\(\int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{(4+20 e^x+20 x+5 x^2) \log ^2(\frac {1}{20} e^{x+x^2} (4+20 e^x+20 x+5 x^2))} \, dx\) [4723]

Optimal result

Integrand size = 71, antiderivative size = 26 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log \left (e^{x (1+x)} \left (\frac {1}{5}+e^x+x+\frac {x^2}{4}\right )\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx \]

[In]

Int[(-24 + E^x*(-40 - 40*x) - 38*x - 45*x^2 - 10*x^3)/((4 + 20*E^x + 20*x + 5*x^2)*Log[(E^(x + x^2)*(4 + 20*E^
x + 20*x + 5*x^2))/20]^2),x]

[Out]

-2*Defer[Int][Log[(E^(x + x^2)*(4 + 20*E^x + 20*x + 5*x^2))/20]^(-2), x] - 2*Defer[Int][x/Log[(E^(x + x^2)*(4
+ 20*E^x + 20*x + 5*x^2))/20]^2, x] - 16*Defer[Int][1/((4 + 20*E^x + 20*x + 5*x^2)*Log[(E^(x + x^2)*(4 + 20*E^
x + 20*x + 5*x^2))/20]^2), x] + 10*Defer[Int][x/((4 + 20*E^x + 20*x + 5*x^2)*Log[(E^(x + x^2)*(4 + 20*E^x + 20
*x + 5*x^2))/20]^2), x] + 5*Defer[Int][x^2/((4 + 20*E^x + 20*x + 5*x^2)*Log[(E^(x + x^2)*(4 + 20*E^x + 20*x +
5*x^2))/20]^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 (1+x)}{\log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )}+\frac {-16+10 x+5 x^2}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )}\right ) \, dx \\ & = -\left (2 \int \frac {1+x}{\log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx\right )+\int \frac {-16+10 x+5 x^2}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx \\ & = -\left (2 \int \left (\frac {1}{\log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )}+\frac {x}{\log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )}\right ) \, dx\right )+\int \left (-\frac {16}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )}+\frac {10 x}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )}+\frac {5 x^2}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )}\right ) \, dx \\ & = -\left (2 \int \frac {1}{\log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx\right )-2 \int \frac {x}{\log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx+5 \int \frac {x^2}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx+10 \int \frac {x}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx-16 \int \frac {1}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=-\frac {1}{\log (20)-\log \left (e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \]

[In]

Integrate[(-24 + E^x*(-40 - 40*x) - 38*x - 45*x^2 - 10*x^3)/((4 + 20*E^x + 20*x + 5*x^2)*Log[(E^(x + x^2)*(4 +
20*E^x + 20*x + 5*x^2))/20]^2),x]

[Out]

-(Log[20] - Log[E^(x + x^2)*(4 + 20*E^x + 20*x + 5*x^2)])^(-1)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {1}{\ln \left (\frac {\left (20 \,{\mathrm e}^{x}+5 x^{2}+20 x +4\right ) {\mathrm e}^{x^{2}+x}}{20}\right )}\)	\(26\)
risch	\(-\frac {2}{4 \ln \left (2\right )-2 \ln \left ({\mathrm e}^{\left (1+x \right ) x}\right )-2 \ln \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )+i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )-i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{3}}\)	\(196\)

[In]

int(((-40*x-40)*exp(x)-10*x^3-45*x^2-38*x-24)/(20*exp(x)+5*x^2+20*x+4)/ln(1/20*(20*exp(x)+5*x^2+20*x+4)*exp(x^
2+x))^2,x,method=_RETURNVERBOSE)

[Out]

1/ln(1/20*(20*exp(x)+5*x^2+20*x+4)*exp(x^2+x))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log \left (\frac {1}{20} \, {\left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} e^{\left (x^{2} + x\right )}\right )} \]

[In]

integrate(((-40*x-40)*exp(x)-10*x^3-45*x^2-38*x-24)/(20*exp(x)+5*x^2+20*x+4)/log(1/20*(20*exp(x)+5*x^2+20*x+4)
*exp(x^2+x))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log {\left (\left (\frac {x^{2}}{4} + x + e^{x} + \frac {1}{5}\right ) e^{x^{2} + x} \right )}} \]

[In]

integrate(((-40*x-40)*exp(x)-10*x**3-45*x**2-38*x-24)/(20*exp(x)+5*x**2+20*x+4)/ln(1/20*(20*exp(x)+5*x**2+20*x
+4)*exp(x**2+x))**2,x)

[Out]

1/log((x**2/4 + x + exp(x) + 1/5)*exp(x**2 + x))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x^{2} + x - \log \left (5\right ) - 2 \, \log \left (2\right ) + \log \left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} \]

[In]

integrate(((-40*x-40)*exp(x)-10*x^3-45*x^2-38*x-24)/(20*exp(x)+5*x^2+20*x+4)/log(1/20*(20*exp(x)+5*x^2+20*x+4)
*exp(x^2+x))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x^{2} + x - \log \left (20\right ) + \log \left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} \]

[In]

integrate(((-40*x-40)*exp(x)-10*x^3-45*x^2-38*x-24)/(20*exp(x)+5*x^2+20*x+4)/log(1/20*(20*exp(x)+5*x^2+20*x+4)
*exp(x^2+x))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x+\ln \left (x+{\mathrm {e}}^x+\frac {x^2}{4}+\frac {1}{5}\right )+x^2} \]

[In]

int(-(38*x + exp(x)*(40*x + 40) + 45*x^2 + 10*x^3 + 24)/(log((exp(x + x^2)*(20*x + 20*exp(x) + 5*x^2 + 4))/20)
^2*(20*x + 20*exp(x) + 5*x^2 + 4)),x)

[Out]

1/(x + log(x + exp(x) + x^2/4 + 1/5) + x^2)