3.12.37 \(\int \frac {e^{\frac {5 x-\log (\frac {1}{9} (-100+9 \log (\log (-2+x))))}{x}} (-9 e^4 x+(e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))) \log (\frac {1}{9} (-100+9 \log (\log (-2+x)))))}{(200 x^2-100 x^3) \log (-2+x)+(-18 x^2+9 x^3) \log (-2+x) \log (\log (-2+x))} \, dx\)

Optimal. Leaf size=19 \[ e^{9-\frac {\log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x}} \]

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Rubi [F]  time = 6.05, 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 {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}\right ) \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((5*x - Log[(-100 + 9*Log[Log[-2 + x]])/9])/x)*(-9*E^4*x + (E^4*(200 - 100*x)*Log[-2 + x] + E^4*(-18 +
9*x)*Log[-2 + x]*Log[Log[-2 + x]])*Log[(-100 + 9*Log[Log[-2 + x]])/9]))/((200*x^2 - 100*x^3)*Log[-2 + x] + (-1
8*x^2 + 9*x^3)*Log[-2 + x]*Log[Log[-2 + x]]),x]

[Out]

-1/2*(E^9*Defer[Int][(-100/9 + Log[Log[-2 + x]])^(-1 - x^(-1))/((-2 + x)*Log[-2 + x]), x]) + (E^9*Defer[Int][(
-100/9 + Log[Log[-2 + x]])^(-1 - x^(-1))/(x*Log[-2 + x]), x])/2 + E^9*Defer[Int][Log[-100/9 + Log[Log[-2 + x]]
]/(x^2*(-100/9 + Log[Log[-2 + x]])^x^(-1)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9^{\frac {1}{x}} e^9 (-100+9 \log (\log (-2+x)))^{-1-\frac {1}{x}} \left (9 x-(-2+x) \log (-2+x) (-100+9 \log (\log (-2+x))) \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )\right )}{(2-x) x^2 \log (-2+x)} \, dx\\ &=e^9 \int \frac {9^{\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1-\frac {1}{x}} \left (9 x-(-2+x) \log (-2+x) (-100+9 \log (\log (-2+x))) \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )\right )}{(2-x) x^2 \log (-2+x)} \, dx\\ &=e^9 \int \left (-\frac {9^{1+\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1-\frac {1}{x}}}{(-2+x) x \log (-2+x)}+\frac {9^{\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2}\right ) \, dx\\ &=-\left (e^9 \int \frac {9^{1+\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1-\frac {1}{x}}}{(-2+x) x \log (-2+x)} \, dx\right )+e^9 \int \frac {9^{\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2} \, dx\\ &=-\left (e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{(-2+x) x \log (-2+x)} \, dx\right )+e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2} \, dx\\ &=-\left (e^9 \int \left (\frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{2 (-2+x) \log (-2+x)}-\frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{2 x \log (-2+x)}\right ) \, dx\right )+e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2} \, dx\\ &=-\left (\frac {1}{2} e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{(-2+x) \log (-2+x)} \, dx\right )+\frac {1}{2} e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{x \log (-2+x)} \, dx+e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.68, size = 19, normalized size = 1.00 \begin {gather*} e^9 \left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5*x - Log[(-100 + 9*Log[Log[-2 + x]])/9])/x)*(-9*E^4*x + (E^4*(200 - 100*x)*Log[-2 + x] + E^4*(
-18 + 9*x)*Log[-2 + x]*Log[Log[-2 + x]])*Log[(-100 + 9*Log[Log[-2 + x]])/9]))/((200*x^2 - 100*x^3)*Log[-2 + x]
+ (-18*x^2 + 9*x^3)*Log[-2 + x]*Log[Log[-2 + x]]),x]

[Out]

E^9/(-100/9 + Log[Log[-2 + x]])^x^(-1)

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fricas [A]  time = 0.52, size = 21, normalized size = 1.11 \begin {gather*} e^{\left (\frac {5 \, x - \log \left (\log \left (\log \left (x - 2\right )\right ) - \frac {100}{9}\right )}{x} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x-18)*exp(4)*log(x-2)*log(log(x-2))+(-100*x+200)*exp(4)*log(x-2))*log(log(log(x-2))-100/9)-9*x*
exp(4))*exp((-log(log(log(x-2))-100/9)+5*x)/x)/((9*x^3-18*x^2)*log(x-2)*log(log(x-2))+(-100*x^3+200*x^2)*log(x
-2)),x, algorithm="fricas")

[Out]

e^((5*x - log(log(log(x - 2)) - 100/9))/x + 4)

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giac [A]  time = 3.51, size = 16, normalized size = 0.84 \begin {gather*} e^{\left (-\frac {\log \left (\log \left (\log \left (x - 2\right )\right ) - \frac {100}{9}\right )}{x} + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x-18)*exp(4)*log(x-2)*log(log(x-2))+(-100*x+200)*exp(4)*log(x-2))*log(log(log(x-2))-100/9)-9*x*
exp(4))*exp((-log(log(log(x-2))-100/9)+5*x)/x)/((9*x^3-18*x^2)*log(x-2)*log(log(x-2))+(-100*x^3+200*x^2)*log(x
-2)),x, algorithm="giac")

[Out]

e^(-log(log(log(x - 2)) - 100/9)/x + 9)

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maple [A]  time = 0.05, size = 17, normalized size = 0.89

method	result	size
risch	\(\left (\ln \left (\ln \left (x -2\right )\right )-\frac {100}{9}\right )^{-\frac {1}{x}} {\mathrm e}^{9}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((9*x-18)*exp(4)*ln(x-2)*ln(ln(x-2))+(-100*x+200)*exp(4)*ln(x-2))*ln(ln(ln(x-2))-100/9)-9*x*exp(4))*exp((
-ln(ln(ln(x-2))-100/9)+5*x)/x)/((9*x^3-18*x^2)*ln(x-2)*ln(ln(x-2))+(-100*x^3+200*x^2)*ln(x-2)),x,method=_RETUR
NVERBOSE)

[Out]

(ln(ln(x-2))-100/9)^(-1/x)*exp(9)

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maxima [A]  time = 1.22, size = 25, normalized size = 1.32 \begin {gather*} e^{\left (\frac {2 \, \log \relax (3)}{x} - \frac {\log \left (9 \, \log \left (\log \left (x - 2\right )\right ) - 100\right )}{x} + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x-18)*exp(4)*log(x-2)*log(log(x-2))+(-100*x+200)*exp(4)*log(x-2))*log(log(log(x-2))-100/9)-9*x*
exp(4))*exp((-log(log(log(x-2))-100/9)+5*x)/x)/((9*x^3-18*x^2)*log(x-2)*log(log(x-2))+(-100*x^3+200*x^2)*log(x
-2)),x, algorithm="maxima")

[Out]

e^(2*log(3)/x - log(9*log(log(x - 2)) - 100)/x + 9)

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mupad [B]  time = 1.15, size = 16, normalized size = 0.84 \begin {gather*} \frac {{\mathrm {e}}^9}{{\left (\ln \left (\ln \left (x-2\right )\right )-\frac {100}{9}\right )}^{1/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((5*x - log(log(log(x - 2)) - 100/9))/x)*(9*x*exp(4) + log(log(log(x - 2)) - 100/9)*(log(x - 2)*exp(4
)*(100*x - 200) - log(x - 2)*exp(4)*log(log(x - 2))*(9*x - 18))))/(log(x - 2)*(200*x^2 - 100*x^3) - log(x - 2)
*log(log(x - 2))*(18*x^2 - 9*x^3)),x)

[Out]

exp(9)/(log(log(x - 2)) - 100/9)^(1/x)

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sympy [A]  time = 3.37, size = 20, normalized size = 1.05 \begin {gather*} e^{4} e^{\frac {5 x - \log {\left (\log {\left (\log {\left (x - 2 \right )} \right )} - \frac {100}{9} \right )}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x-18)*exp(4)*ln(x-2)*ln(ln(x-2))+(-100*x+200)*exp(4)*ln(x-2))*ln(ln(ln(x-2))-100/9)-9*x*exp(4))
*exp((-ln(ln(ln(x-2))-100/9)+5*x)/x)/((9*x**3-18*x**2)*ln(x-2)*ln(ln(x-2))+(-100*x**3+200*x**2)*ln(x-2)),x)

[Out]

exp(4)*exp((5*x - log(log(log(x - 2)) - 100/9))/x)

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