3.6.28 \(\int \frac {e^{-4 e^3+x} (-32 x+16 x^2-2 x^3)+(-16+20 x-4 x^2) \log (e^{-x} x)-2 x \log ^2(e^{-x} x)}{16 x-8 x^2+x^3} \, dx\)

Optimal. Leaf size=30 \[ 2 \left (-e^{-4 e^3+x}+\frac {\log ^2\left (e^{-x} x\right )}{-4+x}\right ) \]

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Rubi [F]  time = 1.36, 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^{-4 e^3+x} \left (-32 x+16 x^2-2 x^3\right )+\left (-16+20 x-4 x^2\right ) \log \left (e^{-x} x\right )-2 x \log ^2\left (e^{-x} x\right )}{16 x-8 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-4*E^3 + x)*(-32*x + 16*x^2 - 2*x^3) + (-16 + 20*x - 4*x^2)*Log[x/E^x] - 2*x*Log[x/E^x]^2)/(16*x - 8*x
^2 + x^3),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-4 e^3+x} \left (-32 x+16 x^2-2 x^3\right )+\left (-16+20 x-4 x^2\right ) \log \left (e^{-x} x\right )-2 x \log ^2\left (e^{-x} x\right )}{x \left (16-8 x+x^2\right )} \, dx\\ &=\int \frac {e^{-4 e^3+x} \left (-32 x+16 x^2-2 x^3\right )+\left (-16+20 x-4 x^2\right ) \log \left (e^{-x} x\right )-2 x \log ^2\left (e^{-x} x\right )}{(-4+x)^2 x} \, dx\\ &=\int \left (-2 e^{-4 e^3+x}-\frac {2 \log \left (e^{-x} x\right ) \left (8-10 x+2 x^2+x \log \left (e^{-x} x\right )\right )}{(-4+x)^2 x}\right ) \, dx\\ &=-\left (2 \int e^{-4 e^3+x} \, dx\right )-2 \int \frac {\log \left (e^{-x} x\right ) \left (8-10 x+2 x^2+x \log \left (e^{-x} x\right )\right )}{(-4+x)^2 x} \, dx\\ &=-2 e^{-4 e^3+x}-2 \int \left (\frac {2 (-1+x) \log \left (e^{-x} x\right )}{(-4+x) x}+\frac {\log ^2\left (e^{-x} x\right )}{(-4+x)^2}\right ) \, dx\\ &=-2 e^{-4 e^3+x}-2 \int \frac {\log ^2\left (e^{-x} x\right )}{(-4+x)^2} \, dx-4 \int \frac {(-1+x) \log \left (e^{-x} x\right )}{(-4+x) x} \, dx\\ &=-2 e^{-4 e^3+x}-2 \int \frac {\log ^2\left (e^{-x} x\right )}{(-4+x)^2} \, dx-4 \int \left (\frac {3 \log \left (e^{-x} x\right )}{4 (-4+x)}+\frac {\log \left (e^{-x} x\right )}{4 x}\right ) \, dx\\ &=-2 e^{-4 e^3+x}-2 \int \frac {\log ^2\left (e^{-x} x\right )}{(-4+x)^2} \, dx-3 \int \frac {\log \left (e^{-x} x\right )}{-4+x} \, dx-\int \frac {\log \left (e^{-x} x\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.23, size = 100, normalized size = 3.33 \begin {gather*} -\frac {32+8 e^{-4 e^3+x}-24 x-2 e^{-4 e^3+x} x+4 x^2+x^2 \log (4)-x \log (256)+(16-x (4+\log (4))+\log (256)) \log (x)+(-16+x (4+\log (4))-\log (256)) \log \left (e^{-x} x\right )+2 \log ^2\left (e^{-x} x\right )}{4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4*E^3 + x)*(-32*x + 16*x^2 - 2*x^3) + (-16 + 20*x - 4*x^2)*Log[x/E^x] - 2*x*Log[x/E^x]^2)/(16*x
- 8*x^2 + x^3),x]

[Out]

-((32 + 8*E^(-4*E^3 + x) - 24*x - 2*E^(-4*E^3 + x)*x + 4*x^2 + x^2*Log[4] - x*Log[256] + (16 - x*(4 + Log[4])
+ Log[256])*Log[x] + (-16 + x*(4 + Log[4]) - Log[256])*Log[x/E^x] + 2*Log[x/E^x]^2)/(4 - x))

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fricas [A]  time = 0.69, size = 35, normalized size = 1.17 \begin {gather*} -\frac {2 \, {\left ({\left (x - 4\right )} e^{\left (x - e^{\left (2 \, \log \relax (2) + 3\right )}\right )} - \log \left (x e^{\left (-x\right )}\right )^{2}\right )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x/exp(x))^2+(-4*x^2+20*x-16)*log(x/exp(x))+(-2*x^3+16*x^2-32*x)*exp(-exp(2*log(2)+3)+x))/(
x^3-8*x^2+16*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.61, size = 43, normalized size = 1.43 \begin {gather*} \frac {2 \, {\left (x^{2} - x e^{\left (x - 4 \, e^{3}\right )} - 2 \, x \log \relax (x) + \log \relax (x)^{2} - 4 \, x + 4 \, e^{\left (x - 4 \, e^{3}\right )} + 16\right )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x/exp(x))^2+(-4*x^2+20*x-16)*log(x/exp(x))+(-2*x^3+16*x^2-32*x)*exp(-exp(2*log(2)+3)+x))/(
x^3-8*x^2+16*x),x, algorithm="giac")

[Out]

2*(x^2 - x*e^(x - 4*e^3) - 2*x*log(x) + log(x)^2 - 4*x + 4*e^(x - 4*e^3) + 16)/(x - 4)

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maple [C]  time = 0.31, size = 533, normalized size = 17.77

method	result	size
risch	\(\frac {2 \ln \left ({\mathrm e}^{x}\right )^{2}}{x -4}-\frac {2 \left (-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+2 \ln \relax (x )\right ) \ln \left ({\mathrm e}^{x}\right )}{x -4}-\frac {\left (\pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-2 \pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+\pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{4}-2 \pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+4 \pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{4}-2 \pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{5}+\pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{4}-2 \pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{5}+\pi ^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{6}+4 i \pi \,{\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} \ln \relax (x )+4 i \pi \,{\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \ln \relax (x )-4 i \pi \,{\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \ln \relax (x )-4 i \pi \,{\mathrm e}^{4 \,{\mathrm e}^{3}} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \ln \relax (x )-4 \,{\mathrm e}^{4 \,{\mathrm e}^{3}} \ln \relax (x )^{2}+4 \,{\mathrm e}^{x} x -16 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-4 \,{\mathrm e}^{3}}}{2 \left (x -4\right )}\)	\(533\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(x/exp(x))^2+(-4*x^2+20*x-16)*ln(x/exp(x))+(-2*x^3+16*x^2-32*x)*exp(-exp(2*ln(2)+3)+x))/(x^3-8*x^2
+16*x),x,method=_RETURNVERBOSE)

[Out]

2/(x-4)*ln(exp(x))^2-2*(-I*Pi*csgn(I*x)*csgn(I*exp(-x))*csgn(I*x*exp(-x))+I*Pi*csgn(I*x)*csgn(I*x*exp(-x))^2+I
*Pi*csgn(I*exp(-x))*csgn(I*x*exp(-x))^2-I*Pi*csgn(I*x*exp(-x))^3+2*ln(x))/(x-4)*ln(exp(x))-1/2*(Pi^2*exp(4*exp
(3))*csgn(I*x)^2*csgn(I*exp(-x))^2*csgn(I*x*exp(-x))^2-2*Pi^2*exp(4*exp(3))*csgn(I*x)^2*csgn(I*exp(-x))*csgn(I
*x*exp(-x))^3+Pi^2*exp(4*exp(3))*csgn(I*x)^2*csgn(I*x*exp(-x))^4-2*Pi^2*exp(4*exp(3))*csgn(I*x)*csgn(I*exp(-x)
)^2*csgn(I*x*exp(-x))^3+4*Pi^2*exp(4*exp(3))*csgn(I*x)*csgn(I*exp(-x))*csgn(I*x*exp(-x))^4-2*Pi^2*exp(4*exp(3)
)*csgn(I*x)*csgn(I*x*exp(-x))^5+Pi^2*exp(4*exp(3))*csgn(I*exp(-x))^2*csgn(I*x*exp(-x))^4-2*Pi^2*exp(4*exp(3))*
csgn(I*exp(-x))*csgn(I*x*exp(-x))^5+Pi^2*exp(4*exp(3))*csgn(I*x*exp(-x))^6+4*I*Pi*exp(4*exp(3))*csgn(I*x*exp(-
x))^3*ln(x)+4*I*Pi*exp(4*exp(3))*csgn(I*x)*csgn(I*exp(-x))*csgn(I*x*exp(-x))*ln(x)-4*I*Pi*exp(4*exp(3))*csgn(I
*exp(-x))*csgn(I*x*exp(-x))^2*ln(x)-4*I*Pi*exp(4*exp(3))*csgn(I*x)*csgn(I*x*exp(-x))^2*ln(x)-4*exp(4*exp(3))*l
n(x)^2+4*exp(x)*x-16*exp(x))/(x-4)*exp(-4*exp(3))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {32 \, e^{\left (-4 \, e^{3} + 4\right )} E_{2}\left (-x + 4\right )}{x - 4} + \frac {2 \, {\left (x^{2} - 2 \, x \log \relax (x) + \log \relax (x)^{2} - 4 \, x + 16\right )}}{x - 4} - 2 \, \int \frac {{\left (x^{2} - 8 \, x\right )} e^{x}}{x^{2} e^{\left (4 \, e^{3}\right )} - 8 \, x e^{\left (4 \, e^{3}\right )} + 16 \, e^{\left (4 \, e^{3}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x/exp(x))^2+(-4*x^2+20*x-16)*log(x/exp(x))+(-2*x^3+16*x^2-32*x)*exp(-exp(2*log(2)+3)+x))/(
x^3-8*x^2+16*x),x, algorithm="maxima")

[Out]

32*e^(-4*e^3 + 4)*exp_integral_e(2, -x + 4)/(x - 4) + 2*(x^2 - 2*x*log(x) + log(x)^2 - 4*x + 16)/(x - 4) - 2*i
ntegrate((x^2 - 8*x)*e^x/(x^2*e^(4*e^3) - 8*x*e^(4*e^3) + 16*e^(4*e^3)), x)

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mupad [B]  time = 0.72, size = 41, normalized size = 1.37 \begin {gather*} \frac {2\,x^2}{x-4}-2\,{\mathrm {e}}^{-4\,{\mathrm {e}}^3}\,{\mathrm {e}}^x+\frac {2\,{\ln \relax (x)}^2}{x-4}-\frac {4\,x\,\ln \relax (x)}{x-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - exp(2*log(2) + 3))*(32*x - 16*x^2 + 2*x^3) + log(x*exp(-x))*(4*x^2 - 20*x + 16) + 2*x*log(x*exp(
-x))^2)/(16*x - 8*x^2 + x^3),x)

[Out]

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

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sympy [A]  time = 0.29, size = 24, normalized size = 0.80 \begin {gather*} - \frac {2 e^{x}}{e^{4 e^{3}}} + \frac {2 \log {\left (x e^{- x} \right )}^{2}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(x/exp(x))**2+(-4*x**2+20*x-16)*ln(x/exp(x))+(-2*x**3+16*x**2-32*x)*exp(-exp(2*ln(2)+3)+x))/
(x**3-8*x**2+16*x),x)

[Out]

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

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