3.49.82 \(\int \frac {-2 e^x x+(16+8 e^x) \log (18+9 e^x)+(4+2 e^x) \log (18+9 e^x) \log (\log (18+9 e^x))}{(160+80 x+10 x^2+e^x (80+40 x+5 x^2)) \log (18+9 e^x)+(80+20 x+e^x (40+10 x)) \log (18+9 e^x) \log (\log (18+9 e^x))+(10+5 e^x) \log (18+9 e^x) \log ^2(\log (18+9 e^x))} \, dx\)

Optimal. Leaf size=19 \[ \frac {2 x}{5 \left (4+x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )\right )} \]

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Rubi [F]  time = 1.99, 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 {-2 e^x x+\left (16+8 e^x\right ) \log \left (18+9 e^x\right )+\left (4+2 e^x\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )}{\left (160+80 x+10 x^2+e^x \left (80+40 x+5 x^2\right )\right ) \log \left (18+9 e^x\right )+\left (80+20 x+e^x (40+10 x)\right ) \log \left (18+9 e^x\right ) \log \left (\log \left (18+9 e^x\right )\right )+\left (10+5 e^x\right ) \log \left (18+9 e^x\right ) \log ^2\left (\log \left (18+9 e^x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*E^x*x + (16 + 8*E^x)*Log[18 + 9*E^x] + (4 + 2*E^x)*Log[18 + 9*E^x]*Log[Log[18 + 9*E^x]])/((160 + 80*x
+ 10*x^2 + E^x*(80 + 40*x + 5*x^2))*Log[18 + 9*E^x] + (80 + 20*x + E^x*(40 + 10*x))*Log[18 + 9*E^x]*Log[Log[18
+ 9*E^x]] + (10 + 5*E^x)*Log[18 + 9*E^x]*Log[Log[18 + 9*E^x]]^2),x]

[Out]

(-2*Defer[Int][x/(4 + x + Log[Log[9*(2 + E^x)]])^2, x])/5 - (2*Defer[Int][x/(Log[9*(2 + E^x)]*(4 + x + Log[Log
[9*(2 + E^x)]])^2), x])/5 + (4*Defer[Int][x/((2 + E^x)*Log[9*(2 + E^x)]*(4 + x + Log[Log[9*(2 + E^x)]])^2), x]
)/5 + (2*Defer[Int][(4 + x + Log[Log[9*(2 + E^x)]])^(-1), x])/5

Rubi steps

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

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Mathematica [A]  time = 0.49, size = 19, normalized size = 1.00 \begin {gather*} \frac {2 x}{5 \left (4+x+\log \left (\log \left (9 \left (2+e^x\right )\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^x*x + (16 + 8*E^x)*Log[18 + 9*E^x] + (4 + 2*E^x)*Log[18 + 9*E^x]*Log[Log[18 + 9*E^x]])/((160 +
80*x + 10*x^2 + E^x*(80 + 40*x + 5*x^2))*Log[18 + 9*E^x] + (80 + 20*x + E^x*(40 + 10*x))*Log[18 + 9*E^x]*Log[
Log[18 + 9*E^x]] + (10 + 5*E^x)*Log[18 + 9*E^x]*Log[Log[18 + 9*E^x]]^2),x]

[Out]

(2*x)/(5*(4 + x + Log[Log[9*(2 + E^x)]]))

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fricas [A]  time = 0.46, size = 16, normalized size = 0.84 \begin {gather*} \frac {2 \, x}{5 \, {\left (x + \log \left (\log \left (9 \, e^{x} + 18\right )\right ) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)+4)*log(9*exp(x)+18)*log(log(9*exp(x)+18))+(8*exp(x)+16)*log(9*exp(x)+18)-2*exp(x)*x)/((5*
exp(x)+10)*log(9*exp(x)+18)*log(log(9*exp(x)+18))^2+((10*x+40)*exp(x)+20*x+80)*log(9*exp(x)+18)*log(log(9*exp(
x)+18))+((5*x^2+40*x+80)*exp(x)+10*x^2+80*x+160)*log(9*exp(x)+18)),x, algorithm="fricas")

[Out]

2/5*x/(x + log(log(9*e^x + 18)) + 4)

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giac [A]  time = 0.47, size = 16, normalized size = 0.84 \begin {gather*} \frac {2 \, x}{5 \, {\left (x + \log \left (\log \left (9 \, e^{x} + 18\right )\right ) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)+4)*log(9*exp(x)+18)*log(log(9*exp(x)+18))+(8*exp(x)+16)*log(9*exp(x)+18)-2*exp(x)*x)/((5*
exp(x)+10)*log(9*exp(x)+18)*log(log(9*exp(x)+18))^2+((10*x+40)*exp(x)+20*x+80)*log(9*exp(x)+18)*log(log(9*exp(
x)+18))+((5*x^2+40*x+80)*exp(x)+10*x^2+80*x+160)*log(9*exp(x)+18)),x, algorithm="giac")

[Out]

2/5*x/(x + log(log(9*e^x + 18)) + 4)

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maple [A]  time = 0.04, size = 17, normalized size = 0.89

method	result	size
risch	\(\frac {2 x}{5 \left (\ln \left (\ln \left (9 \,{\mathrm e}^{x}+18\right )\right )+4+x \right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)+4)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))+(8*exp(x)+16)*ln(9*exp(x)+18)-2*exp(x)*x)/((5*exp(x)+10)
*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))^2+((10*x+40)*exp(x)+20*x+80)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))+((5*x^2+
40*x+80)*exp(x)+10*x^2+80*x+160)*ln(9*exp(x)+18)),x,method=_RETURNVERBOSE)

[Out]

2/5*x/(ln(ln(9*exp(x)+18))+4+x)

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maxima [A]  time = 0.55, size = 19, normalized size = 1.00 \begin {gather*} \frac {2 \, x}{5 \, {\left (x + \log \left (2 \, \log \relax (3) + \log \left (e^{x} + 2\right )\right ) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)+4)*log(9*exp(x)+18)*log(log(9*exp(x)+18))+(8*exp(x)+16)*log(9*exp(x)+18)-2*exp(x)*x)/((5*
exp(x)+10)*log(9*exp(x)+18)*log(log(9*exp(x)+18))^2+((10*x+40)*exp(x)+20*x+80)*log(9*exp(x)+18)*log(log(9*exp(
x)+18))+((5*x^2+40*x+80)*exp(x)+10*x^2+80*x+160)*log(9*exp(x)+18)),x, algorithm="maxima")

[Out]

2/5*x/(x + log(2*log(3) + log(e^x + 2)) + 4)

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mupad [B]  time = 3.51, size = 20, normalized size = 1.05 \begin {gather*} \frac {2\,x}{5\,\left (x+\ln \left (\ln \left (9\,{\mathrm {e}}^x+18\right )\right )+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(9*exp(x) + 18)*(8*exp(x) + 16) - 2*x*exp(x) + log(9*exp(x) + 18)*log(log(9*exp(x) + 18))*(2*exp(x) +
4))/(log(9*exp(x) + 18)*(80*x + exp(x)*(40*x + 5*x^2 + 80) + 10*x^2 + 160) + log(9*exp(x) + 18)*log(log(9*exp(
x) + 18))*(20*x + exp(x)*(10*x + 40) + 80) + log(9*exp(x) + 18)*log(log(9*exp(x) + 18))^2*(5*exp(x) + 10)),x)

[Out]

(2*x)/(5*(x + log(log(9*exp(x) + 18)) + 4))

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sympy [A]  time = 0.52, size = 19, normalized size = 1.00 \begin {gather*} \frac {2 x}{5 x + 5 \log {\left (\log {\left (9 e^{x} + 18 \right )} \right )} + 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)+4)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))+(8*exp(x)+16)*ln(9*exp(x)+18)-2*exp(x)*x)/((5*exp(
x)+10)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))**2+((10*x+40)*exp(x)+20*x+80)*ln(9*exp(x)+18)*ln(ln(9*exp(x)+18))+(
(5*x**2+40*x+80)*exp(x)+10*x**2+80*x+160)*ln(9*exp(x)+18)),x)

[Out]

2*x/(5*x + 5*log(log(9*exp(x) + 18)) + 20)

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