Integrand size = 81, antiderivative size = 33 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log \left (\frac {e^{2 e^{2 e^3}} \left (e^3-e^{-2 x} x\right )}{\log (x)}\right )} \]
Time = 0.43 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log \left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \]
Integrate[(5*E^(3 + 2*x) - 5*x + (5*x - 10*x^2)*Log[x])/((E^(3 + 2*x)*x - x^2)*Log[x]*Log[(E^(2*E^(2*E^3) - 2*x)*(E^(3 + 2*x) - x))/Log[x]]^2),x]
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.
\(\displaystyle \int \frac {\left (5 x-10 x^2\right ) \log (x)-5 x+5 e^{2 x+3}}{\left (e^{2 x+3} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{2 x+3}-x\right )}{\log (x)}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 \left (-2 x^2 \log (x)-x+e^{2 x+3}+x \log (x)\right )}{\left (e^{2 x+3} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{2 x+3}-x\right )}{\log (x)}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int \frac {-2 \log (x) x^2+\log (x) x-x+e^{2 x+3}}{\left (e^{2 x+3} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{2 x+3}-x\right )}{\log (x)}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 5 \int \left (\frac {2 x-1}{\left (x-e^{2 x+3}\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{2 x+3}-x\right )}{\log (x)}\right )}+\frac {1}{x \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{2 x+3}-x\right )}{\log (x)}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \left (\int \frac {1}{\left (e^{2 x+3}-x\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{2 x+3}-x\right )}{\log (x)}\right )}dx-2 \int \frac {x}{\left (e^{2 x+3}-x\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{2 x+3}-x\right )}{\log (x)}\right )}dx+\int \frac {1}{x \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{2 x+3}-x\right )}{\log (x)}\right )}dx\right )\) |
Int[(5*E^(3 + 2*x) - 5*x + (5*x - 10*x^2)*Log[x])/((E^(3 + 2*x)*x - x^2)*L og[x]*Log[(E^(2*E^(2*E^3) - 2*x)*(E^(3 + 2*x) - x))/Log[x]]^2),x]
3.24.89.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 12.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\frac {5}{\ln \left (\frac {\left ({\mathrm e}^{3} {\mathrm e}^{2 x}-x \right ) {\mathrm e}^{2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}} {\mathrm e}^{-2 x}}{\ln \left (x \right )}\right )}\) | \(36\) |
risch | \(\frac {10 i}{\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{3+2 x}+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{3+2 x}+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )}^{2}-\pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )}^{3}+\pi \,\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )^{3}+2 i \ln \left ({\mathrm e}^{3+2 x}-x \right )-2 i \ln \left (\ln \left (x \right )\right )-2 i \ln \left ({\mathrm e}^{2 x}\right )+4 i {\mathrm e}^{2 \,{\mathrm e}^{3}}}\) | \(331\) |
int(((-10*x^2+5*x)*ln(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)-x^2)/ln (x)/ln((exp(3)*exp(2*x)-x)*exp(exp(exp(3))^2)^2/exp(2*x)/ln(x))^2,x,method =_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log \left (-\frac {x e^{\left (-2 \, x + 2 \, e^{\left (2 \, e^{3}\right )}\right )} - e^{\left (2 \, e^{\left (2 \, e^{3}\right )} + 3\right )}}{\log \left (x\right )}\right )} \]
integrate(((-10*x^2+5*x)*log(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)- x^2)/log(x)/log((exp(3)*exp(2*x)-x)*exp(exp(exp(3))^2)^2/exp(2*x)/log(x))^ 2,x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log {\left (\frac {\left (- x + e^{3} e^{2 x}\right ) e^{- 2 x} e^{2 e^{2 e^{3}}}}{\log {\left (x \right )}} \right )}} \]
integrate(((-10*x**2+5*x)*ln(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)- x**2)/ln(x)/ln((exp(3)*exp(2*x)-x)*exp(exp(exp(3))**2)**2/exp(2*x)/ln(x))* *2,x)
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=-\frac {5}{2 \, x - 2 \, e^{\left (2 \, e^{3}\right )} - \log \left (-x + e^{\left (2 \, x + 3\right )}\right ) + \log \left (\log \left (x\right )\right )} \]
integrate(((-10*x^2+5*x)*log(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)- x^2)/log(x)/log((exp(3)*exp(2*x)-x)*exp(exp(exp(3))^2)^2/exp(2*x)/log(x))^ 2,x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log \left (-{\left (x e^{\left (2 \, e^{\left (2 \, e^{3}\right )}\right )} - e^{\left (2 \, x + 2 \, e^{\left (2 \, e^{3}\right )} + 3\right )}\right )} e^{\left (-2 \, x\right )}\right ) - \log \left (\log \left (x\right )\right )} \]
integrate(((-10*x^2+5*x)*log(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)- x^2)/log(x)/log((exp(3)*exp(2*x)-x)*exp(exp(exp(3))^2)^2/exp(2*x)/log(x))^ 2,x, algorithm=\
Time = 13.69 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\ln \left (-\frac {{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}}\,\left (x-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3\right )}{\ln \left (x\right )}\right )} \]