\(\int \frac {(-12 x-24 x^2-12 x^3) \log (5)+(9 x^2+36 x^3+27 x^4) \log (x)+((8+16 x+8 x^2) \log ^2(5)+(-24 x^2-24 x^3) \log (5) \log (x)) \log (\log (x))+(-4+4 x^2) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx\) [8469]

Optimal result

Integrand size = 106, antiderivative size = 25 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {(1+x)^2 \left (x-\frac {2}{3} \log (5) \log (\log (x))\right )^2}{e^5 x} \]

[Out]

1/exp(5)/x*(1+x)^2*(x-2/3*ln(ln(x))*ln(5))^2

Rubi [F]

\[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx \]

[In]

Int[((-12*x - 24*x^2 - 12*x^3)*Log[5] + (9*x^2 + 36*x^3 + 27*x^4)*Log[x] + ((8 + 16*x + 8*x^2)*Log[5]^2 + (-24
*x^2 - 24*x^3)*Log[5]*Log[x])*Log[Log[x]] + (-4 + 4*x^2)*Log[5]^2*Log[x]*Log[Log[x]]^2)/(9*E^5*x^2*Log[x]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{x^2 \log (x)} \, dx}{9 e^5} \\ & = \frac {\int \frac {(1+x) (3 x-2 \log (5) \log (\log (x))) (-4 (1+x) \log (5)+\log (x) (3 x (1+3 x)-2 (-1+x) \log (5) \log (\log (x))))}{x^2 \log (x)} \, dx}{9 e^5} \\ & = \frac {\int \left (\frac {3 (1+x) \left (-4 \log (5)-4 x \log (5)+3 x \log (x)+9 x^2 \log (x)\right )}{x \log (x)}-\frac {8 (1+x) \log (5) \left (-\log (5)-x \log (5)+3 x^2 \log (x)\right ) \log (\log (x))}{x^2 \log (x)}+\frac {4 (-1+x) (1+x) \log ^2(5) \log ^2(\log (x))}{x^2}\right ) \, dx}{9 e^5} \\ & = \frac {\int \frac {(1+x) \left (-4 \log (5)-4 x \log (5)+3 x \log (x)+9 x^2 \log (x)\right )}{x \log (x)} \, dx}{3 e^5}-\frac {(8 \log (5)) \int \frac {(1+x) \left (-\log (5)-x \log (5)+3 x^2 \log (x)\right ) \log (\log (x))}{x^2 \log (x)} \, dx}{9 e^5}+\frac {\left (4 \log ^2(5)\right ) \int \frac {(-1+x) (1+x) \log ^2(\log (x))}{x^2} \, dx}{9 e^5} \\ & = \frac {\int \left (3 \left (1+4 x+3 x^2\right )-\frac {4 (1+x)^2 \log (5)}{x \log (x)}\right ) \, dx}{3 e^5}-\frac {(8 \log (5)) \int \left (\frac {\left (-\log (5)-x \log (5)+3 x^2 \log (x)\right ) \log (\log (x))}{x^2 \log (x)}+\frac {\left (-\log (5)-x \log (5)+3 x^2 \log (x)\right ) \log (\log (x))}{x \log (x)}\right ) \, dx}{9 e^5}+\frac {\left (4 \log ^2(5)\right ) \int \left (\log ^2(\log (x))-\frac {\log ^2(\log (x))}{x^2}\right ) \, dx}{9 e^5} \\ & = \frac {\int \left (1+4 x+3 x^2\right ) \, dx}{e^5}-\frac {(8 \log (5)) \int \frac {\left (-\log (5)-x \log (5)+3 x^2 \log (x)\right ) \log (\log (x))}{x^2 \log (x)} \, dx}{9 e^5}-\frac {(8 \log (5)) \int \frac {\left (-\log (5)-x \log (5)+3 x^2 \log (x)\right ) \log (\log (x))}{x \log (x)} \, dx}{9 e^5}-\frac {(4 \log (5)) \int \frac {(1+x)^2}{x \log (x)} \, dx}{3 e^5}+\frac {\left (4 \log ^2(5)\right ) \int \log ^2(\log (x)) \, dx}{9 e^5}-\frac {\left (4 \log ^2(5)\right ) \int \frac {\log ^2(\log (x))}{x^2} \, dx}{9 e^5} \\ & = \frac {x}{e^5}+\frac {2 x^2}{e^5}+\frac {x^3}{e^5}-\frac {(8 \log (5)) \int \left (3-\frac {(1+x) \log (5)}{x^2 \log (x)}\right ) \log (\log (x)) \, dx}{9 e^5}-\frac {(8 \log (5)) \int \left (3 x \log (\log (x))-\frac {\log (5) \log (\log (x))}{\log (x)}-\frac {\log (5) \log (\log (x))}{x \log (x)}\right ) \, dx}{9 e^5}-\frac {(4 \log (5)) \int \left (\frac {2}{\log (x)}+\frac {1}{x \log (x)}+\frac {x}{\log (x)}\right ) \, dx}{3 e^5}+\frac {\left (4 \log ^2(5)\right ) \int \log ^2(\log (x)) \, dx}{9 e^5}-\frac {\left (4 \log ^2(5)\right ) \int \frac {\log ^2(\log (x))}{x^2} \, dx}{9 e^5} \\ & = \frac {x}{e^5}+\frac {2 x^2}{e^5}+\frac {x^3}{e^5}-\frac {(8 \log (5)) \int \left (3 \log (\log (x))-\frac {\log (5) \log (\log (x))}{x^2 \log (x)}-\frac {\log (5) \log (\log (x))}{x \log (x)}\right ) \, dx}{9 e^5}-\frac {(4 \log (5)) \int \frac {1}{x \log (x)} \, dx}{3 e^5}-\frac {(4 \log (5)) \int \frac {x}{\log (x)} \, dx}{3 e^5}-\frac {(8 \log (5)) \int \frac {1}{\log (x)} \, dx}{3 e^5}-\frac {(8 \log (5)) \int x \log (\log (x)) \, dx}{3 e^5}+\frac {\left (4 \log ^2(5)\right ) \int \log ^2(\log (x)) \, dx}{9 e^5}-\frac {\left (4 \log ^2(5)\right ) \int \frac {\log ^2(\log (x))}{x^2} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{\log (x)} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{x \log (x)} \, dx}{9 e^5} \\ & = \frac {x}{e^5}+\frac {2 x^2}{e^5}+\frac {x^3}{e^5}-\frac {4 x^2 \log (5) \log (\log (x))}{3 e^5}-\frac {8 \log (5) \operatorname {LogIntegral}(x)}{3 e^5}+\frac {(4 \log (5)) \int \frac {x}{\log (x)} \, dx}{3 e^5}-\frac {(4 \log (5)) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )}{3 e^5}-\frac {(4 \log (5)) \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )}{3 e^5}-\frac {(8 \log (5)) \int \log (\log (x)) \, dx}{3 e^5}+\frac {\left (4 \log ^2(5)\right ) \int \log ^2(\log (x)) \, dx}{9 e^5}-\frac {\left (4 \log ^2(5)\right ) \int \frac {\log ^2(\log (x))}{x^2} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{\log (x)} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{x^2 \log (x)} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{x \log (x)} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,\log (x)\right )}{9 e^5} \\ & = \frac {x}{e^5}+\frac {2 x^2}{e^5}+\frac {x^3}{e^5}-\frac {4 \operatorname {ExpIntegralEi}(2 \log (x)) \log (5)}{3 e^5}-\frac {4 \log (5) \log (\log (x))}{3 e^5}-\frac {8 x \log (5) \log (\log (x))}{3 e^5}-\frac {4 x^2 \log (5) \log (\log (x))}{3 e^5}+\frac {4 \log ^2(5) \log ^2(\log (x))}{9 e^5}-\frac {8 \log (5) \operatorname {LogIntegral}(x)}{3 e^5}+\frac {(4 \log (5)) \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )}{3 e^5}+\frac {(8 \log (5)) \int \frac {1}{\log (x)} \, dx}{3 e^5}+\frac {\left (4 \log ^2(5)\right ) \int \log ^2(\log (x)) \, dx}{9 e^5}-\frac {\left (4 \log ^2(5)\right ) \int \frac {\log ^2(\log (x))}{x^2} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{\log (x)} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{x^2 \log (x)} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,\log (x)\right )}{9 e^5} \\ & = \frac {x}{e^5}+\frac {2 x^2}{e^5}+\frac {x^3}{e^5}-\frac {4 \log (5) \log (\log (x))}{3 e^5}-\frac {8 x \log (5) \log (\log (x))}{3 e^5}-\frac {4 x^2 \log (5) \log (\log (x))}{3 e^5}+\frac {8 \log ^2(5) \log ^2(\log (x))}{9 e^5}+\frac {\left (4 \log ^2(5)\right ) \int \log ^2(\log (x)) \, dx}{9 e^5}-\frac {\left (4 \log ^2(5)\right ) \int \frac {\log ^2(\log (x))}{x^2} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{\log (x)} \, dx}{9 e^5}+\frac {\left (8 \log ^2(5)\right ) \int \frac {\log (\log (x))}{x^2 \log (x)} \, dx}{9 e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {(1+x)^2 (3 x-2 \log (5) \log (\log (x)))^2}{9 e^5 x} \]

[In]

Integrate[((-12*x - 24*x^2 - 12*x^3)*Log[5] + (9*x^2 + 36*x^3 + 27*x^4)*Log[x] + ((8 + 16*x + 8*x^2)*Log[5]^2
+ (-24*x^2 - 24*x^3)*Log[5]*Log[x])*Log[Log[x]] + (-4 + 4*x^2)*Log[5]^2*Log[x]*Log[Log[x]]^2)/(9*E^5*x^2*Log[x
]),x]

[Out]

((1 + x)^2*(3*x - 2*Log[5]*Log[Log[x]])^2)/(9*E^5*x)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(64\) vs. \(2(24)=48\).

Time = 1.66 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.60

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-5} \ln \left (5\right )^{2} \left (x^{2}+2 x +1\right ) \ln \left (\ln \left (x \right )\right )^{2}}{9 x}-\frac {4 \,{\mathrm e}^{-5} \ln \left (5\right ) x \left (2+x \right ) \ln \left (\ln \left (x \right )\right )}{3}+{\mathrm e}^{-5} x^{3}+2 x^{2} {\mathrm e}^{-5}+x \,{\mathrm e}^{-5}-\frac {4 \,{\mathrm e}^{-5} \ln \left (\ln \left (x \right )\right ) \ln \left (5\right )}{3}\)	\(65\)
parallelrisch	\(\frac {{\mathrm e}^{-5} \left (4 \ln \left (5\right )^{2} \ln \left (\ln \left (x \right )\right )^{2} x^{2}+8 \ln \left (5\right )^{2} \ln \left (\ln \left (x \right )\right )^{2} x -12 \ln \left (5\right ) \ln \left (\ln \left (x \right )\right ) x^{3}+4 \ln \left (5\right )^{2} \ln \left (\ln \left (x \right )\right )^{2}-24 \ln \left (5\right ) x^{2} \ln \left (\ln \left (x \right )\right )+9 x^{4}-12 x \ln \left (5\right ) \ln \left (\ln \left (x \right )\right )+18 x^{3}+9 x^{2}\right )}{9 x}\)	\(91\)

[In]

int(1/9*((4*x^2-4)*ln(5)^2*ln(x)*ln(ln(x))^2+((-24*x^3-24*x^2)*ln(5)*ln(x)+(8*x^2+16*x+8)*ln(5)^2)*ln(ln(x))+(
27*x^4+36*x^3+9*x^2)*ln(x)+(-12*x^3-24*x^2-12*x)*ln(5))/x^2/exp(5)/ln(x),x,method=_RETURNVERBOSE)

[Out]

4/9*exp(-5)*ln(5)^2*(x^2+2*x+1)/x*ln(ln(x))^2-4/3*exp(-5)*ln(5)*x*(2+x)*ln(ln(x))+exp(-5)*x^3+2*x^2*exp(-5)+x*
exp(-5)-4/3*exp(-5)*ln(ln(x))*ln(5)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).

Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.36 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {{\left (4 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (5\right )^{2} \log \left (\log \left (x\right )\right )^{2} + 9 \, x^{4} + 18 \, x^{3} - 12 \, {\left (x^{3} + 2 \, x^{2} + x\right )} \log \left (5\right ) \log \left (\log \left (x\right )\right ) + 9 \, x^{2}\right )} e^{\left (-5\right )}}{9 \, x} \]

[In]

integrate(1/9*((4*x^2-4)*log(5)^2*log(x)*log(log(x))^2+((-24*x^3-24*x^2)*log(5)*log(x)+(8*x^2+16*x+8)*log(5)^2
)*log(log(x))+(27*x^4+36*x^3+9*x^2)*log(x)+(-12*x^3-24*x^2-12*x)*log(5))/x^2/exp(5)/log(x),x, algorithm="frica
s")

[Out]

1/9*(4*(x^2 + 2*x + 1)*log(5)^2*log(log(x))^2 + 9*x^4 + 18*x^3 - 12*(x^3 + 2*x^2 + x)*log(5)*log(log(x)) + 9*x
^2)*e^(-5)/x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (24) = 48\).

Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.00 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {x^{3}}{e^{5}} + \frac {2 x^{2}}{e^{5}} + \frac {x}{e^{5}} + \frac {\left (- 4 x^{2} \log {\left (5 \right )} - 8 x \log {\left (5 \right )}\right ) \log {\left (\log {\left (x \right )} \right )}}{3 e^{5}} - \frac {4 \log {\left (5 \right )} \log {\left (\log {\left (x \right )} \right )}}{3 e^{5}} + \frac {\left (4 x^{2} \log {\left (5 \right )}^{2} + 8 x \log {\left (5 \right )}^{2} + 4 \log {\left (5 \right )}^{2}\right ) \log {\left (\log {\left (x \right )} \right )}^{2}}{9 x e^{5}} \]

[In]

integrate(1/9*((4*x**2-4)*ln(5)**2*ln(x)*ln(ln(x))**2+((-24*x**3-24*x**2)*ln(5)*ln(x)+(8*x**2+16*x+8)*ln(5)**2
)*ln(ln(x))+(27*x**4+36*x**3+9*x**2)*ln(x)+(-12*x**3-24*x**2-12*x)*ln(5))/x**2/exp(5)/ln(x),x)

[Out]

x**3*exp(-5) + 2*x**2*exp(-5) + x*exp(-5) + (-4*x**2*log(5) - 8*x*log(5))*exp(-5)*log(log(x))/3 - 4*exp(-5)*lo
g(5)*log(log(x))/3 + (4*x**2*log(5)**2 + 8*x*log(5)**2 + 4*log(5)**2)*exp(-5)*log(log(x))**2/(9*x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.40 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.20 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {1}{9} \, {\left (9 \, x^{3} + 18 \, x^{2} - 12 \, {\left (x^{2} \log \left (\log \left (x\right )\right ) - {\rm Ei}\left (2 \, \log \left (x\right )\right )\right )} \log \left (5\right ) - 24 \, {\left (x \log \left (\log \left (x\right )\right ) - {\rm Ei}\left (\log \left (x\right )\right )\right )} \log \left (5\right ) - 12 \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) \log \left (5\right ) - 24 \, {\rm Ei}\left (\log \left (x\right )\right ) \log \left (5\right ) - 12 \, \log \left (5\right ) \log \left (\log \left (x\right )\right ) + \frac {4 \, {\left (x^{2} \log \left (5\right )^{2} + 2 \, x \log \left (5\right )^{2} + \log \left (5\right )^{2}\right )} \log \left (\log \left (x\right )\right )^{2}}{x} + 9 \, x\right )} e^{\left (-5\right )} \]

[In]

integrate(1/9*((4*x^2-4)*log(5)^2*log(x)*log(log(x))^2+((-24*x^3-24*x^2)*log(5)*log(x)+(8*x^2+16*x+8)*log(5)^2
)*log(log(x))+(27*x^4+36*x^3+9*x^2)*log(x)+(-12*x^3-24*x^2-12*x)*log(5))/x^2/exp(5)/log(x),x, algorithm="maxim
a")

[Out]

1/9*(9*x^3 + 18*x^2 - 12*(x^2*log(log(x)) - Ei(2*log(x)))*log(5) - 24*(x*log(log(x)) - Ei(log(x)))*log(5) - 12
*Ei(2*log(x))*log(5) - 24*Ei(log(x))*log(5) - 12*log(5)*log(log(x)) + 4*(x^2*log(5)^2 + 2*x*log(5)^2 + log(5)^
2)*log(log(x))^2/x + 9*x)*e^(-5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.80 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {1}{9} \, {\left (9 \, x^{3} + 4 \, {\left (x \log \left (5\right )^{2} + 2 \, \log \left (5\right )^{2} + \frac {\log \left (5\right )^{2}}{x}\right )} \log \left (\log \left (x\right )\right )^{2} + 18 \, x^{2} - 12 \, {\left (x^{2} \log \left (5\right ) + 2 \, x \log \left (5\right )\right )} \log \left (\log \left (x\right )\right ) - 12 \, \log \left (5\right ) \log \left (\log \left (x\right )\right ) + 9 \, x\right )} e^{\left (-5\right )} \]

[In]

integrate(1/9*((4*x^2-4)*log(5)^2*log(x)*log(log(x))^2+((-24*x^3-24*x^2)*log(5)*log(x)+(8*x^2+16*x+8)*log(5)^2
)*log(log(x))+(27*x^4+36*x^3+9*x^2)*log(x)+(-12*x^3-24*x^2-12*x)*log(5))/x^2/exp(5)/log(x),x, algorithm="giac"
)

[Out]

1/9*(9*x^3 + 4*(x*log(5)^2 + 2*log(5)^2 + log(5)^2/x)*log(log(x))^2 + 18*x^2 - 12*(x^2*log(5) + 2*x*log(5))*lo
g(log(x)) - 12*log(5)*log(log(x)) + 9*x)*e^(-5)

Mupad [B] (verification not implemented)

Time = 14.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.76 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx={\ln \left (\ln \left (x\right )\right )}^2\,\left (\frac {8\,{\mathrm {e}}^{-5}\,{\ln \left (5\right )}^2}{9}+\frac {8\,x\,{\mathrm {e}}^{-5}\,{\ln \left (5\right )}^2}{9}-\frac {{\mathrm {e}}^{-5}\,\left (4\,x^2\,{\ln \left (5\right )}^2-4\,{\ln \left (5\right )}^2\right )}{9\,x}\right )+x\,{\mathrm {e}}^{-5}+2\,x^2\,{\mathrm {e}}^{-5}+x^3\,{\mathrm {e}}^{-5}-\frac {\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-5}\,\left (4\,\ln \left (5\right )\,x^2+8\,\ln \left (5\right )\,x\right )}{3}-\frac {4\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-5}\,\ln \left (5\right )}{3} \]

[In]

int((exp(-5)*((log(log(x))*(log(5)^2*(16*x + 8*x^2 + 8) - log(5)*log(x)*(24*x^2 + 24*x^3)))/9 + (log(x)*(9*x^2
+ 36*x^3 + 27*x^4))/9 - (log(5)*(12*x + 24*x^2 + 12*x^3))/9 + (log(log(x))^2*log(5)^2*log(x)*(4*x^2 - 4))/9))
/(x^2*log(x)),x)

[Out]

log(log(x))^2*((8*exp(-5)*log(5)^2)/9 + (8*x*exp(-5)*log(5)^2)/9 - (exp(-5)*(4*x^2*log(5)^2 - 4*log(5)^2))/(9*
x)) + x*exp(-5) + 2*x^2*exp(-5) + x^3*exp(-5) - (log(log(x))*exp(-5)*(8*x*log(5) + 4*x^2*log(5)))/3 - (4*log(l
og(x))*exp(-5)*log(5))/3