3.46.1 \(\int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} (-4+4 x-x^2+(-4+6 x-x^2) \log (x))}{(20 x^2-20 x^3+5 x^4) \log (x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {e^{-\frac {x}{-2+x}}}{45 x \log (x)} \]

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Rubi [F]  time = 2.20, 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 {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+4 x-x^2+\left (-4+6 x-x^2\right ) \log (x)\right )}{\left (20 x^2-20 x^3+5 x^4\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4 + 4*x - x^2 + (-4 + 6*x - x^2)*Log[x])/(E^((x + (-2 + x)*Log[9*Log[x]])/(-2 + x))*(20*x^2 - 20*x^3 + 5
*x^4)*Log[x]),x]

[Out]

Defer[Int][1/(E^((x + (-2 + x)*Log[9*Log[x]])/(-2 + x))*(-2 + x)^2), x]/5 - Defer[Int][1/(E^((x + (-2 + x)*Log
[9*Log[x]])/(-2 + x))*(-2 + x)), x]/10 - Defer[Int][1/(E^((x + (-2 + x)*Log[9*Log[x]])/(-2 + x))*x^2), x]/5 +
Defer[Int][1/(E^((x + (-2 + x)*Log[9*Log[x]])/(-2 + x))*x), x]/10 - Defer[Int][1/(E^(x/(-2 + x))*x^2*Log[x]^2)
, x]/45

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+4 x-x^2+\left (-4+6 x-x^2\right ) \log (x)\right )}{x^2 \left (20-20 x+5 x^2\right ) \log (x)} \, dx\\ &=\int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+4 x-x^2+\left (-4+6 x-x^2\right ) \log (x)\right )}{5 (-2+x)^2 x^2 \log (x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+4 x-x^2+\left (-4+6 x-x^2\right ) \log (x)\right )}{(-2+x)^2 x^2 \log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+6 x-x^2\right )}{(-2+x)^2 x^2}-\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x^2 \log (x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+6 x-x^2\right )}{(-2+x)^2 x^2} \, dx-\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x^2 \log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{(-2+x)^2}-\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{2 (-2+x)}-\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x^2}+\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{2 x}\right ) \, dx-\frac {1}{5} \int \frac {e^{-\frac {x}{-2+x}}}{9 x^2 \log ^2(x)} \, dx\\ &=-\left (\frac {1}{45} \int \frac {e^{-\frac {x}{-2+x}}}{x^2 \log ^2(x)} \, dx\right )-\frac {1}{10} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{-2+x} \, dx+\frac {1}{10} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x} \, dx+\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{(-2+x)^2} \, dx-\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 22, normalized size = 1.05 \begin {gather*} \frac {e^{-1-\frac {2}{-2+x}}}{45 x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 4*x - x^2 + (-4 + 6*x - x^2)*Log[x])/(E^((x + (-2 + x)*Log[9*Log[x]])/(-2 + x))*(20*x^2 - 20*x
^3 + 5*x^4)*Log[x]),x]

[Out]

E^(-1 - 2/(-2 + x))/(45*x*Log[x])

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fricas [A]  time = 0.60, size = 24, normalized size = 1.14 \begin {gather*} \frac {e^{\left (-\frac {{\left (x - 2\right )} \log \left (9 \, \log \relax (x)\right ) + x}{x - 2}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+6*x-4)*log(x)-x^2+4*x-4)/(5*x^4-20*x^3+20*x^2)/log(x)/exp(((x-2)*log(9*log(x))+x)/(x-2)),x, a
lgorithm="fricas")

[Out]

1/5*e^(-((x - 2)*log(9*log(x)) + x)/(x - 2))/x

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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+6*x-4)*log(x)-x^2+4*x-4)/(5*x^4-20*x^3+20*x^2)/log(x)/exp(((x-2)*log(9*log(x))+x)/(x-2)),x, a
lgorithm="giac")

[Out]

undef

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maple [A]  time = 0.04, size = 30, normalized size = 1.43

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {\ln \left (9 \ln \relax (x )\right ) x -2 \ln \left (9 \ln \relax (x )\right )+x}{x -2}}}{5 x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+6*x-4)*ln(x)-x^2+4*x-4)/(5*x^4-20*x^3+20*x^2)/ln(x)/exp(((x-2)*ln(9*ln(x))+x)/(x-2)),x,method=_RETU
RNVERBOSE)

[Out]

1/5/x*exp(-(ln(9*ln(x))*x-2*ln(9*ln(x))+x)/(x-2))

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maxima [A]  time = 0.56, size = 19, normalized size = 0.90 \begin {gather*} \frac {e^{\left (-\frac {2}{x - 2} - 1\right )}}{45 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+6*x-4)*log(x)-x^2+4*x-4)/(5*x^4-20*x^3+20*x^2)/log(x)/exp(((x-2)*log(9*log(x))+x)/(x-2)),x, a
lgorithm="maxima")

[Out]

1/45*e^(-2/(x - 2) - 1)/(x*log(x))

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mupad [B]  time = 3.42, size = 18, normalized size = 0.86 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {x}{x-2}}}{45\,x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x + log(9*log(x))*(x - 2))/(x - 2))*(log(x)*(x^2 - 6*x + 4) - 4*x + x^2 + 4))/(log(x)*(20*x^2 - 20
*x^3 + 5*x^4)),x)

[Out]

exp(-x/(x - 2))/(45*x*log(x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+6*x-4)*ln(x)-x**2+4*x-4)/(5*x**4-20*x**3+20*x**2)/ln(x)/exp(((x-2)*ln(9*ln(x))+x)/(x-2)),x)

[Out]

Timed out

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