3.94.78 \(\int \frac {e^{\frac {-9+6 x+x^2}{9 x}} (9+x^2)+(63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x) \log (7-e^{\frac {-9+6 x+x^2}{9 x}}) \log (\log (7-e^{\frac {-9+6 x+x^2}{9 x}}))}{(-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3) \log (7-e^{\frac {-9+6 x+x^2}{9 x}})} \, dx\)

Optimal. Leaf size=25 \[ \frac {\log \left (\log \left (7-e^{-\frac {1}{x}+\frac {6+x}{9}}\right )\right )}{x} \]

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Rubi [F]  time = 4.16, 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 {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-9 + 6*x + x^2)/(9*x))*(9 + x^2) + (63*x - 9*E^((-9 + 6*x + x^2)/(9*x))*x)*Log[7 - E^((-9 + 6*x + x^2
)/(9*x))]*Log[Log[7 - E^((-9 + 6*x + x^2)/(9*x))]])/((-63*x^3 + 9*E^((-9 + 6*x + x^2)/(9*x))*x^3)*Log[7 - E^((
-9 + 6*x + x^2)/(9*x))]),x]

[Out]

Defer[Int][E^(2/3 + x/9)/((E^(2/3 + x/9) - 7*E^x^(-1))*x^3*Log[7 - E^((6 - 9/x + x)/9)]), x] + Defer[Int][E^(2
/3 + x/9)/((E^(2/3 + x/9) - 7*E^x^(-1))*x*Log[7 - E^((6 - 9/x + x)/9)]), x]/9 - Defer[Int][Log[Log[7 - E^((6 -
9/x + x)/9)]]/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^{\frac {2}{3}+\frac {x}{9}} \left (9+x^2\right )}{9 \left (-7 e^{\frac {1}{x}}+e^{\frac {6+x}{9}}\right ) x^3 \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )}-\frac {\log \left (\log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )\right )}{x^2}\right ) \, dx\\ &=\frac {1}{9} \int \frac {e^{\frac {2}{3}+\frac {x}{9}} \left (9+x^2\right )}{\left (-7 e^{\frac {1}{x}}+e^{\frac {6+x}{9}}\right ) x^3 \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )} \, dx-\int \frac {\log \left (\log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )\right )}{x^2} \, dx\\ &=\frac {1}{9} \int \left (\frac {9 e^{\frac {2}{3}+\frac {x}{9}}}{\left (e^{\frac {2}{3}+\frac {x}{9}}-7 e^{\frac {1}{x}}\right ) x^3 \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )}+\frac {e^{\frac {2}{3}+\frac {x}{9}}}{\left (e^{\frac {2}{3}+\frac {x}{9}}-7 e^{\frac {1}{x}}\right ) x \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )}\right ) \, dx-\int \frac {\log \left (\log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )\right )}{x^2} \, dx\\ &=\frac {1}{9} \int \frac {e^{\frac {2}{3}+\frac {x}{9}}}{\left (e^{\frac {2}{3}+\frac {x}{9}}-7 e^{\frac {1}{x}}\right ) x \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )} \, dx+\int \frac {e^{\frac {2}{3}+\frac {x}{9}}}{\left (e^{\frac {2}{3}+\frac {x}{9}}-7 e^{\frac {1}{x}}\right ) x^3 \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )} \, dx-\int \frac {\log \left (\log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.84, size = 26, normalized size = 1.04 \begin {gather*} \frac {\log \left (\log \left (7-e^{\frac {2}{3}-\frac {1}{x}+\frac {x}{9}}\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-9 + 6*x + x^2)/(9*x))*(9 + x^2) + (63*x - 9*E^((-9 + 6*x + x^2)/(9*x))*x)*Log[7 - E^((-9 + 6*x
+ x^2)/(9*x))]*Log[Log[7 - E^((-9 + 6*x + x^2)/(9*x))]])/((-63*x^3 + 9*E^((-9 + 6*x + x^2)/(9*x))*x^3)*Log[7
- E^((-9 + 6*x + x^2)/(9*x))]),x]

[Out]

Log[Log[7 - E^(2/3 - x^(-1) + x/9)]]/x

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fricas [A]  time = 0.78, size = 24, normalized size = 0.96 \begin {gather*} \frac {\log \left (\log \left (-e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} + 7\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x*exp(1/9*(x^2+6*x-9)/x)+63*x)*log(-exp(1/9*(x^2+6*x-9)/x)+7)*log(log(-exp(1/9*(x^2+6*x-9)/x)+7
))+(x^2+9)*exp(1/9*(x^2+6*x-9)/x))/(9*x^3*exp(1/9*(x^2+6*x-9)/x)-63*x^3)/log(-exp(1/9*(x^2+6*x-9)/x)+7),x, alg
orithm="fricas")

[Out]

log(log(-e^(1/9*(x^2 + 6*x - 9)/x) + 7))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {9 \, {\left (x e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} - 7 \, x\right )} \log \left (-e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} + 7\right ) \log \left (\log \left (-e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} + 7\right )\right ) - {\left (x^{2} + 9\right )} e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )}}{9 \, {\left (x^{3} e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} - 7 \, x^{3}\right )} \log \left (-e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} + 7\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x*exp(1/9*(x^2+6*x-9)/x)+63*x)*log(-exp(1/9*(x^2+6*x-9)/x)+7)*log(log(-exp(1/9*(x^2+6*x-9)/x)+7
))+(x^2+9)*exp(1/9*(x^2+6*x-9)/x))/(9*x^3*exp(1/9*(x^2+6*x-9)/x)-63*x^3)/log(-exp(1/9*(x^2+6*x-9)/x)+7),x, alg
orithm="giac")

[Out]

integrate(-1/9*(9*(x*e^(1/9*(x^2 + 6*x - 9)/x) - 7*x)*log(-e^(1/9*(x^2 + 6*x - 9)/x) + 7)*log(log(-e^(1/9*(x^2
+ 6*x - 9)/x) + 7)) - (x^2 + 9)*e^(1/9*(x^2 + 6*x - 9)/x))/((x^3*e^(1/9*(x^2 + 6*x - 9)/x) - 7*x^3)*log(-e^(1
/9*(x^2 + 6*x - 9)/x) + 7)), x)

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maple [A]  time = 0.03, size = 25, normalized size = 1.00

method	result	size
risch	\(\frac {\ln \left (\ln \left (-{\mathrm e}^{\frac {x^{2}+6 x -9}{9 x}}+7\right )\right )}{x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*x*exp(1/9*(x^2+6*x-9)/x)+63*x)*ln(-exp(1/9*(x^2+6*x-9)/x)+7)*ln(ln(-exp(1/9*(x^2+6*x-9)/x)+7))+(x^2+9
)*exp(1/9*(x^2+6*x-9)/x))/(9*x^3*exp(1/9*(x^2+6*x-9)/x)-63*x^3)/ln(-exp(1/9*(x^2+6*x-9)/x)+7),x,method=_RETURN
VERBOSE)

[Out]

1/x*ln(ln(-exp(1/9*(x^2+6*x-9)/x)+7))

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maxima [A]  time = 0.45, size = 30, normalized size = 1.20 \begin {gather*} \frac {\log \left (x \log \left (-e^{\left (\frac {1}{9} \, x + \frac {2}{3}\right )} + 7 \, e^{\frac {1}{x}}\right ) - 1\right ) - \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x*exp(1/9*(x^2+6*x-9)/x)+63*x)*log(-exp(1/9*(x^2+6*x-9)/x)+7)*log(log(-exp(1/9*(x^2+6*x-9)/x)+7
))+(x^2+9)*exp(1/9*(x^2+6*x-9)/x))/(9*x^3*exp(1/9*(x^2+6*x-9)/x)-63*x^3)/log(-exp(1/9*(x^2+6*x-9)/x)+7),x, alg
orithm="maxima")

[Out]

(log(x*log(-e^(1/9*x + 2/3) + 7*e^(1/x)) - 1) - log(x))/x

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mupad [B]  time = 6.29, size = 22, normalized size = 0.88 \begin {gather*} \frac {\ln \left (\ln \left (7-{\mathrm {e}}^{x/9}\,{\mathrm {e}}^{2/3}\,{\mathrm {e}}^{-\frac {1}{x}}\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(((2*x)/3 + x^2/9 - 1)/x)*(x^2 + 9) + log(7 - exp(((2*x)/3 + x^2/9 - 1)/x))*log(log(7 - exp(((2*x)/3 +
x^2/9 - 1)/x)))*(63*x - 9*x*exp(((2*x)/3 + x^2/9 - 1)/x)))/(log(7 - exp(((2*x)/3 + x^2/9 - 1)/x))*(9*x^3*exp(
((2*x)/3 + x^2/9 - 1)/x) - 63*x^3)),x)

[Out]

log(log(7 - exp(x/9)*exp(2/3)*exp(-1/x)))/x

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sympy [A]  time = 0.68, size = 20, normalized size = 0.80 \begin {gather*} \frac {\log {\left (\log {\left (7 - e^{\frac {\frac {x^{2}}{9} + \frac {2 x}{3} - 1}{x}} \right )} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x*exp(1/9*(x**2+6*x-9)/x)+63*x)*ln(-exp(1/9*(x**2+6*x-9)/x)+7)*ln(ln(-exp(1/9*(x**2+6*x-9)/x)+7
))+(x**2+9)*exp(1/9*(x**2+6*x-9)/x))/(9*x**3*exp(1/9*(x**2+6*x-9)/x)-63*x**3)/ln(-exp(1/9*(x**2+6*x-9)/x)+7),x
)

[Out]

log(log(7 - exp((x**2/9 + 2*x/3 - 1)/x)))/x

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