3.23.72 \(\int \frac {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+(-12 x^2+4 x^3+12 e^2 x^3) \log (x)+(9-6 x+x^2+9 e^4 x^2+e^2 (-18 x+6 x^2)) \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {1}{2 x-\frac {\left (3-x-3 e^2 x\right ) \log (x)}{x}} \]

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Rubi [F]  time = 1.09, 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 {3-x-3 e^2 x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3 - x - 3*E^2*x - 2*x^2 - 3*Log[x])/(4*x^4 + (-12*x^2 + 4*x^3 + 12*E^2*x^3)*Log[x] + (9 - 6*x + x^2 + 9*E
^4*x^2 + E^2*(-18*x + 6*x^2))*Log[x]^2),x]

[Out]

(3*(7 + 6*E^2 + 9*E^4)*Defer[Int][(2*x^2 - 3*Log[x] + (1 + 3*E^2)*x*Log[x])^(-2), x])/(1 + 3*E^2)^2 + ((5 - 6*
E^2 - 9*E^4)*Defer[Int][x/(2*x^2 - 3*Log[x] + (1 + 3*E^2)*x*Log[x])^2, x])/(1 + 3*E^2) - 2*Defer[Int][x^2/(2*x
^2 - 3*Log[x] + (1 + 3*E^2)*x*Log[x])^2, x] + (54*Defer[Int][1/((-3 + (1 + 3*E^2)*x)*(2*x^2 - 3*Log[x] + (1 +
3*E^2)*x*Log[x])^2), x])/(1 + 3*E^2)^2 + 3*Defer[Int][1/((3 - (1 + 3*E^2)*x)*(2*x^2 - 3*Log[x] + (1 + 3*E^2)*x
*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3+\left (-1-3 e^2\right ) x-2 x^2-3 \log (x)}{4 x^4+\left (-12 x^2+4 x^3+12 e^2 x^3\right ) \log (x)+\left (9-6 x+x^2+9 e^4 x^2+e^2 \left (-18 x+6 x^2\right )\right ) \log ^2(x)} \, dx\\ &=\int \frac {3-\left (1+3 e^2\right ) x-2 x^2-3 \log (x)}{\left (2 x^2+\left (-3+x+3 e^2 x\right ) \log (x)\right )^2} \, dx\\ &=\int \left (\frac {9-6 \left (1+3 e^2\right ) x-\left (11-6 e^2-9 e^4\right ) x^2+2 \left (1+3 e^2\right ) x^3}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2}+\frac {3}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )}\right ) \, dx\\ &=3 \int \frac {1}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )} \, dx+\int \frac {9-6 \left (1+3 e^2\right ) x-\left (11-6 e^2-9 e^4\right ) x^2+2 \left (1+3 e^2\right ) x^3}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2} \, dx\\ &=3 \int \frac {1}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )} \, dx+\int \left (\frac {3 \left (7+6 e^2+9 e^4\right )}{\left (1+3 e^2\right )^2 \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2}+\frac {\left (5-6 e^2-9 e^4\right ) x}{\left (1+3 e^2\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2}-\frac {2 x^2}{\left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2}+\frac {54}{\left (1+3 e^2\right )^2 \left (-3+\left (1+3 e^2\right ) x\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {x^2}{\left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2} \, dx\right )+3 \int \frac {1}{\left (3-\left (1+3 e^2\right ) x\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )} \, dx+\frac {54 \int \frac {1}{\left (-3+\left (1+3 e^2\right ) x\right ) \left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2} \, dx}{\left (1+3 e^2\right )^2}+\frac {\left (5-6 e^2-9 e^4\right ) \int \frac {x}{\left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2} \, dx}{1+3 e^2}+\frac {\left (3 \left (7+6 e^2+9 e^4\right )\right ) \int \frac {1}{\left (2 x^2-3 \log (x)+\left (1+3 e^2\right ) x \log (x)\right )^2} \, dx}{\left (1+3 e^2\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 22, normalized size = 0.92 \begin {gather*} \frac {x}{2 x^2+\left (-3+x+3 e^2 x\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - x - 3*E^2*x - 2*x^2 - 3*Log[x])/(4*x^4 + (-12*x^2 + 4*x^3 + 12*E^2*x^3)*Log[x] + (9 - 6*x + x^2
+ 9*E^4*x^2 + E^2*(-18*x + 6*x^2))*Log[x]^2),x]

[Out]

x/(2*x^2 + (-3 + x + 3*E^2*x)*Log[x])

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fricas [A]  time = 0.74, size = 21, normalized size = 0.88 \begin {gather*} \frac {x}{2 \, x^{2} + {\left (3 \, x e^{2} + x - 3\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*exp(2)+x^2-6*x+9)*log(x)^2+(12*x^3*ex
p(2)+4*x^3-12*x^2)*log(x)+4*x^4),x, algorithm="fricas")

[Out]

x/(2*x^2 + (3*x*e^2 + x - 3)*log(x))

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giac [A]  time = 0.44, size = 25, normalized size = 1.04 \begin {gather*} \frac {x}{3 \, x e^{2} \log \relax (x) + 2 \, x^{2} + x \log \relax (x) - 3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*exp(2)+x^2-6*x+9)*log(x)^2+(12*x^3*ex
p(2)+4*x^3-12*x^2)*log(x)+4*x^4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 26, normalized size = 1.08

method	result	size
norman	\(\frac {x}{3 x \,{\mathrm e}^{2} \ln \relax (x )+x \ln \relax (x )+2 x^{2}-3 \ln \relax (x )}\)	\(26\)
risch	\(\frac {x}{3 x \,{\mathrm e}^{2} \ln \relax (x )+x \ln \relax (x )+2 x^{2}-3 \ln \relax (x )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*ln(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*exp(2)+x^2-6*x+9)*ln(x)^2+(12*x^3*exp(2)+4*x
^3-12*x^2)*ln(x)+4*x^4),x,method=_RETURNVERBOSE)

[Out]

x/(3*x*exp(2)*ln(x)+x*ln(x)+2*x^2-3*ln(x))

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maxima [A]  time = 0.49, size = 23, normalized size = 0.96 \begin {gather*} \frac {x}{2 \, x^{2} + {\left (x {\left (3 \, e^{2} + 1\right )} - 3\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)-3*exp(2)*x-2*x^2-x+3)/((9*x^2*exp(2)^2+(6*x^2-18*x)*exp(2)+x^2-6*x+9)*log(x)^2+(12*x^3*ex
p(2)+4*x^3-12*x^2)*log(x)+4*x^4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.98, size = 21, normalized size = 0.88 \begin {gather*} \frac {x}{\ln \relax (x)\,\left (x+3\,x\,{\mathrm {e}}^2-3\right )+2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + 3*log(x) + 3*x*exp(2) + 2*x^2 - 3)/(log(x)^2*(9*x^2*exp(4) - exp(2)*(18*x - 6*x^2) - 6*x + x^2 + 9)
+ log(x)*(12*x^3*exp(2) - 12*x^2 + 4*x^3) + 4*x^4),x)

[Out]

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

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sympy [A]  time = 0.25, size = 19, normalized size = 0.79 \begin {gather*} \frac {x}{2 x^{2} + \left (x + 3 x e^{2} - 3\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*ln(x)-3*exp(2)*x-2*x**2-x+3)/((9*x**2*exp(2)**2+(6*x**2-18*x)*exp(2)+x**2-6*x+9)*ln(x)**2+(12*x*
*3*exp(2)+4*x**3-12*x**2)*ln(x)+4*x**4),x)

[Out]

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

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