3.82.94 \(\int \frac {(-2 x^2+e^2 (250 x-250 x^2+2 x^3-2 x^4+e (50 x-50 x^2))) \log (x)+(e^2 (125-250 x+126 x^2-2 x^3+x^4+e (25-50 x+25 x^2))+(125+25 e+x^2) \log (\frac {1}{25} (125+25 e+x^2))) \log (e^2 (1-2 x+x^2)+\log (\frac {1}{25} (125+25 e+x^2)))}{(e^2 (125 x-250 x^2+126 x^3-2 x^4+x^5+e (25 x-50 x^2+25 x^3))+(125 x+25 e x+x^3) \log (\frac {1}{25} (125+25 e+x^2))) \log ^2(e^2 (1-2 x+x^2)+\log (\frac {1}{25} (125+25 e+x^2)))} \, dx\)

Optimal. Leaf size=27 \[ \frac {\log (x)}{\log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \]

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Rubi [F]  time = 11.19, 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 (-2 x^2+e^2 \left (250 x-250 x^2+2 x^3-2 x^4+e \left (50 x-50 x^2\right )\right )\right ) \log (x)+\left (e^2 \left (125-250 x+126 x^2-2 x^3+x^4+e \left (25-50 x+25 x^2\right )\right )+\left (125+25 e+x^2\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log \left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )}{\left (e^2 \left (125 x-250 x^2+126 x^3-2 x^4+x^5+e \left (25 x-50 x^2+25 x^3\right )\right )+\left (125 x+25 e x+x^3\right ) \log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right ) \log ^2\left (e^2 \left (1-2 x+x^2\right )+\log \left (\frac {1}{25} \left (125+25 e+x^2\right )\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-2*x^2 + E^2*(250*x - 250*x^2 + 2*x^3 - 2*x^4 + E*(50*x - 50*x^2)))*Log[x] + (E^2*(125 - 250*x + 126*x^2
- 2*x^3 + x^4 + E*(25 - 50*x + 25*x^2)) + (125 + 25*E + x^2)*Log[(125 + 25*E + x^2)/25])*Log[E^2*(1 - 2*x + x
^2) + Log[(125 + 25*E + x^2)/25]])/((E^2*(125*x - 250*x^2 + 126*x^3 - 2*x^4 + x^5 + E*(25*x - 50*x^2 + 25*x^3)
) + (125*x + 25*E*x + x^3)*Log[(125 + 25*E + x^2)/25])*Log[E^2*(1 - 2*x + x^2) + Log[(125 + 25*E + x^2)/25]]^2
),x]

[Out]

2*E^2*Defer[Int][Log[x]/((E^2 - 2*E^2*x + E^2*x^2 + Log[5 + E + x^2/25])*Log[E^2*(-1 + x)^2 + Log[5 + E + x^2/
25]]^2), x] + Defer[Int][Log[x]/(((5*I)*Sqrt[5 + E] - x)*(E^2 - 2*E^2*x + E^2*x^2 + Log[5 + E + x^2/25])*Log[E
^2*(-1 + x)^2 + Log[5 + E + x^2/25]]^2), x] - 2*E^2*Defer[Int][(x*Log[x])/((E^2 - 2*E^2*x + E^2*x^2 + Log[5 +
E + x^2/25])*Log[E^2*(-1 + x)^2 + Log[5 + E + x^2/25]]^2), x] - Defer[Int][Log[x]/(((5*I)*Sqrt[5 + E] + x)*(E^
2 - 2*E^2*x + E^2*x^2 + Log[5 + E + x^2/25])*Log[E^2*(-1 + x)^2 + Log[5 + E + x^2/25]]^2), x] + Defer[Int][1/(
x*Log[E^2*(-1 + x)^2 + Log[5 + E + x^2/25]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {\left (-2 x^2-2 e^2 (-1+x) x \left (125+25 e+x^2\right )\right ) \log (x)}{\left (125+25 e+x^2\right ) \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}+\log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}{x \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx\\ &=\int \left (\frac {2 \left (25 e^2 (5+e)-\left (1+125 e^2+25 e^3\right ) x+e^2 x^2-e^2 x^3\right ) \log (x)}{\left (125+25 e+x^2\right ) \left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}+\frac {1}{x \log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}\right ) \, dx\\ &=2 \int \frac {\left (25 e^2 (5+e)-\left (1+125 e^2+25 e^3\right ) x+e^2 x^2-e^2 x^3\right ) \log (x)}{\left (125+25 e+x^2\right ) \left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx+\int \frac {1}{x \log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx\\ &=2 \int \left (\frac {e^2 \log (x)}{\left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}-\frac {e^2 x \log (x)}{\left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}-\frac {x \log (x)}{\left (125+25 e+x^2\right ) \left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}\right ) \, dx+\int \frac {1}{x \log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx\\ &=-\left (2 \int \frac {x \log (x)}{\left (125+25 e+x^2\right ) \left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx\right )+\left (2 e^2\right ) \int \frac {\log (x)}{\left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx-\left (2 e^2\right ) \int \frac {x \log (x)}{\left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx+\int \frac {1}{x \log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx\\ &=-\left (2 \int \left (-\frac {\log (x)}{2 \left (5 i \sqrt {5+e}-x\right ) \left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}+\frac {\log (x)}{2 \left (5 i \sqrt {5+e}+x\right ) \left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )}\right ) \, dx\right )+\left (2 e^2\right ) \int \frac {\log (x)}{\left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx-\left (2 e^2\right ) \int \frac {x \log (x)}{\left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx+\int \frac {1}{x \log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx\\ &=\left (2 e^2\right ) \int \frac {\log (x)}{\left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx-\left (2 e^2\right ) \int \frac {x \log (x)}{\left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx+\int \frac {\log (x)}{\left (5 i \sqrt {5+e}-x\right ) \left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx-\int \frac {\log (x)}{\left (5 i \sqrt {5+e}+x\right ) \left (e^2-2 e^2 x+e^2 x^2+\log \left (5+e+\frac {x^2}{25}\right )\right ) \log ^2\left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx+\int \frac {1}{x \log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 27, normalized size = 1.00 \begin {gather*} \frac {\log (x)}{\log \left (e^2 (-1+x)^2+\log \left (5+e+\frac {x^2}{25}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2*x^2 + E^2*(250*x - 250*x^2 + 2*x^3 - 2*x^4 + E*(50*x - 50*x^2)))*Log[x] + (E^2*(125 - 250*x + 1
26*x^2 - 2*x^3 + x^4 + E*(25 - 50*x + 25*x^2)) + (125 + 25*E + x^2)*Log[(125 + 25*E + x^2)/25])*Log[E^2*(1 - 2
*x + x^2) + Log[(125 + 25*E + x^2)/25]])/((E^2*(125*x - 250*x^2 + 126*x^3 - 2*x^4 + x^5 + E*(25*x - 50*x^2 + 2
5*x^3)) + (125*x + 25*E*x + x^3)*Log[(125 + 25*E + x^2)/25])*Log[E^2*(1 - 2*x + x^2) + Log[(125 + 25*E + x^2)/
25]]^2),x]

[Out]

Log[x]/Log[E^2*(-1 + x)^2 + Log[5 + E + x^2/25]]

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fricas [A]  time = 1.24, size = 28, normalized size = 1.04 \begin {gather*} \frac {\log \relax (x)}{\log \left ({\left (x^{2} - 2 \, x + 1\right )} e^{2} + \log \left (\frac {1}{25} \, x^{2} + e + 5\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((25*exp(1)+x^2+125)*log(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*exp(1)+x^4-2*x^3+126*x^2-250*x+125)*e
xp(2))*log(log(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2
)-2*x^2)*log(x))/((25*x*exp(1)+x^3+125*x)*log(exp(1)+1/25*x^2+5)+((25*x^3-50*x^2+25*x)*exp(1)+x^5-2*x^4+126*x^
3-250*x^2+125*x)*exp(2))/log(log(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))^2,x, algorithm="fricas")

[Out]

log(x)/log((x^2 - 2*x + 1)*e^2 + log(1/25*x^2 + e + 5))

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giac [B]  time = 3.37, size = 1134, normalized size = 42.00 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((25*exp(1)+x^2+125)*log(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*exp(1)+x^4-2*x^3+126*x^2-250*x+125)*e
xp(2))*log(log(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2
)-2*x^2)*log(x))/((25*x*exp(1)+x^3+125*x)*log(exp(1)+1/25*x^2+5)+((25*x^3-50*x^2+25*x)*exp(1)+x^5-2*x^4+126*x^
3-250*x^2+125*x)*exp(2))/log(log(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))^2,x, algorithm="giac")

[Out]

(x^5*e^4*log(x) - 3*x^4*e^4*log(x) + x^3*e^2*log(1/25*x^2 + e + 5)*log(x) + 25*x^3*e^5*log(x) + 128*x^3*e^4*lo
g(x) + x^3*e^2*log(x) - x^2*e^2*log(1/25*x^2 + e + 5)*log(x) - 75*x^2*e^5*log(x) - 376*x^2*e^4*log(x) - 2*x^2*
e^2*log(x) + 25*x*e^3*log(1/25*x^2 + e + 5)*log(x) + 125*x*e^2*log(1/25*x^2 + e + 5)*log(x) + 75*x*e^5*log(x)
+ 375*x*e^4*log(x) + x*e^2*log(x) - 2*x*log(5)*log(x) + x*log(x^2 + 25*e + 125)*log(x) - 25*e^3*log(1/25*x^2 +
e + 5)*log(x) - 125*e^2*log(1/25*x^2 + e + 5)*log(x) - 25*e^5*log(x) - 125*e^4*log(x))/(x^5*e^4*log(x^2*e^2 -
2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 3*x^4*e^4*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 2*x
^3*e^2*log(5)*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + x^3*e^2*log(x^2*e^2 - 2*x*e^2 + e^2 + log
(1/25*x^2 + e + 5))*log(x^2 + 25*e + 125) + 25*x^3*e^5*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) +
128*x^3*e^4*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + x^3*e^2*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1
/25*x^2 + e + 5)) + 2*x^2*e^2*log(5)*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - x^2*e^2*log(x^2*e^
2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5))*log(x^2 + 25*e + 125) - 75*x^2*e^5*log(x^2*e^2 - 2*x*e^2 + e^2 + lo
g(1/25*x^2 + e + 5)) - 376*x^2*e^4*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 2*x^2*e^2*log(x^2*e^
2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 50*x*e^3*log(5)*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e +
5)) - 250*x*e^2*log(5)*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + 25*x*e^3*log(x^2*e^2 - 2*x*e^2 +
e^2 + log(1/25*x^2 + e + 5))*log(x^2 + 25*e + 125) + 125*x*e^2*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e
+ 5))*log(x^2 + 25*e + 125) + 75*x*e^5*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + 375*x*e^4*log(x
^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + x*e^2*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) -
2*x*log(5)*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + 50*e^3*log(5)*log(x^2*e^2 - 2*x*e^2 + e^2 +
log(1/25*x^2 + e + 5)) + 250*e^2*log(5)*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) + x*log(x^2*e^2
- 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5))*log(x^2 + 25*e + 125) - 25*e^3*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25
*x^2 + e + 5))*log(x^2 + 25*e + 125) - 125*e^2*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5))*log(x^2 +
25*e + 125) - 25*e^5*log(x^2*e^2 - 2*x*e^2 + e^2 + log(1/25*x^2 + e + 5)) - 125*e^4*log(x^2*e^2 - 2*x*e^2 + e^
2 + log(1/25*x^2 + e + 5)))

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maple [A]  time = 0.10, size = 29, normalized size = 1.07

method	result	size
risch	\(\frac {\ln \relax (x )}{\ln \left (\ln \left ({\mathrm e}+\frac {x^{2}}{25}+5\right )+\left (x^{2}-2 x +1\right ) {\mathrm e}^{2}\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((25*exp(1)+x^2+125)*ln(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*exp(1)+x^4-2*x^3+126*x^2-250*x+125)*exp(2))*
ln(ln(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2)-2*x^2)*
ln(x))/((25*x*exp(1)+x^3+125*x)*ln(exp(1)+1/25*x^2+5)+((25*x^3-50*x^2+25*x)*exp(1)+x^5-2*x^4+126*x^3-250*x^2+1
25*x)*exp(2))/ln(ln(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))^2,x,method=_RETURNVERBOSE)

[Out]

ln(x)/ln(ln(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))

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maxima [A]  time = 0.57, size = 34, normalized size = 1.26 \begin {gather*} \frac {\log \relax (x)}{\log \left (x^{2} e^{2} - 2 \, x e^{2} + e^{2} - 2 \, \log \relax (5) + \log \left (x^{2} + 25 \, e + 125\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((25*exp(1)+x^2+125)*log(exp(1)+1/25*x^2+5)+((25*x^2-50*x+25)*exp(1)+x^4-2*x^3+126*x^2-250*x+125)*e
xp(2))*log(log(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))+(((-50*x^2+50*x)*exp(1)-2*x^4+2*x^3-250*x^2+250*x)*exp(2
)-2*x^2)*log(x))/((25*x*exp(1)+x^3+125*x)*log(exp(1)+1/25*x^2+5)+((25*x^3-50*x^2+25*x)*exp(1)+x^5-2*x^4+126*x^
3-250*x^2+125*x)*exp(2))/log(log(exp(1)+1/25*x^2+5)+(x^2-2*x+1)*exp(2))^2,x, algorithm="maxima")

[Out]

log(x)/log(x^2*e^2 - 2*x*e^2 + e^2 - 2*log(5) + log(x^2 + 25*e + 125))

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mupad [B]  time = 9.68, size = 28, normalized size = 1.04 \begin {gather*} \frac {\ln \relax (x)}{\ln \left (\ln \left (\frac {x^2}{25}+\mathrm {e}+5\right )+{\mathrm {e}}^2\,\left (x^2-2\,x+1\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(exp(2)*(250*x + exp(1)*(50*x - 50*x^2) - 250*x^2 + 2*x^3 - 2*x^4) - 2*x^2) + log(log(exp(1) + x^2
/25 + 5) + exp(2)*(x^2 - 2*x + 1))*(exp(2)*(exp(1)*(25*x^2 - 50*x + 25) - 250*x + 126*x^2 - 2*x^3 + x^4 + 125)
+ log(exp(1) + x^2/25 + 5)*(25*exp(1) + x^2 + 125)))/(log(log(exp(1) + x^2/25 + 5) + exp(2)*(x^2 - 2*x + 1))^
2*(log(exp(1) + x^2/25 + 5)*(125*x + 25*x*exp(1) + x^3) + exp(2)*(125*x + exp(1)*(25*x - 50*x^2 + 25*x^3) - 25
0*x^2 + 126*x^3 - 2*x^4 + x^5))),x)

[Out]

log(x)/log(log(exp(1) + x^2/25 + 5) + exp(2)*(x^2 - 2*x + 1))

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sympy [A]  time = 7.84, size = 27, normalized size = 1.00 \begin {gather*} \frac {\log {\relax (x )}}{\log {\left (\left (x^{2} - 2 x + 1\right ) e^{2} + \log {\left (\frac {x^{2}}{25} + e + 5 \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((25*exp(1)+x**2+125)*ln(exp(1)+1/25*x**2+5)+((25*x**2-50*x+25)*exp(1)+x**4-2*x**3+126*x**2-250*x+1
25)*exp(2))*ln(ln(exp(1)+1/25*x**2+5)+(x**2-2*x+1)*exp(2))+(((-50*x**2+50*x)*exp(1)-2*x**4+2*x**3-250*x**2+250
*x)*exp(2)-2*x**2)*ln(x))/((25*x*exp(1)+x**3+125*x)*ln(exp(1)+1/25*x**2+5)+((25*x**3-50*x**2+25*x)*exp(1)+x**5
-2*x**4+126*x**3-250*x**2+125*x)*exp(2))/ln(ln(exp(1)+1/25*x**2+5)+(x**2-2*x+1)*exp(2))**2,x)

[Out]

log(x)/log((x**2 - 2*x + 1)*exp(2) + log(x**2/25 + E + 5))

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