Integrand size = 94, antiderivative size = 26 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 (x-\log (4)) \log \left (\frac {15 \left (-5-e^{x/5}+x\right )}{x}\right ) \]
\[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=\int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx \]
Integrate[(-1800*x + 1800*Log[4] + E^(x/5)*(-360*x + 72*x^2 + (360 - 72*x) *Log[4]) + (1800*x + 360*E^(x/5)*x - 360*x^2)*Log[(-75 - 15*E^(x/5) + 15*x )/x])/(25*x + 5*E^(x/5)*x - 5*x^2),x]
Integrate[(-1800*x + 1800*Log[4] + E^(x/5)*(-360*x + 72*x^2 + (360 - 72*x) *Log[4]) + (1800*x + 360*E^(x/5)*x - 360*x^2)*Log[(-75 - 15*E^(x/5) + 15*x )/x])/(25*x + 5*E^(x/5)*x - 5*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.
\(\displaystyle \int \frac {e^{x/5} \left (72 x^2-360 x+(360-72 x) \log (4)\right )+\left (-360 x^2+360 e^{x/5} x+1800 x\right ) \log \left (\frac {15 x-15 e^{x/5}-75}{x}\right )-1800 x+1800 \log (4)}{-5 x^2+5 e^{x/5} x+25 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x/5} \left (72 x^2-360 x+(360-72 x) \log (4)\right )+\left (-360 x^2+360 e^{x/5} x+1800 x\right ) \log \left (\frac {15 x-15 e^{x/5}-75}{x}\right )-1800 x+1800 \log (4)}{5 \left (-x+e^{x/5}+5\right ) x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {72 \left (25 x+e^{x/5} \left (-x^2+5 x-(5-x) \log (4)\right )-5 \left (-x^2+e^{x/5} x+5 x\right ) \log \left (-\frac {15 \left (-x+e^{x/5}+5\right )}{x}\right )-25 \log (4)\right )}{\left (-x+e^{x/5}+5\right ) x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {72}{5} \int \frac {25 x+e^{x/5} \left (-x^2+5 x-(5-x) \log (4)\right )-5 \left (-x^2+e^{x/5} x+5 x\right ) \log \left (-\frac {15 \left (-x+e^{x/5}+5\right )}{x}\right )-25 \log (4)}{\left (-x+e^{x/5}+5\right ) x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {72}{5} \int \left (\frac {-x^2-5 \log \left (-\frac {15 \left (-x+e^{x/5}+5\right )}{x}\right ) x+5 \left (1+\frac {2 \log (2)}{5}\right ) x-5 \log (4)}{x}-\frac {(x-10) (x-\log (4))}{-x+e^{x/5}+5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {72}{5} \left (125 \text {Subst}\left (\int \frac {x^2}{-5 x+e^x+5}dx,x,\frac {x}{5}\right )-250 \text {Subst}\left (\int \frac {x}{-5 x+e^x+5}dx,x,\frac {x}{5}\right )-50 \log (4) \text {Subst}\left (\int \frac {1}{-5 x+e^x+5}dx,x,\frac {x}{5}\right )-\int \frac {x^2}{-x+e^{x/5}+5}dx+(10+\log (4)) \int \frac {x}{-x+e^{x/5}+5}dx-\frac {x^2}{2}+\frac {1}{2} (5-x)^2-5 x \log \left (-\frac {15 \left (-x+e^{x/5}+5\right )}{x}\right )-5 \log (4) \log (x)+x (5+\log (4))\right )\) |
Int[(-1800*x + 1800*Log[4] + E^(x/5)*(-360*x + 72*x^2 + (360 - 72*x)*Log[4 ]) + (1800*x + 360*E^(x/5)*x - 360*x^2)*Log[(-75 - 15*E^(x/5) + 15*x)/x])/ (25*x + 5*E^(x/5)*x - 5*x^2),x]
3.14.34.3.1 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 = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50
method | result | size |
parallelrisch | \(-144 \ln \left (-\frac {15 \left ({\mathrm e}^{\frac {x}{5}}-x +5\right )}{x}\right ) \ln \left (2\right )+72 \ln \left (-\frac {15 \left ({\mathrm e}^{\frac {x}{5}}-x +5\right )}{x}\right ) x\) | \(39\) |
norman | \(-144 \ln \left (2\right ) \ln \left (\frac {-15 \,{\mathrm e}^{\frac {x}{5}}+15 x -75}{x}\right )+72 x \ln \left (\frac {-15 \,{\mathrm e}^{\frac {x}{5}}+15 x -75}{x}\right )\) | \(41\) |
risch | \(72 x \ln \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )-72 x \ln \left (x \right )-36 i \pi x \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )+36 i \pi x \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )}^{2}+36 i \pi x \,\operatorname {csgn}\left (i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )}^{2}-36 i \pi x {\operatorname {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )}^{3}+72 x \ln \left (5\right )+72 x \ln \left (3\right )+144 \ln \left (2\right ) \ln \left (x \right )-144 \ln \left (2\right ) \ln \left ({\mathrm e}^{\frac {x}{5}}-x +5\right )\) | \(180\) |
int(((360*x*exp(1/5*x)-360*x^2+1800*x)*ln((-15*exp(1/5*x)+15*x-75)/x)+(2*( -72*x+360)*ln(2)+72*x^2-360*x)*exp(1/5*x)+3600*ln(2)-1800*x)/(5*x*exp(1/5* x)-5*x^2+25*x),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 \, {\left (x - 2 \, \log \left (2\right )\right )} \log \left (\frac {15 \, {\left (x - e^{\left (\frac {1}{5} \, x\right )} - 5\right )}}{x}\right ) \]
integrate(((360*x*exp(1/5*x)-360*x^2+1800*x)*log((-15*exp(1/5*x)+15*x-75)/ x)+(2*(-72*x+360)*log(2)+72*x^2-360*x)*exp(1/5*x)+3600*log(2)-1800*x)/(5*x *exp(1/5*x)-5*x^2+25*x),x, algorithm=\
Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 x \log {\left (\frac {15 x - 15 e^{\frac {x}{5}} - 75}{x} \right )} + 144 \log {\left (2 \right )} \log {\left (x \right )} - 144 \log {\left (2 \right )} \log {\left (- x + e^{\frac {x}{5}} + 5 \right )} \]
integrate(((360*x*exp(1/5*x)-360*x**2+1800*x)*ln((-15*exp(1/5*x)+15*x-75)/ x)+(2*(-72*x+360)*ln(2)+72*x**2-360*x)*exp(1/5*x)+3600*ln(2)-1800*x)/(5*x* exp(1/5*x)-5*x**2+25*x),x)
72*x*log((15*x - 15*exp(x/5) - 75)/x) + 144*log(2)*log(x) - 144*log(2)*log (-x + exp(x/5) + 5)
Result contains complex when optimal does not.
Time = 0.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 \, {\left (i \, \pi + \log \left (5\right ) + \log \left (3\right )\right )} x - 72 \, {\left (x - 2 \, \log \left (2\right )\right )} \log \left (x\right ) + 72 \, {\left (x - 2 \, \log \left (2\right )\right )} \log \left (-x + e^{\left (\frac {1}{5} \, x\right )} + 5\right ) \]
integrate(((360*x*exp(1/5*x)-360*x^2+1800*x)*log((-15*exp(1/5*x)+15*x-75)/ x)+(2*(-72*x+360)*log(2)+72*x^2-360*x)*exp(1/5*x)+3600*log(2)-1800*x)/(5*x *exp(1/5*x)-5*x^2+25*x),x, algorithm=\
72*(I*pi + log(5) + log(3))*x - 72*(x - 2*log(2))*log(x) + 72*(x - 2*log(2 ))*log(-x + e^(1/5*x) + 5)
Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 \, x \log \left (15 \, x - 15 \, e^{\left (\frac {1}{5} \, x\right )} - 75\right ) - 72 \, x \log \left (x\right ) + 144 \, \log \left (2\right ) \log \left (x\right ) - 144 \, \log \left (2\right ) \log \left (-x + e^{\left (\frac {1}{5} \, x\right )} + 5\right ) \]
integrate(((360*x*exp(1/5*x)-360*x^2+1800*x)*log((-15*exp(1/5*x)+15*x-75)/ x)+(2*(-72*x+360)*log(2)+72*x^2-360*x)*exp(1/5*x)+3600*log(2)-1800*x)/(5*x *exp(1/5*x)-5*x^2+25*x),x, algorithm=\
72*x*log(15*x - 15*e^(1/5*x) - 75) - 72*x*log(x) + 144*log(2)*log(x) - 144 *log(2)*log(-x + e^(1/5*x) + 5)
Time = 10.84 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72\,x\,\ln \left (-\frac {15\,{\left ({\mathrm {e}}^x\right )}^{1/5}-15\,x+75}{x}\right )-72\,\ln \left (4\right )\,\ln \left ({\left ({\mathrm {e}}^x\right )}^{1/5}-x+5\right )+72\,\ln \left (4\right )\,\ln \left (x\right ) \]
int(-(1800*x - 3600*log(2) + exp(x/5)*(360*x + 2*log(2)*(72*x - 360) - 72* x^2) - log(-(15*exp(x/5) - 15*x + 75)/x)*(1800*x + 360*x*exp(x/5) - 360*x^ 2))/(25*x + 5*x*exp(x/5) - 5*x^2),x)