3.3.3 \(\int \frac {-4-3 x+(7+7 x) \log (x)+(-4-4 x) \log ^2(x)}{5 e^x x^2-10 e^x x^2 \log (x)+5 e^x x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {3 e^{-x} \left (x+\frac {x}{3-\frac {3}{\log (x)}}\right )}{5 x^2} \]

________________________________________________________________________________________

Rubi [F]  time = 0.96, 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 {-4-3 x+(7+7 x) \log (x)+(-4-4 x) \log ^2(x)}{5 e^x x^2-10 e^x x^2 \log (x)+5 e^x x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4 - 3*x + (7 + 7*x)*Log[x] + (-4 - 4*x)*Log[x]^2)/(5*E^x*x^2 - 10*E^x*x^2*Log[x] + 5*E^x*x^2*Log[x]^2),x
]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (-4-3 x+7 (1+x) \log (x)-4 (1+x) \log ^2(x)\right )}{5 x^2 (1-\log (x))^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-x} \left (-4-3 x+7 (1+x) \log (x)-4 (1+x) \log ^2(x)\right )}{x^2 (1-\log (x))^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {4 e^{-x} (1+x)}{x^2}-\frac {e^{-x}}{x^2 (-1+\log (x))^2}+\frac {e^{-x} (-1-x)}{x^2 (-1+\log (x))}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{-x}}{x^2 (-1+\log (x))^2} \, dx\right )+\frac {1}{5} \int \frac {e^{-x} (-1-x)}{x^2 (-1+\log (x))} \, dx-\frac {4}{5} \int \frac {e^{-x} (1+x)}{x^2} \, dx\\ &=\frac {4 e^{-x}}{5 x}+\frac {1}{5} \int \left (-\frac {e^{-x}}{x^2 (-1+\log (x))}-\frac {e^{-x}}{x (-1+\log (x))}\right ) \, dx-\frac {1}{5} \int \frac {e^{-x}}{x^2 (-1+\log (x))^2} \, dx\\ &=\frac {4 e^{-x}}{5 x}-\frac {1}{5} \int \frac {e^{-x}}{x^2 (-1+\log (x))^2} \, dx-\frac {1}{5} \int \frac {e^{-x}}{x^2 (-1+\log (x))} \, dx-\frac {1}{5} \int \frac {e^{-x}}{x (-1+\log (x))} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 24, normalized size = 0.92 \begin {gather*} \frac {e^{-x} (-3+4 \log (x))}{5 x (-1+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 3*x + (7 + 7*x)*Log[x] + (-4 - 4*x)*Log[x]^2)/(5*E^x*x^2 - 10*E^x*x^2*Log[x] + 5*E^x*x^2*Log[x
]^2),x]

[Out]

(-3 + 4*Log[x])/(5*E^x*x*(-1 + Log[x]))

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 22, normalized size = 0.85 \begin {gather*} \frac {4 \, \log \relax (x) - 3}{5 \, {\left (x e^{x} \log \relax (x) - x e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-4)*log(x)^2+(7*x+7)*log(x)-3*x-4)/(5*x^2*exp(x)*log(x)^2-10*x^2*exp(x)*log(x)+5*exp(x)*x^2),x
, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.24, size = 27, normalized size = 1.04 \begin {gather*} \frac {4 \, e^{\left (-x\right )} \log \relax (x) - 3 \, e^{\left (-x\right )}}{5 \, {\left (x \log \relax (x) - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-4)*log(x)^2+(7*x+7)*log(x)-3*x-4)/(5*x^2*exp(x)*log(x)^2-10*x^2*exp(x)*log(x)+5*exp(x)*x^2),x
, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 21, normalized size = 0.81

method	result	size
norman	\(\frac {\left (-\frac {3}{5}+\frac {4 \ln \relax (x )}{5}\right ) {\mathrm e}^{-x}}{x \left (\ln \relax (x )-1\right )}\)	\(21\)
risch	\(\frac {4 \,{\mathrm e}^{-x}}{5 x}+\frac {{\mathrm e}^{-x}}{5 x \left (\ln \relax (x )-1\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x-4)*ln(x)^2+(7*x+7)*ln(x)-3*x-4)/(5*x^2*exp(x)*ln(x)^2-10*x^2*exp(x)*ln(x)+5*exp(x)*x^2),x,method=_R
ETURNVERBOSE)

[Out]

(-3/5+4/5*ln(x))/exp(x)/x/(ln(x)-1)

________________________________________________________________________________________

maxima [A]  time = 0.84, size = 22, normalized size = 0.85 \begin {gather*} \frac {{\left (4 \, \log \relax (x) - 3\right )} e^{\left (-x\right )}}{5 \, {\left (x \log \relax (x) - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-4)*log(x)^2+(7*x+7)*log(x)-3*x-4)/(5*x^2*exp(x)*log(x)^2-10*x^2*exp(x)*log(x)+5*exp(x)*x^2),x
, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.59, size = 28, normalized size = 1.08 \begin {gather*} -\frac {x\,\left (4\,\ln \relax (x)-3\right )}{5\,\left (x^2\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x\,\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x - log(x)*(7*x + 7) + log(x)^2*(4*x + 4) + 4)/(5*x^2*exp(x) + 5*x^2*exp(x)*log(x)^2 - 10*x^2*exp(x)*l
og(x)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.31, size = 19, normalized size = 0.73 \begin {gather*} \frac {\left (4 \log {\relax (x )} - 3\right ) e^{- x}}{5 x \log {\relax (x )} - 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-4)*ln(x)**2+(7*x+7)*ln(x)-3*x-4)/(5*x**2*exp(x)*ln(x)**2-10*x**2*exp(x)*ln(x)+5*exp(x)*x**2),
x)

[Out]

(4*log(x) - 3)*exp(-x)/(5*x*log(x) - 5*x)

________________________________________________________________________________________