∫ −405x log(x)+90x log(x) log(log(x))−5x log(x) log2(log(x))+e− 4ex ________________−9+log (log (x)) (−20ex−180exx log(x)+20exx log(x) log(log(x)))_____________________________________________________________ (729e2x−486ex2+81x3) log(x)+(−162e2x+108ex2−18x3) log(x) log(log(x))+(9e2x−6ex2+x3) log(x) log2(log(x))+e− 8ex______ −9+log(log(x)) (81x log(x)−18x log(x) log(log(x))+x log(x) log2(log(x)))+e− 4ex______ −9+log(log(x)) ((−486ex+162x2) log(x)+(108ex−36x2) log(x) log(log(x))+(−6ex+2x2) log(x) log2(log(x))) dx

3.86.29 $\int \frac {-405 x \log (x)+90 x \log (x) \log (\log (x))-5 x \log (x) \log ^2(\log (x))+e^{-\frac {4 e^x}{-9+\log (\log (x))}} (-20 e^x-180 e^x x \log (x)+20 e^x x \log (x) \log (\log (x)))}{(729 e^2 x-486 e x^2+81 x^3) \log (x)+(-162 e^2 x+108 e x^2-18 x^3) \log (x) \log (\log (x))+(9 e^2 x-6 e x^2+x^3) \log (x) \log ^2(\log (x))+e^{-\frac {8 e^x}{-9+\log (\log (x))}} (81 x \log (x)-18 x \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x)))+e^{-\frac {4 e^x}{-9+\log (\log (x))}} ((-486 e x+162 x^2) \log (x)+(108 e x-36 x^2) \log (x) \log (\log (x))+(-6 e x+2 x^2) \log (x) \log ^2(\log (x)))} \, dx$

Optimal. Leaf size=25 \[ \frac {5}{-3 e+e^{\frac {4 e^x}{9-\log (\log (x))}}+x} \]

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Rubi [F] time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Int[(-405*x*Log[x] + 90*x*Log[x]*Log[Log[x]] - 5*x*Log[x]*Log[Log[x]]^2 + (-20*E^x - 180*E^x*x*Log[x] + 20*E^x

*x*Log[x]*Log[Log[x]])/E^((4*E^x)/(-9 + Log[Log[x]])))/((729*E^2*x - 486*E*x^2 + 81*x^3)*Log[x] + (-162*E^2*x

+ 108*E*x^2 - 18*x^3)*Log[x]*Log[Log[x]] + (9*E^2*x - 6*E*x^2 + x^3)*Log[x]*Log[Log[x]]^2 + (81*x*Log[x] - 18*

x*Log[x]*Log[Log[x]] + x*Log[x]*Log[Log[x]]^2)/E^((8*E^x)/(-9 + Log[Log[x]])) + ((-486*E*x + 162*x^2)*Log[x] +

(108*E*x - 36*x^2)*Log[x]*Log[Log[x]] + (-6*E*x + 2*x^2)*Log[x]*Log[Log[x]]^2)/E^((4*E^x)/(-9 + Log[Log[x]]))

),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [B] time = 0.61, size = 55, normalized size = 2.20 \begin {gather*} -\frac {5 e^{\frac {4 e^x}{-9+\log (\log (x))}}}{-1+3 e^{1+\frac {4 e^x}{-9+\log (\log (x))}}-e^{\frac {4 e^x}{-9+\log (\log (x))}} x} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-405*x*Log[x] + 90*x*Log[x]*Log[Log[x]] - 5*x*Log[x]*Log[Log[x]]^2 + (-20*E^x - 180*E^x*x*Log[x] +

20*E^x*x*Log[x]*Log[Log[x]])/E^((4*E^x)/(-9 + Log[Log[x]])))/((729*E^2*x - 486*E*x^2 + 81*x^3)*Log[x] + (-162*

E^2*x + 108*E*x^2 - 18*x^3)*Log[x]*Log[Log[x]] + (9*E^2*x - 6*E*x^2 + x^3)*Log[x]*Log[Log[x]]^2 + (81*x*Log[x]

- 18*x*Log[x]*Log[Log[x]] + x*Log[x]*Log[Log[x]]^2)/E^((8*E^x)/(-9 + Log[Log[x]])) + ((-486*E*x + 162*x^2)*Lo

g[x] + (108*E*x - 36*x^2)*Log[x]*Log[Log[x]] + (-6*E*x + 2*x^2)*Log[x]*Log[Log[x]]^2)/E^((4*E^x)/(-9 + Log[Log

[x]]))),x]

[Out]

(-5*E^((4*E^x)/(-9 + Log[Log[x]])))/(-1 + 3*E^(1 + (4*E^x)/(-9 + Log[Log[x]])) - E^((4*E^x)/(-9 + Log[Log[x]])

)*x)

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fricas [A] time = 1.01, size = 22, normalized size = 0.88 \begin {gather*} \frac {5}{x - 3 \, e + e^{\left (-\frac {4 \, e^{x}}{\log \left (\log \relax (x)\right ) - 9}\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*exp(x)*log(x)*log(log(x))-180*x*exp(x)*log(x)-20*exp(x))*exp(-4*exp(x)/(log(log(x))-9))-5*x*l

og(x)*log(log(x))^2+90*x*log(x)*log(log(x))-405*x*log(x))/((x*log(x)*log(log(x))^2-18*x*log(x)*log(log(x))+81*

x*log(x))*exp(-4*exp(x)/(log(log(x))-9))^2+((-6*x*exp(1)+2*x^2)*log(x)*log(log(x))^2+(108*x*exp(1)-36*x^2)*log

(x)*log(log(x))+(-486*x*exp(1)+162*x^2)*log(x))*exp(-4*exp(x)/(log(log(x))-9))+(9*x*exp(1)^2-6*x^2*exp(1)+x^3)

*log(x)*log(log(x))^2+(-162*x*exp(1)^2+108*x^2*exp(1)-18*x^3)*log(x)*log(log(x))+(729*x*exp(1)^2-486*x^2*exp(1

)+81*x^3)*log(x)),x, algorithm="fricas")

[Out]

5/(x - 3*e + e^(-4*e^x/(log(log(x)) - 9)))

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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(((20*x*exp(x)*log(x)*log(log(x))-180*x*exp(x)*log(x)-20*exp(x))*exp(-4*exp(x)/(log(log(x))-9))-5*x*l

og(x)*log(log(x))^2+90*x*log(x)*log(log(x))-405*x*log(x))/((x*log(x)*log(log(x))^2-18*x*log(x)*log(log(x))+81*

x*log(x))*exp(-4*exp(x)/(log(log(x))-9))^2+((-6*x*exp(1)+2*x^2)*log(x)*log(log(x))^2+(108*x*exp(1)-36*x^2)*log

(x)*log(log(x))+(-486*x*exp(1)+162*x^2)*log(x))*exp(-4*exp(x)/(log(log(x))-9))+(9*x*exp(1)^2-6*x^2*exp(1)+x^3)

*log(x)*log(log(x))^2+(-162*x*exp(1)^2+108*x^2*exp(1)-18*x^3)*log(x)*log(log(x))+(729*x*exp(1)^2-486*x^2*exp(1

)+81*x^3)*log(x)),x, algorithm="giac")

[Out]

undef

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maple [A] time = 0.16, size = 27, normalized size = 1.08


method	result	size

risch	$-\frac {5}{-{\mathrm e}^{-\frac {4 \,{\mathrm e}^{x}}{\ln \left (\ln \relax (x )\right )-9}}-x +3 \,{\mathrm e}}$	$27$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x*exp(x)*ln(x)*ln(ln(x))-180*x*exp(x)*ln(x)-20*exp(x))*exp(-4*exp(x)/(ln(ln(x))-9))-5*x*ln(x)*ln(ln(x

))^2+90*x*ln(x)*ln(ln(x))-405*x*ln(x))/((x*ln(x)*ln(ln(x))^2-18*x*ln(x)*ln(ln(x))+81*x*ln(x))*exp(-4*exp(x)/(l

n(ln(x))-9))^2+((-6*x*exp(1)+2*x^2)*ln(x)*ln(ln(x))^2+(108*x*exp(1)-36*x^2)*ln(x)*ln(ln(x))+(-486*x*exp(1)+162

*x^2)*ln(x))*exp(-4*exp(x)/(ln(ln(x))-9))+(9*x*exp(1)^2-6*x^2*exp(1)+x^3)*ln(x)*ln(ln(x))^2+(-162*x*exp(1)^2+1

08*x^2*exp(1)-18*x^3)*ln(x)*ln(ln(x))+(729*x*exp(1)^2-486*x^2*exp(1)+81*x^3)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-5/(-exp(-4*exp(x)/(ln(ln(x))-9))-x+3*exp(1))

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maxima [A] time = 0.59, size = 37, normalized size = 1.48 \begin {gather*} \frac {5 \, e^{\left (\frac {4 \, e^{x}}{\log \left (\log \relax (x)\right ) - 9}\right )}}{{\left (x - 3 \, e\right )} e^{\left (\frac {4 \, e^{x}}{\log \left (\log \relax (x)\right ) - 9}\right )} + 1} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*exp(x)*log(x)*log(log(x))-180*x*exp(x)*log(x)-20*exp(x))*exp(-4*exp(x)/(log(log(x))-9))-5*x*l

og(x)*log(log(x))^2+90*x*log(x)*log(log(x))-405*x*log(x))/((x*log(x)*log(log(x))^2-18*x*log(x)*log(log(x))+81*

x*log(x))*exp(-4*exp(x)/(log(log(x))-9))^2+((-6*x*exp(1)+2*x^2)*log(x)*log(log(x))^2+(108*x*exp(1)-36*x^2)*log

(x)*log(log(x))+(-486*x*exp(1)+162*x^2)*log(x))*exp(-4*exp(x)/(log(log(x))-9))+(9*x*exp(1)^2-6*x^2*exp(1)+x^3)

*log(x)*log(log(x))^2+(-162*x*exp(1)^2+108*x^2*exp(1)-18*x^3)*log(x)*log(log(x))+(729*x*exp(1)^2-486*x^2*exp(1

)+81*x^3)*log(x)),x, algorithm="maxima")

[Out]

5*e^(4*e^x/(log(log(x)) - 9))/((x - 3*e)*e^(4*e^x/(log(log(x)) - 9)) + 1)

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mupad [B] time = 5.71, size = 22, normalized size = 0.88 \begin {gather*} \frac {5}{x-3\,\mathrm {e}+{\mathrm {e}}^{-\frac {4\,{\mathrm {e}}^x}{\ln \left (\ln \relax (x)\right )-9}}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(4*exp(x))/(log(log(x)) - 9))*(20*exp(x) + 180*x*exp(x)*log(x) - 20*x*log(log(x))*exp(x)*log(x)) +

405*x*log(x) - 90*x*log(log(x))*log(x) + 5*x*log(log(x))^2*log(x))/(exp(-(8*exp(x))/(log(log(x)) - 9))*(81*x*l

og(x) - 18*x*log(log(x))*log(x) + x*log(log(x))^2*log(x)) + log(x)*(729*x*exp(2) - 486*x^2*exp(1) + 81*x^3) -

exp(-(4*exp(x))/(log(log(x)) - 9))*(log(x)*(486*x*exp(1) - 162*x^2) - log(log(x))*log(x)*(108*x*exp(1) - 36*x^

2) + log(log(x))^2*log(x)*(6*x*exp(1) - 2*x^2)) - log(log(x))*log(x)*(162*x*exp(2) - 108*x^2*exp(1) + 18*x^3)

+ log(log(x))^2*log(x)*(9*x*exp(2) - 6*x^2*exp(1) + x^3)),x)

[Out]

5/(x - 3*exp(1) + exp(-(4*exp(x))/(log(log(x)) - 9)))

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sympy [A] time = 2.68, size = 22, normalized size = 0.88 \begin {gather*} \frac {5}{x - 3 e + e^{- \frac {4 e^{x}}{\log {\left (\log {\relax (x )} \right )} - 9}}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*exp(x)*ln(x)*ln(ln(x))-180*x*exp(x)*ln(x)-20*exp(x))*exp(-4*exp(x)/(ln(ln(x))-9))-5*x*ln(x)*l

n(ln(x))**2+90*x*ln(x)*ln(ln(x))-405*x*ln(x))/((x*ln(x)*ln(ln(x))**2-18*x*ln(x)*ln(ln(x))+81*x*ln(x))*exp(-4*e

xp(x)/(ln(ln(x))-9))**2+((-6*x*exp(1)+2*x**2)*ln(x)*ln(ln(x))**2+(108*x*exp(1)-36*x**2)*ln(x)*ln(ln(x))+(-486*

x*exp(1)+162*x**2)*ln(x))*exp(-4*exp(x)/(ln(ln(x))-9))+(9*x*exp(1)**2-6*x**2*exp(1)+x**3)*ln(x)*ln(ln(x))**2+(

-162*x*exp(1)**2+108*x**2*exp(1)-18*x**3)*ln(x)*ln(ln(x))+(729*x*exp(1)**2-486*x**2*exp(1)+81*x**3)*ln(x)),x)

[Out]

5/(x - 3*E + exp(-4*exp(x)/(log(log(x)) - 9)))

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