3.21.48 \(\int \frac {e^{-x} (48-48 x+(-48+48 x-48 x^2) \log (x)+(-32 x+32 x^2-16 x^3+e^2 (-16+16 x-16 x^2)) \log ^2(x))}{(15-30 x+15 x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=33 \[ \frac {4 e^{-x} x \left (e^2+x+\frac {3}{\log (x)}\right )}{3 \left (-2+x+\frac {3+x}{4}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 1.07, 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 (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(48 - 48*x + (-48 + 48*x - 48*x^2)*Log[x] + (-32*x + 32*x^2 - 16*x^3 + E^2*(-16 + 16*x - 16*x^2))*Log[x]^2
)/(E^x*(15 - 30*x + 15*x^2)*Log[x]^2),x]

[Out]

(16*E^(2 - x))/15 + 16/(15*E^x) - (16*(1 + E^2))/(15*E^x*(1 - x)) + (16*x)/(15*E^x) - (16*Defer[Int][1/(E^x*(-
1 + x)*Log[x]^2), x])/5 - (16*Defer[Int][1/(E^x*Log[x]), x])/5 - (16*Defer[Int][1/(E^x*(-1 + x)^2*Log[x]), x])
/5 - (16*Defer[Int][1/(E^x*(-1 + x)*Log[x]), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{15 (-1+x)^2 \log ^2(x)} \, dx\\ &=\frac {1}{15} \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{(-1+x)^2 \log ^2(x)} \, dx\\ &=\frac {1}{15} \int \left (\frac {16 e^{-x} \left (-e^2-\left (2-e^2\right ) x+\left (2-e^2\right ) x^2-x^3\right )}{(1-x)^2}-\frac {48 e^{-x}}{(-1+x) \log ^2(x)}-\frac {48 e^{-x} \left (1-x+x^2\right )}{(-1+x)^2 \log (x)}\right ) \, dx\\ &=\frac {16}{15} \int \frac {e^{-x} \left (-e^2-\left (2-e^2\right ) x+\left (2-e^2\right ) x^2-x^3\right )}{(1-x)^2} \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x) \log ^2(x)} \, dx-\frac {16}{5} \int \frac {e^{-x} \left (1-x+x^2\right )}{(-1+x)^2 \log (x)} \, dx\\ &=\frac {16}{15} \int \left (-e^{2-x}+\frac {e^{-x} \left (-1-e^2\right )}{(-1+x)^2}+\frac {e^{-x} \left (-1-e^2\right )}{-1+x}-e^{-x} x\right ) \, dx-\frac {16}{5} \int \left (\frac {e^{-x}}{\log (x)}+\frac {e^{-x}}{(-1+x)^2 \log (x)}+\frac {e^{-x}}{(-1+x) \log (x)}\right ) \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x) \log ^2(x)} \, dx\\ &=-\left (\frac {16}{15} \int e^{2-x} \, dx\right )-\frac {16}{15} \int e^{-x} x \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x) \log ^2(x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{\log (x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x)^2 \log (x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x) \log (x)} \, dx-\frac {1}{15} \left (16 \left (1+e^2\right )\right ) \int \frac {e^{-x}}{(-1+x)^2} \, dx-\frac {1}{15} \left (16 \left (1+e^2\right )\right ) \int \frac {e^{-x}}{-1+x} \, dx\\ &=\frac {16 e^{2-x}}{15}-\frac {16 e^{-x} \left (1+e^2\right )}{15 (1-x)}+\frac {16 e^{-x} x}{15}-\frac {16 \left (1+e^2\right ) \text {Ei}(1-x)}{15 e}-\frac {16}{15} \int e^{-x} \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x) \log ^2(x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{\log (x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x)^2 \log (x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x) \log (x)} \, dx+\frac {1}{15} \left (16 \left (1+e^2\right )\right ) \int \frac {e^{-x}}{-1+x} \, dx\\ &=\frac {16 e^{2-x}}{15}+\frac {16 e^{-x}}{15}-\frac {16 e^{-x} \left (1+e^2\right )}{15 (1-x)}+\frac {16 e^{-x} x}{15}-\frac {16}{5} \int \frac {e^{-x}}{(-1+x) \log ^2(x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{\log (x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x)^2 \log (x)} \, dx-\frac {16}{5} \int \frac {e^{-x}}{(-1+x) \log (x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.30, size = 29, normalized size = 0.88 \begin {gather*} \frac {16 e^{-x} x \left (3+\left (e^2+x\right ) \log (x)\right )}{15 (-1+x) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48 - 48*x + (-48 + 48*x - 48*x^2)*Log[x] + (-32*x + 32*x^2 - 16*x^3 + E^2*(-16 + 16*x - 16*x^2))*Lo
g[x]^2)/(E^x*(15 - 30*x + 15*x^2)*Log[x]^2),x]

[Out]

(16*x*(3 + (E^2 + x)*Log[x]))/(15*E^x*(-1 + x)*Log[x])

________________________________________________________________________________________

fricas [A]  time = 0.78, size = 34, normalized size = 1.03 \begin {gather*} \frac {16 \, {\left ({\left (x^{2} + x e^{2}\right )} e^{\left (-x\right )} \log \relax (x) + 3 \, x e^{\left (-x\right )}\right )}}{15 \, {\left (x - 1\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*log(x)^2+(-48*x^2+48*x-48)*log(x)-48*x+48)/(15*x^2-30
*x+15)/exp(x)/log(x)^2,x, algorithm="fricas")

[Out]

16/15*((x^2 + x*e^2)*e^(-x)*log(x) + 3*x*e^(-x))/((x - 1)*log(x))

________________________________________________________________________________________

giac [A]  time = 0.19, size = 41, normalized size = 1.24 \begin {gather*} \frac {16 \, {\left (x^{2} e^{\left (-x\right )} \log \relax (x) + x e^{\left (-x + 2\right )} \log \relax (x) + 3 \, x e^{\left (-x\right )}\right )}}{15 \, {\left (x \log \relax (x) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*log(x)^2+(-48*x^2+48*x-48)*log(x)-48*x+48)/(15*x^2-30
*x+15)/exp(x)/log(x)^2,x, algorithm="giac")

[Out]

16/15*(x^2*e^(-x)*log(x) + x*e^(-x + 2)*log(x) + 3*x*e^(-x))/(x*log(x) - log(x))

________________________________________________________________________________________

maple [A]  time = 0.41, size = 34, normalized size = 1.03

method	result	size
risch	\(\frac {16 x \left (x +{\mathrm e}^{2}\right ) {\mathrm e}^{-x}}{15 \left (x -1\right )}+\frac {16 x \,{\mathrm e}^{-x}}{5 \left (x -1\right ) \ln \relax (x )}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*ln(x)^2+(-48*x^2+48*x-48)*ln(x)-48*x+48)/(15*x^2-30*x+15)/e
xp(x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

16/15*x*(x+exp(2))/(x-1)*exp(-x)+16/5*x*exp(-x)/(x-1)/ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.47, size = 30, normalized size = 0.91 \begin {gather*} \frac {16 \, {\left ({\left (x^{2} + x e^{2}\right )} \log \relax (x) + 3 \, x\right )} e^{\left (-x\right )}}{15 \, {\left (x - 1\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*log(x)^2+(-48*x^2+48*x-48)*log(x)-48*x+48)/(15*x^2-30
*x+15)/exp(x)/log(x)^2,x, algorithm="maxima")

[Out]

16/15*((x^2 + x*e^2)*log(x) + 3*x)*e^(-x)/((x - 1)*log(x))

________________________________________________________________________________________

mupad [B]  time = 1.49, size = 35, normalized size = 1.06 \begin {gather*} \frac {16\,x\,{\mathrm {e}}^{-x}\,\left (x+{\mathrm {e}}^2\right )}{15\,\left (x-1\right )}+\frac {16\,x\,{\mathrm {e}}^{-x}}{5\,\ln \relax (x)\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(48*x + log(x)^2*(32*x + exp(2)*(16*x^2 - 16*x + 16) - 32*x^2 + 16*x^3) + log(x)*(48*x^2 - 48*x
+ 48) - 48))/(log(x)^2*(15*x^2 - 30*x + 15)),x)

[Out]

(16*x*exp(-x)*(x + exp(2)))/(15*(x - 1)) + (16*x*exp(-x))/(5*log(x)*(x - 1))

________________________________________________________________________________________

sympy [A]  time = 0.40, size = 36, normalized size = 1.09 \begin {gather*} \frac {\left (16 x^{2} \log {\relax (x )} + 16 x e^{2} \log {\relax (x )} + 48 x\right ) e^{- x}}{15 x \log {\relax (x )} - 15 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x**2+16*x-16)*exp(2)-16*x**3+32*x**2-32*x)*ln(x)**2+(-48*x**2+48*x-48)*ln(x)-48*x+48)/(15*x**
2-30*x+15)/exp(x)/ln(x)**2,x)

[Out]

(16*x**2*log(x) + 16*x*exp(2)*log(x) + 48*x)*exp(-x)/(15*x*log(x) - 15*log(x))

________________________________________________________________________________________