3.89.77 \(\int \frac {4-12 x+e^x (8+4 x)+8 \log (x)}{e^{2 x} x^3-2 e^x x^4+x^5+(2 e^x x^3-2 x^4) \log (x)+x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {4}{x^2 \left (-e^x+x-\log (x)\right )} \]

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Rubi [F]  time = 2.09, 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 {4-12 x+e^x (8+4 x)+8 \log (x)}{e^{2 x} x^3-2 e^x x^4+x^5+\left (2 e^x x^3-2 x^4\right ) \log (x)+x^3 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4 - 12*x + E^x*(8 + 4*x) + 8*Log[x])/(E^(2*x)*x^3 - 2*E^x*x^4 + x^5 + (2*E^x*x^3 - 2*x^4)*Log[x] + x^3*Lo
g[x]^2),x]

[Out]

4*Defer[Int][1/(x^3*(-E^x + x - Log[x])^2), x] - 4*Defer[Int][1/(x^2*(-E^x + x - Log[x])^2), x] + 4*Defer[Int]
[1/(x*(-E^x + x - Log[x])^2), x] - 8*Defer[Int][1/(x^3*(-E^x + x - Log[x])), x] - 4*Defer[Int][1/(x^2*(-E^x +
x - Log[x])), x] - 4*Defer[Int][Log[x]/(x^2*(-E^x + x - Log[x])^2), x]

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 16, normalized size = 0.89 \begin {gather*} -\frac {4}{x^2 \left (e^x-x+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 12*x + E^x*(8 + 4*x) + 8*Log[x])/(E^(2*x)*x^3 - 2*E^x*x^4 + x^5 + (2*E^x*x^3 - 2*x^4)*Log[x] +
x^3*Log[x]^2),x]

[Out]

-4/(x^2*(E^x - x + Log[x]))

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fricas [A]  time = 0.53, size = 22, normalized size = 1.22 \begin {gather*} \frac {4}{x^{3} - x^{2} e^{x} - x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(x)+(4*x+8)*exp(x)-12*x+4)/(x^3*log(x)^2+(2*exp(x)*x^3-2*x^4)*log(x)+exp(x)^2*x^3-2*exp(x)*x^4
+x^5),x, algorithm="fricas")

[Out]

4/(x^3 - x^2*e^x - x^2*log(x))

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giac [A]  time = 0.16, size = 22, normalized size = 1.22 \begin {gather*} \frac {4}{x^{3} - x^{2} e^{x} - x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(x)+(4*x+8)*exp(x)-12*x+4)/(x^3*log(x)^2+(2*exp(x)*x^3-2*x^4)*log(x)+exp(x)^2*x^3-2*exp(x)*x^4
+x^5),x, algorithm="giac")

[Out]

4/(x^3 - x^2*e^x - x^2*log(x))

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maple [A]  time = 0.02, size = 18, normalized size = 1.00

method	result	size
risch	\(\frac {4}{x^{2} \left (-\ln \relax (x )-{\mathrm e}^{x}+x \right )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*ln(x)+(4*x+8)*exp(x)-12*x+4)/(x^3*ln(x)^2+(2*exp(x)*x^3-2*x^4)*ln(x)+exp(x)^2*x^3-2*exp(x)*x^4+x^5),x,m
ethod=_RETURNVERBOSE)

[Out]

4/x^2/(-ln(x)-exp(x)+x)

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maxima [A]  time = 0.41, size = 22, normalized size = 1.22 \begin {gather*} \frac {4}{x^{3} - x^{2} e^{x} - x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(x)+(4*x+8)*exp(x)-12*x+4)/(x^3*log(x)^2+(2*exp(x)*x^3-2*x^4)*log(x)+exp(x)^2*x^3-2*exp(x)*x^4
+x^5),x, algorithm="maxima")

[Out]

4/(x^3 - x^2*e^x - x^2*log(x))

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mupad [B]  time = 5.28, size = 15, normalized size = 0.83 \begin {gather*} -\frac {4}{x^2\,\left ({\mathrm {e}}^x-x+\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*log(x) - 12*x + exp(x)*(4*x + 8) + 4)/(x^3*exp(2*x) - 2*x^4*exp(x) + x^3*log(x)^2 + x^5 + log(x)*(2*x^3
*exp(x) - 2*x^4)),x)

[Out]

-4/(x^2*(exp(x) - x + log(x)))

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sympy [A]  time = 0.26, size = 19, normalized size = 1.06 \begin {gather*} - \frac {4}{- x^{3} + x^{2} e^{x} + x^{2} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*ln(x)+(4*x+8)*exp(x)-12*x+4)/(x**3*ln(x)**2+(2*exp(x)*x**3-2*x**4)*ln(x)+exp(x)**2*x**3-2*exp(x)*
x**4+x**5),x)

[Out]

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

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