3.42.81 \(\int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log (\frac {x}{5}))+e^{e^{\frac {-5+10 x}{2 x}}} (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+(30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)) \log (\frac {x}{5})-75 e^{\frac {-5+10 x}{2 x}} \log ^2(\frac {x}{5}))}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log (\frac {x}{5})-3 x^2 \log (4) \log ^2(\frac {x}{5})+x^2 \log ^3(\frac {x}{5})} \, dx\)

Optimal. Leaf size=32 \[ \left (15-\frac {e^{e^{5-\frac {5}{2 x}}}}{-\log (4)+\log \left (\frac {x}{5}\right )}\right )^2 \]

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Rubi [F]  time = 6.30, 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 {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log[4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/5]) +
E^E^((-5 + 10*x)/(2*x))*(-30*x*Log[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(2*x))*
Log[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4]^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*x^2*Lo
g[4]*Log[x/5]^2 + x^2*Log[x/5]^3),x]

[Out]

-2*Defer[Int][E^(2*E^(5 - 5/(2*x)))/(x*Log[x/20]^3), x] + 5*Defer[Int][E^(5 + 2*E^(5 - 5/(2*x)) - 5/(2*x))/(x^
2*Log[x/20]^2), x] + 30*Defer[Int][E^E^(5 - 5/(2*x))/(x*Log[x/20]^2), x] - 75*Defer[Int][E^(5 + E^(5 - 5/(2*x)
) - 5/(2*x))/(x^2*Log[x/20]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^{5-\frac {5}{2 x}}-\frac {5}{2 x}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right ) \left (-2 e^{\left .\frac {5}{2}\right /x} x-5 e^5 \log (4)+5 e^5 \log \left (\frac {x}{5}\right )\right )}{x^2 \log ^3\left (\frac {x}{20}\right )} \, dx\\ &=\int \left (-\frac {2 e^{e^{5-\frac {5}{2 x}}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right )}{x \log ^3\left (\frac {x}{20}\right )}-\frac {5 e^{5+e^{5-\frac {5}{2 x}}-\frac {5}{2 x}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right ) (\log (20)-\log (x))}{x^2 \log ^3\left (\frac {x}{20}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{e^{5-\frac {5}{2 x}}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right )}{x \log ^3\left (\frac {x}{20}\right )} \, dx\right )-5 \int \frac {e^{5+e^{5-\frac {5}{2 x}}-\frac {5}{2 x}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right ) (\log (20)-\log (x))}{x^2 \log ^3\left (\frac {x}{20}\right )} \, dx\\ &=-\left (2 \int \left (\frac {e^{2 e^{5-\frac {5}{2 x}}}}{x \log ^3\left (\frac {x}{20}\right )}-\frac {15 e^{e^{5-\frac {5}{2 x}}}}{x \log ^2\left (\frac {x}{20}\right )}\right ) \, dx\right )-5 \int \left (-\frac {e^{5+2 e^{5-\frac {5}{2 x}}-\frac {5}{2 x}}}{x^2 \log ^2\left (\frac {x}{20}\right )}+\frac {15 e^{5+e^{5-\frac {5}{2 x}}-\frac {5}{2 x}}}{x^2 \log \left (\frac {x}{20}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2 e^{5-\frac {5}{2 x}}}}{x \log ^3\left (\frac {x}{20}\right )} \, dx\right )+5 \int \frac {e^{5+2 e^{5-\frac {5}{2 x}}-\frac {5}{2 x}}}{x^2 \log ^2\left (\frac {x}{20}\right )} \, dx+30 \int \frac {e^{e^{5-\frac {5}{2 x}}}}{x \log ^2\left (\frac {x}{20}\right )} \, dx-75 \int \frac {e^{5+e^{5-\frac {5}{2 x}}-\frac {5}{2 x}}}{x^2 \log \left (\frac {x}{20}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.11, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log[4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/
5]) + E^E^((-5 + 10*x)/(2*x))*(-30*x*Log[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(
2*x))*Log[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4]^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*
x^2*Log[4]*Log[x/5]^2 + x^2*Log[x/5]^3),x]

[Out]

Integrate[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log[4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/
5]) + E^E^((-5 + 10*x)/(2*x))*(-30*x*Log[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(
2*x))*Log[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4]^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*
x^2*Log[4]*Log[x/5]^2 + x^2*Log[x/5]^3), x]

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fricas [B]  time = 2.53, size = 64, normalized size = 2.00 \begin {gather*} \frac {30 \, {\left (2 \, \log \relax (2) - \log \left (\frac {1}{5} \, x\right )\right )} e^{\left (e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )} + e^{\left (2 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) \log \left (\frac {1}{5} \, x\right ) + \log \left (\frac {1}{5} \, x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-7
5*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(300*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x
-5)/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2)*log(1/5*x)^2+12*x^2*log(2)^2*log(
1/5*x)-8*x^2*log(2)^3),x, algorithm="fricas")

[Out]

(30*(2*log(2) - log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)) + e^(2*e^(5/2*(2*x - 1)/x)))/(4*log(2)^2 - 4*log(2)*log(1/
5*x) + log(1/5*x)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (10 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \relax (2) - 5 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (\frac {1}{5} \, x\right ) + 2 \, x\right )} e^{\left (2 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )} + 15 \, {\left (20 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \relax (2)^{2} + 5 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (\frac {1}{5} \, x\right )^{2} + 4 \, x \log \relax (2) - 2 \, {\left (10 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \relax (2) + x\right )} \log \left (\frac {1}{5} \, x\right )\right )} e^{\left (e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )}}{8 \, x^{2} \log \relax (2)^{3} - 12 \, x^{2} \log \relax (2)^{2} \log \left (\frac {1}{5} \, x\right ) + 6 \, x^{2} \log \relax (2) \log \left (\frac {1}{5} \, x\right )^{2} - x^{2} \log \left (\frac {1}{5} \, x\right )^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-7
5*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(300*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x
-5)/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2)*log(1/5*x)^2+12*x^2*log(2)^2*log(
1/5*x)-8*x^2*log(2)^3),x, algorithm="giac")

[Out]

integrate(((10*e^(5/2*(2*x - 1)/x)*log(2) - 5*e^(5/2*(2*x - 1)/x)*log(1/5*x) + 2*x)*e^(2*e^(5/2*(2*x - 1)/x))
+ 15*(20*e^(5/2*(2*x - 1)/x)*log(2)^2 + 5*e^(5/2*(2*x - 1)/x)*log(1/5*x)^2 + 4*x*log(2) - 2*(10*e^(5/2*(2*x -
1)/x)*log(2) + x)*log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)))/(8*x^2*log(2)^3 - 12*x^2*log(2)^2*log(1/5*x) + 6*x^2*lo
g(2)*log(1/5*x)^2 - x^2*log(1/5*x)^3), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (5 \,{\mathrm e}^{\frac {10 x -5}{2 x}} \ln \left (\frac {x}{5}\right )-10 \ln \relax (2) {\mathrm e}^{\frac {10 x -5}{2 x}}-2 x \right ) {\mathrm e}^{2 \,{\mathrm e}^{\frac {10 x -5}{2 x}}}+\left (-75 \,{\mathrm e}^{\frac {10 x -5}{2 x}} \ln \left (\frac {x}{5}\right )^{2}+\left (300 \ln \relax (2) {\mathrm e}^{\frac {10 x -5}{2 x}}+30 x \right ) \ln \left (\frac {x}{5}\right )-300 \ln \relax (2)^{2} {\mathrm e}^{\frac {10 x -5}{2 x}}-60 x \ln \relax (2)\right ) {\mathrm e}^{{\mathrm e}^{\frac {10 x -5}{2 x}}}}{x^{2} \ln \left (\frac {x}{5}\right )^{3}-6 x^{2} \ln \relax (2) \ln \left (\frac {x}{5}\right )^{2}+12 x^{2} \ln \relax (2)^{2} \ln \left (\frac {x}{5}\right )-8 x^{2} \ln \relax (2)^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(1/2*(10*x-5)/x)*ln(1/5*x)-10*ln(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/
2*(10*x-5)/x)*ln(1/5*x)^2+(300*ln(2)*exp(1/2*(10*x-5)/x)+30*x)*ln(1/5*x)-300*ln(2)^2*exp(1/2*(10*x-5)/x)-60*x*
ln(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*ln(1/5*x)^3-6*x^2*ln(2)*ln(1/5*x)^2+12*x^2*ln(2)^2*ln(1/5*x)-8*x^2*ln(2)
^3),x)

[Out]

int(((5*exp(1/2*(10*x-5)/x)*ln(1/5*x)-10*ln(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/
2*(10*x-5)/x)*ln(1/5*x)^2+(300*ln(2)*exp(1/2*(10*x-5)/x)+30*x)*ln(1/5*x)-300*ln(2)^2*exp(1/2*(10*x-5)/x)-60*x*
ln(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*ln(1/5*x)^3-6*x^2*ln(2)*ln(1/5*x)^2+12*x^2*ln(2)^2*ln(1/5*x)-8*x^2*ln(2)
^3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (10 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \relax (2) - 5 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (\frac {1}{5} \, x\right ) + 2 \, x\right )} e^{\left (2 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )} + 15 \, {\left (20 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \relax (2)^{2} + 5 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (\frac {1}{5} \, x\right )^{2} + 4 \, x \log \relax (2) - 2 \, {\left (10 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \relax (2) + x\right )} \log \left (\frac {1}{5} \, x\right )\right )} e^{\left (e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )}}{8 \, x^{2} \log \relax (2)^{3} - 12 \, x^{2} \log \relax (2)^{2} \log \left (\frac {1}{5} \, x\right ) + 6 \, x^{2} \log \relax (2) \log \left (\frac {1}{5} \, x\right )^{2} - x^{2} \log \left (\frac {1}{5} \, x\right )^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-7
5*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(300*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x
-5)/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2)*log(1/5*x)^2+12*x^2*log(2)^2*log(
1/5*x)-8*x^2*log(2)^3),x, algorithm="maxima")

[Out]

integrate(((10*e^(5/2*(2*x - 1)/x)*log(2) - 5*e^(5/2*(2*x - 1)/x)*log(1/5*x) + 2*x)*e^(2*e^(5/2*(2*x - 1)/x))
+ 15*(20*e^(5/2*(2*x - 1)/x)*log(2)^2 + 5*e^(5/2*(2*x - 1)/x)*log(1/5*x)^2 + 4*x*log(2) - 2*(10*e^(5/2*(2*x -
1)/x)*log(2) + x)*log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)))/(8*x^2*log(2)^3 - 12*x^2*log(2)^2*log(1/5*x) + 6*x^2*lo
g(2)*log(1/5*x)^2 - x^2*log(1/5*x)^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int -\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}}\,\left (2\,x-5\,\ln \left (\frac {x}{5}\right )\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}+10\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,\ln \relax (2)\right )+{\mathrm {e}}^{{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}}\,\left (75\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,{\ln \left (\frac {x}{5}\right )}^2+\left (-30\,x-300\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,\ln \relax (2)\right )\,\ln \left (\frac {x}{5}\right )+60\,x\,\ln \relax (2)+300\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,{\ln \relax (2)}^2\right )}{-x^2\,{\ln \left (\frac {x}{5}\right )}^3+6\,\ln \relax (2)\,x^2\,{\ln \left (\frac {x}{5}\right )}^2-12\,{\ln \relax (2)}^2\,x^2\,\ln \left (\frac {x}{5}\right )+8\,{\ln \relax (2)}^3\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp((5*x - 5/2)/x))*(2*x - 5*log(x/5)*exp((5*x - 5/2)/x) + 10*exp((5*x - 5/2)/x)*log(2)) + exp(exp(
(5*x - 5/2)/x))*(60*x*log(2) - log(x/5)*(30*x + 300*exp((5*x - 5/2)/x)*log(2)) + 75*log(x/5)^2*exp((5*x - 5/2)
/x) + 300*exp((5*x - 5/2)/x)*log(2)^2))/(8*x^2*log(2)^3 - x^2*log(x/5)^3 - 12*x^2*log(x/5)*log(2)^2 + 6*x^2*lo
g(x/5)^2*log(2)),x)

[Out]

-int(-(exp(2*exp((5*x - 5/2)/x))*(2*x - 5*log(x/5)*exp((5*x - 5/2)/x) + 10*exp((5*x - 5/2)/x)*log(2)) + exp(ex
p((5*x - 5/2)/x))*(60*x*log(2) - log(x/5)*(30*x + 300*exp((5*x - 5/2)/x)*log(2)) + 75*log(x/5)^2*exp((5*x - 5/
2)/x) + 300*exp((5*x - 5/2)/x)*log(2)^2))/(8*x^2*log(2)^3 - x^2*log(x/5)^3 - 12*x^2*log(x/5)*log(2)^2 + 6*x^2*
log(x/5)^2*log(2)), x)

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sympy [B]  time = 0.60, size = 97, normalized size = 3.03 \begin {gather*} \frac {\left (\log {\left (\frac {x}{5} \right )} - 2 \log {\relax (2 )}\right ) e^{2 e^{\frac {5 x - \frac {5}{2}}{x}}} + \left (- 30 \log {\left (\frac {x}{5} \right )}^{2} + 120 \log {\relax (2 )} \log {\left (\frac {x}{5} \right )} - 120 \log {\relax (2 )}^{2}\right ) e^{e^{\frac {5 x - \frac {5}{2}}{x}}}}{\log {\left (\frac {x}{5} \right )}^{3} - 6 \log {\relax (2 )} \log {\left (\frac {x}{5} \right )}^{2} + 12 \log {\relax (2 )}^{2} \log {\left (\frac {x}{5} \right )} - 8 \log {\relax (2 )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1/2*(10*x-5)/x)*ln(1/5*x)-10*ln(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))**2+(-75
*exp(1/2*(10*x-5)/x)*ln(1/5*x)**2+(300*ln(2)*exp(1/2*(10*x-5)/x)+30*x)*ln(1/5*x)-300*ln(2)**2*exp(1/2*(10*x-5)
/x)-60*x*ln(2))*exp(exp(1/2*(10*x-5)/x)))/(x**2*ln(1/5*x)**3-6*x**2*ln(2)*ln(1/5*x)**2+12*x**2*ln(2)**2*ln(1/5
*x)-8*x**2*ln(2)**3),x)

[Out]

((log(x/5) - 2*log(2))*exp(2*exp((5*x - 5/2)/x)) + (-30*log(x/5)**2 + 120*log(2)*log(x/5) - 120*log(2)**2)*exp
(exp((5*x - 5/2)/x)))/(log(x/5)**3 - 6*log(2)*log(x/5)**2 + 12*log(2)**2*log(x/5) - 8*log(2)**3)

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