3.62.42 \(\int \frac {e^{-1+x} (-2+3 x-x^2)+e^{-1+x} (-2+6 x-4 x^2+x^3) \log (x)}{e^3 (x^2-2 x^3+x^4) \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ -2+\frac {e^{-4+x}}{\left (1+\frac {1}{-2+x}\right ) x \log (x)} \]

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Rubi [F]  time = 3.46, 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^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-1 + x)*(-2 + 3*x - x^2) + E^(-1 + x)*(-2 + 6*x - 4*x^2 + x^3)*Log[x])/(E^3*(x^2 - 2*x^3 + x^4)*Log[x]
^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{x^2 \left (1-2 x+x^2\right ) \log ^2(x)} \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{(-1+x)^2 x^2 \log ^2(x)} \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x} \left (-2+3 x-x^2-2 \log (x)+6 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right )}{(1-x)^2 x^2 \log ^2(x)} \, dx}{e^3}\\ &=\frac {\int \left (\frac {e^{-1+x} (2-x)}{(-1+x) x^2 \log ^2(x)}+\frac {e^{-1+x} \left (-2+6 x-4 x^2+x^3\right )}{(-1+x)^2 x^2 \log (x)}\right ) \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x} (2-x)}{(-1+x) x^2 \log ^2(x)} \, dx}{e^3}+\frac {\int \frac {e^{-1+x} \left (-2+6 x-4 x^2+x^3\right )}{(-1+x)^2 x^2 \log (x)} \, dx}{e^3}\\ &=\frac {\int \left (\frac {e^{-1+x}}{(-1+x) \log ^2(x)}-\frac {2 e^{-1+x}}{x^2 \log ^2(x)}-\frac {e^{-1+x}}{x \log ^2(x)}\right ) \, dx}{e^3}+\frac {\int \left (\frac {e^{-1+x}}{(-1+x)^2 \log (x)}-\frac {e^{-1+x}}{(-1+x) \log (x)}-\frac {2 e^{-1+x}}{x^2 \log (x)}+\frac {2 e^{-1+x}}{x \log (x)}\right ) \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x}}{(-1+x) \log ^2(x)} \, dx}{e^3}-\frac {\int \frac {e^{-1+x}}{x \log ^2(x)} \, dx}{e^3}+\frac {\int \frac {e^{-1+x}}{(-1+x)^2 \log (x)} \, dx}{e^3}-\frac {\int \frac {e^{-1+x}}{(-1+x) \log (x)} \, dx}{e^3}-\frac {2 \int \frac {e^{-1+x}}{x^2 \log ^2(x)} \, dx}{e^3}-\frac {2 \int \frac {e^{-1+x}}{x^2 \log (x)} \, dx}{e^3}+\frac {2 \int \frac {e^{-1+x}}{x \log (x)} \, dx}{e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 21, normalized size = 0.88 \begin {gather*} \frac {e^{-4+x} (-2+x)}{(-1+x) x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 + x)*(-2 + 3*x - x^2) + E^(-1 + x)*(-2 + 6*x - 4*x^2 + x^3)*Log[x])/(E^3*(x^2 - 2*x^3 + x^4)*
Log[x]^2),x]

[Out]

(E^(-4 + x)*(-2 + x))/((-1 + x)*x*Log[x])

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fricas [A]  time = 0.57, size = 21, normalized size = 0.88 \begin {gather*} \frac {{\left (x - 2\right )} e^{\left (x - 4\right )}}{{\left (x^{2} - x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+6*x-2)*exp(x-1)*log(x)+(-x^2+3*x-2)*exp(x-1))/(x^4-2*x^3+x^2)/exp(3)/log(x)^2,x, algorit
hm="fricas")

[Out]

(x - 2)*e^(x - 4)/((x^2 - x)*log(x))

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giac [A]  time = 0.20, size = 30, normalized size = 1.25 \begin {gather*} \frac {{\left (x e^{x} - 2 \, e^{x}\right )} e^{\left (-3\right )}}{x^{2} e \log \relax (x) - x e \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+6*x-2)*exp(x-1)*log(x)+(-x^2+3*x-2)*exp(x-1))/(x^4-2*x^3+x^2)/exp(3)/log(x)^2,x, algorit
hm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 21, normalized size = 0.88

method	result	size
risch	\(\frac {\left (x -2\right ) {\mathrm e}^{x -4}}{\ln \relax (x ) \left (x -1\right ) x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-4*x^2+6*x-2)*exp(x-1)*ln(x)+(-x^2+3*x-2)*exp(x-1))/(x^4-2*x^3+x^2)/exp(3)/ln(x)^2,x,method=_RETURNVE
RBOSE)

[Out]

1/ln(x)/(x-1)*(x-2)/x*exp(x-4)

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maxima [A]  time = 0.40, size = 26, normalized size = 1.08 \begin {gather*} \frac {{\left (x - 2\right )} e^{\left (x - 3\right )}}{{\left (x^{2} e - x e\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+6*x-2)*exp(x-1)*log(x)+(-x^2+3*x-2)*exp(x-1))/(x^4-2*x^3+x^2)/exp(3)/log(x)^2,x, algorit
hm="maxima")

[Out]

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

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mupad [B]  time = 4.60, size = 20, normalized size = 0.83 \begin {gather*} \frac {{\mathrm {e}}^{x-4}\,\left (x-2\right )}{x\,\ln \relax (x)\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-3)*(exp(x - 1)*(x^2 - 3*x + 2) - exp(x - 1)*log(x)*(6*x - 4*x^2 + x^3 - 2)))/(log(x)^2*(x^2 - 2*x^3
+ x^4)),x)

[Out]

(exp(x - 4)*(x - 2))/(x*log(x)*(x - 1))

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sympy [A]  time = 0.33, size = 26, normalized size = 1.08 \begin {gather*} \frac {\left (x - 2\right ) e^{x - 1}}{x^{2} e^{3} \log {\relax (x )} - x e^{3} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-4*x**2+6*x-2)*exp(x-1)*ln(x)+(-x**2+3*x-2)*exp(x-1))/(x**4-2*x**3+x**2)/exp(3)/ln(x)**2,x)

[Out]

(x - 2)*exp(x - 1)/(x**2*exp(3)*log(x) - x*exp(3)*log(x))

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