3.3.41 \(\int \frac {e^{-x} (-x-x^2-5 x \log ^2(3)+(2 x+2 x^2-x^3+(10 x-5 x^2) \log ^2(3)) \log (x)+(1-x-x^2+(-5-5 x) \log ^2(3)) \log ^2(x))}{\log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ 5 e^{-x} \left (\frac {1+x}{5}+\log ^2(3)\right ) \left (2+x+\frac {x^2}{\log (x)}\right ) \]

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Rubi [F]  time = 0.97, 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 {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-x - x^2 - 5*x*Log[3]^2 + (2*x + 2*x^2 - x^3 + (10*x - 5*x^2)*Log[3]^2)*Log[x] + (1 - x - x^2 + (-5 - 5*x
)*Log[3]^2)*Log[x]^2)/(E^x*Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (-x^2+x \left (-1-5 \log ^2(3)\right )+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=\int \left (-e^{-x} x^2+e^{-x} \left (1-5 \log ^2(3)\right )-e^{-x} x \left (1+5 \log ^2(3)\right )-\frac {e^{-x} x \left (1+x+5 \log ^2(3)\right )}{\log ^2(x)}+\frac {e^{-x} x \left (-x^2+x \left (2-5 \log ^2(3)\right )+2 \left (1+5 \log ^2(3)\right )\right )}{\log (x)}\right ) \, dx\\ &=\left (-1-5 \log ^2(3)\right ) \int e^{-x} x \, dx+\left (1-5 \log ^2(3)\right ) \int e^{-x} \, dx-\int e^{-x} x^2 \, dx-\int \frac {e^{-x} x \left (1+x+5 \log ^2(3)\right )}{\log ^2(x)} \, dx+\int \frac {e^{-x} x \left (-x^2+x \left (2-5 \log ^2(3)\right )+2 \left (1+5 \log ^2(3)\right )\right )}{\log (x)} \, dx\\ &=e^{-x} x^2-e^{-x} \left (1-5 \log ^2(3)\right )+e^{-x} x \left (1+5 \log ^2(3)\right )-2 \int e^{-x} x \, dx+\left (-1-5 \log ^2(3)\right ) \int e^{-x} \, dx-\int \left (\frac {e^{-x} x^2}{\log ^2(x)}+\frac {e^{-x} x \left (1+5 \log ^2(3)\right )}{\log ^2(x)}\right ) \, dx+\int \left (-\frac {e^{-x} x^3}{\log (x)}+\frac {e^{-x} x^2 \left (2-5 \log ^2(3)\right )}{\log (x)}+\frac {2 e^{-x} x \left (1+5 \log ^2(3)\right )}{\log (x)}\right ) \, dx\\ &=2 e^{-x} x+e^{-x} x^2-e^{-x} \left (1-5 \log ^2(3)\right )+e^{-x} \left (1+5 \log ^2(3)\right )+e^{-x} x \left (1+5 \log ^2(3)\right )-2 \int e^{-x} \, dx+\left (2-5 \log ^2(3)\right ) \int \frac {e^{-x} x^2}{\log (x)} \, dx-\left (1+5 \log ^2(3)\right ) \int \frac {e^{-x} x}{\log ^2(x)} \, dx+\left (2 \left (1+5 \log ^2(3)\right )\right ) \int \frac {e^{-x} x}{\log (x)} \, dx-\int \frac {e^{-x} x^2}{\log ^2(x)} \, dx-\int \frac {e^{-x} x^3}{\log (x)} \, dx\\ &=2 e^{-x}+2 e^{-x} x+e^{-x} x^2-e^{-x} \left (1-5 \log ^2(3)\right )+e^{-x} \left (1+5 \log ^2(3)\right )+e^{-x} x \left (1+5 \log ^2(3)\right )+\left (2-5 \log ^2(3)\right ) \int \frac {e^{-x} x^2}{\log (x)} \, dx-\left (1+5 \log ^2(3)\right ) \int \frac {e^{-x} x}{\log ^2(x)} \, dx+\left (2 \left (1+5 \log ^2(3)\right )\right ) \int \frac {e^{-x} x}{\log (x)} \, dx-\int \frac {e^{-x} x^2}{\log ^2(x)} \, dx-\int \frac {e^{-x} x^3}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 29, normalized size = 0.97 \begin {gather*} \frac {e^{-x} \left (1+x+5 \log ^2(3)\right ) \left (x^2+(2+x) \log (x)\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x - x^2 - 5*x*Log[3]^2 + (2*x + 2*x^2 - x^3 + (10*x - 5*x^2)*Log[3]^2)*Log[x] + (1 - x - x^2 + (-5
- 5*x)*Log[3]^2)*Log[x]^2)/(E^x*Log[x]^2),x]

[Out]

((1 + x + 5*Log[3]^2)*(x^2 + (2 + x)*Log[x]))/(E^x*Log[x])

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fricas [B]  time = 0.90, size = 51, normalized size = 1.70 \begin {gather*} \frac {{\left (5 \, {\left (x + 2\right )} \log \relax (3)^{2} + x^{2} + 3 \, x + 2\right )} e^{\left (-x\right )} \log \relax (x) + {\left (5 \, x^{2} \log \relax (3)^{2} + x^{3} + x^{2}\right )} e^{\left (-x\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x-5)*log(3)^2-x^2-x+1)*log(x)^2+((-5*x^2+10*x)*log(3)^2-x^3+2*x^2+2*x)*log(x)-5*x*log(3)^2-x^2
-x)/exp(x)/log(x)^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x-5)*log(3)^2-x^2-x+1)*log(x)^2+((-5*x^2+10*x)*log(3)^2-x^3+2*x^2+2*x)*log(x)-5*x*log(3)^2-x^2
-x)/exp(x)/log(x)^2,x, algorithm="giac")

[Out]

integrate(-(5*x*log(3)^2 + (5*(x + 1)*log(3)^2 + x^2 + x - 1)*log(x)^2 + x^2 + (x^3 + 5*(x^2 - 2*x)*log(3)^2 -
2*x^2 - 2*x)*log(x) + x)*e^(-x)/log(x)^2, x)

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maple [A]  time = 0.08, size = 49, normalized size = 1.63

method	result	size
risch	\(\left (5 x \ln \relax (3)^{2}+10 \ln \relax (3)^{2}+x^{2}+3 x +2\right ) {\mathrm e}^{-x}+\frac {x^{2} {\mathrm e}^{-x} \left (5 \ln \relax (3)^{2}+x +1\right )}{\ln \relax (x )}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x-5)*ln(3)^2-x^2-x+1)*ln(x)^2+((-5*x^2+10*x)*ln(3)^2-x^3+2*x^2+2*x)*ln(x)-5*x*ln(3)^2-x^2-x)/exp(x)/
ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

(5*x*ln(3)^2+10*ln(3)^2+x^2+3*x+2)*exp(-x)+x^2*exp(-x)*(5*ln(3)^2+x+1)/ln(x)

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maxima [B]  time = 0.78, size = 76, normalized size = 2.53 \begin {gather*} 5 \, {\left (x + 1\right )} e^{\left (-x\right )} \log \relax (3)^{2} + 5 \, e^{\left (-x\right )} \log \relax (3)^{2} + {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} + {\left (x + 1\right )} e^{\left (-x\right )} + \frac {{\left ({\left (5 \, \log \relax (3)^{2} + 1\right )} x^{2} + x^{3}\right )} e^{\left (-x\right )}}{\log \relax (x)} - e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x-5)*log(3)^2-x^2-x+1)*log(x)^2+((-5*x^2+10*x)*log(3)^2-x^3+2*x^2+2*x)*log(x)-5*x*log(3)^2-x^2
-x)/exp(x)/log(x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.51, size = 39, normalized size = 1.30 \begin {gather*} {\mathrm {e}}^{-x}\,\left (x+2\right )\,\left (x+5\,{\ln \relax (3)}^2+1\right )+\frac {x^2\,{\mathrm {e}}^{-x}\,\left (x+5\,{\ln \relax (3)}^2+1\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(x - log(x)*(2*x + log(3)^2*(10*x - 5*x^2) + 2*x^2 - x^3) + 5*x*log(3)^2 + log(x)^2*(x + log(3)^
2*(5*x + 5) + x^2 - 1) + x^2))/log(x)^2,x)

[Out]

exp(-x)*(x + 2)*(x + 5*log(3)^2 + 1) + (x^2*exp(-x)*(x + 5*log(3)^2 + 1))/log(x)

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sympy [B]  time = 0.33, size = 63, normalized size = 2.10 \begin {gather*} \frac {\left (x^{3} + x^{2} \log {\relax (x )} + x^{2} + 5 x^{2} \log {\relax (3 )}^{2} + 3 x \log {\relax (x )} + 5 x \log {\relax (3 )}^{2} \log {\relax (x )} + 2 \log {\relax (x )} + 10 \log {\relax (3 )}^{2} \log {\relax (x )}\right ) e^{- x}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x-5)*ln(3)**2-x**2-x+1)*ln(x)**2+((-5*x**2+10*x)*ln(3)**2-x**3+2*x**2+2*x)*ln(x)-5*x*ln(3)**2-
x**2-x)/exp(x)/ln(x)**2,x)

[Out]

(x**3 + x**2*log(x) + x**2 + 5*x**2*log(3)**2 + 3*x*log(x) + 5*x*log(3)**2*log(x) + 2*log(x) + 10*log(3)**2*lo
g(x))*exp(-x)/log(x)

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