3.80.55 \(\int \frac {35-16 x+x^2+(-5+2 x) \log (\frac {1}{100} e^{-x} x^3)}{(20 x-4 x^2+(-5 x+x^2) \log (\frac {1}{100} e^{-x} x^3)) \log (\frac {5 x-x^2}{-4+\log (\frac {1}{100} e^{-x} x^3)})} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\log \left (\frac {(-5+x) x}{4-\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )\right ) \]

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Rubi [F]  time = 5.00, 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 {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(35 - 16*x + x^2 + (-5 + 2*x)*Log[x^3/(100*E^x)])/((20*x - 4*x^2 + (-5*x + x^2)*Log[x^3/(100*E^x)])*Log[(5
*x - x^2)/(-4 + Log[x^3/(100*E^x)])]),x]

[Out]

Defer[Int][1/((-4 + Log[x^3/(100*E^x)])*Log[-(((-5 + x)*x)/(-4 + Log[x^3/(100*E^x)]))]), x] - 4*Defer[Int][1/(
(-5 + x)*(-4 + Log[x^3/(100*E^x)])*Log[-(((-5 + x)*x)/(-4 + Log[x^3/(100*E^x)]))]), x] - 7*Defer[Int][1/(x*(-4
+ Log[x^3/(100*E^x)])*Log[-(((-5 + x)*x)/(-4 + Log[x^3/(100*E^x)]))]), x] + Defer[Int][Log[x^3/(100*E^x)]/((-
5 + x)*(-4 + Log[x^3/(100*E^x)])*Log[-(((-5 + x)*x)/(-4 + Log[x^3/(100*E^x)]))]), x] + Defer[Int][Log[x^3/(100
*E^x)]/(x*(-4 + Log[x^3/(100*E^x)])*Log[-(((-5 + x)*x)/(-4 + Log[x^3/(100*E^x)]))]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{(5-x) x \left (4-\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {(5-x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx\\ &=\int \left (\frac {-35+16 x-x^2+5 \log \left (\frac {1}{100} e^{-x} x^3\right )-2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{5 x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {35-16 x+x^2-5 \log \left (\frac {1}{100} e^{-x} x^3\right )+2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{5 (-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-35+16 x-x^2+5 \log \left (\frac {1}{100} e^{-x} x^3\right )-2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\frac {1}{5} \int \frac {35-16 x+x^2-5 \log \left (\frac {1}{100} e^{-x} x^3\right )+2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx\\ &=\frac {1}{5} \int \left (\frac {16}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {35}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {x}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {2 \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {5 \log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx+\frac {1}{5} \int \left (\frac {35}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {16 x}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {x^2}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {5 \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {x}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx\right )+\frac {1}{5} \int \frac {x^2}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\frac {2}{5} \int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\frac {2}{5} \int \frac {x \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\frac {16}{5} \int \frac {1}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\frac {16}{5} \int \frac {x}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+7 \int \frac {1}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-7 \int \frac {1}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 25, normalized size = 0.96 \begin {gather*} \log \left (\log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(35 - 16*x + x^2 + (-5 + 2*x)*Log[x^3/(100*E^x)])/((20*x - 4*x^2 + (-5*x + x^2)*Log[x^3/(100*E^x)])*
Log[(5*x - x^2)/(-4 + Log[x^3/(100*E^x)])]),x]

[Out]

Log[Log[-(((-5 + x)*x)/(-4 + Log[x^3/(100*E^x)]))]]

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fricas [A]  time = 1.32, size = 25, normalized size = 0.96 \begin {gather*} \log \left (\log \left (-\frac {x^{2} - 5 \, x}{\log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-5)*log(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*log(1/100*x^3/exp(x))-4*x^2+20*x)/log((-x^2+5
*x)/(log(1/100*x^3/exp(x))-4)),x, algorithm="fricas")

[Out]

log(log(-(x^2 - 5*x)/(log(1/100*x^3*e^(-x)) - 4)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {x^{2} + {\left (2 \, x - 5\right )} \log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 16 \, x + 35}{{\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} \log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 20 \, x\right )} \log \left (-\frac {x^{2} - 5 \, x}{\log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 4}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-5)*log(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*log(1/100*x^3/exp(x))-4*x^2+20*x)/log((-x^2+5
*x)/(log(1/100*x^3/exp(x))-4)),x, algorithm="giac")

[Out]

integrate(-(x^2 + (2*x - 5)*log(1/100*x^3*e^(-x)) - 16*x + 35)/((4*x^2 - (x^2 - 5*x)*log(1/100*x^3*e^(-x)) - 2
0*x)*log(-(x^2 - 5*x)/(log(1/100*x^3*e^(-x)) - 4))), x)

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[\int \frac {\left (2 x -5\right ) \ln \left (\frac {x^{3} {\mathrm e}^{-x}}{100}\right )+x^{2}-16 x +35}{\left (\left (x^{2}-5 x \right ) \ln \left (\frac {x^{3} {\mathrm e}^{-x}}{100}\right )-4 x^{2}+20 x \right ) \ln \left (\frac {-x^{2}+5 x}{\ln \left (\frac {x^{3} {\mathrm e}^{-x}}{100}\right )-4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x-5)*ln(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*ln(1/100*x^3/exp(x))-4*x^2+20*x)/ln((-x^2+5*x)/(ln(1
/100*x^3/exp(x))-4)),x)

[Out]

int(((2*x-5)*ln(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*ln(1/100*x^3/exp(x))-4*x^2+20*x)/ln((-x^2+5*x)/(ln(1
/100*x^3/exp(x))-4)),x)

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maxima [A]  time = 0.53, size = 28, normalized size = 1.08 \begin {gather*} \log \left (\log \left (x + 2 \, \log \relax (5) + 2 \, \log \relax (2) - 3 \, \log \relax (x) + 4\right ) - \log \left (x - 5\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-5)*log(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*log(1/100*x^3/exp(x))-4*x^2+20*x)/log((-x^2+5
*x)/(log(1/100*x^3/exp(x))-4)),x, algorithm="maxima")

[Out]

log(log(x + 2*log(5) + 2*log(2) - 3*log(x) + 4) - log(x - 5) - log(x))

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mupad [B]  time = 8.63, size = 28, normalized size = 1.08 \begin {gather*} \ln \left (\ln \left (-\frac {5\,x-x^2}{x-\ln \left (x^3\right )+2\,\ln \left (10\right )+4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((x^3*exp(-x))/100)*(2*x - 5) - 16*x + x^2 + 35)/(log((5*x - x^2)/(log((x^3*exp(-x))/100) - 4))*(log(
(x^3*exp(-x))/100)*(5*x - x^2) - 20*x + 4*x^2)),x)

[Out]

log(log(-(5*x - x^2)/(x - log(x^3) + 2*log(10) + 4)))

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sympy [A]  time = 0.97, size = 20, normalized size = 0.77 \begin {gather*} \log {\left (\log {\left (\frac {- x^{2} + 5 x}{\log {\left (\frac {x^{3} e^{- x}}{100} \right )} - 4} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-5)*ln(1/100*x**3/exp(x))+x**2-16*x+35)/((x**2-5*x)*ln(1/100*x**3/exp(x))-4*x**2+20*x)/ln((-x**
2+5*x)/(ln(1/100*x**3/exp(x))-4)),x)

[Out]

log(log((-x**2 + 5*x)/(log(x**3*exp(-x)/100) - 4)))

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