\(\int \frac {12 x^2+e^x (-60 x-48 x^2+12 x^3+e^{4/x} (-240-12 x+12 x^2))+(-60 x+12 x^2+e^{4/x} (-240-12 x+12 x^2)) \log (5-x)+(e^x (60 x-12 x^2)+(60 x-12 x^2) \log (5-x)) \log (e^x+\log (5-x))}{e^x (-5 x+x^2+e^{8/x} (-5 x+x^2)+e^{4/x} (-10 x+2 x^2))+(-5 x+x^2+e^{8/x} (-5 x+x^2)+e^{4/x} (-10 x+2 x^2)) \log (5-x)+(e^x (10 x-2 x^2+e^{4/x} (10 x-2 x^2))+(10 x-2 x^2+e^{4/x} (10 x-2 x^2)) \log (5-x)) \log (e^x+\log (5-x))+(e^x (-5 x+x^2)+(-5 x+x^2) \log (5-x)) \log ^2(e^x+\log (5-x))} \, dx\) [3272]

Optimal result

Integrand size = 327, antiderivative size = 27 \[ \int \frac {12 x^2+e^x \left (-60 x-48 x^2+12 x^3+e^{4/x} \left (-240-12 x+12 x^2\right )\right )+\left (-60 x+12 x^2+e^{4/x} \left (-240-12 x+12 x^2\right )\right ) \log (5-x)+\left (e^x \left (60 x-12 x^2\right )+\left (60 x-12 x^2\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )}{e^x \left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right )+\left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right ) \log (5-x)+\left (e^x \left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right )+\left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )+\left (e^x \left (-5 x+x^2\right )+\left (-5 x+x^2\right ) \log (5-x)\right ) \log ^2\left (e^x+\log (5-x)\right )} \, dx=\frac {12 x}{1+e^{4/x}-\log \left (e^x+\log (5-x)\right )} \]

[Out]

3*x/(1/4*exp(4/x)-1/4*ln(ln(5-x)+exp(x))+1/4)

Rubi [A] (verified)

Time = 1.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6820, 12, 6819} \[ \int \frac {12 x^2+e^x \left (-60 x-48 x^2+12 x^3+e^{4/x} \left (-240-12 x+12 x^2\right )\right )+\left (-60 x+12 x^2+e^{4/x} \left (-240-12 x+12 x^2\right )\right ) \log (5-x)+\left (e^x \left (60 x-12 x^2\right )+\left (60 x-12 x^2\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )}{e^x \left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right )+\left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right ) \log (5-x)+\left (e^x \left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right )+\left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )+\left (e^x \left (-5 x+x^2\right )+\left (-5 x+x^2\right ) \log (5-x)\right ) \log ^2\left (e^x+\log (5-x)\right )} \, dx=\frac {12 x}{e^{4/x}-\log \left (e^x+\log (5-x)\right )+1} \]

[In]

Int[(12*x^2 + E^x*(-60*x - 48*x^2 + 12*x^3 + E^(4/x)*(-240 - 12*x + 12*x^2)) + (-60*x + 12*x^2 + E^(4/x)*(-240
- 12*x + 12*x^2))*Log[5 - x] + (E^x*(60*x - 12*x^2) + (60*x - 12*x^2)*Log[5 - x])*Log[E^x + Log[5 - x]])/(E^x
*(-5*x + x^2 + E^(8/x)*(-5*x + x^2) + E^(4/x)*(-10*x + 2*x^2)) + (-5*x + x^2 + E^(8/x)*(-5*x + x^2) + E^(4/x)*
(-10*x + 2*x^2))*Log[5 - x] + (E^x*(10*x - 2*x^2 + E^(4/x)*(10*x - 2*x^2)) + (10*x - 2*x^2 + E^(4/x)*(10*x - 2
*x^2))*Log[5 - x])*Log[E^x + Log[5 - x]] + (E^x*(-5*x + x^2) + (-5*x + x^2)*Log[5 - x])*Log[E^x + Log[5 - x]]^
2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6819

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[q*y^(m +
1)*(z^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {12 \left (20 e^{\frac {4}{x}+x}+5 e^x x+e^{\frac {4}{x}+x} x-x^2+4 e^x x^2-e^{\frac {4}{x}+x} x^2-e^x x^3+e^x (-5+x) x \log \left (e^x+\log (5-x)\right )-(-5+x) \log (5-x) \left (x+e^{4/x} (4+x)-x \log \left (e^x+\log (5-x)\right )\right )\right )}{(5-x) x \left (e^x+\log (5-x)\right ) \left (1+e^{4/x}-\log \left (e^x+\log (5-x)\right )\right )^2} \, dx \\ & = 12 \int \frac {20 e^{\frac {4}{x}+x}+5 e^x x+e^{\frac {4}{x}+x} x-x^2+4 e^x x^2-e^{\frac {4}{x}+x} x^2-e^x x^3+e^x (-5+x) x \log \left (e^x+\log (5-x)\right )-(-5+x) \log (5-x) \left (x+e^{4/x} (4+x)-x \log \left (e^x+\log (5-x)\right )\right )}{(5-x) x \left (e^x+\log (5-x)\right ) \left (1+e^{4/x}-\log \left (e^x+\log (5-x)\right )\right )^2} \, dx \\ & = \frac {12 x}{1+e^{4/x}-\log \left (e^x+\log (5-x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {12 x^2+e^x \left (-60 x-48 x^2+12 x^3+e^{4/x} \left (-240-12 x+12 x^2\right )\right )+\left (-60 x+12 x^2+e^{4/x} \left (-240-12 x+12 x^2\right )\right ) \log (5-x)+\left (e^x \left (60 x-12 x^2\right )+\left (60 x-12 x^2\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )}{e^x \left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right )+\left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right ) \log (5-x)+\left (e^x \left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right )+\left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )+\left (e^x \left (-5 x+x^2\right )+\left (-5 x+x^2\right ) \log (5-x)\right ) \log ^2\left (e^x+\log (5-x)\right )} \, dx=-\frac {12 x}{-1-e^{4/x}+\log \left (e^x+\log (5-x)\right )} \]

[In]

Integrate[(12*x^2 + E^x*(-60*x - 48*x^2 + 12*x^3 + E^(4/x)*(-240 - 12*x + 12*x^2)) + (-60*x + 12*x^2 + E^(4/x)
*(-240 - 12*x + 12*x^2))*Log[5 - x] + (E^x*(60*x - 12*x^2) + (60*x - 12*x^2)*Log[5 - x])*Log[E^x + Log[5 - x]]
)/(E^x*(-5*x + x^2 + E^(8/x)*(-5*x + x^2) + E^(4/x)*(-10*x + 2*x^2)) + (-5*x + x^2 + E^(8/x)*(-5*x + x^2) + E^
(4/x)*(-10*x + 2*x^2))*Log[5 - x] + (E^x*(10*x - 2*x^2 + E^(4/x)*(10*x - 2*x^2)) + (10*x - 2*x^2 + E^(4/x)*(10
*x - 2*x^2))*Log[5 - x])*Log[E^x + Log[5 - x]] + (E^x*(-5*x + x^2) + (-5*x + x^2)*Log[5 - x])*Log[E^x + Log[5
- x]]^2),x]

[Out]

(-12*x)/(-1 - E^(4/x) + Log[E^x + Log[5 - x]])

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96

\[\frac {12 x}{{\mathrm e}^{\frac {4}{x}}-\ln \left (\ln \left (5-x \right )+{\mathrm e}^{x}\right )+1}\]

[In]

int((((-12*x^2+60*x)*ln(5-x)+(-12*x^2+60*x)*exp(x))*ln(ln(5-x)+exp(x))+((12*x^2-12*x-240)*exp(4/x)+12*x^2-60*x
)*ln(5-x)+((12*x^2-12*x-240)*exp(4/x)+12*x^3-48*x^2-60*x)*exp(x)+12*x^2)/(((x^2-5*x)*ln(5-x)+(x^2-5*x)*exp(x))
*ln(ln(5-x)+exp(x))^2+(((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*ln(5-x)+((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*exp(x))
*ln(ln(5-x)+exp(x))+((x^2-5*x)*exp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*ln(5-x)+((x^2-5*x)*exp(4/x)^2+(2*x^2-
10*x)*exp(4/x)+x^2-5*x)*exp(x)),x)

[Out]

12*x/(exp(4/x)-ln(ln(5-x)+exp(x))+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {12 x^2+e^x \left (-60 x-48 x^2+12 x^3+e^{4/x} \left (-240-12 x+12 x^2\right )\right )+\left (-60 x+12 x^2+e^{4/x} \left (-240-12 x+12 x^2\right )\right ) \log (5-x)+\left (e^x \left (60 x-12 x^2\right )+\left (60 x-12 x^2\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )}{e^x \left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right )+\left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right ) \log (5-x)+\left (e^x \left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right )+\left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )+\left (e^x \left (-5 x+x^2\right )+\left (-5 x+x^2\right ) \log (5-x)\right ) \log ^2\left (e^x+\log (5-x)\right )} \, dx=\frac {12 \, x}{e^{\frac {4}{x}} - \log \left (e^{x} + \log \left (-x + 5\right )\right ) + 1} \]

[In]

integrate((((-12*x^2+60*x)*log(5-x)+(-12*x^2+60*x)*exp(x))*log(log(5-x)+exp(x))+((12*x^2-12*x-240)*exp(4/x)+12
*x^2-60*x)*log(5-x)+((12*x^2-12*x-240)*exp(4/x)+12*x^3-48*x^2-60*x)*exp(x)+12*x^2)/(((x^2-5*x)*log(5-x)+(x^2-5
*x)*exp(x))*log(log(5-x)+exp(x))^2+(((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*log(5-x)+((-2*x^2+10*x)*exp(4/x)-2*x^2
+10*x)*exp(x))*log(log(5-x)+exp(x))+((x^2-5*x)*exp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*log(5-x)+((x^2-5*x)*e
xp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*exp(x)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {12 x^2+e^x \left (-60 x-48 x^2+12 x^3+e^{4/x} \left (-240-12 x+12 x^2\right )\right )+\left (-60 x+12 x^2+e^{4/x} \left (-240-12 x+12 x^2\right )\right ) \log (5-x)+\left (e^x \left (60 x-12 x^2\right )+\left (60 x-12 x^2\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )}{e^x \left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right )+\left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right ) \log (5-x)+\left (e^x \left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right )+\left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )+\left (e^x \left (-5 x+x^2\right )+\left (-5 x+x^2\right ) \log (5-x)\right ) \log ^2\left (e^x+\log (5-x)\right )} \, dx=- \frac {12 x}{- e^{\frac {4}{x}} + \log {\left (e^{x} + \log {\left (5 - x \right )} \right )} - 1} \]

[In]

integrate((((-12*x**2+60*x)*ln(5-x)+(-12*x**2+60*x)*exp(x))*ln(ln(5-x)+exp(x))+((12*x**2-12*x-240)*exp(4/x)+12
*x**2-60*x)*ln(5-x)+((12*x**2-12*x-240)*exp(4/x)+12*x**3-48*x**2-60*x)*exp(x)+12*x**2)/(((x**2-5*x)*ln(5-x)+(x
**2-5*x)*exp(x))*ln(ln(5-x)+exp(x))**2+(((-2*x**2+10*x)*exp(4/x)-2*x**2+10*x)*ln(5-x)+((-2*x**2+10*x)*exp(4/x)
-2*x**2+10*x)*exp(x))*ln(ln(5-x)+exp(x))+((x**2-5*x)*exp(4/x)**2+(2*x**2-10*x)*exp(4/x)+x**2-5*x)*ln(5-x)+((x*
*2-5*x)*exp(4/x)**2+(2*x**2-10*x)*exp(4/x)+x**2-5*x)*exp(x)),x)

[Out]

-12*x/(-exp(4/x) + log(exp(x) + log(5 - x)) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {12 x^2+e^x \left (-60 x-48 x^2+12 x^3+e^{4/x} \left (-240-12 x+12 x^2\right )\right )+\left (-60 x+12 x^2+e^{4/x} \left (-240-12 x+12 x^2\right )\right ) \log (5-x)+\left (e^x \left (60 x-12 x^2\right )+\left (60 x-12 x^2\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )}{e^x \left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right )+\left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right ) \log (5-x)+\left (e^x \left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right )+\left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )+\left (e^x \left (-5 x+x^2\right )+\left (-5 x+x^2\right ) \log (5-x)\right ) \log ^2\left (e^x+\log (5-x)\right )} \, dx=\frac {12 \, x}{e^{\frac {4}{x}} - \log \left (e^{x} + \log \left (-x + 5\right )\right ) + 1} \]

[In]

integrate((((-12*x^2+60*x)*log(5-x)+(-12*x^2+60*x)*exp(x))*log(log(5-x)+exp(x))+((12*x^2-12*x-240)*exp(4/x)+12
*x^2-60*x)*log(5-x)+((12*x^2-12*x-240)*exp(4/x)+12*x^3-48*x^2-60*x)*exp(x)+12*x^2)/(((x^2-5*x)*log(5-x)+(x^2-5
*x)*exp(x))*log(log(5-x)+exp(x))^2+(((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*log(5-x)+((-2*x^2+10*x)*exp(4/x)-2*x^2
+10*x)*exp(x))*log(log(5-x)+exp(x))+((x^2-5*x)*exp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*log(5-x)+((x^2-5*x)*e
xp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*exp(x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {12 x^2+e^x \left (-60 x-48 x^2+12 x^3+e^{4/x} \left (-240-12 x+12 x^2\right )\right )+\left (-60 x+12 x^2+e^{4/x} \left (-240-12 x+12 x^2\right )\right ) \log (5-x)+\left (e^x \left (60 x-12 x^2\right )+\left (60 x-12 x^2\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )}{e^x \left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right )+\left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right ) \log (5-x)+\left (e^x \left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right )+\left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )+\left (e^x \left (-5 x+x^2\right )+\left (-5 x+x^2\right ) \log (5-x)\right ) \log ^2\left (e^x+\log (5-x)\right )} \, dx=\frac {12 \, x}{e^{\frac {4}{x}} - \log \left (e^{x} + \log \left (-x + 5\right )\right ) + 1} \]

[In]

integrate((((-12*x^2+60*x)*log(5-x)+(-12*x^2+60*x)*exp(x))*log(log(5-x)+exp(x))+((12*x^2-12*x-240)*exp(4/x)+12
*x^2-60*x)*log(5-x)+((12*x^2-12*x-240)*exp(4/x)+12*x^3-48*x^2-60*x)*exp(x)+12*x^2)/(((x^2-5*x)*log(5-x)+(x^2-5
*x)*exp(x))*log(log(5-x)+exp(x))^2+(((-2*x^2+10*x)*exp(4/x)-2*x^2+10*x)*log(5-x)+((-2*x^2+10*x)*exp(4/x)-2*x^2
+10*x)*exp(x))*log(log(5-x)+exp(x))+((x^2-5*x)*exp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*log(5-x)+((x^2-5*x)*e
xp(4/x)^2+(2*x^2-10*x)*exp(4/x)+x^2-5*x)*exp(x)),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {12 x^2+e^x \left (-60 x-48 x^2+12 x^3+e^{4/x} \left (-240-12 x+12 x^2\right )\right )+\left (-60 x+12 x^2+e^{4/x} \left (-240-12 x+12 x^2\right )\right ) \log (5-x)+\left (e^x \left (60 x-12 x^2\right )+\left (60 x-12 x^2\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )}{e^x \left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right )+\left (-5 x+x^2+e^{8/x} \left (-5 x+x^2\right )+e^{4/x} \left (-10 x+2 x^2\right )\right ) \log (5-x)+\left (e^x \left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right )+\left (10 x-2 x^2+e^{4/x} \left (10 x-2 x^2\right )\right ) \log (5-x)\right ) \log \left (e^x+\log (5-x)\right )+\left (e^x \left (-5 x+x^2\right )+\left (-5 x+x^2\right ) \log (5-x)\right ) \log ^2\left (e^x+\log (5-x)\right )} \, dx=\int -\frac {\ln \left (\ln \left (5-x\right )+{\mathrm {e}}^x\right )\,\left (\ln \left (5-x\right )\,\left (60\,x-12\,x^2\right )+{\mathrm {e}}^x\,\left (60\,x-12\,x^2\right )\right )-{\mathrm {e}}^x\,\left (60\,x+{\mathrm {e}}^{4/x}\,\left (-12\,x^2+12\,x+240\right )+48\,x^2-12\,x^3\right )+12\,x^2-\ln \left (5-x\right )\,\left (60\,x+{\mathrm {e}}^{4/x}\,\left (-12\,x^2+12\,x+240\right )-12\,x^2\right )}{\left (\ln \left (5-x\right )\,\left (5\,x-x^2\right )+{\mathrm {e}}^x\,\left (5\,x-x^2\right )\right )\,{\ln \left (\ln \left (5-x\right )+{\mathrm {e}}^x\right )}^2+\left (-{\mathrm {e}}^x\,\left (10\,x+{\mathrm {e}}^{4/x}\,\left (10\,x-2\,x^2\right )-2\,x^2\right )-\ln \left (5-x\right )\,\left (10\,x+{\mathrm {e}}^{4/x}\,\left (10\,x-2\,x^2\right )-2\,x^2\right )\right )\,\ln \left (\ln \left (5-x\right )+{\mathrm {e}}^x\right )+\ln \left (5-x\right )\,\left (5\,x+{\mathrm {e}}^{8/x}\,\left (5\,x-x^2\right )+{\mathrm {e}}^{4/x}\,\left (10\,x-2\,x^2\right )-x^2\right )+{\mathrm {e}}^x\,\left (5\,x+{\mathrm {e}}^{8/x}\,\left (5\,x-x^2\right )+{\mathrm {e}}^{4/x}\,\left (10\,x-2\,x^2\right )-x^2\right )} \,d x \]

[In]

int(-(log(log(5 - x) + exp(x))*(log(5 - x)*(60*x - 12*x^2) + exp(x)*(60*x - 12*x^2)) - exp(x)*(60*x + exp(4/x)
*(12*x - 12*x^2 + 240) + 48*x^2 - 12*x^3) + 12*x^2 - log(5 - x)*(60*x + exp(4/x)*(12*x - 12*x^2 + 240) - 12*x^
2))/(log(5 - x)*(5*x + exp(8/x)*(5*x - x^2) + exp(4/x)*(10*x - 2*x^2) - x^2) + exp(x)*(5*x + exp(8/x)*(5*x - x
^2) + exp(4/x)*(10*x - 2*x^2) - x^2) + log(log(5 - x) + exp(x))^2*(log(5 - x)*(5*x - x^2) + exp(x)*(5*x - x^2)
) - log(log(5 - x) + exp(x))*(exp(x)*(10*x + exp(4/x)*(10*x - 2*x^2) - 2*x^2) + log(5 - x)*(10*x + exp(4/x)*(1
0*x - 2*x^2) - 2*x^2))),x)

[Out]

int(-(log(log(5 - x) + exp(x))*(log(5 - x)*(60*x - 12*x^2) + exp(x)*(60*x - 12*x^2)) - exp(x)*(60*x + exp(4/x)
*(12*x - 12*x^2 + 240) + 48*x^2 - 12*x^3) + 12*x^2 - log(5 - x)*(60*x + exp(4/x)*(12*x - 12*x^2 + 240) - 12*x^
2))/(log(5 - x)*(5*x + exp(8/x)*(5*x - x^2) + exp(4/x)*(10*x - 2*x^2) - x^2) + exp(x)*(5*x + exp(8/x)*(5*x - x
^2) + exp(4/x)*(10*x - 2*x^2) - x^2) + log(log(5 - x) + exp(x))^2*(log(5 - x)*(5*x - x^2) + exp(x)*(5*x - x^2)
) - log(log(5 - x) + exp(x))*(exp(x)*(10*x + exp(4/x)*(10*x - 2*x^2) - 2*x^2) + log(5 - x)*(10*x + exp(4/x)*(1
0*x - 2*x^2) - 2*x^2))), x)