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

Optimal. Leaf size=26 \[ \frac {1}{\log \left (e^{x (1+x)} \left (\frac {1}{5}+e^x+x+\frac {x^2}{4}\right )\right )} \]

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Rubi [F]  time = 1.33, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 33, normalized size = 1.27 \begin {gather*} -\frac {1}{\log (20)-\log \left (e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.50, size = 25, normalized size = 0.96 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.18, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{x^{2} + x - \log \left (20\right ) + \log \left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 0.13, size = 194, normalized size = 7.46

method	result	size
risch	\(\frac {2 i}{\pi \,\mathrm {csgn}\left (i \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x +1\right ) x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x +1\right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )-\pi \,\mathrm {csgn}\left (i \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x +1\right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{\left (x +1\right ) x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x +1\right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{\left (x +1\right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{3}-4 i \ln \relax (2)+2 i \ln \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )+2 i \ln \left ({\mathrm e}^{\left (x +1\right ) x}\right )}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[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]

2*I/(Pi*csgn(I*(x^2+4*x+4*exp(x)+4/5))*csgn(I*exp((x+1)*x))*csgn(I*exp((x+1)*x)*(x^2+4*x+4*exp(x)+4/5))-Pi*csg
n(I*(x^2+4*x+4*exp(x)+4/5))*csgn(I*exp((x+1)*x)*(x^2+4*x+4*exp(x)+4/5))^2-Pi*csgn(I*exp((x+1)*x))*csgn(I*exp((
x+1)*x)*(x^2+4*x+4*exp(x)+4/5))^2+Pi*csgn(I*exp((x+1)*x)*(x^2+4*x+4*exp(x)+4/5))^3-4*I*ln(2)+2*I*ln(x^2+4*x+4*
exp(x)+4/5)+2*I*ln(exp((x+1)*x)))

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maxima [A]  time = 0.51, size = 30, normalized size = 1.15 \begin {gather*} \frac {1}{x^{2} + x - \log \relax (5) - 2 \, \log \relax (2) + \log \left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 3.52, size = 18, normalized size = 0.69 \begin {gather*} \frac {1}{x+\ln \left (x+{\mathrm {e}}^x+\frac {x^2}{4}+\frac {1}{5}\right )+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.45, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{\log {\left (\left (\frac {x^{2}}{4} + x + e^{x} + \frac {1}{5}\right ) e^{x^{2} + x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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