\(\int \frac {e^{-\frac {1}{-x+(-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))) \log (x)}} (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+(9 x+e^x (2 x-3 x^2+x \log (4))) \log (x))}{x^3+(30 x^2-18 x^3+6 x^2 \log (4)+e^x (-10 x^2+6 x^3-2 x^2 \log (4))) \log (x)+(225 x-270 x^2+81 x^3+(90 x-54 x^2) \log (4)+9 x \log ^2(4)+e^x (-150 x+180 x^2-54 x^3+(-60 x+36 x^2) \log (4)-6 x \log ^2(4))+e^{2 x} (25 x-30 x^2+9 x^3+(10 x-6 x^2) \log (4)+x \log ^2(4))) \log ^2(x)} \, dx\) [1379]

Optimal result

Integrand size = 240, antiderivative size = 26 \[ \int \frac {e^{-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}} \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx=e^{\frac {1}{x-\left (3-e^x\right ) (-5+3 x-\log (4)) \log (x)}} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}} \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx=\int \frac {\exp \left (-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}\right ) \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx \]

[In]

Int[(-15 + 8*x - 3*Log[4] + E^x*(5 - 3*x + Log[4]) + (9*x + E^x*(2*x - 3*x^2 + x*Log[4]))*Log[x])/(E^(-x + (-1
5 + 9*x - 3*Log[4] + E^x*(5 - 3*x + Log[4]))*Log[x])^(-1)*(x^3 + (30*x^2 - 18*x^3 + 6*x^2*Log[4] + E^x*(-10*x^
2 + 6*x^3 - 2*x^2*Log[4]))*Log[x] + (225*x - 270*x^2 + 81*x^3 + (90*x - 54*x^2)*Log[4] + 9*x*Log[4]^2 + E^x*(-
150*x + 180*x^2 - 54*x^3 + (-60*x + 36*x^2)*Log[4] - 6*x*Log[4]^2) + E^(2*x)*(25*x - 30*x^2 + 9*x^3 + (10*x -
6*x^2)*Log[4] + x*Log[4]^2))*Log[x]^2)),x]

[Out]

(-2*(1 + Log[2])*Defer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)/(-x - (-3 + E^x)*(-5 + 3*x - Lo
g[4])*Log[x]), x])/(5 + Log[4]) + (3*Defer[Int][(E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)*x)/(-x - (
-3 + E^x)*(-5 + 3*x - Log[4])*Log[x]), x])/(5 + Log[4]) + (3*Defer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*
Log[x])^(-1)/(Log[x]*(-x - (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])), x])/(5 + Log[4]) + Defer[Int][(E^(x + (-3
+ E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)*x)/(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^2, x] + (5 + Log[4])*De
fer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)/((-5 + 3*x - Log[4])*(x + (-3 + E^x)*(-5 + 3*x - L
og[4])*Log[x])^2), x] + ((5 + Log[4])^2*Defer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)/((-5 + 3
*x - Log[4])*(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^2), x])/3 + ((5 + Log[4])^2*Defer[Int][E^(x + (-3 + E
^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)/((5 - 3*x + Log[4])*(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^2), x])/3
+ Defer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)/(Log[x]*(x + (-3 + E^x)*(-5 + 3*x - Log[4])*L
og[x])^2), x] + 3*(5 + Log[4])*Defer[Int][(E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)*Log[x])/(x + (-3
+ E^x)*(-5 + 3*x - Log[4])*Log[x])^2, x] - 9*Defer[Int][(E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)*x
*Log[x])/(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^2, x] - 3*(5 + Log[4])^2*Defer[Int][(E^(x + (-3 + E^x)*(-
5 + 3*x - Log[4])*Log[x])^(-1)*Log[x])/((5 - 3*x + Log[4])*(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^2), x]
+ (75 + Log[4]*(30 + Log[64]))*Defer[Int][(E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)*Log[x])/((5 - 3*
x + Log[4])*(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^2), x] - Defer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[
4])*Log[x])^(-1)/(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x]), x] - (2*(1 + Log[2])*Defer[Int][E^(x + (-3 + E^x
)*(-5 + 3*x - Log[4])*Log[x])^(-1)/(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x]), x])/(5 + Log[4]) + (3*Defer[In
t][(E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)*x)/(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x]), x])/(5
+ Log[4]) - 2*(1 + Log[2])*Defer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)/((5 - 3*x + Log[4])*(
x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])), x] + (5 + Log[4])*Defer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[4]
)*Log[x])^(-1)/((5 - 3*x + Log[4])*(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])), x] + (3*Defer[Int][E^(x + (-3
+ E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)/(Log[x]*(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])), x])/(5 + Log[4]
) - Defer[Int][E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)/(x*Log[x]*(x + (-3 + E^x)*(-5 + 3*x - Log[4]
)*Log[x])), x]

Rubi steps Aborted

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}} \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx=e^{\frac {1}{x+\left (-3+e^x\right ) (-5+3 x-\log (4)) \log (x)}} \]

[In]

Integrate[(-15 + 8*x - 3*Log[4] + E^x*(5 - 3*x + Log[4]) + (9*x + E^x*(2*x - 3*x^2 + x*Log[4]))*Log[x])/(E^(-x
+ (-15 + 9*x - 3*Log[4] + E^x*(5 - 3*x + Log[4]))*Log[x])^(-1)*(x^3 + (30*x^2 - 18*x^3 + 6*x^2*Log[4] + E^x*(
-10*x^2 + 6*x^3 - 2*x^2*Log[4]))*Log[x] + (225*x - 270*x^2 + 81*x^3 + (90*x - 54*x^2)*Log[4] + 9*x*Log[4]^2 +
E^x*(-150*x + 180*x^2 - 54*x^3 + (-60*x + 36*x^2)*Log[4] - 6*x*Log[4]^2) + E^(2*x)*(25*x - 30*x^2 + 9*x^3 + (1
0*x - 6*x^2)*Log[4] + x*Log[4]^2))*Log[x]^2)),x]

[Out]

E^(x + (-3 + E^x)*(-5 + 3*x - Log[4])*Log[x])^(-1)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77

\[{\mathrm e}^{-\frac {1}{2 \ln \left (2\right ) \ln \left (x \right ) {\mathrm e}^{x}-3 x \,{\mathrm e}^{x} \ln \left (x \right )-6 \ln \left (2\right ) \ln \left (x \right )+5 \,{\mathrm e}^{x} \ln \left (x \right )+9 x \ln \left (x \right )-15 \ln \left (x \right )-x}}\]

[In]

int((((2*x*ln(2)-3*x^2+2*x)*exp(x)+9*x)*ln(x)+(2*ln(2)-3*x+5)*exp(x)-6*ln(2)+8*x-15)*exp(-1/(((2*ln(2)-3*x+5)*
exp(x)-6*ln(2)+9*x-15)*ln(x)-x))/(((4*x*ln(2)^2+2*(-6*x^2+10*x)*ln(2)+9*x^3-30*x^2+25*x)*exp(x)^2+(-24*x*ln(2)
^2+2*(36*x^2-60*x)*ln(2)-54*x^3+180*x^2-150*x)*exp(x)+36*x*ln(2)^2+2*(-54*x^2+90*x)*ln(2)+81*x^3-270*x^2+225*x
)*ln(x)^2+((-4*x^2*ln(2)+6*x^3-10*x^2)*exp(x)+12*x^2*ln(2)-18*x^3+30*x^2)*ln(x)+x^3),x)

[Out]

exp(-1/(2*ln(2)*ln(x)*exp(x)-3*x*exp(x)*ln(x)-6*ln(2)*ln(x)+5*exp(x)*ln(x)+9*x*ln(x)-15*ln(x)-x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}} \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx=e^{\left (\frac {1}{{\left ({\left (3 \, x - 2 \, \log \left (2\right ) - 5\right )} e^{x} - 9 \, x + 6 \, \log \left (2\right ) + 15\right )} \log \left (x\right ) + x}\right )} \]

[In]

integrate((((2*x*log(2)-3*x^2+2*x)*exp(x)+9*x)*log(x)+(2*log(2)-3*x+5)*exp(x)-6*log(2)+8*x-15)*exp(-1/(((2*log
(2)-3*x+5)*exp(x)-6*log(2)+9*x-15)*log(x)-x))/(((4*x*log(2)^2+2*(-6*x^2+10*x)*log(2)+9*x^3-30*x^2+25*x)*exp(x)
^2+(-24*x*log(2)^2+2*(36*x^2-60*x)*log(2)-54*x^3+180*x^2-150*x)*exp(x)+36*x*log(2)^2+2*(-54*x^2+90*x)*log(2)+8
1*x^3-270*x^2+225*x)*log(x)^2+((-4*x^2*log(2)+6*x^3-10*x^2)*exp(x)+12*x^2*log(2)-18*x^3+30*x^2)*log(x)+x^3),x,
algorithm="fricas")

[Out]

e^(1/(((3*x - 2*log(2) - 5)*e^x - 9*x + 6*log(2) + 15)*log(x) + x))

Sympy [A] (verification not implemented)

Time = 16.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}} \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx=e^{- \frac {1}{- x + \left (9 x + \left (- 3 x + 2 \log {\left (2 \right )} + 5\right ) e^{x} - 15 - 6 \log {\left (2 \right )}\right ) \log {\left (x \right )}}} \]

[In]

integrate((((2*x*ln(2)-3*x**2+2*x)*exp(x)+9*x)*ln(x)+(2*ln(2)-3*x+5)*exp(x)-6*ln(2)+8*x-15)*exp(-1/(((2*ln(2)-
3*x+5)*exp(x)-6*ln(2)+9*x-15)*ln(x)-x))/(((4*x*ln(2)**2+2*(-6*x**2+10*x)*ln(2)+9*x**3-30*x**2+25*x)*exp(x)**2+
(-24*x*ln(2)**2+2*(36*x**2-60*x)*ln(2)-54*x**3+180*x**2-150*x)*exp(x)+36*x*ln(2)**2+2*(-54*x**2+90*x)*ln(2)+81
*x**3-270*x**2+225*x)*ln(x)**2+((-4*x**2*ln(2)+6*x**3-10*x**2)*exp(x)+12*x**2*ln(2)-18*x**3+30*x**2)*ln(x)+x**
3),x)

[Out]

exp(-1/(-x + (9*x + (-3*x + 2*log(2) + 5)*exp(x) - 15 - 6*log(2))*log(x)))

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}} \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx=e^{\left (\frac {1}{{\left ({\left (3 \, x - 2 \, \log \left (2\right ) - 5\right )} e^{x} - 9 \, x + 6 \, \log \left (2\right ) + 15\right )} \log \left (x\right ) + x}\right )} \]

[In]

integrate((((2*x*log(2)-3*x^2+2*x)*exp(x)+9*x)*log(x)+(2*log(2)-3*x+5)*exp(x)-6*log(2)+8*x-15)*exp(-1/(((2*log
(2)-3*x+5)*exp(x)-6*log(2)+9*x-15)*log(x)-x))/(((4*x*log(2)^2+2*(-6*x^2+10*x)*log(2)+9*x^3-30*x^2+25*x)*exp(x)
^2+(-24*x*log(2)^2+2*(36*x^2-60*x)*log(2)-54*x^3+180*x^2-150*x)*exp(x)+36*x*log(2)^2+2*(-54*x^2+90*x)*log(2)+8
1*x^3-270*x^2+225*x)*log(x)^2+((-4*x^2*log(2)+6*x^3-10*x^2)*exp(x)+12*x^2*log(2)-18*x^3+30*x^2)*log(x)+x^3),x,
algorithm="maxima")

[Out]

e^(1/(((3*x - 2*log(2) - 5)*e^x - 9*x + 6*log(2) + 15)*log(x) + x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}} \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx=e^{\left (\frac {1}{3 \, x e^{x} \log \left (x\right ) - 2 \, e^{x} \log \left (2\right ) \log \left (x\right ) - 9 \, x \log \left (x\right ) - 5 \, e^{x} \log \left (x\right ) + 6 \, \log \left (2\right ) \log \left (x\right ) + x + 15 \, \log \left (x\right )}\right )} \]

[In]

integrate((((2*x*log(2)-3*x^2+2*x)*exp(x)+9*x)*log(x)+(2*log(2)-3*x+5)*exp(x)-6*log(2)+8*x-15)*exp(-1/(((2*log
(2)-3*x+5)*exp(x)-6*log(2)+9*x-15)*log(x)-x))/(((4*x*log(2)^2+2*(-6*x^2+10*x)*log(2)+9*x^3-30*x^2+25*x)*exp(x)
^2+(-24*x*log(2)^2+2*(36*x^2-60*x)*log(2)-54*x^3+180*x^2-150*x)*exp(x)+36*x*log(2)^2+2*(-54*x^2+90*x)*log(2)+8
1*x^3-270*x^2+225*x)*log(x)^2+((-4*x^2*log(2)+6*x^3-10*x^2)*exp(x)+12*x^2*log(2)-18*x^3+30*x^2)*log(x)+x^3),x,
algorithm="giac")

[Out]

e^(1/(3*x*e^x*log(x) - 2*e^x*log(2)*log(x) - 9*x*log(x) - 5*e^x*log(x) + 6*log(2)*log(x) + x + 15*log(x)))

Mupad [B] (verification not implemented)

Time = 10.61 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-\frac {1}{-x+\left (-15+9 x-3 \log (4)+e^x (5-3 x+\log (4))\right ) \log (x)}} \left (-15+8 x-3 \log (4)+e^x (5-3 x+\log (4))+\left (9 x+e^x \left (2 x-3 x^2+x \log (4)\right )\right ) \log (x)\right )}{x^3+\left (30 x^2-18 x^3+6 x^2 \log (4)+e^x \left (-10 x^2+6 x^3-2 x^2 \log (4)\right )\right ) \log (x)+\left (225 x-270 x^2+81 x^3+\left (90 x-54 x^2\right ) \log (4)+9 x \log ^2(4)+e^x \left (-150 x+180 x^2-54 x^3+\left (-60 x+36 x^2\right ) \log (4)-6 x \log ^2(4)\right )+e^{2 x} \left (25 x-30 x^2+9 x^3+\left (10 x-6 x^2\right ) \log (4)+x \log ^2(4)\right )\right ) \log ^2(x)} \, dx={\mathrm {e}}^{\frac {1}{x+15\,\ln \left (x\right )-5\,{\mathrm {e}}^x\,\ln \left (x\right )+6\,\ln \left (2\right )\,\ln \left (x\right )-9\,x\,\ln \left (x\right )-2\,{\mathrm {e}}^x\,\ln \left (2\right )\,\ln \left (x\right )+3\,x\,{\mathrm {e}}^x\,\ln \left (x\right )}} \]

[In]

int((exp(1/(x - log(x)*(9*x - 6*log(2) + exp(x)*(2*log(2) - 3*x + 5) - 15)))*(8*x - 6*log(2) + exp(x)*(2*log(2
) - 3*x + 5) + log(x)*(9*x + exp(x)*(2*x + 2*x*log(2) - 3*x^2)) - 15))/(log(x)^2*(225*x + exp(2*x)*(25*x + 2*l
og(2)*(10*x - 6*x^2) + 4*x*log(2)^2 - 30*x^2 + 9*x^3) + 2*log(2)*(90*x - 54*x^2) + 36*x*log(2)^2 - exp(x)*(150
*x + 2*log(2)*(60*x - 36*x^2) + 24*x*log(2)^2 - 180*x^2 + 54*x^3) - 270*x^2 + 81*x^3) + log(x)*(12*x^2*log(2)
- exp(x)*(4*x^2*log(2) + 10*x^2 - 6*x^3) + 30*x^2 - 18*x^3) + x^3),x)

[Out]

exp(1/(x + 15*log(x) - 5*exp(x)*log(x) + 6*log(2)*log(x) - 9*x*log(x) - 2*exp(x)*log(2)*log(x) + 3*x*exp(x)*lo
g(x)))