3.30.84 \(\int \frac {e^{\frac {-6-3 x+x^2+(-2-x) \log (4 x-x^2)}{2+x}} (16+8 x-20 x^2-2 x^3+x^4)}{-16 x-12 x^2+x^4} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^{-3+\frac {4}{-2-\frac {4}{x}}+x}}{(4-x) x} \]

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Rubi [F]  time = 2.14, 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 {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right ) \left (16+8 x-20 x^2-2 x^3+x^4\right )}{-16 x-12 x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-6 - 3*x + x^2 + (-2 - x)*Log[4*x - x^2])/(2 + x))*(16 + 8*x - 20*x^2 - 2*x^3 + x^4))/(-16*x - 12*x^2
+ x^4),x]

[Out]

Defer[Int][E^((-6 - 3*x + x^2 + (-2 - x)*Log[4*x - x^2])/(2 + x)), x] + Defer[Int][E^((-6 - 3*x + x^2)/(2 + x)
)/(-4 + x)^2, x]/4 - Defer[Int][E^((-6 - 3*x + x^2)/(2 + x))/(-4 + x), x]/16 + Defer[Int][E^((-6 - 3*x + x^2)/
(2 + x))/x, x]/16 - Defer[Int][E^((-6 - 3*x + x^2 + (-2 - x)*Log[4*x - x^2])/(2 + x))/x, x] - 4*Defer[Int][E^(
(-6 - 3*x + x^2 + (-2 - x)*Log[4*x - x^2])/(2 + x))/(2 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right ) \left (16+8 x-20 x^2-2 x^3+x^4\right )}{x \left (-16-12 x+x^3\right )} \, dx\\ &=\int \left (\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )+\frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{4-x}-\frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{x}-\frac {4 \exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{(2+x)^2}\right ) \, dx\\ &=-\left (4 \int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{(2+x)^2} \, dx\right )+\int \exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right ) \, dx+\int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{4-x} \, dx-\int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{x} \, dx\\ &=-\left (4 \int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{(2+x)^2} \, dx\right )+\int \exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right ) \, dx-\int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{x} \, dx+\int \frac {e^{\frac {-6-3 x+x^2}{2+x}}}{(4-x)^2 x} \, dx\\ &=-\left (4 \int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{(2+x)^2} \, dx\right )+\int \exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right ) \, dx+\int \left (\frac {e^{\frac {-6-3 x+x^2}{2+x}}}{4 (-4+x)^2}-\frac {e^{\frac {-6-3 x+x^2}{2+x}}}{16 (-4+x)}+\frac {e^{\frac {-6-3 x+x^2}{2+x}}}{16 x}\right ) \, dx-\int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{x} \, dx\\ &=-\left (\frac {1}{16} \int \frac {e^{\frac {-6-3 x+x^2}{2+x}}}{-4+x} \, dx\right )+\frac {1}{16} \int \frac {e^{\frac {-6-3 x+x^2}{2+x}}}{x} \, dx+\frac {1}{4} \int \frac {e^{\frac {-6-3 x+x^2}{2+x}}}{(-4+x)^2} \, dx-4 \int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{(2+x)^2} \, dx+\int \exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right ) \, dx-\int \frac {\exp \left (\frac {-6-3 x+x^2+(-2-x) \log \left (4 x-x^2\right )}{2+x}\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 22, normalized size = 0.81 \begin {gather*} -\frac {e^{-5+x+\frac {4}{2+x}}}{(-4+x) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-6 - 3*x + x^2 + (-2 - x)*Log[4*x - x^2])/(2 + x))*(16 + 8*x - 20*x^2 - 2*x^3 + x^4))/(-16*x -
12*x^2 + x^4),x]

[Out]

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

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fricas [A]  time = 1.01, size = 30, normalized size = 1.11 \begin {gather*} e^{\left (\frac {x^{2} - {\left (x + 2\right )} \log \left (-x^{2} + 4 \, x\right ) - 3 \, x - 6}{x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-2*x^3-20*x^2+8*x+16)*exp(((-x-2)*log(-x^2+4*x)+x^2-3*x-6)/(2+x))/(x^4-12*x^2-16*x),x, algorithm
="fricas")

[Out]

e^((x^2 - (x + 2)*log(-x^2 + 4*x) - 3*x - 6)/(x + 2))

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giac [B]  time = 0.24, size = 61, normalized size = 2.26 \begin {gather*} e^{\left (\frac {x^{2}}{x + 2} - \frac {x \log \left (-x^{2} + 4 \, x\right )}{x + 2} - \frac {3 \, x}{x + 2} - \frac {2 \, \log \left (-x^{2} + 4 \, x\right )}{x + 2} - \frac {6}{x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-2*x^3-20*x^2+8*x+16)*exp(((-x-2)*log(-x^2+4*x)+x^2-3*x-6)/(2+x))/(x^4-12*x^2-16*x),x, algorithm
="giac")

[Out]

e^(x^2/(x + 2) - x*log(-x^2 + 4*x)/(x + 2) - 3*x/(x + 2) - 2*log(-x^2 + 4*x)/(x + 2) - 6/(x + 2))

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maple [A]  time = 0.05, size = 41, normalized size = 1.52

method	result	size
risch	\({\mathrm e}^{\frac {-\ln \left (-x^{2}+4 x \right ) x +x^{2}-2 \ln \left (-x^{2}+4 x \right )-3 x -6}{2+x}}\)	\(41\)
gosper	\({\mathrm e}^{-\frac {\ln \left (-x^{2}+4 x \right ) x -x^{2}+2 \ln \left (-x^{2}+4 x \right )+3 x +6}{2+x}}\)	\(43\)
norman	\(\frac {x \,{\mathrm e}^{\frac {\left (-x -2\right ) \ln \left (-x^{2}+4 x \right )+x^{2}-3 x -6}{2+x}}+2 \,{\mathrm e}^{\frac {\left (-x -2\right ) \ln \left (-x^{2}+4 x \right )+x^{2}-3 x -6}{2+x}}}{2+x}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-2*x^3-20*x^2+8*x+16)*exp(((-x-2)*ln(-x^2+4*x)+x^2-3*x-6)/(2+x))/(x^4-12*x^2-16*x),x,method=_RETURNVER
BOSE)

[Out]

exp((-ln(-x^2+4*x)*x+x^2-2*ln(-x^2+4*x)-3*x-6)/(2+x))

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maxima [A]  time = 0.64, size = 26, normalized size = 0.96 \begin {gather*} -\frac {e^{\left (x + \frac {4}{x + 2}\right )}}{x^{2} e^{5} - 4 \, x e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-2*x^3-20*x^2+8*x+16)*exp(((-x-2)*log(-x^2+4*x)+x^2-3*x-6)/(2+x))/(x^4-12*x^2-16*x),x, algorithm
="maxima")

[Out]

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

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mupad [B]  time = 2.08, size = 39, normalized size = 1.44 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {3\,x}{x+2}}\,{\mathrm {e}}^{\frac {x^2}{x+2}}\,{\mathrm {e}}^{-\frac {6}{x+2}}}{4\,x-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(3*x + log(4*x - x^2)*(x + 2) - x^2 + 6)/(x + 2))*(8*x - 20*x^2 - 2*x^3 + x^4 + 16))/(16*x + 12*x^2
- x^4),x)

[Out]

(exp(-(3*x)/(x + 2))*exp(x^2/(x + 2))*exp(-6/(x + 2)))/(4*x - x^2)

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sympy [A]  time = 0.55, size = 26, normalized size = 0.96 \begin {gather*} e^{\frac {x^{2} - 3 x + \left (- x - 2\right ) \log {\left (- x^{2} + 4 x \right )} - 6}{x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-2*x**3-20*x**2+8*x+16)*exp(((-x-2)*ln(-x**2+4*x)+x**2-3*x-6)/(2+x))/(x**4-12*x**2-16*x),x)

[Out]

exp((x**2 - 3*x + (-x - 2)*log(-x**2 + 4*x) - 6)/(x + 2))

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