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

Optimal. Leaf size=25 \[ e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \]

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Rubi [F]  time = 4.34, 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 {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2 - Log[((-4 + x)*Log[4])/(16*x)])/(-2 + x))*(8 + 4*x - 2*x^2 + (-4*x + x^2)*Log[((-4 + x)*Log[4])/(1
6*x)]))/(-16*x + 20*x^2 - 8*x^3 + x^4),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 5.08, size = 37, normalized size = 1.48 \begin {gather*} e^{\frac {2}{-2+x}} \left (\frac {-4+x}{x}\right )^{-\frac {1}{-2+x}} \left (\frac {16}{\log (4)}\right )^{\frac {1}{-2+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(2/(-2 + x))*(16/Log[4])^(-2 + x)^(-1))/((-4 + x)/x)^(-2 + x)^(-1)

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fricas [A]  time = 0.63, size = 21, normalized size = 0.84 \begin {gather*} e^{\left (-\frac {\log \left (\frac {{\left (x - 4\right )} \log \relax (2)}{8 \, x}\right ) - 2}{x - 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-4*x)*log(1/8*(x-4)*log(2)/x)-2*x^2+4*x+8)*exp((-log(1/8*(x-4)*log(2)/x)+2)/(x-2))/(x^4-8*x^3+2
0*x^2-16*x),x, algorithm="fricas")

[Out]

e^(-(log(1/8*(x - 4)*log(2)/x) - 2)/(x - 2))

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giac [A]  time = 0.25, size = 29, normalized size = 1.16 \begin {gather*} e^{\left (-\frac {\log \left (-\frac {\log \relax (2)}{2 \, x} + \frac {1}{8} \, \log \relax (2)\right )}{x - 2} + \frac {2}{x - 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-4*x)*log(1/8*(x-4)*log(2)/x)-2*x^2+4*x+8)*exp((-log(1/8*(x-4)*log(2)/x)+2)/(x-2))/(x^4-8*x^3+2
0*x^2-16*x),x, algorithm="giac")

[Out]

e^(-log(-1/2*log(2)/x + 1/8*log(2))/(x - 2) + 2/(x - 2))

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maple [A]  time = 0.23, size = 22, normalized size = 0.88

method	result	size
risch	\({\mathrm e}^{-\frac {\ln \left (\frac {\left (x -4\right ) \ln \relax (2)}{8 x}\right )-2}{x -2}}\)	\(22\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-\ln \left (\frac {\left (x -4\right ) \ln \relax (2)}{8 x}\right )+2}{x -2}}-2 \,{\mathrm e}^{\frac {-\ln \left (\frac {\left (x -4\right ) \ln \relax (2)}{8 x}\right )+2}{x -2}}}{x -2}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-4*x)*ln(1/8*(x-4)*ln(2)/x)-2*x^2+4*x+8)*exp((-ln(1/8*(x-4)*ln(2)/x)+2)/(x-2))/(x^4-8*x^3+20*x^2-16*x
),x,method=_RETURNVERBOSE)

[Out]

exp(-(ln(1/8*(x-4)*ln(2)/x)-2)/(x-2))

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maxima [B]  time = 0.55, size = 47, normalized size = 1.88 \begin {gather*} e^{\left (\frac {3 \, \log \relax (2)}{x - 2} - \frac {\log \left (x - 4\right )}{x - 2} + \frac {\log \relax (x)}{x - 2} - \frac {\log \left (\log \relax (2)\right )}{x - 2} + \frac {2}{x - 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-4*x)*log(1/8*(x-4)*log(2)/x)-2*x^2+4*x+8)*exp((-log(1/8*(x-4)*log(2)/x)+2)/(x-2))/(x^4-8*x^3+2
0*x^2-16*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.61, size = 29, normalized size = 1.16 \begin {gather*} \frac {{\mathrm {e}}^{\frac {2}{x-2}}}{{\left (\frac {\ln \relax (2)}{8}-\frac {\ln \relax (2)}{2\,x}\right )}^{\frac {1}{x-2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2/(x - 2))/(log(2)/8 - log(2)/(2*x))^(1/(x - 2))

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sympy [A]  time = 0.51, size = 19, normalized size = 0.76 \begin {gather*} e^{\frac {2 - \log {\left (\frac {\left (\frac {x}{8} - \frac {1}{2}\right ) \log {\relax (2 )}}{x} \right )}}{x - 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-4*x)*ln(1/8*(x-4)*ln(2)/x)-2*x**2+4*x+8)*exp((-ln(1/8*(x-4)*ln(2)/x)+2)/(x-2))/(x**4-8*x**3+2
0*x**2-16*x),x)

[Out]

exp((2 - log((x/8 - 1/2)*log(2)/x))/(x - 2))

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