\(\int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 (-40+72 x-2 x^2)+e^x (-40+8 e^5+72 x-2 x^2+x^3)}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 (-40 x+72 x^2)+e^x (-40 x+8 e^5 x+72 x^2)} \, dx\) [8956]

Optimal result

Integrand size = 126, antiderivative size = 28 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=2+\log \left (e^{-\frac {x}{4 \left (9+\frac {-5+e^5+e^x}{x}\right )}} x\right ) \]

[Out]

ln(x/exp(1/4*x/((exp(5)-5+exp(x))/x+9)))+2

Rubi [F]

\[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=\int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx \]

[In]

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]

[Out]

Log[x] + Defer[Int][x/(-E^x + 5*(1 - E^5/5) - 9*x), x]/2 + ((14 - E^5)*Defer[Int][x^2/(E^x - 5*(1 - E^5/5) + 9
*x)^2, x])/4 - (9*Defer[Int][x^3/(E^x - 5*(1 - E^5/5) + 9*x)^2, x])/4 + Defer[Int][x^2/(E^x - 5*(1 - E^5/5) +
9*x), x]/4

Rubi steps \begin{align*} \text {integral}& = \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 )}{4 e^{2 x} x+\left (100+4 e^{10}\right ) 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 \\ & = \int \frac {4 e^{2 x}+100 \left (1+\frac {e^{10}}{25}\right )-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 )}{4 x \left (e^x-5 \left (1-\frac {e^5}{5}\right )+9 x\right )^2} \, dx \\ & = \frac {1}{4} \int \frac {4 e^{2 x}+100 \left (1+\frac {e^{10}}{25}\right )-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 )}{x \left (e^x-5 \left (1-\frac {e^5}{5}\right )+9 x\right )^2} \, dx \\ & = \frac {1}{4} \int \left (\frac {4}{x}+\frac {\left (14-e^5-9 x\right ) x^2}{\left (e^x-5 \left (1-\frac {e^5}{5}\right )+9 x\right )^2}+\frac {(-2+x) x}{e^x-5 \left (1-\frac {e^5}{5}\right )+9 x}\right ) \, dx \\ & = \log (x)+\frac {1}{4} \int \frac {\left (14-e^5-9 x\right ) x^2}{\left (e^x-5 \left (1-\frac {e^5}{5}\right )+9 x\right )^2} \, dx+\frac {1}{4} \int \frac {(-2+x) x}{e^x-5 \left (1-\frac {e^5}{5}\right )+9 x} \, dx \\ & = \log (x)+\frac {1}{4} \int \left (\frac {\left (14-e^5\right ) x^2}{\left (e^x-5 \left (1-\frac {e^5}{5}\right )+9 x\right )^2}-\frac {9 x^3}{\left (e^x-5 \left (1-\frac {e^5}{5}\right )+9 x\right )^2}\right ) \, dx+\frac {1}{4} \int \left (\frac {2 x}{-e^x+5 \left (1-\frac {e^5}{5}\right )-9 x}+\frac {x^2}{e^x-5 \left (1-\frac {e^5}{5}\right )+9 x}\right ) \, dx \\ & = \log (x)+\frac {1}{4} \int \frac {x^2}{e^x-5 \left (1-\frac {e^5}{5}\right )+9 x} \, dx+\frac {1}{2} \int \frac {x}{-e^x+5 \left (1-\frac {e^5}{5}\right )-9 x} \, dx-\frac {9}{4} \int \frac {x^3}{\left (e^x-5 \left (1-\frac {e^5}{5}\right )+9 x\right )^2} \, dx+\frac {1}{4} \left (14-e^5\right ) \int \frac {x^2}{\left (e^x-5 \left (1-\frac {e^5}{5}\right )+9 x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=\frac {1}{4} \left (-\frac {x^2}{-5+e^5+e^x+9 x}+4 \log (x)\right ) \]

[In]

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]

[Out]

(-(x^2/(-5 + E^5 + E^x + 9*x)) + 4*Log[x])/4

Maple [A] (verified)

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	\(\frac {4 \,{\mathrm e}^{5} \ln \left (x \right )+4 \,{\mathrm e}^{x} \ln \left (x \right )+36 x \ln \left (x \right )-x^{2}-20 \ln \left (x \right )}{36 x -20+4 \,{\mathrm e}^{5}+4 \,{\mathrm e}^{x}}\)	\(41\)

[In]

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,met
hod=_RETURNVERBOSE)

[Out]

-1/4*x^2/(9*x-5+exp(5)+exp(x))+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=-\frac {x^{2} - 4 \, {\left (9 \, x + e^{5} + e^{x} - 5\right )} \log \left (x\right )}{4 \, {\left (9 \, x + e^{5} + e^{x} - 5\right )}} \]

[In]

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="fricas")

[Out]

-1/4*(x^2 - 4*(9*x + e^5 + e^x - 5)*log(x))/(9*x + e^5 + e^x - 5)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=- \frac {x^{2}}{36 x + 4 e^{x} - 20 + 4 e^{5}} + \log {\left (x \right )} \]

[In]

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**3-36
0*x**2+100*x),x)

[Out]

-x**2/(36*x + 4*exp(x) - 20 + 4*exp(5)) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=-\frac {x^{2}}{4 \, {\left (9 \, x + e^{5} + e^{x} - 5\right )}} + \log \left (x\right ) \]

[In]

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="maxima")

[Out]

-1/4*x^2/(9*x + e^5 + e^x - 5) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=-\frac {x^{2} - 36 \, x \log \left (x\right ) - 4 \, e^{5} \log \left (x\right ) - 4 \, e^{x} \log \left (x\right ) + 20 \, \log \left (x\right )}{4 \, {\left (9 \, x + e^{5} + e^{x} - 5\right )}} \]

[In]

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="giac")

[Out]

-1/4*(x^2 - 36*x*log(x) - 4*e^5*log(x) - 4*e^x*log(x) + 20*log(x))/(9*x + e^5 + e^x - 5)

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {100+4 e^{10}+4 e^{2 x}-360 x+334 x^2-9 x^3+e^5 \left (-40+72 x-2 x^2\right )+e^x \left (-40+8 e^5+72 x-2 x^2+x^3\right )}{100 x+4 e^{10} x+4 e^{2 x} x-360 x^2+324 x^3+e^5 \left (-40 x+72 x^2\right )+e^x \left (-40 x+8 e^5 x+72 x^2\right )} \, dx=\ln \left (x\right )-\frac {x^2\,{\mathrm {e}}^5-14\,x^2+9\,x^3}{4\,\left (9\,x+{\mathrm {e}}^5-14\right )\,\left (9\,x+{\mathrm {e}}^5+{\mathrm {e}}^x-5\right )} \]

[In]

int((4*exp(2*x) - 360*x + 4*exp(10) - exp(5)*(2*x^2 - 72*x + 40) + exp(x)*(72*x + 8*exp(5) - 2*x^2 + x^3 - 40)
+ 334*x^2 - 9*x^3 + 100)/(100*x + 4*x*exp(2*x) - exp(5)*(40*x - 72*x^2) + 4*x*exp(10) - 360*x^2 + 324*x^3 + e
xp(x)*(8*x*exp(5) - 40*x + 72*x^2)),x)

[Out]

log(x) - (x^2*exp(5) - 14*x^2 + 9*x^3)/(4*(9*x + exp(5) - 14)*(9*x + exp(5) + exp(x) - 5))