3.8.54 \(\int \frac {-5 e^4+5 x^2-5 x^3-5 x^4+(10 e^4-5 x^3-10 x^4) \log (x) \log (\log (x))+(e^4 (18-3 x)-8 x^3-18 x^4+x^5) \log (x) \log ^2(\log (x))}{25 x^3 \log (x)+(90 x^3-10 x^4) \log (x) \log (\log (x))+(81 x^3-18 x^4+x^5) \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=33 \[ \frac {1-\frac {e^4}{x^2}-x-x^2}{9-x+\frac {5}{\log (\log (x))}} \]

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Rubi [F]  time = 8.12, 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 {-5 e^4+5 x^2-5 x^3-5 x^4+\left (10 e^4-5 x^3-10 x^4\right ) \log (x) \log (\log (x))+\left (e^4 (18-3 x)-8 x^3-18 x^4+x^5\right ) \log (x) \log ^2(\log (x))}{25 x^3 \log (x)+\left (90 x^3-10 x^4\right ) \log (x) \log (\log (x))+\left (81 x^3-18 x^4+x^5\right ) \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5*E^4 + 5*x^2 - 5*x^3 - 5*x^4 + (10*E^4 - 5*x^3 - 10*x^4)*Log[x]*Log[Log[x]] + (E^4*(18 - 3*x) - 8*x^3 -
18*x^4 + x^5)*Log[x]*Log[Log[x]]^2)/(25*x^3*Log[x] + (90*x^3 - 10*x^4)*Log[x]*Log[Log[x]] + (81*x^3 - 18*x^4
+ x^5)*Log[x]*Log[Log[x]]^2),x]

[Out]

-1/81*(7209 + E^4)/(9 - x) - E^4/(9*x^2) - E^4/(81*x) + x + (25*(27 - E^4)*Defer[Int][(-5 - 9*Log[Log[x]] + x*
Log[Log[x]])^(-2), x])/2187 - (25*(2214 - E^4)*Defer[Int][(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^(-2), x])/2187
- (25*(7209 + E^4)*Defer[Int][1/((-9 + x)^2*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/81 - (25*(2214 - E^4)
*Defer[Int][1/((-9 + x)*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/243 - (25*(7209 + E^4)*Defer[Int][1/((-9
+ x)*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/729 - (25*E^4*Defer[Int][1/(x^2*(-5 - 9*Log[Log[x]] + x*Log[
Log[x]])^2), x])/81 - (50*E^4*Defer[Int][1/(x*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/729 - (10*E^4*Defer
[Int][1/(Log[x]*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/729 - (10*(27 - E^4)*Defer[Int][1/(Log[x]*(-5 - 9
*Log[Log[x]] + x*Log[Log[x]])^2), x])/243 + (5*(2214 - E^4)*Defer[Int][1/(Log[x]*(-5 - 9*Log[Log[x]] + x*Log[L
og[x]])^2), x])/243 - (5*(7209 + E^4)*Defer[Int][1/(Log[x]*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/729 -
5*E^4*Defer[Int][1/(x^3*Log[x]*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x] + (5*E^4*Defer[Int][1/(x*Log[x]*(-5
- 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/27 + (5*(27 - E^4)*Defer[Int][1/(x*Log[x]*(-5 - 9*Log[Log[x]] + x*Lo
g[Log[x]])^2), x])/27 + (5*(27 - E^4)*Defer[Int][x/(Log[x]*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/2187 -
(5*(2214 - E^4)*Defer[Int][x/(Log[x]*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])^2), x])/2187 - (10*(7209 + E^4)*Def
er[Int][1/((-9 + x)^2*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])), x])/81 - (5*(13851 - 2*E^4)*Defer[Int][1/((-9 + x
)*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])), x])/729 + (10*E^4*Defer[Int][1/(x^3*(-5 - 9*Log[Log[x]] + x*Log[Log[x
]])), x])/9 - (10*E^4*Defer[Int][1/(x*(-5 - 9*Log[Log[x]] + x*Log[Log[x]])), x])/729

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5 \left (e^4+x^2 \left (-1+x+x^2\right )\right )+\log (x) \log (\log (x)) \left (10 e^4-5 x^3 (1+2 x)+\left (-3 e^4 (-6+x)+x^3 \left (-8-18 x+x^2\right )\right ) \log (\log (x))\right )}{x^3 \log (x) (5-(-9+x) \log (\log (x)))^2} \, dx\\ &=\int \left (\frac {18 e^4-3 e^4 x-8 x^3-18 x^4+x^5}{(-9+x)^2 x^3}-\frac {5 \left (e^4-x^2+x^3+x^4\right ) \left (81-18 x+x^2+5 x \log (x)\right )}{(-9+x)^2 x^3 \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2}-\frac {5 \left (-18 e^4+4 e^4 x+7 x^3+19 x^4\right )}{(-9+x)^2 x^3 (-5-9 \log (\log (x))+x \log (\log (x)))}\right ) \, dx\\ &=-\left (5 \int \frac {\left (e^4-x^2+x^3+x^4\right ) \left (81-18 x+x^2+5 x \log (x)\right )}{(-9+x)^2 x^3 \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2} \, dx\right )-5 \int \frac {-18 e^4+4 e^4 x+7 x^3+19 x^4}{(-9+x)^2 x^3 (-5-9 \log (\log (x))+x \log (\log (x)))} \, dx+\int \frac {18 e^4-3 e^4 x-8 x^3-18 x^4+x^5}{(-9+x)^2 x^3} \, dx\\ &=-\left (5 \int \left (\frac {\left (7209+e^4\right ) \left (81-18 x+x^2+5 x \log (x)\right )}{729 (-9+x)^2 \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2}-\frac {\left (-2214+e^4\right ) \left (81-18 x+x^2+5 x \log (x)\right )}{2187 (-9+x) \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2}+\frac {e^4 \left (81-18 x+x^2+5 x \log (x)\right )}{81 x^3 \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2}+\frac {2 e^4 \left (81-18 x+x^2+5 x \log (x)\right )}{729 x^2 \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2}+\frac {\left (-27+e^4\right ) \left (81-18 x+x^2+5 x \log (x)\right )}{2187 x \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2}\right ) \, dx\right )-5 \int \left (\frac {2 \left (7209+e^4\right )}{81 (-9+x)^2 (-5-9 \log (\log (x))+x \log (\log (x)))}+\frac {13851-2 e^4}{729 (-9+x) (-5-9 \log (\log (x))+x \log (\log (x)))}-\frac {2 e^4}{9 x^3 (-5-9 \log (\log (x))+x \log (\log (x)))}+\frac {2 e^4}{729 x (-5-9 \log (\log (x))+x \log (\log (x)))}\right ) \, dx+\int \left (1+\frac {-7209-e^4}{81 (-9+x)^2}+\frac {2 e^4}{9 x^3}+\frac {e^4}{81 x^2}\right ) \, dx\\ &=-\frac {7209+e^4}{81 (9-x)}-\frac {e^4}{9 x^2}-\frac {e^4}{81 x}+x-\frac {1}{729} \left (10 e^4\right ) \int \frac {81-18 x+x^2+5 x \log (x)}{x^2 \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2} \, dx-\frac {1}{729} \left (10 e^4\right ) \int \frac {1}{x (-5-9 \log (\log (x))+x \log (\log (x)))} \, dx-\frac {1}{81} \left (5 e^4\right ) \int \frac {81-18 x+x^2+5 x \log (x)}{x^3 \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2} \, dx+\frac {1}{9} \left (10 e^4\right ) \int \frac {1}{x^3 (-5-9 \log (\log (x))+x \log (\log (x)))} \, dx-\frac {1}{729} \left (5 \left (13851-2 e^4\right )\right ) \int \frac {1}{(-9+x) (-5-9 \log (\log (x))+x \log (\log (x)))} \, dx+\frac {\left (5 \left (27-e^4\right )\right ) \int \frac {81-18 x+x^2+5 x \log (x)}{x \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2} \, dx}{2187}-\frac {\left (5 \left (2214-e^4\right )\right ) \int \frac {81-18 x+x^2+5 x \log (x)}{(-9+x) \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2} \, dx}{2187}-\frac {1}{729} \left (5 \left (7209+e^4\right )\right ) \int \frac {81-18 x+x^2+5 x \log (x)}{(-9+x)^2 \log (x) (-5-9 \log (\log (x))+x \log (\log (x)))^2} \, dx-\frac {1}{81} \left (10 \left (7209+e^4\right )\right ) \int \frac {1}{(-9+x)^2 (-5-9 \log (\log (x))+x \log (\log (x)))} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 41, normalized size = 1.24 \begin {gather*} \frac {50 x^2+\left (e^4+x^2 \left (89-9 x+x^2\right )\right ) \log (\log (x))}{x^2 (-5+(-9+x) \log (\log (x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*E^4 + 5*x^2 - 5*x^3 - 5*x^4 + (10*E^4 - 5*x^3 - 10*x^4)*Log[x]*Log[Log[x]] + (E^4*(18 - 3*x) - 8
*x^3 - 18*x^4 + x^5)*Log[x]*Log[Log[x]]^2)/(25*x^3*Log[x] + (90*x^3 - 10*x^4)*Log[x]*Log[Log[x]] + (81*x^3 - 1
8*x^4 + x^5)*Log[x]*Log[Log[x]]^2),x]

[Out]

(50*x^2 + (E^4 + x^2*(89 - 9*x + x^2))*Log[Log[x]])/(x^2*(-5 + (-9 + x)*Log[Log[x]]))

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fricas [A]  time = 0.70, size = 50, normalized size = 1.52 \begin {gather*} -\frac {50 \, x^{2} + {\left (x^{4} - 9 \, x^{3} + 89 \, x^{2} + e^{4}\right )} \log \left (\log \relax (x)\right )}{5 \, x^{2} - {\left (x^{3} - 9 \, x^{2}\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x+18)*exp(4)+x^5-18*x^4-8*x^3)*log(x)*log(log(x))^2+(10*exp(4)-10*x^4-5*x^3)*log(x)*log(log(x)
)-5*exp(4)-5*x^4-5*x^3+5*x^2)/((x^5-18*x^4+81*x^3)*log(x)*log(log(x))^2+(-10*x^4+90*x^3)*log(x)*log(log(x))+25
*x^3*log(x)),x, algorithm="fricas")

[Out]

-(50*x^2 + (x^4 - 9*x^3 + 89*x^2 + e^4)*log(log(x)))/(5*x^2 - (x^3 - 9*x^2)*log(log(x)))

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giac [B]  time = 0.65, size = 59, normalized size = 1.79 \begin {gather*} \frac {x^{4} \log \left (\log \relax (x)\right ) - 9 \, x^{3} \log \left (\log \relax (x)\right ) + 89 \, x^{2} \log \left (\log \relax (x)\right ) + 50 \, x^{2} + e^{4} \log \left (\log \relax (x)\right )}{x^{3} \log \left (\log \relax (x)\right ) - 9 \, x^{2} \log \left (\log \relax (x)\right ) - 5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x+18)*exp(4)+x^5-18*x^4-8*x^3)*log(x)*log(log(x))^2+(10*exp(4)-10*x^4-5*x^3)*log(x)*log(log(x)
)-5*exp(4)-5*x^4-5*x^3+5*x^2)/((x^5-18*x^4+81*x^3)*log(x)*log(log(x))^2+(-10*x^4+90*x^3)*log(x)*log(log(x))+25
*x^3*log(x)),x, algorithm="giac")

[Out]

(x^4*log(log(x)) - 9*x^3*log(log(x)) + 89*x^2*log(log(x)) + 50*x^2 + e^4*log(log(x)))/(x^3*log(log(x)) - 9*x^2
*log(log(x)) - 5*x^2)

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maple [A]  time = 0.06, size = 65, normalized size = 1.97

method	result	size
risch	\(\frac {x^{4}-9 x^{3}+89 x^{2}+{\mathrm e}^{4}}{x^{2} \left (x -9\right )}+\frac {5 x^{4}+5 x^{3}-5 x^{2}+5 \,{\mathrm e}^{4}}{x^{2} \left (x -9\right ) \left (x \ln \left (\ln \relax (x )\right )-9 \ln \left (\ln \relax (x )\right )-5\right )}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x+18)*exp(4)+x^5-18*x^4-8*x^3)*ln(x)*ln(ln(x))^2+(10*exp(4)-10*x^4-5*x^3)*ln(x)*ln(ln(x))-5*exp(4)-5
*x^4-5*x^3+5*x^2)/((x^5-18*x^4+81*x^3)*ln(x)*ln(ln(x))^2+(-10*x^4+90*x^3)*ln(x)*ln(ln(x))+25*x^3*ln(x)),x,meth
od=_RETURNVERBOSE)

[Out]

(x^4-9*x^3+89*x^2+exp(4))/x^2/(x-9)+5*(x^4+x^3-x^2+exp(4))/x^2/(x-9)/(x*ln(ln(x))-9*ln(ln(x))-5)

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maxima [A]  time = 0.74, size = 50, normalized size = 1.52 \begin {gather*} -\frac {50 \, x^{2} + {\left (x^{4} - 9 \, x^{3} + 89 \, x^{2} + e^{4}\right )} \log \left (\log \relax (x)\right )}{5 \, x^{2} - {\left (x^{3} - 9 \, x^{2}\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x+18)*exp(4)+x^5-18*x^4-8*x^3)*log(x)*log(log(x))^2+(10*exp(4)-10*x^4-5*x^3)*log(x)*log(log(x)
)-5*exp(4)-5*x^4-5*x^3+5*x^2)/((x^5-18*x^4+81*x^3)*log(x)*log(log(x))^2+(-10*x^4+90*x^3)*log(x)*log(log(x))+25
*x^3*log(x)),x, algorithm="maxima")

[Out]

-(50*x^2 + (x^4 - 9*x^3 + 89*x^2 + e^4)*log(log(x)))/(5*x^2 - (x^3 - 9*x^2)*log(log(x)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {5\,{\mathrm {e}}^4-5\,x^2+5\,x^3+5\,x^4+\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (10\,x^4+5\,x^3-10\,{\mathrm {e}}^4\right )+{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)\,\left (8\,x^3+18\,x^4-x^5+{\mathrm {e}}^4\,\left (3\,x-18\right )\right )}{25\,x^3\,\ln \relax (x)+\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (90\,x^3-10\,x^4\right )+{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)\,\left (x^5-18\,x^4+81\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(4) - 5*x^2 + 5*x^3 + 5*x^4 + log(log(x))*log(x)*(5*x^3 - 10*exp(4) + 10*x^4) + log(log(x))^2*log(x
)*(8*x^3 + 18*x^4 - x^5 + exp(4)*(3*x - 18)))/(25*x^3*log(x) + log(log(x))*log(x)*(90*x^3 - 10*x^4) + log(log(
x))^2*log(x)*(81*x^3 - 18*x^4 + x^5)),x)

[Out]

int(-(5*exp(4) - 5*x^2 + 5*x^3 + 5*x^4 + log(log(x))*log(x)*(5*x^3 - 10*exp(4) + 10*x^4) + log(log(x))^2*log(x
)*(8*x^3 + 18*x^4 - x^5 + exp(4)*(3*x - 18)))/(25*x^3*log(x) + log(log(x))*log(x)*(90*x^3 - 10*x^4) + log(log(
x))^2*log(x)*(81*x^3 - 18*x^4 + x^5)), x)

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sympy [B]  time = 0.66, size = 66, normalized size = 2.00 \begin {gather*} x + \frac {89 x^{2} + e^{4}}{x^{3} - 9 x^{2}} + \frac {5 x^{4} + 5 x^{3} - 5 x^{2} + 5 e^{4}}{- 5 x^{3} + 45 x^{2} + \left (x^{4} - 18 x^{3} + 81 x^{2}\right ) \log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x+18)*exp(4)+x**5-18*x**4-8*x**3)*ln(x)*ln(ln(x))**2+(10*exp(4)-10*x**4-5*x**3)*ln(x)*ln(ln(x)
)-5*exp(4)-5*x**4-5*x**3+5*x**2)/((x**5-18*x**4+81*x**3)*ln(x)*ln(ln(x))**2+(-10*x**4+90*x**3)*ln(x)*ln(ln(x))
+25*x**3*ln(x)),x)

[Out]

x + (89*x**2 + exp(4))/(x**3 - 9*x**2) + (5*x**4 + 5*x**3 - 5*x**2 + 5*exp(4))/(-5*x**3 + 45*x**2 + (x**4 - 18
*x**3 + 81*x**2)*log(log(x)))

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