Integrand size = 110, antiderivative size = 29 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \log (x) \log \left (e^e \left (3 e^{4 x}-\frac {2}{5-x}\right ) x\right ) \]
Time = 0.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \log (x) \log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right ) \]
Integrate[((-10 + E^(4*x)*(75 + 270*x - 117*x^2 + 12*x^3))*Log[x] + (-10 + 2*x + E^(4*x)*(75 - 30*x + 3*x^2))*Log[(E^E*(2*x + E^(4*x)*(-15*x + 3*x^2 )))/(-5 + x)])/(-50*x + 10*x^2 + E^(4*x)*(375*x - 150*x^2 + 15*x^3)),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.
| \(\displaystyle \int \frac {\left (e^{4 x} \left (3 x^2-30 x+75\right )+2 x-10\right ) \log \left (\frac {e^e \left (e^{4 x} \left (3 x^2-15 x\right )+2 x\right )}{x-5}\right )+\left (e^{4 x} \left (12 x^3-117 x^2+270 x+75\right )-10\right ) \log (x)}{10 x^2+e^{4 x} \left (15 x^3-150 x^2+375 x\right )-50 x} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-\left (e^{4 x} \left (3 x^2-30 x+75\right )+2 x-10\right ) \log \left (\frac {e^e \left (e^{4 x} \left (3 x^2-15 x\right )+2 x\right )}{x-5}\right )-\left (\left (e^{4 x} \left (12 x^3-117 x^2+270 x+75\right )-10\right ) \log (x)\right )}{5 (5-x) x \left (3 e^{4 x} x-15 e^{4 x}+2\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (10-3 e^{4 x} \left (4 x^3-39 x^2+90 x+25\right )\right ) \log (x)+\left (-2 x-3 e^{4 x} \left (x^2-10 x+25\right )+10\right ) \log \left (-\frac {e^e \left (2 x-3 e^{4 x} \left (5 x-x^2\right )\right )}{5-x}\right )}{(5-x) x \left (3 e^{4 x} x-15 e^{4 x}+2\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \frac {1}{5} \int \frac {\frac {\left (3 e^{4 x} (x-5)^2 (4 x+1)-10\right ) \log (x)}{\left (3 e^{4 x} (x-5)+2\right ) (x-5)}+\log \left (e^e \left (3 e^{4 x}+\frac {2}{x-5}\right ) x\right )}{x}dx\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \frac {1}{5} \int \left (\frac {4 x \log (x)+\log (x)+\log \left (e^e \left (3 e^{4 x}+\frac {2}{x-5}\right ) x\right )}{x}-\frac {2 (4 x-19) \log (x)}{(x-5) \left (3 e^{4 x} x-15 e^{4 x}+2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (8 \int \frac {\int \frac {1}{3 e^{4 x} (x-5)+2}dx}{x}dx+2 \int \frac {\int \frac {1}{\left (3 e^{4 x} (x-5)+2\right ) (x-5)}dx}{x}dx-8 \log (x) \int \frac {1}{3 e^{4 x} x-15 e^{4 x}+2}dx-2 \log (x) \int \frac {1}{(x-5) \left (3 e^{4 x} x-15 e^{4 x}+2\right )}dx+\int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{x-5}\right ) x\right )}{x}dx-4 x+\frac {\log ^2(x)}{2}+4 x \log (x)\right )\) |
Int[((-10 + E^(4*x)*(75 + 270*x - 117*x^2 + 12*x^3))*Log[x] + (-10 + 2*x + E^(4*x)*(75 - 30*x + 3*x^2))*Log[(E^E*(2*x + E^(4*x)*(-15*x + 3*x^2)))/(- 5 + x)])/(-50*x + 10*x^2 + E^(4*x)*(375*x - 150*x^2 + 15*x^3)),x]
3.23.5.3.1 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]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 10.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
| method | result | size |
| parallelrisch | \(\frac {\ln \left (x \right ) \ln \left (\frac {\left (\left (3 x^{2}-15 x \right ) {\mathrm e}^{4 x}+2 x \right ) {\mathrm e}^{{\mathrm e}}}{-5+x}\right )}{5}\) | \(33\) |
| risch | \(\frac {\ln \left (x \right ) \ln \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{5}-\frac {\ln \left (x \right ) \ln \left (-5+x \right )}{5}+\frac {\ln \left (x \right )^{2}}{5}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right ) {\operatorname {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{2}}{10}-\frac {i \pi \ln \left (x \right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{3}}{10}-\frac {i \pi \ln \left (x \right ) {\operatorname {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{3}}{10}-\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}{10}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{2}}{10}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{2}}{10}-\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}{10}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i}{-5+x}\right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{2}}{10}+\frac {\ln \left (3\right ) \ln \left (x \right )}{5}+\frac {{\mathrm e} \ln \left (x \right )}{5}\) | \(405\) |
int((((3*x^2-30*x+75)*exp(4*x)+2*x-10)*ln(((3*x^2-15*x)*exp(4*x)+2*x)*exp( exp(1))/(-5+x))+((12*x^3-117*x^2+270*x+75)*exp(4*x)-10)*ln(x))/((15*x^3-15 0*x^2+375*x)*exp(4*x)+10*x^2-50*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \, \log \left (x\right ) \log \left (\frac {{\left (3 \, {\left (x^{2} - 5 \, x\right )} e^{\left (4 \, x\right )} + 2 \, x\right )} e^{e}}{x - 5}\right ) \]
integrate((((3*x^2-30*x+75)*exp(4*x)+2*x-10)*log(((3*x^2-15*x)*exp(4*x)+2* x)*exp(exp(1))/(-5+x))+((12*x^3-117*x^2+270*x+75)*exp(4*x)-10)*log(x))/((1 5*x^3-150*x^2+375*x)*exp(4*x)+10*x^2-50*x),x, algorithm=\
Time = 0.51 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {\log {\left (x \right )} \log {\left (\frac {\left (2 x + \left (3 x^{2} - 15 x\right ) e^{4 x}\right ) e^{e}}{x - 5} \right )}}{5} \]
integrate((((3*x**2-30*x+75)*exp(4*x)+2*x-10)*ln(((3*x**2-15*x)*exp(4*x)+2 *x)*exp(exp(1))/(-5+x))+((12*x**3-117*x**2+270*x+75)*exp(4*x)-10)*ln(x))/( (15*x**3-150*x**2+375*x)*exp(4*x)+10*x**2-50*x),x)
Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \, e \log \left (x\right ) + \frac {1}{5} \, \log \left (3 \, {\left (x - 5\right )} e^{\left (4 \, x\right )} + 2\right ) \log \left (x\right ) - \frac {1}{5} \, \log \left (x - 5\right ) \log \left (x\right ) + \frac {1}{5} \, \log \left (x\right )^{2} \]
integrate((((3*x^2-30*x+75)*exp(4*x)+2*x-10)*log(((3*x^2-15*x)*exp(4*x)+2* x)*exp(exp(1))/(-5+x))+((12*x^3-117*x^2+270*x+75)*exp(4*x)-10)*log(x))/((1 5*x^3-150*x^2+375*x)*exp(4*x)+10*x^2-50*x),x, algorithm=\
Time = 0.37 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \, e \log \left (x\right ) + \frac {1}{5} \, \log \left (3 \, x e^{\left (4 \, x\right )} - 15 \, e^{\left (4 \, x\right )} + 2\right ) \log \left (x\right ) - \frac {1}{5} \, \log \left (x - 5\right ) \log \left (x\right ) + \frac {1}{5} \, \log \left (x\right )^{2} \]
integrate((((3*x^2-30*x+75)*exp(4*x)+2*x-10)*log(((3*x^2-15*x)*exp(4*x)+2* x)*exp(exp(1))/(-5+x))+((12*x^3-117*x^2+270*x+75)*exp(4*x)-10)*log(x))/((1 5*x^3-150*x^2+375*x)*exp(4*x)+10*x^2-50*x),x, algorithm=\
1/5*e*log(x) + 1/5*log(3*x*e^(4*x) - 15*e^(4*x) + 2)*log(x) - 1/5*log(x - 5)*log(x) + 1/5*log(x)^2
Time = 12.57 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {\ln \left (x\right )\,\left (\ln \left (\frac {2\,x-{\mathrm {e}}^{4\,x}\,\left (15\,x-3\,x^2\right )}{x-5}\right )+\mathrm {e}\right )}{5} \]