3.92.96 \(\int \frac {1}{4} e^{-x} (4 e^{2 x}+e^x (10 x-x^3)+(2 x+e^x (20 x-4 x^3)) \log (x)+(2 x-x^2) \log ^2(x)) \, dx\)

Optimal. Leaf size=28 \[ e^x+\frac {1}{4} x^2 \log (x) \left (10-x^2+e^{-x} \log (x)\right ) \]

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Rubi [F]  time = 0.66, 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 {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4*E^(2*x) + E^x*(10*x - x^3) + (2*x + E^x*(20*x - 4*x^3))*Log[x] + (2*x - x^2)*Log[x]^2)/(4*E^x),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx\\ &=\frac {1}{4} \int \left (4 e^x-x \left (-10+x^2\right )-2 e^{-x} x \left (-1-10 e^x+2 e^x x^2\right ) \log (x)-e^{-x} (-2+x) x \log ^2(x)\right ) \, dx\\ &=-\left (\frac {1}{4} \int x \left (-10+x^2\right ) \, dx\right )-\frac {1}{4} \int e^{-x} (-2+x) x \log ^2(x) \, dx-\frac {1}{2} \int e^{-x} x \left (-1-10 e^x+2 e^x x^2\right ) \log (x) \, dx+\int e^x \, dx\\ &=e^x-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)-\frac {1}{4} \int \left (-10 x+x^3\right ) \, dx-\frac {1}{4} \int \left (-2 e^{-x} x \log ^2(x)+e^{-x} x^2 \log ^2(x)\right ) \, dx+\frac {1}{2} \int \left (e^{-x} \left (1+\frac {1}{x}\right )+\frac {1}{2} x \left (-10+x^2\right )\right ) \, dx\\ &=e^x+\frac {5 x^2}{4}-\frac {x^4}{16}-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)+\frac {1}{4} \int x \left (-10+x^2\right ) \, dx-\frac {1}{4} \int e^{-x} x^2 \log ^2(x) \, dx+\frac {1}{2} \int e^{-x} \left (1+\frac {1}{x}\right ) \, dx+\frac {1}{2} \int e^{-x} x \log ^2(x) \, dx\\ &=e^x+\frac {5 x^2}{4}-\frac {x^4}{16}-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)+\frac {1}{4} \int \left (-10 x+x^3\right ) \, dx-\frac {1}{4} \int e^{-x} x^2 \log ^2(x) \, dx+\frac {1}{2} \int \left (e^{-x}+\frac {e^{-x}}{x}\right ) \, dx+\frac {1}{2} \int e^{-x} x \log ^2(x) \, dx\\ &=e^x-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)-\frac {1}{4} \int e^{-x} x^2 \log ^2(x) \, dx+\frac {1}{2} \int e^{-x} \, dx+\frac {1}{2} \int \frac {e^{-x}}{x} \, dx+\frac {1}{2} \int e^{-x} x \log ^2(x) \, dx\\ &=-\frac {e^{-x}}{2}+e^x+\frac {\text {Ei}(-x)}{2}-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)-\frac {1}{4} \int e^{-x} x^2 \log ^2(x) \, dx+\frac {1}{2} \int e^{-x} x \log ^2(x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 37, normalized size = 1.32 \begin {gather*} \frac {1}{4} \left (4 e^x+10 x^2 \log (x)-x^4 \log (x)+e^{-x} x^2 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^(2*x) + E^x*(10*x - x^3) + (2*x + E^x*(20*x - 4*x^3))*Log[x] + (2*x - x^2)*Log[x]^2)/(4*E^x),x]

[Out]

(4*E^x + 10*x^2*Log[x] - x^4*Log[x] + (x^2*Log[x]^2)/E^x)/4

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fricas [A]  time = 1.00, size = 36, normalized size = 1.29 \begin {gather*} \frac {1}{4} \, {\left (x^{2} \log \relax (x)^{2} - {\left (x^{4} - 10 \, x^{2}\right )} e^{x} \log \relax (x) + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^2*log(x)^2 - (x^4 - 10*x^2)*e^x*log(x) + 4*e^(2*x))*e^(-x)

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giac [A]  time = 0.21, size = 30, normalized size = 1.07 \begin {gather*} -\frac {1}{4} \, x^{4} \log \relax (x) + \frac {1}{4} \, x^{2} e^{\left (-x\right )} \log \relax (x)^{2} + \frac {5}{2} \, x^{2} \log \relax (x) + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 31, normalized size = 1.11

method	result	size
default	\(\frac {\ln \relax (x )^{2} x^{2} {\mathrm e}^{-x}}{4}-\frac {x^{4} \ln \relax (x )}{4}+\frac {5 x^{2} \ln \relax (x )}{2}+{\mathrm e}^{x}\)	\(31\)
risch	\(\frac {\ln \relax (x )^{2} x^{2} {\mathrm e}^{-x}}{4}+\frac {\left (-x^{4}+10 x^{2}\right ) \ln \relax (x )}{4}+{\mathrm e}^{x}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-x^2+2*x)*ln(x)^2+((-4*x^3+20*x)*exp(x)+2*x)*ln(x)+4*exp(x)^2+(-x^3+10*x)*exp(x))/exp(x),x,method=_R
ETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 30, normalized size = 1.07 \begin {gather*} -\frac {1}{4} \, x^{4} \log \relax (x) + \frac {1}{4} \, x^{2} e^{\left (-x\right )} \log \relax (x)^{2} + \frac {5}{2} \, x^{2} \log \relax (x) + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.04, size = 109, normalized size = 3.89 \begin {gather*} {\mathrm {e}}^x+\frac {2\,x\,{\mathrm {e}}^{-x}+2\,x^2\,{\mathrm {e}}^{-x}\,\ln \relax (x)+x^3\,{\mathrm {e}}^{-x}\,{\ln \relax (x)}^2+2\,x\,{\mathrm {e}}^{-x}\,\ln \relax (x)}{4\,x}+\frac {x^2\,\left (40\,\ln \relax (x)-4\,x^2\,\ln \relax (x)+x^2-20\right )}{16}+\frac {5\,x^2}{4}-\frac {x^4}{16}-\frac {x\,{\mathrm {e}}^{-x}+x^2\,{\mathrm {e}}^{-x}\,\ln \relax (x)+x\,{\mathrm {e}}^{-x}\,\ln \relax (x)}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-x)*(exp(2*x) + (log(x)^2*(2*x - x^2))/4 + (log(x)*(2*x + exp(x)*(20*x - 4*x^3)))/4 + (exp(x)*(10*x -
x^3))/4),x)

[Out]

exp(x) + (2*x*exp(-x) + 2*x^2*exp(-x)*log(x) + x^3*exp(-x)*log(x)^2 + 2*x*exp(-x)*log(x))/(4*x) + (x^2*(40*log
(x) - 4*x^2*log(x) + x^2 - 20))/16 + (5*x^2)/4 - x^4/16 - (x*exp(-x) + x^2*exp(-x)*log(x) + x*exp(-x)*log(x))/
(2*x)

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sympy [A]  time = 0.40, size = 31, normalized size = 1.11 \begin {gather*} \frac {x^{2} e^{- x} \log {\relax (x )}^{2}}{4} + \left (- \frac {x^{4}}{4} + \frac {5 x^{2}}{2}\right ) \log {\relax (x )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-x**2+2*x)*ln(x)**2+((-4*x**3+20*x)*exp(x)+2*x)*ln(x)+4*exp(x)**2+(-x**3+10*x)*exp(x))/exp(x),
x)

[Out]

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

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