3.86.98 \(\int \frac {(-100 e^{5/x}+e^{e^{-5/x} (-e^{5/x} x+\log (x))} (10 x+e^{5/x} (-10 x-10 x^2)+50 \log (x))) \log (\frac {10+2 e^{e^{-5/x} (-e^{5/x} x+\log (x))} x}{x^2})}{5 e^{5/x} x+e^{\frac {5}{x}+e^{-5/x} (-e^{5/x} x+\log (x))} x^2} \, dx\)

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

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Rubi [F]  time = 76.46, 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 (-100 e^{5/x}+e^{e^{-5/x} \left (-e^{5/x} x+\log (x)\right )} \left (10 x+e^{5/x} \left (-10 x-10 x^2\right )+50 \log (x)\right )\right ) \log \left (\frac {10+2 e^{e^{-5/x} \left (-e^{5/x} x+\log (x)\right )} x}{x^2}\right )}{5 e^{5/x} x+e^{\frac {5}{x}+e^{-5/x} \left (-e^{5/x} x+\log (x)\right )} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-100*E^(5/x) + E^((-(E^(5/x)*x) + Log[x])/E^(5/x))*(10*x + E^(5/x)*(-10*x - 10*x^2) + 50*Log[x]))*Log[(1
0 + 2*E^((-(E^(5/x)*x) + Log[x])/E^(5/x))*x)/x^2])/(5*E^(5/x)*x + E^(5/x + (-(E^(5/x)*x) + Log[x])/E^(5/x))*x^
2),x]

[Out]

-100*Log[(2*(5 + x^(1 + E^(-5/x))/E^x))/x^2]*Defer[Int][E^x/(x*(5*E^x + x^(1 + E^(-5/x)))), x] - 10*Log[(2*(5
+ x^(1 + E^(-5/x))/E^x))/x^2]*Defer[Int][x^E^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x] + 10*Log[(2*(5 + x^(1 + E^(
-5/x))/E^x))/x^2]*Defer[Int][x^E^(-5/x)/(E^(5/x)*(5*E^x + x^(1 + E^(-5/x)))), x] + 50*Log[x]*Log[(2*(5 + x^(1
+ E^(-5/x))/E^x))/x^2]*Defer[Int][x^(-1 + E^(-5/x))/(E^(5/x)*(5*E^x + x^(1 + E^(-5/x)))), x] - 10*Log[(2*(5 +
x^(1 + E^(-5/x))/E^x))/x^2]*Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x] + 50*Log[x]*Defer[Int][
Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x], x] + 50*Log[x]*Defer[Int][Defer[Int][1/(E^(5/x)*x*(x +
(5*E^x)/x^E^(-5/x))), x]/x, x] - 50*Log[(2*(5 + x^(1 + E^(-5/x))/E^x))/x^2]*Defer[Int][Defer[Int][1/(E^(5/x)*x
*(x + (5*E^x)/x^E^(-5/x))), x]/x, x] - 50*Log[x]*Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))),
x]/(E^(5/x)*x), x] - 250*Log[x]*Defer[Int][(E^x*Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x])/(5*E^x
+ x^(1 + E^(-5/x))), x] + 250*Log[x]*Defer[Int][(E^x*Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x])/(
x*(5*E^x + x^(1 + E^(-5/x)))), x] + 250*Log[x]*Defer[Int][(E^(-5/x + x)*Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x
^E^(-5/x))), x])/(x*(5*E^x + x^(1 + E^(-5/x)))), x] - 250*Defer[Int][(Log[x]^2*Defer[Int][1/(E^(5/x)*x*(x + (5
*E^x)/x^E^(-5/x))), x])/(E^(5/x)*x^2), x] + 1250*Defer[Int][(E^(-5/x + x)*Log[x]^2*Defer[Int][1/(E^(5/x)*x*(x
+ (5*E^x)/x^E^(-5/x))), x])/(x^2*(5*E^x + x^(1 + E^(-5/x)))), x] - 10*Defer[Int][Defer[Int][x^E^(-5/x)/(5*E^x
+ x^(1 + E^(-5/x))), x], x] + 50*Log[x]*Defer[Int][Defer[Int][x^E^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x]/(E^(5/
x)*x^2), x] - 10*Defer[Int][Defer[Int][x^E^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x]/x, x] + 10*Defer[Int][Defer[I
nt][x^E^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x]/(E^(5/x)*x), x] + 50*Defer[Int][(E^x*Defer[Int][x^E^(-5/x)/(5*E^
x + x^(1 + E^(-5/x))), x])/(5*E^x + x^(1 + E^(-5/x))), x] - 250*Log[x]*Defer[Int][(E^(-5/x + x)*Defer[Int][x^E
^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x])/(x^2*(5*E^x + x^(1 + E^(-5/x)))), x] - 50*Defer[Int][(E^x*Defer[Int][x
^E^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x])/(x*(5*E^x + x^(1 + E^(-5/x)))), x] - 50*Defer[Int][(E^(-5/x + x)*Def
er[Int][x^E^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x])/(x*(5*E^x + x^(1 + E^(-5/x)))), x] + 10*Defer[Int][Defer[In
t][x^E^(-5/x)/(E^(5/x)*(5*E^x + x^(1 + E^(-5/x)))), x], x] - 50*Log[x]*Defer[Int][Defer[Int][x^E^(-5/x)/(E^(5/
x)*(5*E^x + x^(1 + E^(-5/x)))), x]/(E^(5/x)*x^2), x] + 10*Defer[Int][Defer[Int][x^E^(-5/x)/(E^(5/x)*(5*E^x + x
^(1 + E^(-5/x)))), x]/x, x] - 10*Defer[Int][Defer[Int][x^E^(-5/x)/(E^(5/x)*(5*E^x + x^(1 + E^(-5/x)))), x]/(E^
(5/x)*x), x] - 50*Defer[Int][(E^x*Defer[Int][x^E^(-5/x)/(E^(5/x)*(5*E^x + x^(1 + E^(-5/x)))), x])/(5*E^x + x^(
1 + E^(-5/x))), x] + 250*Log[x]*Defer[Int][(E^(-5/x + x)*Defer[Int][x^E^(-5/x)/(E^(5/x)*(5*E^x + x^(1 + E^(-5/
x)))), x])/(x^2*(5*E^x + x^(1 + E^(-5/x)))), x] + 50*Defer[Int][(E^x*Defer[Int][x^E^(-5/x)/(E^(5/x)*(5*E^x + x
^(1 + E^(-5/x)))), x])/(x*(5*E^x + x^(1 + E^(-5/x)))), x] + 50*Defer[Int][(E^(-5/x + x)*Defer[Int][x^E^(-5/x)/
(E^(5/x)*(5*E^x + x^(1 + E^(-5/x)))), x])/(x*(5*E^x + x^(1 + E^(-5/x)))), x] - 10*Defer[Int][Defer[Int][x^(1 +
E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x], x] + 50*Log[x]*Defer[Int][Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1
+ E^(-5/x))), x]/(E^(5/x)*x^2), x] - 10*Defer[Int][Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x]
/x, x] + 10*Defer[Int][Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x]/(E^(5/x)*x), x] + 50*Defer[I
nt][(E^x*Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x])/(5*E^x + x^(1 + E^(-5/x))), x] - 250*Log[
x]*Defer[Int][(E^(-5/x + x)*Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x])/(x^2*(5*E^x + x^(1 + E
^(-5/x)))), x] - 50*Defer[Int][(E^x*Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x])/(x*(5*E^x + x^
(1 + E^(-5/x)))), x] - 50*Defer[Int][(E^(-5/x + x)*Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x])
/(x*(5*E^x + x^(1 + E^(-5/x)))), x] - 100*Defer[Int][Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x], x] + 500
*Log[x]*Defer[Int][Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x]/(E^(5/x)*x^2), x] - 100*Defer[Int][Defer[In
t][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x]/x, x] + 100*Defer[Int][Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x]
/(E^(5/x)*x), x] + 500*Defer[Int][(E^x*Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x])/(5*E^x + x^(1 + E^(-5/
x))), x] - 2500*Log[x]*Defer[Int][(E^(-5/x + x)*Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x])/(x^2*(5*E^x +
x^(1 + E^(-5/x)))), x] - 500*Defer[Int][(E^x*Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x])/(x*(5*E^x + x^(
1 + E^(-5/x)))), x] - 500*Defer[Int][(E^(-5/x + x)*Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x])/(x*(5*E^x
+ x^(1 + E^(-5/x)))), x] - 50*Defer[Int][Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x], x]/
x, x] - 50*Defer[Int][Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/x, x], x] + 250*Log[x]*
Defer[Int][Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/x, x]/(E^(5/x)*x^2), x] - 100*Defe
r[Int][Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/x, x]/x, x] + 50*Defer[Int][Defer[Int]
[Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/x, x]/(E^(5/x)*x), x] + 250*Defer[Int][(E^x*Defer[Int][
Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/x, x])/(5*E^x + x^(1 + E^(-5/x))), x] - 1250*Log[x]*Defe
r[Int][(E^(-5/x + x)*Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/x, x])/(x^2*(5*E^x + x^(
1 + E^(-5/x)))), x] - 250*Defer[Int][(E^x*Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/x,
x])/(x*(5*E^x + x^(1 + E^(-5/x)))), x] - 250*Defer[Int][(E^(-5/x + x)*Defer[Int][Defer[Int][1/(E^(5/x)*x*(x +
(5*E^x)/x^E^(-5/x))), x]/x, x])/(x*(5*E^x + x^(1 + E^(-5/x)))), x] + 50*Defer[Int][Defer[Int][Defer[Int][1/(E^
(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/(E^(5/x)*x), x]/x, x] + 250*Defer[Int][Defer[Int][(E^x*Defer[Int][1/(E^(
5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x])/(5*E^x + x^(1 + E^(-5/x))), x]/x, x] - 250*Defer[Int][Defer[Int][(E^x*De
fer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x])/(5*E^x*x + x^(2 + E^(-5/x))), x]/x, x] - 250*Defer[Int][D
efer[Int][(E^(-5/x + x)*Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x])/(5*E^x*x + x^(2 + E^(-5/x))), x
]/x, x] - 50*Defer[Int][Defer[Int][Defer[Int][x^E^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x]/(E^(5/x)*x^2), x]/x, x
] + 250*Defer[Int][Defer[Int][(E^(-5/x + x)*Defer[Int][x^E^(-5/x)/(5*E^x + x^(1 + E^(-5/x))), x])/(5*E^x*x^2 +
x^(3 + E^(-5/x))), x]/x, x] + 50*Defer[Int][Defer[Int][Defer[Int][x^E^(-5/x)/(E^(5/x)*(5*E^x + x^(1 + E^(-5/x
)))), x]/(E^(5/x)*x^2), x]/x, x] - 250*Defer[Int][Defer[Int][(E^(-5/x + x)*Defer[Int][x^E^(-5/x)/(E^(5/x)*(5*E
^x + x^(1 + E^(-5/x)))), x])/(5*E^x*x^2 + x^(3 + E^(-5/x))), x]/x, x] - 50*Defer[Int][Defer[Int][Defer[Int][x^
(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x]/(E^(5/x)*x^2), x]/x, x] + 250*Defer[Int][Defer[Int][(E^(-5/x + x
)*Defer[Int][x^(1 + E^(-5/x))/(5*E^x + x^(1 + E^(-5/x))), x])/(5*E^x*x^2 + x^(3 + E^(-5/x))), x]/x, x] - 500*D
efer[Int][Defer[Int][Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x]/(E^(5/x)*x^2), x]/x, x] + 2500*Defer[Int]
[Defer[Int][(E^(-5/x + x)*Defer[Int][E^x/(5*E^x*x + x^(2 + E^(-5/x))), x])/(5*E^x*x^2 + x^(3 + E^(-5/x))), x]/
x, x] - 250*Defer[Int][Defer[Int][Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/x^E^(-5/x))), x]/x, x]/(E^(5
/x)*x^2), x]/x, x] + 1250*Defer[Int][Defer[Int][(E^(-5/x + x)*Defer[Int][Defer[Int][1/(E^(5/x)*x*(x + (5*E^x)/
x^E^(-5/x))), x]/x, x])/(x^2*(5*E^x + x^(1 + E^(-5/x)))), x]/x, x]

Rubi steps

Aborted

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Mathematica [A]  time = 0.13, size = 29, normalized size = 0.91 \begin {gather*} 5 \log ^2\left (\frac {2 \left (5+e^{-x} x^{1+e^{-5/x}}\right )}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-100*E^(5/x) + E^((-(E^(5/x)*x) + Log[x])/E^(5/x))*(10*x + E^(5/x)*(-10*x - 10*x^2) + 50*Log[x]))*
Log[(10 + 2*E^((-(E^(5/x)*x) + Log[x])/E^(5/x))*x)/x^2])/(5*E^(5/x)*x + E^(5/x + (-(E^(5/x)*x) + Log[x])/E^(5/
x))*x^2),x]

[Out]

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

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fricas [A]  time = 0.55, size = 57, normalized size = 1.78 \begin {gather*} 5 \, \log \left (\frac {2 \, {\left (x e^{\left (-\frac {{\left ({\left (x^{2} - 5\right )} e^{\frac {5}{x}} - x \log \relax (x)\right )} e^{\left (-\frac {5}{x}\right )}}{x}\right )} + 5 \, e^{\frac {5}{x}}\right )} e^{\left (-\frac {5}{x}\right )}}{x^{2}}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*log(x)+(-10*x^2-10*x)*exp(5/x)+10*x)*exp((log(x)-x*exp(5/x))/exp(5/x))-100*exp(5/x))*log((2*x*e
xp((log(x)-x*exp(5/x))/exp(5/x))+10)/x^2)/(x^2*exp(5/x)*exp((log(x)-x*exp(5/x))/exp(5/x))+5*x*exp(5/x)),x, alg
orithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*log(x)+(-10*x^2-10*x)*exp(5/x)+10*x)*exp((log(x)-x*exp(5/x))/exp(5/x))-100*exp(5/x))*log((2*x*e
xp((log(x)-x*exp(5/x))/exp(5/x))+10)/x^2)/(x^2*exp(5/x)*exp((log(x)-x*exp(5/x))/exp(5/x))+5*x*exp(5/x)),x, alg
orithm="giac")

[Out]

integrate(-10*(((x^2 + x)*e^(5/x) - x - 5*log(x))*e^(-(x*e^(5/x) - log(x))*e^(-5/x)) + 10*e^(5/x))*log(2*(x*e^
(-(x*e^(5/x) - log(x))*e^(-5/x)) + 5)/x^2)/(x^2*e^(-(x*e^(5/x) - log(x))*e^(-5/x) + 5/x) + 5*x*e^(5/x)), x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (50 \ln \relax (x )+\left (-10 x^{2}-10 x \right ) {\mathrm e}^{\frac {5}{x}}+10 x \right ) {\mathrm e}^{\left (\ln \relax (x )-x \,{\mathrm e}^{\frac {5}{x}}\right ) {\mathrm e}^{-\frac {5}{x}}}-100 \,{\mathrm e}^{\frac {5}{x}}\right ) \ln \left (\frac {2 x \,{\mathrm e}^{\left (\ln \relax (x )-x \,{\mathrm e}^{\frac {5}{x}}\right ) {\mathrm e}^{-\frac {5}{x}}}+10}{x^{2}}\right )}{x^{2} {\mathrm e}^{\frac {5}{x}} {\mathrm e}^{\left (\ln \relax (x )-x \,{\mathrm e}^{\frac {5}{x}}\right ) {\mathrm e}^{-\frac {5}{x}}}+5 x \,{\mathrm e}^{\frac {5}{x}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*ln(x)+(-10*x^2-10*x)*exp(5/x)+10*x)*exp((ln(x)-x*exp(5/x))/exp(5/x))-100*exp(5/x))*ln((2*x*exp((ln(x)
-x*exp(5/x))/exp(5/x))+10)/x^2)/(x^2*exp(5/x)*exp((ln(x)-x*exp(5/x))/exp(5/x))+5*x*exp(5/x)),x)

[Out]

int(((50*ln(x)+(-10*x^2-10*x)*exp(5/x)+10*x)*exp((ln(x)-x*exp(5/x))/exp(5/x))-100*exp(5/x))*ln((2*x*exp((ln(x)
-x*exp(5/x))/exp(5/x))+10)/x^2)/(x^2*exp(5/x)*exp((ln(x)-x*exp(5/x))/exp(5/x))+5*x*exp(5/x)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*log(x)+(-10*x^2-10*x)*exp(5/x)+10*x)*exp((log(x)-x*exp(5/x))/exp(5/x))-100*exp(5/x))*log((2*x*e
xp((log(x)-x*exp(5/x))/exp(5/x))+10)/x^2)/(x^2*exp(5/x)*exp((log(x)-x*exp(5/x))/exp(5/x))+5*x*exp(5/x)),x, alg
orithm="maxima")

[Out]

-10*integrate((((x^2 + x)*e^(5/x) - x - 5*log(x))*e^(-(x*e^(5/x) - log(x))*e^(-5/x)) + 10*e^(5/x))*log(2*(x*e^
(-(x*e^(5/x) - log(x))*e^(-5/x)) + 5)/x^2)/(x^2*e^(-(x*e^(5/x) - log(x))*e^(-5/x) + 5/x) + 5*x*e^(5/x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\ln \left (\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {5}{x}}\,\left (\ln \relax (x)-x\,{\mathrm {e}}^{5/x}\right )}+10}{x^2}\right )\,\left (100\,{\mathrm {e}}^{5/x}-{\mathrm {e}}^{{\mathrm {e}}^{-\frac {5}{x}}\,\left (\ln \relax (x)-x\,{\mathrm {e}}^{5/x}\right )}\,\left (10\,x+50\,\ln \relax (x)-{\mathrm {e}}^{5/x}\,\left (10\,x^2+10\,x\right )\right )\right )}{5\,x\,{\mathrm {e}}^{5/x}+x^2\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {5}{x}}\,\left (\ln \relax (x)-x\,{\mathrm {e}}^{5/x}\right )}\,{\mathrm {e}}^{5/x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((2*x*exp(exp(-5/x)*(log(x) - x*exp(5/x))) + 10)/x^2)*(100*exp(5/x) - exp(exp(-5/x)*(log(x) - x*exp(5
/x)))*(10*x + 50*log(x) - exp(5/x)*(10*x + 10*x^2))))/(5*x*exp(5/x) + x^2*exp(exp(-5/x)*(log(x) - x*exp(5/x)))
*exp(5/x)),x)

[Out]

int(-(log((2*x*exp(exp(-5/x)*(log(x) - x*exp(5/x))) + 10)/x^2)*(100*exp(5/x) - exp(exp(-5/x)*(log(x) - x*exp(5
/x)))*(10*x + 50*log(x) - exp(5/x)*(10*x + 10*x^2))))/(5*x*exp(5/x) + x^2*exp(exp(-5/x)*(log(x) - x*exp(5/x)))
*exp(5/x)), x)

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sympy [A]  time = 50.51, size = 29, normalized size = 0.91 \begin {gather*} 5 \log {\left (\frac {2 x e^{\left (- x e^{\frac {5}{x}} + \log {\relax (x )}\right ) e^{- \frac {5}{x}}} + 10}{x^{2}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*ln(x)+(-10*x**2-10*x)*exp(5/x)+10*x)*exp((ln(x)-x*exp(5/x))/exp(5/x))-100*exp(5/x))*ln((2*x*exp
((ln(x)-x*exp(5/x))/exp(5/x))+10)/x**2)/(x**2*exp(5/x)*exp((ln(x)-x*exp(5/x))/exp(5/x))+5*x*exp(5/x)),x)

[Out]

5*log((2*x*exp((-x*exp(5/x) + log(x))*exp(-5/x)) + 10)/x**2)**2

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