3.11.85 \(\int \frac {e^{\frac {(-16 x+4 x^2) \log ^2(4)-8 \log ^2(4) \log (\log (4-x))}{-4 x+x^2}} (-8 x \log ^2(4)+(-32+16 x) \log ^2(4) \log (4-x) \log (\log (4-x)))}{(16 x^2-8 x^3+x^4) \log (4-x)} \, dx\)

Optimal. Leaf size=29 \[ e^{\frac {4 \log ^2(4) \left (x+\frac {2 \log (\log (4-x))}{4-x}\right )}{x}} \]

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Rubi [F]  time = 5.39, 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 {\left (-16 x+4 x^2\right ) \log ^2(4)-8 \log ^2(4) \log (\log (4-x))}{-4 x+x^2}\right ) \left (-8 x \log ^2(4)+(-32+16 x) \log ^2(4) \log (4-x) \log (\log (4-x))\right )}{\left (16 x^2-8 x^3+x^4\right ) \log (4-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(((-16*x + 4*x^2)*Log[4]^2 - 8*Log[4]^2*Log[Log[4 - x]])/(-4*x + x^2))*(-8*x*Log[4]^2 + (-32 + 16*x)*Lo
g[4]^2*Log[4 - x]*Log[Log[4 - x]]))/((16*x^2 - 8*x^3 + x^4)*Log[4 - x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {\left (-16 x+4 x^2\right ) \log ^2(4)-8 \log ^2(4) \log (\log (4-x))}{-4 x+x^2}\right ) \left (-8 x \log ^2(4)+(-32+16 x) \log ^2(4) \log (4-x) \log (\log (4-x))\right )}{x^2 \left (16-8 x+x^2\right ) \log (4-x)} \, dx\\ &=\int \frac {\exp \left (\frac {\left (-16 x+4 x^2\right ) \log ^2(4)-8 \log ^2(4) \log (\log (4-x))}{-4 x+x^2}\right ) \left (-8 x \log ^2(4)+(-32+16 x) \log ^2(4) \log (4-x) \log (\log (4-x))\right )}{(-4+x)^2 x^2 \log (4-x)} \, dx\\ &=\int \frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right ) \left (-8 x \log ^2(4)+(-32+16 x) \log ^2(4) \log (4-x) \log (\log (4-x))\right )}{(4-x)^2 x^2 \log (4-x)} \, dx\\ &=\int \left (-\frac {8 \exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right ) \log ^2(4)}{(-4+x)^2 x \log (4-x)}+\frac {16 \exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right ) (-2+x) \log ^2(4) \log (\log (4-x))}{(-4+x)^2 x^2}\right ) \, dx\\ &=-\left (\left (8 \log ^2(4)\right ) \int \frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right )}{(-4+x)^2 x \log (4-x)} \, dx\right )+\left (16 \log ^2(4)\right ) \int \frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right ) (-2+x) \log (\log (4-x))}{(-4+x)^2 x^2} \, dx\\ &=-\left (\left (8 \log ^2(4)\right ) \int \left (\frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right )}{4 (-4+x)^2 \log (4-x)}-\frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right )}{16 (-4+x) \log (4-x)}+\frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right )}{16 x \log (4-x)}\right ) \, dx\right )+\left (16 \log ^2(4)\right ) \int \left (\frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right ) \log (\log (4-x))}{8 (-4+x)^2}-\frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right ) \log (\log (4-x))}{8 x^2}\right ) \, dx\\ &=\frac {1}{2} \log ^2(4) \int \frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right )}{(-4+x) \log (4-x)} \, dx-\frac {1}{2} \log ^2(4) \int \frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right )}{x \log (4-x)} \, dx-\left (2 \log ^2(4)\right ) \int \frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right )}{(-4+x)^2 \log (4-x)} \, dx+\left (2 \log ^2(4)\right ) \int \frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right ) \log (\log (4-x))}{(-4+x)^2} \, dx-\left (2 \log ^2(4)\right ) \int \frac {\exp \left (\frac {4 \log ^2(4) \left (-4 x+x^2-2 \log (\log (4-x))\right )}{(-4+x) x}\right ) \log (\log (4-x))}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 30, normalized size = 1.03 \begin {gather*} e^{4 \log ^2(4)} \log ^{-\frac {8 \log ^2(4)}{(-4+x) x}}(4-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(((-16*x + 4*x^2)*Log[4]^2 - 8*Log[4]^2*Log[Log[4 - x]])/(-4*x + x^2))*(-8*x*Log[4]^2 + (-32 + 16
*x)*Log[4]^2*Log[4 - x]*Log[Log[4 - x]]))/((16*x^2 - 8*x^3 + x^4)*Log[4 - x]),x]

[Out]

E^(4*Log[4]^2)/Log[4 - x]^((8*Log[4]^2)/((-4 + x)*x))

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fricas [A]  time = 0.75, size = 38, normalized size = 1.31 \begin {gather*} e^{\left (\frac {16 \, {\left ({\left (x^{2} - 4 \, x\right )} \log \relax (2)^{2} - 2 \, \log \relax (2)^{2} \log \left (\log \left (-x + 4\right )\right )\right )}}{x^{2} - 4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(16*x-32)*log(2)^2*log(-x+4)*log(log(-x+4))-32*x*log(2)^2)*exp((-32*log(2)^2*log(log(-x+4))+4*(4*
x^2-16*x)*log(2)^2)/(x^2-4*x))/(x^4-8*x^3+16*x^2)/log(-x+4),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.67, size = 58, normalized size = 2.00 \begin {gather*} e^{\left (\frac {16 \, x^{2} \log \relax (2)^{2}}{x^{2} - 4 \, x} - \frac {64 \, x \log \relax (2)^{2}}{x^{2} - 4 \, x} - \frac {32 \, \log \relax (2)^{2} \log \left (\log \left (-x + 4\right )\right )}{x^{2} - 4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(16*x-32)*log(2)^2*log(-x+4)*log(log(-x+4))-32*x*log(2)^2)*exp((-32*log(2)^2*log(log(-x+4))+4*(4*
x^2-16*x)*log(2)^2)/(x^2-4*x))/(x^4-8*x^3+16*x^2)/log(-x+4),x, algorithm="giac")

[Out]

e^(16*x^2*log(2)^2/(x^2 - 4*x) - 64*x*log(2)^2/(x^2 - 4*x) - 32*log(2)^2*log(log(-x + 4))/(x^2 - 4*x))

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maple [A]  time = 0.04, size = 32, normalized size = 1.10

method	result	size
risch	\({\mathrm e}^{\frac {16 \ln \relax (2)^{2} \left (x^{2}-2 \ln \left (\ln \left (-x +4\right )\right )-4 x \right )}{\left (x -4\right ) x}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*(16*x-32)*ln(2)^2*ln(-x+4)*ln(ln(-x+4))-32*x*ln(2)^2)*exp((-32*ln(2)^2*ln(ln(-x+4))+4*(4*x^2-16*x)*ln(2
)^2)/(x^2-4*x))/(x^4-8*x^3+16*x^2)/ln(-x+4),x,method=_RETURNVERBOSE)

[Out]

exp(16*ln(2)^2*(x^2-2*ln(ln(-x+4))-4*x)/(x-4)/x)

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maxima [A]  time = 1.20, size = 42, normalized size = 1.45 \begin {gather*} e^{\left (16 \, \log \relax (2)^{2} - \frac {8 \, \log \relax (2)^{2} \log \left (\log \left (-x + 4\right )\right )}{x - 4} + \frac {8 \, \log \relax (2)^{2} \log \left (\log \left (-x + 4\right )\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(16*x-32)*log(2)^2*log(-x+4)*log(log(-x+4))-32*x*log(2)^2)*exp((-32*log(2)^2*log(log(-x+4))+4*(4*
x^2-16*x)*log(2)^2)/(x^2-4*x))/(x^4-8*x^3+16*x^2)/log(-x+4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.02, size = 50, normalized size = 1.72 \begin {gather*} {\mathrm {e}}^{\frac {16\,x\,{\ln \relax (2)}^2}{x-4}}\,{\mathrm {e}}^{-\frac {64\,{\ln \relax (2)}^2}{x-4}}\,{\ln \left (4-x\right )}^{\frac {32\,{\ln \relax (2)}^2}{4\,x-x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((32*log(log(4 - x))*log(2)^2 + 4*log(2)^2*(16*x - 4*x^2))/(4*x - x^2))*(32*x*log(2)^2 - 4*log(log(4
- x))*log(2)^2*log(4 - x)*(16*x - 32)))/(log(4 - x)*(16*x^2 - 8*x^3 + x^4)),x)

[Out]

exp((16*x*log(2)^2)/(x - 4))*exp(-(64*log(2)^2)/(x - 4))*log(4 - x)^((32*log(2)^2)/(4*x - x^2))

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sympy [A]  time = 0.86, size = 34, normalized size = 1.17 \begin {gather*} e^{\frac {\left (16 x^{2} - 64 x\right ) \log {\relax (2 )}^{2} - 32 \log {\relax (2 )}^{2} \log {\left (\log {\left (4 - x \right )} \right )}}{x^{2} - 4 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(16*x-32)*ln(2)**2*ln(-x+4)*ln(ln(-x+4))-32*x*ln(2)**2)*exp((-32*ln(2)**2*ln(ln(-x+4))+4*(4*x**2-
16*x)*ln(2)**2)/(x**2-4*x))/(x**4-8*x**3+16*x**2)/ln(-x+4),x)

[Out]

exp(((16*x**2 - 64*x)*log(2)**2 - 32*log(2)**2*log(log(4 - x)))/(x**2 - 4*x))

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