)*(I*Pi + Log[9 - E])^2))/((16 + 8*x + x^2)*(I*Pi + Log[9 - E])^2),x]

[Out]

2*x - Defer[Int][E^(-2 + x + (E^(-2 + x)*x)/((4 + x)*(I*Pi + Log[9 - E])^2)), x]/(I*Pi + Log[9 - E])^2 - (4*De
fer[Int][E^(-2 + x + (E^(-2 + x)*x)/((4 + x)*(I*Pi + Log[9 - E])^2))/(4 + x)^2, x])/(I*Pi + Log[9 - E])^2 + (4
*Defer[Int][E^(-2 + x + (E^(-2 + x)*x)/((4 + x)*(I*Pi + Log[9 - E])^2))/(4 + x), x])/(I*Pi + Log[9 - E])^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{-2+x} \left (\exp \left (\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right ) \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{16+8 x+x^2} \, dx}{(i \pi +\log (9-e))^2}\\ &=\frac {\int \frac {e^{-2+x} \left (\exp \left (\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right ) \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{(4+x)^2} \, dx}{(i \pi +\log (9-e))^2}\\ &=\frac {\int \left (-\frac {\exp \left (-2+x+\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right ) (2+x)^2}{(4+x)^2}-2 (\pi -i \log (9-e))^2\right ) \, dx}{(i \pi +\log (9-e))^2}\\ &=2 x-\frac {\int \frac {\exp \left (-2+x+\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right ) (2+x)^2}{(4+x)^2} \, dx}{(i \pi +\log (9-e))^2}\\ &=2 x-\frac {\int \left (\exp \left (-2+x+\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right )+\frac {4 \exp \left (-2+x+\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right )}{(4+x)^2}-\frac {4 \exp \left (-2+x+\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right )}{4+x}\right ) \, dx}{(i \pi +\log (9-e))^2}\\ &=2 x-\frac {\int \exp \left (-2+x+\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right ) \, dx}{(i \pi +\log (9-e))^2}-\frac {4 \int \frac {\exp \left (-2+x+\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right )}{(4+x)^2} \, dx}{(i \pi +\log (9-e))^2}+\frac {4 \int \frac {\exp \left (-2+x+\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}\right )}{4+x} \, dx}{(i \pi +\log (9-e))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.07, size = 35, normalized size = 1.03 \begin {gather*} -e^{-\frac {e^{-2+x} x}{(4+x) (\pi -i \log (9-e))^2}}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 + x)*(E^((E^(-2 + x)*x)/((4 + x)*(I*Pi + Log[9 - E])^2))*(-4 - 4*x - x^2) + E^(2 - x)*(32 + 1
6*x + 2*x^2)*(I*Pi + Log[9 - E])^2))/((16 + 8*x + x^2)*(I*Pi + Log[9 - E])^2),x]

[Out]

-E^(-((E^(-2 + x)*x)/((4 + x)*(Pi - I*Log[9 - E])^2))) + 2*x

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fricas [A]  time = 0.82, size = 25, normalized size = 0.74 \begin {gather*} 2 \, x - e^{\left (\frac {x e^{\left (x - 2\right )}}{{\left (x + 4\right )} \log \left (e - 9\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/log(exp(1)-9)^2)+(2*x^2+16*x+32)*exp(2-x)*log(exp(1)-9)^2)/(x^2+8
*x+16)/exp(2-x)/log(exp(1)-9)^2,x, algorithm="fricas")

[Out]

2*x - e^(x*e^(x - 2)/((x + 4)*log(e - 9)^2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/log(exp(1)-9)^2)+(2*x^2+16*x+32)*exp(2-x)*log(exp(1)-9)^2)/(x^2+8
*x+16)/exp(2-x)/log(exp(1)-9)^2,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.34, size = 26, normalized size = 0.76

method	result	size
risch	\(2 x -{\mathrm e}^{\frac {x \,{\mathrm e}^{x -2}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}}\)	\(26\)
norman	\(\frac {\left (-32 \,{\mathrm e}^{2-x} \ln \left ({\mathrm e}-9\right )-4 \,{\mathrm e}^{2-x} \ln \left ({\mathrm e}-9\right ) {\mathrm e}^{\frac {x \,{\mathrm e}^{x -2}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}}+2 \ln \left ({\mathrm e}-9\right ) x^{2} {\mathrm e}^{2-x}-\ln \left ({\mathrm e}-9\right ) x \,{\mathrm e}^{2-x} {\mathrm e}^{\frac {x \,{\mathrm e}^{x -2}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}}\right ) {\mathrm e}^{x -2}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/ln(exp(1)-9)^2)+(2*x^2+16*x+32)*exp(2-x)*ln(exp(1)-9)^2)/(x^2+8*x+16)/e
xp(2-x)/ln(exp(1)-9)^2,x,method=_RETURNVERBOSE)

[Out]

2*x-exp(x/(4+x)*exp(x-2)/ln(exp(1)-9)^2)

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maxima [B]  time = 0.54, size = 120, normalized size = 3.53 \begin {gather*} \frac {2 \, {\left (x - \frac {16}{x + 4} - 8 \, \log \left (x + 4\right )\right )} \log \left (e - 9\right )^{2} + 16 \, {\left (\frac {4}{x + 4} + \log \left (x + 4\right )\right )} \log \left (e - 9\right )^{2} - e^{\left (-\frac {4 \, e^{x}}{x e^{2} \log \left (e - 9\right )^{2} + 4 \, e^{2} \log \left (e - 9\right )^{2}} + \frac {e^{\left (x - 2\right )}}{\log \left (e - 9\right )^{2}}\right )} \log \left (e - 9\right )^{2} - \frac {32 \, \log \left (e - 9\right )^{2}}{x + 4}}{\log \left (e - 9\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/log(exp(1)-9)^2)+(2*x^2+16*x+32)*exp(2-x)*log(exp(1)-9)^2)/(x^2+8
*x+16)/exp(2-x)/log(exp(1)-9)^2,x, algorithm="maxima")

[Out]

(2*(x - 16/(x + 4) - 8*log(x + 4))*log(e - 9)^2 + 16*(4/(x + 4) + log(x + 4))*log(e - 9)^2 - e^(-4*e^x/(x*e^2*
log(e - 9)^2 + 4*e^2*log(e - 9)^2) + e^(x - 2)/log(e - 9)^2)*log(e - 9)^2 - 32*log(e - 9)^2/(x + 4))/log(e - 9
)^2

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mupad [B]  time = 5.89, size = 34, normalized size = 1.00 \begin {gather*} 2\,x-{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x\,{\ln \left (\mathrm {e}-9\right )}^2+4\,{\ln \left (\mathrm {e}-9\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 2)*(exp((x*exp(x - 2))/(log(exp(1) - 9)^2*(x + 4)))*(4*x + x^2 + 4) - log(exp(1) - 9)^2*exp(2 -
x)*(16*x + 2*x^2 + 32)))/(log(exp(1) - 9)^2*(8*x + x^2 + 16)),x)

[Out]

2*x - exp((x*exp(-2)*exp(x))/(x*log(exp(1) - 9)^2 + 4*log(exp(1) - 9)^2))

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sympy [B]  time = 2.36, size = 85, normalized size = 2.50 \begin {gather*} 2 x - e^{\frac {x e^{x}}{- \pi ^{2} x e^{2} + x e^{2} \log {\left (9 - e \right )}^{2} + 2 i \pi x e^{2} \log {\left (9 - e \right )} - 4 \pi ^{2} e^{2} + 4 e^{2} \log {\left (9 - e \right )}^{2} + 8 i \pi e^{2} \log {\left (9 - e \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2-4*x-4)*exp(x/(4+x)/exp(2-x)/ln(exp(1)-9)**2)+(2*x**2+16*x+32)*exp(2-x)*ln(exp(1)-9)**2)/(x**
2+8*x+16)/exp(2-x)/ln(exp(1)-9)**2,x)

[Out]

2*x - exp(x*exp(x)/(-pi**2*x*exp(2) + x*exp(2)*log(9 - E)**2 + 2*I*pi*x*exp(2)*log(9 - E) - 4*pi**2*exp(2) + 4
*exp(2)*log(9 - E)**2 + 8*I*pi*exp(2)*log(9 - E)))

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