Integrand size = 65, antiderivative size = 28 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{\left (\frac {3}{5}+x\right ) \left (-e^{e^3}-4 (1-x)+x-\log (x)\right )} \]
Time = 1.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{-\frac {1}{5} \left (4+e^{e^3}-5 x\right ) (3+5 x)} x^{-\frac {3}{5}-x} \]
Integrate[(E^((-12 + E^E^3*(-3 - 5*x) - 5*x + 25*x^2 + (-3 - 5*x)*Log[x])/ 5)*(-3 - 10*x - 5*E^E^3*x + 50*x^2 - 5*x*Log[x]))/(5*x),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.
\(\displaystyle \int \frac {\left (50 x^2-5 e^{e^3} x-10 x-5 x \log (x)-3\right ) \exp \left (\frac {1}{5} \left (25 x^2-5 x+e^{e^3} (-5 x-3)+(-5 x-3) \log (x)-12\right )\right )}{5 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (50 x^2+\left (-10-5 e^{e^3}\right ) x-5 x \log (x)-3\right ) \exp \left (\frac {1}{5} \left (25 x^2-5 x+e^{e^3} (-5 x-3)+(-5 x-3) \log (x)-12\right )\right )}{5 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -e^{\frac {1}{5} \left (25 x^2-5 x-e^{e^3} (5 x+3)-12\right )} x^{\frac {1}{5} (-5 x-3)-1} \left (-50 x^2+5 \log (x) x+5 \left (2+e^{e^3}\right ) x+3\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int e^{\frac {1}{5} \left (25 x^2-5 x-e^{e^3} (5 x+3)-12\right )} x^{\frac {1}{5} (-5 x-3)-1} \left (-50 x^2+5 \log (x) x+5 \left (2+e^{e^3}\right ) x+3\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {1}{5} \int \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) x^{-x-\frac {8}{5}} \left (-50 x^2+5 \log (x) x+5 \left (2+e^{e^3}\right ) x+3\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (3 \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) x^{-x-\frac {8}{5}}+5 \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) \log (x) x^{-x-\frac {3}{5}}+5 \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) \left (2+e^{e^3}\right ) x^{-x-\frac {3}{5}}-50 \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) x^{\frac {2}{5}-x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-3 \int \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) x^{-x-\frac {8}{5}}dx-5 \left (2+e^{e^3}\right ) \int \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) x^{-x-\frac {3}{5}}dx+50 \int \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) x^{\frac {2}{5}-x}dx-5 \log (x) \int \exp \left (\frac {1}{5} \left (25 x^2-5 \left (1+e^{e^3}\right ) x-3 \left (4+e^{e^3}\right )\right )\right ) x^{-x-\frac {3}{5}}dx+5 \int \frac {\int e^{-\frac {1}{5} \left (-5 x+e^{e^3}+4\right ) (5 x+3)} x^{-x-\frac {3}{5}}dx}{x}dx\right )\) |
Int[(E^((-12 + E^E^3*(-3 - 5*x) - 5*x + 25*x^2 + (-3 - 5*x)*Log[x])/5)*(-3 - 10*x - 5*E^E^3*x + 50*x^2 - 5*x*Log[x]))/(5*x),x]
3.18.41.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
risch | \(x^{-\frac {3}{5}-x} {\mathrm e}^{\frac {\left (5 x +3\right ) \left (5 x -{\mathrm e}^{{\mathrm e}^{3}}-4\right )}{5}}\) | \(27\) |
norman | \({\mathrm e}^{\frac {\left (-5 x -3\right ) \ln \left (x \right )}{5}+\frac {\left (-5 x -3\right ) {\mathrm e}^{{\mathrm e}^{3}}}{5}+5 x^{2}-x -\frac {12}{5}}\) | \(31\) |
parallelrisch | \({\mathrm e}^{\frac {\left (-5 x -3\right ) \ln \left (x \right )}{5}+\frac {\left (-5 x -3\right ) {\mathrm e}^{{\mathrm e}^{3}}}{5}+5 x^{2}-x -\frac {12}{5}}\) | \(31\) |
int(1/5*(-5*x*ln(x)-5*x*exp(exp(3))+50*x^2-10*x-3)*exp(1/5*(-5*x-3)*ln(x)+ 1/5*(-5*x-3)*exp(exp(3))+5*x^2-x-12/5)/x,x,method=_RETURNVERBOSE)
Time = 0.44 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{\left (5 \, x^{2} - \frac {1}{5} \, {\left (5 \, x + 3\right )} e^{\left (e^{3}\right )} - \frac {1}{5} \, {\left (5 \, x + 3\right )} \log \left (x\right ) - x - \frac {12}{5}\right )} \]
integrate(1/5*(-5*x*log(x)-5*x*exp(exp(3))+50*x^2-10*x-3)*exp(1/5*(-5*x-3) *log(x)+1/5*(-5*x-3)*exp(exp(3))+5*x^2-x-12/5)/x,x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{5 x^{2} - x + \left (- x - \frac {3}{5}\right ) \log {\left (x \right )} + \left (- x - \frac {3}{5}\right ) e^{e^{3}} - \frac {12}{5}} \]
integrate(1/5*(-5*x*ln(x)-5*x*exp(exp(3))+50*x**2-10*x-3)*exp(1/5*(-5*x-3) *ln(x)+1/5*(-5*x-3)*exp(exp(3))+5*x**2-x-12/5)/x,x)
\[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=\int { \frac {{\left (50 \, x^{2} - 5 \, x e^{\left (e^{3}\right )} - 5 \, x \log \left (x\right ) - 10 \, x - 3\right )} e^{\left (5 \, x^{2} - \frac {1}{5} \, {\left (5 \, x + 3\right )} e^{\left (e^{3}\right )} - \frac {1}{5} \, {\left (5 \, x + 3\right )} \log \left (x\right ) - x - \frac {12}{5}\right )}}{5 \, x} \,d x } \]
integrate(1/5*(-5*x*log(x)-5*x*exp(exp(3))+50*x^2-10*x-3)*exp(1/5*(-5*x-3) *log(x)+1/5*(-5*x-3)*exp(exp(3))+5*x^2-x-12/5)/x,x, algorithm=\
1/5*integrate((50*x^2 - 5*x*e^(e^3) - 5*x*log(x) - 10*x - 3)*e^(5*x^2 - 1/ 5*(5*x + 3)*e^(e^3) - 1/5*(5*x + 3)*log(x) - x - 12/5)/x, x)
Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{\left (5 \, x^{2} - x e^{\left (e^{3}\right )} - x \log \left (x\right ) - x - \frac {3}{5} \, e^{\left (e^{3}\right )} - \frac {3}{5} \, \log \left (x\right ) - \frac {12}{5}\right )} \]
integrate(1/5*(-5*x*log(x)-5*x*exp(exp(3))+50*x^2-10*x-3)*exp(1/5*(-5*x-3) *log(x)+1/5*(-5*x-3)*exp(exp(3))+5*x^2-x-12/5)/x,x, algorithm=\
Time = 12.73 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=\frac {{\mathrm {e}}^{-x\,{\mathrm {e}}^{{\mathrm {e}}^3}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-\frac {12}{5}}\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^{{\mathrm {e}}^3}}{5}}\,{\mathrm {e}}^{5\,x^2}}{x^{x+\frac {3}{5}}} \]