Integrand size = 95, antiderivative size = 22 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\frac {x+\log \left (100 \left (-4+e^{10/x}\right )^2\right )}{x}} \] Output:
exp((ln(20*(exp(10/x)-4)*(5*exp(10/x)-20))+x)/x)
\[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx \] Input:
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 - 800*E^(10/x) + 100*E^(20/x)]))/(-4*x^3 + E^(10/x)*x^3),x]
Output:
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 - 800*E^(10/x) + 100*E^(20/x)]))/(-4*x^3 + E^(10/x)*x^3), x]
Time = 0.77 (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 = {7257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{\frac {x+\log \left (-800 e^{10/x}+100 e^{20/x}+1600\right )}{x}} \left (\left (4 x-e^{10/x} x\right ) \log \left (-800 e^{10/x}+100 e^{20/x}+1600\right )-20 e^{10/x}\right )}{e^{10/x} x^3-4 x^3} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e \left (-800 e^{10/x}+100 e^{20/x}+1600\right )^{\frac {1}{x}}\) |
Input:
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^(1 0/x)*x^3),x]
Output:
E*(1600 - 800*E^(10/x) + 100*E^(20/x))^x^(-1)
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 0.90 (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 | \(4^{\frac {1}{x}} 25^{\frac {1}{x}} \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{\frac {2}{x}} {\mathrm e}^{\frac {-i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )}^{3}+2 i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )\right )-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )\right )}^{2}+2 x}{2 x}}\) | \(120\) |
Input:
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(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^3) ,x,method=_RETURNVERBOSE)
Output:
exp((ln(100*exp(10/x)^2-800*exp(10/x)+1600)+x)/x)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\left (\frac {x + \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right )}{x}\right )} \] Input:
integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*e xp(10/x))*exp((log(100*exp(10/x)^2-800*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x )-4*x^3),x, algorithm="fricas")
Output:
e^((x + log(100*e^(20/x) - 800*e^(10/x) + 1600))/x)
Timed out. \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\text {Timed out} \] Input:
integrate(((-x*exp(10/x)+4*x)*ln(100*exp(10/x)**2-800*exp(10/x)+1600)-20*e xp(10/x))*exp((ln(100*exp(10/x)**2-800*exp(10/x)+1600)+x)/x)/(x**3*exp(10/ x)-4*x**3),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\left (\frac {2 \, \log \left (5\right )}{x} + \frac {2 \, \log \left (2\right )}{x} + \frac {2 \, \log \left (e^{\frac {5}{x}} + 2\right )}{x} + \frac {2 \, \log \left (e^{\frac {5}{x}} - 2\right )}{x} + 1\right )} \] Input:
integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*e xp(10/x))*exp((log(100*exp(10/x)^2-800*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x )-4*x^3),x, algorithm="maxima")
Output:
e^(2*log(5)/x + 2*log(2)/x + 2*log(e^(5/x) + 2)/x + 2*log(e^(5/x) - 2)/x + 1)
\[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\int { -\frac {{\left ({\left (x e^{\frac {10}{x}} - 4 \, x\right )} \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right ) + 20 \, e^{\frac {10}{x}}\right )} e^{\left (\frac {x + \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right )}{x}\right )}}{x^{3} e^{\frac {10}{x}} - 4 \, x^{3}} \,d x } \] Input:
integrate(((-x*exp(10/x)+4*x)*log(100*exp(10/x)^2-800*exp(10/x)+1600)-20*e xp(10/x))*exp((log(100*exp(10/x)^2-800*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x )-4*x^3),x, algorithm="giac")
Output:
integrate(-((x*e^(10/x) - 4*x)*log(100*e^(20/x) - 800*e^(10/x) + 1600) + 2 0*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)
Time = 2.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\mathrm {e}\,{\left (100\,{\mathrm {e}}^{20/x}-800\,{\mathrm {e}}^{10/x}+1600\right )}^{1/x} \] Input:
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*ex p(10/x) - 4*x^3),x)
Output:
exp(1)*(100*exp(20/x) - 800*exp(10/x) + 1600)^(1/x)
\[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\int \frac {\left (\left (-x \,{\mathrm e}^{\frac {10}{x}}+4 x \right ) \mathrm {log}\left (100 \left ({\mathrm e}^{\frac {10}{x}}\right )^{2}-800 \,{\mathrm e}^{\frac {10}{x}}+1600\right )-20 \,{\mathrm e}^{\frac {10}{x}}\right ) {\mathrm e}^{\frac {\mathrm {log}\left (100 \left ({\mathrm e}^{\frac {10}{x}}\right )^{2}-800 \,{\mathrm e}^{\frac {10}{x}}+1600\right )+x}{x}}}{x^{3} {\mathrm e}^{\frac {10}{x}}-4 x^{3}}d x \] Input:
int(((-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)^2-800*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^ 3),x)
Output:
int(((-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)^2-800*exp(10/x)+1600)+x)/x)/(x^3*exp(10/x)-4*x^ 3),x)