\(\int \frac {e^{\frac {x+\log (1600-800 e^{10/x}+100 e^{20/x})}{x}} (-20 e^{10/x}+(4 x-e^{10/x} x) \log (1600-800 e^{10/x}+100 e^{20/x}))}{-4 x^3+e^{10/x} x^3} \, dx\) [3436]

Optimal result

Integrand size = 95, antiderivative size = 22 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\frac {x+\log \left (100 \left (-4+e^{10/x}\right )^2\right )}{x}} \]

[Out]

exp((ln(20*(exp(10/x)-4)*(5*exp(10/x)-20))+x)/x)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {6838} \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e \left (-800 e^{10/x}+100 e^{20/x}+1600\right )^{\frac {1}{x}} \]

[In]

Int[(E^((x + Log[1600 - 800*E^(10/x) + 100*E^(20/x)])/x)*(-20*E^(10/x) + (4*x - E^(10/x)*x)*Log[1600 - 800*E^(
10/x) + 100*E^(20/x)]))/(-4*x^3 + E^(10/x)*x^3),x]

[Out]

E*(1600 - 800*E^(10/x) + 100*E^(20/x))^x^(-1)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = e \left (1600-800 e^{10/x}+100 e^{20/x}\right )^{\frac {1}{x}} \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx \]

[In]

Integrate[(E^((x + Log[1600 - 800*E^(10/x) + 100*E^(20/x)])/x)*(-20*E^(10/x) + (4*x - E^(10/x)*x)*Log[1600 - 8
00*E^(10/x) + 100*E^(20/x)]))/(-4*x^3 + E^(10/x)*x^3),x]

[Out]

Integrate[(E^((x + Log[1600 - 800*E^(10/x) + 100*E^(20/x)])/x)*(-20*E^(10/x) + (4*x - E^(10/x)*x)*Log[1600 - 8
00*E^(10/x) + 100*E^(20/x)]))/(-4*x^3 + E^(10/x)*x^3), x]

Maple [A] (verified)

Time = 3.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\ln \left (100 \,{\mathrm e}^{\frac {20}{x}}-800 \,{\mathrm e}^{\frac {10}{x}}+1600\right )+x}{x}}\)	\(29\)
risch	\(4^{\frac {1}{x}} 25^{\frac {1}{x}} \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{\frac {2}{x}} {\mathrm e}^{\frac {-i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )}^{2}-i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )}^{3}+2 x}{2 x}}\)	\(120\)

[In]

int(((-x*exp(10/x)+4*x)*ln(100*exp(10/x)^2-800*exp(10/x)+1600)-20*exp(10/x))*exp((ln(100*exp(10/x)^2-800*exp(1
0/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3),x,method=_RETURNVERBOSE)

[Out]

exp((ln(100*exp(10/x)^2-800*exp(10/x)+1600)+x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\left (\frac {x + \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right )}{x}\right )} \]

[In]

integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*exp(10/x))*exp((log(100*exp(10/x)^2-8
00*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3),x, algorithm="fricas")

[Out]

e^((x + log(100*e^(20/x) - 800*e^(10/x) + 1600))/x)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\text {Timed out} \]

[In]

integrate(((-x*exp(10/x)+4*x)*ln(100*exp(10/x)**2-800*exp(10/x)+1600)-20*exp(10/x))*exp((ln(100*exp(10/x)**2-8
00*exp(10/x)+1600)+x)/x)/(x**3*exp(10/x)-4*x**3),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (20) = 40\).

Time = 0.38 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\left (\frac {2 \, \log \left (5\right )}{x} + \frac {2 \, \log \left (2\right )}{x} + \frac {2 \, \log \left (e^{\frac {5}{x}} + 2\right )}{x} + \frac {2 \, \log \left (e^{\frac {5}{x}} - 2\right )}{x} + 1\right )} \]

[In]

integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*exp(10/x))*exp((log(100*exp(10/x)^2-8
00*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\int { -\frac {{\left ({\left (x e^{\frac {10}{x}} - 4 \, x\right )} \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right ) + 20 \, e^{\frac {10}{x}}\right )} e^{\left (\frac {x + \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right )}{x}\right )}}{x^{3} e^{\frac {10}{x}} - 4 \, x^{3}} \,d x } \]

[In]

integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*exp(10/x))*exp((log(100*exp(10/x)^2-8
00*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3),x, algorithm="giac")

[Out]

integrate(-((x*e^(10/x) - 4*x)*log(100*e^(20/x) - 800*e^(10/x) + 1600) + 20*e^(10/x))*e^((x + log(100*e^(20/x)
- 800*e^(10/x) + 1600))/x)/(x^3*e^(10/x) - 4*x^3), x)

Mupad [B] (verification not implemented)

Time = 8.55 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\mathrm {e}\,{\left (100\,{\mathrm {e}}^{20/x}-800\,{\mathrm {e}}^{10/x}+1600\right )}^{1/x} \]

[In]

int(-(exp((x + log(100*exp(20/x) - 800*exp(10/x) + 1600))/x)*(20*exp(10/x) - log(100*exp(20/x) - 800*exp(10/x)
+ 1600)*(4*x - x*exp(10/x))))/(x^3*exp(10/x) - 4*x^3),x)

[Out]

exp(1)*(100*exp(20/x) - 800*exp(10/x) + 1600)^(1/x)