Integrand size = 126, antiderivative size = 28 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=2+\log \left (e^{-\frac {x}{4 \left (9+\frac {-5+e^5+e^x}{x}\right )}} x\right ) \]
Time = 3.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=\frac {1}{4} \left (-\frac {x^2}{-5+e^5+e^x+9 x}+4 \log (x)\right ) \]
Integrate[(100 + 4*E^10 + 4*E^(2*x) - 360*x + 334*x^2 - 9*x^3 + E^5*(-40 + 72*x - 2*x^2) + E^x*(-40 + 8*E^5 + 72*x - 2*x^2 + x^3))/(100*x + 4*E^10*x + 4*E^(2*x)*x - 360*x^2 + 324*x^3 + E^5*(-40*x + 72*x^2) + E^x*(-40*x + 8 *E^5*x + 72*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.
\(\displaystyle \int \frac {-9 x^3+334 x^2+e^5 \left (-2 x^2+72 x-40\right )+e^x \left (x^3-2 x^2+72 x+8 e^5-40\right )-360 x+4 e^{2 x}+4 e^{10}+100}{324 x^3-360 x^2+e^5 \left (72 x^2-40 x\right )+e^x \left (72 x^2+8 e^5 x-40 x\right )+4 e^{2 x} x+4 e^{10} x+100 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-9 x^3+334 x^2+e^5 \left (-2 x^2+72 x-40\right )+e^x \left (x^3-2 x^2+72 x+8 e^5-40\right )-360 x+4 e^{2 x}+4 e^{10}+100}{324 x^3-360 x^2+e^5 \left (72 x^2-40 x\right )+e^x \left (72 x^2+8 e^5 x-40 x\right )+4 e^{2 x} x+\left (100+4 e^{10}\right ) x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-9 x^3+334 x^2+e^5 \left (-2 x^2+72 x-40\right )+e^x \left (x^3-2 x^2+72 x+8 e^5-40\right )-360 x+4 e^{2 x}+100 \left (1+\frac {e^{10}}{25}\right )}{4 x \left (9 x+e^x-5 \left (1-\frac {e^5}{5}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {-9 x^3+334 x^2-360 x+4 e^{2 x}-2 e^5 \left (x^2-36 x+20\right )-e^x \left (-x^3+2 x^2-72 x+8 \left (5-e^5\right )\right )+4 \left (25+e^{10}\right )}{\left (-9 x-e^x-e^5+5\right )^2 x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{4} \int \frac {-9 x^3+334 x^2-360 x+4 e^{2 x}-2 e^5 \left (x^2-36 x+20\right )-e^x \left (-x^3+2 x^2-72 x+8 \left (5-e^5\right )\right )+4 \left (25+e^{10}\right )}{x \left (9 x+e^x-5 \left (1-\frac {e^5}{5}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (\frac {\left (-9 x-e^5+14\right ) x^2}{\left (9 x+e^x-5 \left (1-\frac {e^5}{5}\right )\right )^2}+\frac {(x-2) x}{9 x+e^x-5 \left (1-\frac {e^5}{5}\right )}+\frac {4}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-9 \int \frac {x^3}{\left (9 x+e^x-5 \left (1-\frac {e^5}{5}\right )\right )^2}dx+\left (14-e^5\right ) \int \frac {x^2}{\left (9 x+e^x-5 \left (1-\frac {e^5}{5}\right )\right )^2}dx+\int \frac {x^2}{9 x+e^x-5 \left (1-\frac {e^5}{5}\right )}dx+2 \int \frac {x}{-9 x-e^x+5 \left (1-\frac {e^5}{5}\right )}dx+4 \log (x)\right )\) |
Int[(100 + 4*E^10 + 4*E^(2*x) - 360*x + 334*x^2 - 9*x^3 + E^5*(-40 + 72*x - 2*x^2) + E^x*(-40 + 8*E^5 + 72*x - 2*x^2 + x^3))/(100*x + 4*E^10*x + 4*E ^(2*x)*x - 360*x^2 + 324*x^3 + E^5*(-40*x + 72*x^2) + E^x*(-40*x + 8*E^5*x + 72*x^2)),x]
3.30.56.3.1 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 = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71
method | result | size |
norman | \(-\frac {x^{2}}{4 \left (9 x -5+{\mathrm e}^{5}+{\mathrm e}^{x}\right )}+\ln \left (x \right )\) | \(20\) |
risch | \(-\frac {x^{2}}{4 \left (9 x -5+{\mathrm e}^{5}+{\mathrm e}^{x}\right )}+\ln \left (x \right )\) | \(20\) |
parallelrisch | \(\frac {4 \,{\mathrm e}^{5} \ln \left (x \right )+4 \,{\mathrm e}^{x} \ln \left (x \right )+36 x \ln \left (x \right )-x^{2}-20 \ln \left (x \right )}{36 x -20+4 \,{\mathrm e}^{5}+4 \,{\mathrm e}^{x}}\) | \(41\) |
int((4*exp(x)^2+(8*exp(5)+x^3-2*x^2+72*x-40)*exp(x)+4*exp(5)^2+(-2*x^2+72* x-40)*exp(5)-9*x^3+334*x^2-360*x+100)/(4*x*exp(x)^2+(8*x*exp(5)+72*x^2-40* x)*exp(x)+4*x*exp(5)^2+(72*x^2-40*x)*exp(5)+324*x^3-360*x^2+100*x),x,metho d=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=-\frac {x^{2} - 4 \, {\left (9 \, x + e^{5} + e^{x} - 5\right )} \log \left (x\right )}{4 \, {\left (9 \, x + e^{5} + e^{x} - 5\right )}} \]
integrate((4*exp(x)^2+(8*exp(5)+x^3-2*x^2+72*x-40)*exp(x)+4*exp(5)^2+(-2*x ^2+72*x-40)*exp(5)-9*x^3+334*x^2-360*x+100)/(4*x*exp(x)^2+(8*x*exp(5)+72*x ^2-40*x)*exp(x)+4*x*exp(5)^2+(72*x^2-40*x)*exp(5)+324*x^3-360*x^2+100*x),x , algorithm=\
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=- \frac {x^{2}}{36 x + 4 e^{x} - 20 + 4 e^{5}} + \log {\left (x \right )} \]
integrate((4*exp(x)**2+(8*exp(5)+x**3-2*x**2+72*x-40)*exp(x)+4*exp(5)**2+( -2*x**2+72*x-40)*exp(5)-9*x**3+334*x**2-360*x+100)/(4*x*exp(x)**2+(8*x*exp (5)+72*x**2-40*x)*exp(x)+4*x*exp(5)**2+(72*x**2-40*x)*exp(5)+324*x**3-360* x**2+100*x),x)
Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=-\frac {x^{2}}{4 \, {\left (9 \, x + e^{5} + e^{x} - 5\right )}} + \log \left (x\right ) \]
integrate((4*exp(x)^2+(8*exp(5)+x^3-2*x^2+72*x-40)*exp(x)+4*exp(5)^2+(-2*x ^2+72*x-40)*exp(5)-9*x^3+334*x^2-360*x+100)/(4*x*exp(x)^2+(8*x*exp(5)+72*x ^2-40*x)*exp(x)+4*x*exp(5)^2+(72*x^2-40*x)*exp(5)+324*x^3-360*x^2+100*x),x , algorithm=\
Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=-\frac {x^{2} - 36 \, x \log \left (x\right ) - 4 \, e^{5} \log \left (x\right ) - 4 \, e^{x} \log \left (x\right ) + 20 \, \log \left (x\right )}{4 \, {\left (9 \, x + e^{5} + e^{x} - 5\right )}} \]
integrate((4*exp(x)^2+(8*exp(5)+x^3-2*x^2+72*x-40)*exp(x)+4*exp(5)^2+(-2*x ^2+72*x-40)*exp(5)-9*x^3+334*x^2-360*x+100)/(4*x*exp(x)^2+(8*x*exp(5)+72*x ^2-40*x)*exp(x)+4*x*exp(5)^2+(72*x^2-40*x)*exp(5)+324*x^3-360*x^2+100*x),x , algorithm=\
Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=\ln \left (x\right )-\frac {x^2\,{\mathrm {e}}^5-14\,x^2+9\,x^3}{4\,\left (9\,x+{\mathrm {e}}^5-14\right )\,\left (9\,x+{\mathrm {e}}^5+{\mathrm {e}}^x-5\right )} \]