3.44.73 \(\int \frac {(-15-3 e^x+3 x) \log (5 x+e^x x-x^2)+(-40 x+16 x^2+e^x (-8 x-8 x^2)) \log (\log (5 x+e^x x-x^2))+(-120-24 e^x+24 x) \log (5 x+e^x x-x^2) \log ^2(\log (5 x+e^x x-x^2))+(-240-48 e^x+48 x) \log (5 x+e^x x-x^2) \log ^4(\log (5 x+e^x x-x^2))}{(15 x^2+3 e^x x^2-3 x^3) \log (5 x+e^x x-x^2)+(120 x^2+24 e^x x^2-24 x^3) \log (5 x+e^x x-x^2) \log ^2(\log (5 x+e^x x-x^2))+(240 x^2+48 e^x x^2-48 x^3) \log (5 x+e^x x-x^2) \log ^4(\log (5 x+e^x x-x^2))} \, dx\)

Optimal. Leaf size=36 \[ -3+\frac {1}{2} \left (\frac {2}{x}-\frac {2}{3+\frac {3}{4 \log ^2\left (\log \left (\left (5+e^x-x\right ) x\right )\right )}}\right ) \]

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Rubi [F]  time = 5.72, 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 {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-15 - 3*E^x + 3*x)*Log[5*x + E^x*x - x^2] + (-40*x + 16*x^2 + E^x*(-8*x - 8*x^2))*Log[Log[5*x + E^x*x -
x^2]] + (-120 - 24*E^x + 24*x)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^2 + (-240 - 48*E^x + 48*x)*L
og[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^4)/((15*x^2 + 3*E^x*x^2 - 3*x^3)*Log[5*x + E^x*x - x^2] + (1
20*x^2 + 24*E^x*x^2 - 24*x^3)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^2 + (240*x^2 + 48*E^x*x^2 - 4
8*x^3)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^4),x]

[Out]

x^(-1) - (8*Defer[Int][Log[Log[5*x + E^x*x - x^2]]/(Log[5*x + E^x*x - x^2]*(1 + 4*Log[Log[-(x*(-5 - E^x + x))]
]^2)^2), x])/3 + 16*Defer[Int][Log[Log[5*x + E^x*x - x^2]]/((5 + E^x - x)*Log[5*x + E^x*x - x^2]*(1 + 4*Log[Lo
g[-(x*(-5 - E^x + x))]]^2)^2), x] - (8*Defer[Int][Log[Log[5*x + E^x*x - x^2]]/(x*Log[5*x + E^x*x - x^2]*(1 + 4
*Log[Log[-(x*(-5 - E^x + x))]]^2)^2), x])/3 + (8*Defer[Int][(x*Log[Log[5*x + E^x*x - x^2]])/((-5 - E^x + x)*Lo
g[5*x + E^x*x - x^2]*(1 + 4*Log[Log[-(x*(-5 - E^x + x))]]^2)^2), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-8 x \left (5-2 x+e^x (1+x)\right ) \log \left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-3 \left (5+e^x-x\right ) \log \left (-x \left (-5-e^x+x\right )\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}{3 \left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (x+4 x \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {-8 x \left (5-2 x+e^x (1+x)\right ) \log \left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-3 \left (5+e^x-x\right ) \log \left (-x \left (-5-e^x+x\right )\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}{\left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (x+4 x \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \left (\frac {-3 \log \left (-x \left (-5-e^x+x\right )\right )-8 x \log \left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-8 x^2 \log \left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-24 \log \left (-x \left (-5-e^x+x\right )\right ) \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-48 \log \left (-x \left (-5-e^x+x\right )\right ) \log ^4\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )}{x^2 \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}+\frac {8 (6-x) \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {-3 \log \left (-x \left (-5-e^x+x\right )\right )-8 x \log \left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-8 x^2 \log \left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-24 \log \left (-x \left (-5-e^x+x\right )\right ) \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-48 \log \left (-x \left (-5-e^x+x\right )\right ) \log ^4\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )}{x^2 \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx+\frac {8}{3} \int \frac {(6-x) \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {-8 x (1+x) \log \left (\log \left (-x \left (-5-e^x+x\right )\right )\right )-3 \log \left (-x \left (-5-e^x+x\right )\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}{\log \left (5 x+e^x x-x^2\right ) \left (x+4 x \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx+\frac {8}{3} \int \left (\frac {6 \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}+\frac {x \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (-5-e^x+x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}\right ) \, dx\\ &=\frac {1}{3} \int \left (-\frac {3}{x^2}+\frac {8 (-1-x) \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{x \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}\right ) \, dx+\frac {8}{3} \int \frac {x \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (-5-e^x+x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx+16 \int \frac {\log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {1}{x}+\frac {8}{3} \int \frac {(-1-x) \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{x \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx+\frac {8}{3} \int \frac {x \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (-5-e^x+x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx+16 \int \frac {\log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {1}{x}+\frac {8}{3} \int \frac {x \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (-5-e^x+x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx+\frac {8}{3} \int \left (-\frac {\log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}-\frac {\log \left (\log \left (5 x+e^x x-x^2\right )\right )}{x \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2}\right ) \, dx+16 \int \frac {\log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {1}{x}-\frac {8}{3} \int \frac {\log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx-\frac {8}{3} \int \frac {\log \left (\log \left (5 x+e^x x-x^2\right )\right )}{x \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx+\frac {8}{3} \int \frac {x \log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (-5-e^x+x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx+16 \int \frac {\log \left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (5+e^x-x\right ) \log \left (5 x+e^x x-x^2\right ) \left (1+4 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 46, normalized size = 1.28 \begin {gather*} \frac {3+x+12 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )}{3 \left (x+4 x \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-15 - 3*E^x + 3*x)*Log[5*x + E^x*x - x^2] + (-40*x + 16*x^2 + E^x*(-8*x - 8*x^2))*Log[Log[5*x + E^
x*x - x^2]] + (-120 - 24*E^x + 24*x)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^2 + (-240 - 48*E^x + 4
8*x)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^4)/((15*x^2 + 3*E^x*x^2 - 3*x^3)*Log[5*x + E^x*x - x^2
] + (120*x^2 + 24*E^x*x^2 - 24*x^3)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^2 + (240*x^2 + 48*E^x*x
^2 - 48*x^3)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^4),x]

[Out]

(3 + x + 12*Log[Log[-(x*(-5 - E^x + x))]]^2)/(3*(x + 4*x*Log[Log[-(x*(-5 - E^x + x))]]^2))

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fricas [A]  time = 0.68, size = 48, normalized size = 1.33 \begin {gather*} \frac {12 \, \log \left (\log \left (-x^{2} + x e^{x} + 5 \, x\right )\right )^{2} + x + 3}{3 \, {\left (4 \, x \log \left (\log \left (-x^{2} + x e^{x} + 5 \, x\right )\right )^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*exp(x)+48*x-240)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^4+(-24*exp(x)+24*x-120)*log(
exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+((-8*x^2-8*x)*exp(x)+16*x^2-40*x)*log(log(exp(x)*x-x^2+5*x))+(-
3*exp(x)+3*x-15)*log(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x
-x^2+5*x))^4+(24*exp(x)*x^2-24*x^3+120*x^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3
*x^3+15*x^2)*log(exp(x)*x-x^2+5*x)),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*exp(x)+48*x-240)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^4+(-24*exp(x)+24*x-120)*log(
exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+((-8*x^2-8*x)*exp(x)+16*x^2-40*x)*log(log(exp(x)*x-x^2+5*x))+(-
3*exp(x)+3*x-15)*log(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x
-x^2+5*x))^4+(24*exp(x)*x^2-24*x^3+120*x^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3
*x^3+15*x^2)*log(exp(x)*x-x^2+5*x)),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.21, size = 125, normalized size = 3.47

method	result	size
risch	\(\frac {1}{x}+\frac {1}{12 \ln \left (i \pi +\ln \relax (x )+\ln \left (-{\mathrm e}^{x}+x -5\right )+\frac {i \pi \,\mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right ) \left (\mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (\mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right )-\mathrm {csgn}\left (i \left ({\mathrm e}^{x}+5-x \right )\right )\right )}{2}+i \pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right )^{2} \left (-\mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right )-1\right )\right )^{2}+3}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-48*exp(x)+48*x-240)*ln(exp(x)*x-x^2+5*x)*ln(ln(exp(x)*x-x^2+5*x))^4+(-24*exp(x)+24*x-120)*ln(exp(x)*x-x
^2+5*x)*ln(ln(exp(x)*x-x^2+5*x))^2+((-8*x^2-8*x)*exp(x)+16*x^2-40*x)*ln(ln(exp(x)*x-x^2+5*x))+(-3*exp(x)+3*x-1
5)*ln(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*ln(exp(x)*x-x^2+5*x)*ln(ln(exp(x)*x-x^2+5*x))^4+(24*e
xp(x)*x^2-24*x^3+120*x^2)*ln(exp(x)*x-x^2+5*x)*ln(ln(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3*x^3+15*x^2)*ln(exp(x
)*x-x^2+5*x)),x,method=_RETURNVERBOSE)

[Out]

1/x+1/3/(4*ln(I*Pi+ln(x)+ln(-exp(x)+x-5)+1/2*I*Pi*csgn(I*x*(exp(x)+5-x))*(csgn(I*x*(exp(x)+5-x))+csgn(I*x))*(c
sgn(I*x*(exp(x)+5-x))-csgn(I*(exp(x)+5-x)))+I*Pi*csgn(I*x*(exp(x)+5-x))^2*(-csgn(I*x*(exp(x)+5-x))-1))^2+1)

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maxima [A]  time = 0.63, size = 42, normalized size = 1.17 \begin {gather*} \frac {12 \, \log \left (\log \relax (x) + \log \left (-x + e^{x} + 5\right )\right )^{2} + x + 3}{3 \, {\left (4 \, x \log \left (\log \relax (x) + \log \left (-x + e^{x} + 5\right )\right )^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*exp(x)+48*x-240)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^4+(-24*exp(x)+24*x-120)*log(
exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+((-8*x^2-8*x)*exp(x)+16*x^2-40*x)*log(log(exp(x)*x-x^2+5*x))+(-
3*exp(x)+3*x-15)*log(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x
-x^2+5*x))^4+(24*exp(x)*x^2-24*x^3+120*x^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3
*x^3+15*x^2)*log(exp(x)*x-x^2+5*x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.51, size = 27, normalized size = 0.75 \begin {gather*} \frac {1}{3\,\left (4\,{\ln \left (\ln \left (5\,x+x\,{\mathrm {e}}^x-x^2\right )\right )}^2+1\right )}+\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5*x + x*exp(x) - x^2)*(3*exp(x) - 3*x + 15) + log(log(5*x + x*exp(x) - x^2))*(40*x + exp(x)*(8*x + 8
*x^2) - 16*x^2) + log(5*x + x*exp(x) - x^2)*log(log(5*x + x*exp(x) - x^2))^2*(24*exp(x) - 24*x + 120) + log(5*
x + x*exp(x) - x^2)*log(log(5*x + x*exp(x) - x^2))^4*(48*exp(x) - 48*x + 240))/(log(5*x + x*exp(x) - x^2)*(3*x
^2*exp(x) + 15*x^2 - 3*x^3) + log(5*x + x*exp(x) - x^2)*log(log(5*x + x*exp(x) - x^2))^2*(24*x^2*exp(x) + 120*
x^2 - 24*x^3) + log(5*x + x*exp(x) - x^2)*log(log(5*x + x*exp(x) - x^2))^4*(48*x^2*exp(x) + 240*x^2 - 48*x^3))
,x)

[Out]

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

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sympy [A]  time = 2.87, size = 24, normalized size = 0.67 \begin {gather*} \frac {1}{12 \log {\left (\log {\left (- x^{2} + x e^{x} + 5 x \right )} \right )}^{2} + 3} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*exp(x)+48*x-240)*ln(exp(x)*x-x**2+5*x)*ln(ln(exp(x)*x-x**2+5*x))**4+(-24*exp(x)+24*x-120)*ln(e
xp(x)*x-x**2+5*x)*ln(ln(exp(x)*x-x**2+5*x))**2+((-8*x**2-8*x)*exp(x)+16*x**2-40*x)*ln(ln(exp(x)*x-x**2+5*x))+(
-3*exp(x)+3*x-15)*ln(exp(x)*x-x**2+5*x))/((48*exp(x)*x**2-48*x**3+240*x**2)*ln(exp(x)*x-x**2+5*x)*ln(ln(exp(x)
*x-x**2+5*x))**4+(24*exp(x)*x**2-24*x**3+120*x**2)*ln(exp(x)*x-x**2+5*x)*ln(ln(exp(x)*x-x**2+5*x))**2+(3*exp(x
)*x**2-3*x**3+15*x**2)*ln(exp(x)*x-x**2+5*x)),x)

[Out]

1/(12*log(log(-x**2 + x*exp(x) + 5*x))**2 + 3) + 1/x

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